mathematics for queensland year 11b

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MATHEMATICSfor QueenslandMATHEMATICSfor QueenslandYEAR 11B

K. Bolger R. Boggs R. Faragher J. Belward

1



3253 Normanby Road, South Melbourne, Victoria, Australia

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© Oxford University Press

First published 2001

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photocopying, recording or otherwise without prior written permission.Enquiries to be made to Oxford University Press.

Copying for educational purposes Where copies of part or the whole of the book are made under Part VBof the Copyright Act, the law requires that prescribed procedures befollowed. For information, contact the Copyright Agency Limited.

National Library of AustraliaCataloguing-in-Publication data:

Mathematics for Queensland year 11B: a graphics calculatorapproach.

ISBN 0 19 550852 1.

1. Mathematics. I. Bolger, Kiddy.

510

Cover design by Sylvia WitteTypeset by Raksar Nominees Pty LtdPrinted through Bookpac Production Services, Singapore



FOREWORD

Mathematics education has undergone revolutionary changes in the last decade. Until the late

1980s mathematics education focused on skill development. The skill development across thevarious areas of mathematics was largely disjointed and skills seemed to be an end in themselves. Such mathematics education always produced the standard question from students, “where arewe ever going to use this?”

Modern mathematics education emphasises modelling, problem solving and applications, andthe use of higher technologies. The QBSSSS Mathematics B syllabus, published in 2001,reflects this change.

This textbook, with the student’s CD, the website and the accompanying mailing list, have been produced to meet the requirements of this syllabus in the four categories of:

Knowledge and procedures

Modelling and problem solvingCommunication and justification

Affective

The text material relating to the first three of these categories incorporates the contexts:, , and .

The fabric of this text has been woven based on five principles and reflects the authors’ philosophyof mathematics education.

A wide variety of mathematical modelling is central to the text. The modelling approach isintroduced once students have encountered linear functions and is further developed withother classes of functions.

Problem solving strategies are developed in all topics. It should be noted that skills are nottreated in isolation, and are introduced in the text as required. Nevertheless, skill developmentis still regarded to be fundamental to the successful study of mathematics. An Appendix con-tains material to enable students to maintain basic knowledge and procedures.

Students are exposed to a catalogue of elementary functions. The functions serve as a bridge between the mathematics and the situations they model. The study of functions is supported by graphics calculator technology. The various ways in which graphics calculators can repre-sent a function help students to develop a deeper understanding of functions. In particular,geometry of functions is used to enhance the algebraic understanding.

The QBSSSS Mathematics B syllabus 2001 has introduced graphics calculators or higher tech-nology throughout the course. The text offers a variety of investigations and applications us-

Application Technology Initiative Complexity

1. Problem solving, mathematical modelling, and applications are

essential to the understanding of mathematics

2. Functions underpin the mathematics

3. Graphics calculators support the understanding and learning of mathematics

ing graphics calculator technology, which have been trialled by students in the classroom.



Although the graphics calculator activities in this book have been trialled by students usingCASIO FX 7400, CASIO 9850, TI 82 and TI 83 calculators, the activities may also be carriedout on any calculator with similar functionality.

In the past, statistics has largely been taught using contrived data. The authors’ view is that

students should study statistics using real data sets wherever possible. Modelling data withelementary functions is introduced using graphics calculators, linking algebra with the notionof regression. To this end, material on the measures of and , together with an explanationof residual plots, has been included for the sake of completeness.

A mathematics text, like any other book, should be very readable. This text has been writtenwith this principle in mind, enabling students to learn independently by studying the text andworking through the numerous examples on their own.

The materials have been trialled for two years with students, working in the classroom and at

home, and feedback from students indicates that the text has met this objective.

The complete textbook with links from the detailed contents list

Exercises linked to the answers

“ ” linked to the glossary

Web page links

End of chapter test worksheets

Chapter revision worksheets

Cumulative review worksheets

The website supports this textbook and contains additional resources including:

Work programs

Assessment instruments

Resources such as grid paper masters, worksheets and teaching ideas

Further graphics calculator activities, investigations, modelling activities and problems

Worked solutions to selected exercises and problems

Web page links

The mailing list allows teachers using this textbook to maintain a dialogue with the authorsand other teachers who are using the text. This dialogue will include syllabus discussion, theexchange of assessment instruments, teaching ideas and professional support.

To subscribe to the mailing list, send an email to:You can also subscribe from the website.

Email address for list members:

4. Statistics and the use of real-life datasets

5. Mathematics books should be readable

Interactive Student’s CD

Website

Mailing list

r r2

Words you should know

.

mathematics-for-queensland.com

[email protected]

[email protected]

mailto:[email protected]

http://mathematics-for-queensland.com/








Key to the Icons in the text and on the CD

+

INVESTIGATION

SPREADSHEET APPLICATION

GRAPHICS CALCULATOR APPLICATION

HISTORICAL NOTE

MODELLING AND PROBLEM SOLVING

WORDS YOU SHOULD KNOW

CHAPTER SUMMARY

CHAPTER TEST, REVISION SETS AND REVISION EXERCISES

Investigations provide opportunities to both develop and apply concepts.

They encourage explorations involving planning, working with others and

decision making.

Activities to make use of computer spreadsheet software (written for MS

Excel) in many applications involving calculation and graphing.

Spreadsheets are linked from the text to the CD.

Another application of technology where appropriate for calculation and

graphing.

Brief discussions of relevant historical facts related to the topic at hand.

The use of mathematical knowledge and procedures in the modelling and

solution of life-related situations.

Important words and terminology used in the text and included in the

glossary. The words and terminology are linked to the glossary on the CD.

An outline of the important ideas and formulae from the chapter.

Are reproduced as student worksheets on the CD and are accessible by

clicking on the icon.



TABLE OF CONTENTS 6

1. INTRODUCTION TO FUNCTIONS 9

2. QUADRATIC FUNCTIONS 47

3. EXPLORING DATA 91

4. MODELLING DATA WITH FUNCTIONS 145

A. Relations 10B. Functions 12C. Exercises, problems, investigations and models 18D. Gradient of a straight line 21E. Linear functions 26F. Equations of linear functions 28G. Communication and justification 34H. Graphing linear functions 36I. Direct variation 38J. Problem solving 43

Chapter 1 Revision Set 44Chapter 1 Test (Knowledge and procedures) 45Extended modelling activity 46

A. A mathematical model of a chook pen 48B. Surds 50C. Review of quadratic expressions 58

D. Solving quadratic equations 64E. Graphing quadratic functions 73F. Transformations of the quadratic function 76G. Transformations of other functions 80H. Modelling using quadratic functions 83I. Problem solving 85

Chapter 2 Revision Set 88Chapter 2 Test (Knowledge and procedures) 89Extended investigation (Absolute value functions) 90Extended modelling activity (Eraser toss) 90

A. Data 92B. Some terminology 95C. Collecting data 99D. Recording data 103E. Visualising data 105F. Quantitative data 113G. Displaying quantitative data 114H. Stemplots 116I. Summarising data - measures of the centre 123J. Summarising data - measures of spread 126K. Boxplots 132L. Problem solving 140

Chapter 3 Revision Set 141

Chapter 3 Test (Mathematical techniques) 143

A. A first model (the regression line) 146B. Least squares regression line 148C. Measuring the 'fit' of a linear model 149D. Residuals and the residual plot 152E. Least squares regression line revisited 159F. Problem solving 165

Chapter 4 Revision Set 166Chapter 4 Test (Knowledge and procedures) 167Extended modelling activity 168



7 TABLE OF CONTENTS

5. TRIGONOMETRY 169

6. REVISION EXERCISES FOR CHAPTERS 1 TO 5 207

7. INDICES AND LOGARITHMS 221

8. POLYNOMIALS 253

9. FURTHER FUNCTIONS AND RELATIONS 281

A. Similar triangles 170B. Trigonometric ratios 171C. Angles of any magnitude 178D. Circular functions 180E. Radian measure 184F. Area of triangles 188G. The sine rule 191H. The cosine rule 195I. Applications of the sine and cosine rules 199J. Problem solving 203

Chapter 5 Revision Set 205Chapter 5 Test (Knowledge and procedures) 206

A. Review of Chapter 1 208B. Review of Chapter 2 210C. Review of Chapter 3 213D. Review of Chapter 4 217

E. Review of Chapter 5 218

A. A review of index notation - a mathematical shorthand 222B. Operations with powers 223C. Indices other than natural numbers 226D. Logarithms 231E. The laws of logarithms 236F. Solving log equations 238G. Change of base of a logarithm 238H. Applications of indices and logarithms 240I. A further application of logarithms 244J. Problem solving 249

Chapter 7 Revision Set 250Chapter 7 Test (Knowledge and procedures) 251Extended modelling activity (Piano keys) 252

A. Definition of a polynomial 255B. Evaluating polynomials 257C. Operations with polynomials 258D. Zeros of a polynomial 263E. Graphs of polynomials 264F. Composition of functions 268G. Transformations of polynomials 271

H. Applications and modelling using polynomials 274I. Problem solving and modelling 278


A. Reciprocal functions 282B. Continuous and discontinuous functions 284C. Inverse variation 286D. Circles as relations 290E. Finding points of intersection of lines, parabolas and circles 296F. More on relations and mapping 299G. Problem solving and modelling 301



TABLE OF CONTENTS 8

Chapter 9 Revision Set 302Chapter 9 Test (Knowledge and procedures) 303Extended modelling activity (The cost of running a car) 304

A. Rates 306B. Average rates of change 310

C. Distance, displacement, speed and velocity 311D. Variable rates of change 317E. Finding the gradient function 321F. Problem solving 321

Chapter 10 Revision Set 322Chapter 10 Test (Knowledge and procedures) 323Extended investigation (Local linearity) 324

A. Instantaneous velocity 327B. The concept of a limit 330C. The gradient of a curve 334D. The gradient of any function at any point 339E. Differentiation by rule 341F. The second (and higher) derivatives 345G. Velocity and acceleration as derivatives 346H. Equations of tangents and normals 348I. Differentiation of composite functions 350J. Differentiation of products 355K. Differentiation of quotients 357L. Functions, derivatives and graphs 360M. Problem solving 365


A. Review of Chapter 7 370B. Review of Chapter 8 372C. Review of Chapter 9 374D. Review of Chapter 10 377E. Review of Chapter 11 380

A. Measurement 386B. Percentage 395

C. Basic algebraic procedures 398D. Rational numbers (fractions) 401E. Linear equations 409F. Formulae 414G. The rule of Pythagoras 418H. Coordinate geometry 422I. Sigma notation 424

10. RATE 305

11. INSTANTANEOUS RATES OF CHANGE– DIFFERENTIAL CALCULUS 325

12. REVISION EXERCISES FOR CHAPTERS 7 TO 11 369

APPENDIX 1 – (Maintaining basic knowledgeand procedures) 385

APPENDIX 2 – (Chapter 3 survey on statistics) 426

APPENDIX 3 – (Glossary) 429

ANSWERS 441



Introduction tofunctions

SUBJECT MATTER concepts of function, domain and range

mappings, tables and graphs as representations of functions and relations

graphs as a representation of the points whosecoordinates satisfy an equation

distinction between continuous and discretefunctions

practical applications of linear functions, includingdirect variation and linear relationships betweenvariables

introduction to modelling, problem solving andinvestigations

CHAPTER 1



10 INTRODUCTION TO FUNCTIONS (Chapter 1)

The season has ended, and it is your job to organise the Sports Presentation Dinner. A caterer tells

you that the cost of serving a three course meal is $200 (for the hire of the hall, staff wages, etc)

plus $10 per head.

The total cost of the dinner depends upon the number of people attending. How can we describe

the relationship between the number of guests and the total cost?

Mathematicians have devised a number of different methods for showing such relationships.

Table

No. of guests Cost

0 2005 250

10 300...

...

30 500...

...

Ordered pairs

(0, 200), (5, 250), (10, 300),

...... (30, 500), ...

A second way of displaying a relation is as a set of ordered

pairs. Notice that the order in which the numbers are writ-

ten is important. (0, 200) means that even if no guests

arrive, the cost is still $200. On the other hand, (200, 0)

implies that there are 200 guests, and the dinner is totally

free! The first number in the ordered pair is always the in-

dependent variable, and the second number is the dependent

variable.

HISTORICAL NOTE +

RELATIONS A

Coordinate geometry allows the power of algebra to be applied to problems ingeometry The person credited with this discovery is

While Descartes was at the Jesuit school of La Fleche in Anjou, he was

of delicate health and hence was allowed to spend his mornings in bed.It is said that he discovered coordinate geometry one morning whilelying in bed watching fly crawl across the ceiling, and wonderinghow the fly’ path could be described mathematically

In honour of Descarte’ discovery the number plane is often referredto as the Cartesian plane. The story is probably not true, but does giveyou good excuse for lying in bed on the weekends!

. .

as .

s ,

a

Rene Descartes

0

5

10

…

30

…

200

250

300

…

500

…

Mappings

The two for this problem are the Number of

Guests and the Cost. The conventional mathematical term

for a relationship between two or more variables is a

. We want to describe the relation between the Number of Guests and the Cost of the Dinner.

One way of presenting the relation is as a table. The logical

order for labelling the columns is to put the Number of

Guests in the first column, and the Cost in the second col-

umn, since you need to know how many guests there will

be before you can determine the cost of the dinner. Getting

the correct order of the columns is important. As the Cost

of the Dinner upon the Number of Guests, the

Cost is said to be the , while the Num-

ber of Guests is called the . The inde-

pendent variable is always the first column of the table (or

first row, if the table is laid out horizontally) and the de-

pendent variable is the second column.

variables

rela-

tion

dependent variable

independent variable

depends



INTRODUCTION TO FUNCTIONS (Chapter 1) 11

A third way to show a relation is as a mapping.

The arrows of a mapping always go from the

independent variable to the dependent variable.

There are two common methods of displaying a

mapping, as shown in the figures alongside.

Plotting the ordered pairs on a number plane, wecan display the relation as a graph. The variables

are:

n = number of guests

c = cost, in dollars

By convention we put the independent variable

(in this case, Number of Guests) on the horizon-

tal axis and the dependent variable (Cost) on the

vertical axis. Note that where possible all graphsshould include a title, and labels on the axes.

For this graph, the points lie in a straight line, so this relation is said to be a linear relation.

Note that the graph was drawn as a continuous straight line and not as a series of points. Do you

think it is appropriate? We will be discussing this question shortly.

We can also represent this relation as an algebraic equation, using the above variables.

The equation for this relation is c = 10n + 200

Can you see that this equation correctly represents the situation? If not, check it for yourself by

substituting the values 0, 5, 10 .., 30, .. for n and calculating the value of c. They should match

those values found in the table.

200 250 300 500

0 5 10 30

500

1000

1500

50 100 150

c

n

cost

Cost of sports dinner

number of guests

EXERCISE 1A

1

i

ii iii iv v vi

a

b

c

d

For each of the following situations

decide which is the independent and which is the dependent variable.

Show the relation as a

table set of ordered pairs mapping graph equation

A plumber charges a $ callout fee, plus $ an hour. What is the relation between thetotal charge and the number of hours charged by the plumber?

A syndicate won $ in Powerball. What is the relation between the number of

people in the syndicate and the amount that each person receives?

A cabinet maker is building a round outdoor table. What is the relation between thediameter of the table and the area of the top of the table?

Which of the representations of a relation (table, ordered pairs, mapping, graph andequation) can you show on a graphics calculator?

60 40

6000000




What is the possible number of guests that may attend the sports dinner? The worst possible case

is that no one turns up, while the maximum number of guests is the capacity of the hall, say 150guests. Any number of guests between 0 and 150 is possible, so the set of all possible values of

the independent variable (Number of Guests) is the set of whole numbers from 0 to 150.

The set only includes whole numbers because we cannot have a fraction of a guest.

The total cost of the evening will range from $200 (if no one turns up) to a maximum of $1700, if

the hall is filled.

In this case, the range is the values that the cost may have, which is the multiples of $10 from $200to $1700 inclusive.

We can also state the domain and range in another way, using set notation, as follows:

domain: f0, 1, 2, ..., 150g range: f200, 210, 220, ..., 1700g.

For this sports dinner, each value of the domain maps to exactly one value in the range. Another

way of saying the same thing is: if you tell me the Number of Guests (a value in the domain), we

can give you the Cost (a value in the range).

The concept of a function is important, so we will explain it yet another way. If you ask a question

such as, ‘What is the time?’, you expect an answer and only one answer. You would be unhappy

with no answer at all, or the answer, ‘The time is now 1:00 pm and 1:20 pm’. Similarly, when

you give a number in the domain, and apply a rule to that number, you should expect exactly one

answer back. That in essence is what is meant by a function.

Most, but not all, of the relations that you will study in this course will be functions. In fact, if

someone described Mathematics B by saying it is the study of functions, they would not be too

far wrong, that is how important functions are in this course.

We will discuss functions in more detail when we meet our first relation that is not a function, in

Chapter 8.

From the many ways to represent a function, three of them; a table, a graph and an equation, are

the most important. When studying a particular group of functions, a significant set of problems is

how to get the table and graph from the equation, the table and equation from the graph, and the

graph and equation from the table. Because a graphics calculator can represent functions in all of

these ways, it is an important tool in the study of mathematics.

DOMAIN AND RANGE

The set of all possible values of the independent variable is called the domain.

The set of all possible values of the dependent variable is called the range.

Relations with the property, that for every value in the domain there is exactly one value in

the range, are called functions.

THREE WAYS TO REPRESENT A FUNCTION

FUNCTIONS B




Mathematicians are able to be concise by carefully choosing how they express ideas in mathematics.

Function notation is a good example of this.

Here is a problem, expressed in words, “I have a rule: take any number and square it, then

multiply the answer by 3, now add on 4 times the number, and finally take 16 away from the lasttotal. Apply this rule to the number ¡2.”.

Here is the same problem expressed in function notation. Note how concise it is.

f (x) = 3x2 + 4x ¡ 16; find f (¡2).

This particular function is given the name ‘f ’, while the independent variable is given the name

‘x’. We use letters of the alphabet to name functions, usually f , g, and h.

This rule is a function, since for any number in the domain, you can apply the rule and you always

get exactly one answer. The notation f (¡2) is a concise way of writing ‘apply the rule named f

to the number ¡2.’

We can evaluate the function when x = ¡2 as follows: f (¡2) = 3(¡2)2 + 4(¡2) ¡ 16, whichequals ¡12. So the value of f (¡2) is ¡12. We write f (¡2) = ¡12. The order of operations that

you learned in the junior school ensures that everyone will do the sequence of operations in the

same order. What is the domain of this function?

When the domain is not given, we assume that the domain is the largest possible set of numbers.

For this function, we can substitute any real number for x and get a unique answer, so the domain

of f is all real numbers.

We write this as x 2 R , where the symbol ‘2’ reads ‘is an element of ’ and the symbol R stands

for the set of all real numbers.

1 a For the function f (x) = 4x ¡ 3, find the value of

i f (¡1) ii f (3) iii f (0) iv f (b)

b If g(x) = 2x2 + x ¡ 12, find

i g(1) ii g(¡2) iii g(1 12 ) iv g(1 + a)

c If p(x) = x2 + 5x + 6, show that p(¡2) = p(¡3)

d If h(x) = x3, show that h(a) =¡

h(¡

a)

For the function f (x) = 3x ¡ 5, find the value of

a f (¡1) b f (3) c f (0) d f (a)

a f (¡1) = 3 £ (¡1) ¡ 5= ¡8

b f (3) = 3 £ 3 ¡ 5= 4

c f (0) = 3 £ 0 ¡ 5= ¡5

d f (a) = 3 £ a ¡ 5= 3a ¡ 5

EXAMPLE 1.1

EXERCISE 1B.1

REPRESENTING A FUNCTION USING AN EQUATION – FUNCTIONNOTATION




2 a Given the function y = 2x ¡ 5, find i y, when x = ¡3 ii x, when y = 3

b Given the function y = ¡3x + 1, find i y, when x = 2 ii x, when y = 0

c Given the function y = 4 ¡ 12 x, find i y, when x = 1

2 ii x, when y = 12

3 For the function f (x) = 2

x + 1, find the value of

a f (2) b f (¡3) c f ( 12 ) d f (0)

4 a Given the function g(x) = x3 ¡ 4x2 ¡ x, find i g(3) ii g(¡3) iii g(0)

b Given the function h(x) = 100 ¡ x4, find i h(0) ii h(¡2) iii h(2)

5 a Is y = 4 a function? Justify your answer.

b Is x = 3 a function? Justify your answer.

6 Which of the following relations is a function? Explain your answers.

a f(1, 2), (1, 3), (2, 4), (3, 5), (4, 6)g b f(2, 3), (3, 3), (4, 3), (5, 3)g

c

7 Geoff was running laps around a track, and timing each one.

a His results are given in the table below. Is this a function? Explain.

Lap number 1 2 3 4 5 6

Time (sec) 71 69 70 72 71 72

b Here is the same data, with time as the independent variable. Is this a function? Explain

Lap number

71 69 70 72 71 72Time (sec)

1 2 3 4 5 6

BHP shares

0

5

10

15

20

25

1 / 0 4 / 0 0

1 / 0 6 / 0 0

1 / 0 8 / 0 0

1 / 1 0 / 0 0

1 / 1 2 / 0 0

1 / 0 2 / 0 1

Price ($)

Date

Many functions cannot be expressed as an

equation. The price of a BHP share on the

share market is one example. For any time

in the past, we can tell you the share price.

We can draw a graph that shows the price

of the share, say for the last thirty days.

However there is certainly not an algebraic

equation that allows us to calculate the

price of the share, given the date and time.

There are too many factors that affect the

share price to allow us to determine the

share price so exactly. In this course,

though, most of the functions that we will

study can be expressed as equations.




CONTINUOUS AND DISCRETE VARIABLES

Let us consider the Sports Dinner graph once again. Did you get the urge to draw a straight line

through the points on the graph?

A line through the points implies that all real numbers from 0 to 150 are in the domain, including

such numbers as fractions, square roots, and even pi.

While you may be happy to have pie for sweets, it is not possible to have pi, or 12 or 1.6 as the

number of guests. So, you should not draw that line, since the number of guests is a discrete

variable.

For example, length is a continuous variable. Given the right measuring instrument, it is possible

to measure lengths to a very high degree of accuracy.

Here is a simple test for determining if a variable is discrete or continuous.

Ask yourself, do I count the variable or do I measure it? I count the number of people on an

aeroplane, therefore that is a discrete variable. I measure the weight of crocodiles (very carefully,

I might add), so weight is a continuous variable.

Why would a mathematician draw a line through those points?

Well, partly for convenience, as it is much easier to draw one line than to plot 151 points. Also

there are some powerful mathematical tools that you will learn to use, that only apply to continuous

variables.

While the mathematician knows the variable is discrete, it useful to treat the variable as if it were continuous.

A discrete variable can only take on certain distinct values, which are usually (but not

always) whole numbers.

A continuous variable, on the other hand, can take on any value between the smallest

and largest possible values.

Which of the following variables are discrete and which are continuous?

a the number of boats in a boat yard

b the height of students in this class

c shoe sizes

a The number of boats can only be a whole number, so the variable is discrete.

We also know the variable is discrete since we count the number of boats.

b

c Although shoes sizes can take on fractional values, such as 12 , they are still

discrete, as they cannot take on every value. Try asking the shoe salesperson

for shoe with sizeap

99, for instance!

EXAMPLE 1.2

The height of person is continuous. During the time that student grows fromcm to cm, his or her height takes on every value in between and eventhough we might round of the height when writing it down. And the variable iscontinuous since we the height of students.

a a,

f

150151 150 151

measure

is often




1 Write in your own words the definitions of

a variable b relation c independent variable

d dependent variable e domain f range

g function h discrete variable i continuous variable

2 Which of the following variables are discrete and which are continuous?

a number of passengers on a bus

b temperature at any time during the day

c number of mangoes picked from a mango tree

d the length of a piece of string

e the volume of water remaining in the bath after the plug has been pulled out

f the time to fly from Brisbane to Rockhampton

g the number of students studying 11 Mathematics B at your school each year

3

The cost of renting a car in Rockhampton is $80 a day plus $0.50 per kilometre. You rent acar for one day, and intend to drive no more than 400 kilometres.

a Which is the dependent variable and which is the independent variable?

b Explain why the relation between rental cost and distance driven is a function.

c Is the independent variable discrete or continuous?

d Represent the function as

i a table ii a set of ordered pairs iii a mapping

iv a graph v an equation

e State the domain and the range of the function.

a The rental cost depends on the distance driven, so rental cost is the dependent

variable. The independent variable is the distance driven.

b The independent variable is distance driven, and the dependent variable is cost of

renting the car. For every distance in the domain, there is exactly one cost in the

range, so the relation is a function.

c Distance driven is a continuous variable. Note that we often round off the distance

to the nearest kilometre, or maybe the nearest 0.1 kilometre.

Note: Rounding off a continuous variable to the nearest whole kilometre does not

make it a discrete variable.

EXERCISE 1B.2

EXAMPLE 1.3

Justin and Candace disagree about whether thetime person is infectious after he or she con-tracts the flu is discrete or continuous. Justinsays “discrete, since we count the number of days person is infectious”. Can

a

a dace says, “con-tinuous, since time is measured”. With whomdo you agree? Why?




d i Table:

Distance driven Rental cost

0 km $80

10 km $85

20 km $90

30 km $95......

400 km $280

ii ordered pairs:

(0, 80), (10, 85),

(20, 90), .... (400, 280)

iv Graph:

iii Partial Mapping: (not every value

has been shown)

v Equation: Let c = rental cost, in dollars and d = distance driven, in km

then c = 0:5d + 80

e domain: 0 6 d 6 400 range: 80 6 c 6 280

If we drive the maximum distance of 400 km, the cost is $(80 + 0:5 £ 400) , o r $280.

1 Each of the following situations can be represented as a function. In each case:

i Which is the dependent variable and which is the independent variable?

ii Are the variables discrete or continuous?

iii Represent the function as: a table, a set of ordered pairs, a graph, an equation.

iv State the domain and the range of the function.

a The weekly rent for a unit is $120 per week. You rent the unit for up to one year.

b The cost of producing ID cards for up to 200 year 11 students is $460 for the hire of the

equipment and the photographer’s fee, plus 25 cents per card.

c The amount of tread on a new tyre is 15 mm. The tread wears away uniformly at a rate

of 1 mm every 3000 km. A tyre becomes illegal once the amount of tread reaches 4 mm.

Date Feb 5 Feb 6 Feb 7 Feb 8 Feb 9

Closing price 3126:7 3178:2 3167:0 3104:8 3137:9

0

$80

100

$130

200

$180

300

$230

400

$280

50

100

150

200

250

100 200 300 400

y

x

EXERCISE 1B.3

d

e

The closing price of the AllOrdinaries Index for the week starting February is given inthe table alongside.

Each point in the graph showsthe length of the metacarpal

bone (a bone in the hand) for one person, and the height of the person.

5

155

160

165

170

175

180

30 40 50 60

height (cm)

metacarp (mm)

Height v metacarpal bone length




Now it is time to step back for a bit, and take a broader look at our study of mathematics.

Suppose we have a mathematical question in front of us. We have seen similar ones before or we

have been taught the standard way to tackle this type of question. We may not know the answer

from looking but we know the procedure to follow to find the solution. This may involve using

textbooks to look up a formula or using a graphics calculator to sketch a graph.

Mathematical questions for which the method of solution is known are called exercises. By doing

exercises we learn and improve our mathematical skills.

There are other types of mathematical questions where we do not know how to answer the question

when we first look. A number of strategies might come to mind which could help us make a start,

but none immediately suggests the solution.

Specific questions where the method of solving them is not immediately known are called prob-lems.

At times we may be studying an interesting topic, but do not yet have specific questions to be

answered. Or we may have a specific question, but we need to gather some data first. This re-

quires an investigation. While investigating, we might draw diagrams, look for patterns and make

conjectures. Often while doing an investigation, new problems arise for which there is no obvious

method.

Alternatively, one step in solving a problem may be investigating an intriguing pattern. Investiga-

tions and problem solving are intertwined.

A mathematical model is just a way of representing a real-life situation mathematically. A func-

tion can be considered to be a mathematical model if it mimics a real situation. Ideally the modelcaptures the important features of the real-life problem, but is simple enough to use for predictions

and further study.

A mathematician once said, “All models are wrong. Some are useful.”. By this he meant that

real-life problems are complex, and it is usually not possible to capture all of this complexity in a

single model. However some models can capture enough of the complexity to be useful for making

predictions.

We will now look at each of these in more detail.

PROBLEM SOLVINGThroughout this course, you will have opportunities to solve problems. You may find it helpful to

rule up a “Thinking Column” to go alongside where you are working. Try to record what you are

doing, what you think you might try, conjectures you have, patterns you spot, when you are stuck

and so on. After a while, you will start to develop a collection of strategies that are helpful. This

column is also useful for recording what you are doing or what you were about to try if you need

to stop work, perhaps at the end of a lesson.

When you begin to solve a problem, you can expect not to know what the solution is. There are

ways you can make a start, though. Many people find trying a few specific examples to be a way

of understanding the problem. After you have tried enough examples, you might begin to spot

patterns or be able to make conjectures about the question.

EXERCISES, PROBLEMS, INVESTIGATIONS AND MODELS

C



GRAPHICS CALCULATOR ACTIVITY

LINEAR FUNCTIONS


Spotting patterns is easier if you have chosen examples in a systematic way. Sometimes this process

is described as “guess and check” but this is not a good name. Guessing suggests you do not know

what you are doing. A much more accurate term is “systematic specialisation”. If you are systematic

in your choice of specific examples and you record your data (results) in a systematic way such as

in a table, you are more likely to spot underlying patterns.

When solving a problem, there are often many ways you can proceed. Do not be surprised if your

path to the solution is different from others in your class.

If you need help, you will need to explain to your group or your teacher what you have already

tried. Otherwise, they will not know what path you are on and therefore how you are seeing the

problem.

Once you have found a solution, you will need to convince yourself that you have solved the

problem. Then try convincing a friend. Finally, you need to refine your argument or modify your

solution until you could convince anyone.

Here is a problem for you to try. It is called a pattern problem.

a How many matchsticks are needed to make the 10th pattern and the 1000th pattern?

b What is the rule for finding the number of matchsticks to make the nth pattern?

c Arrange the matchsticks to demonstrate how you see the pattern building. Compare your

pattern with others in your class. You may be surprised at the variety of ways people see

the patterns.

d Justify your rule in terms of the arrangements of the matchsticks.

A solution to this problem is given at the end of the chapter. But do not read it until you have

tried to solve the problem, and discussed it with your classmates. The best way to learn to solve

problems is to tackle them yourself!

INVESTIGATIONS

Throughout this book are Investigations, similar to the one below. You should do all of the

Investigations in this course, even if your teacher does not require you to do so. Doing them will

help you to think about and understand the mathematics you are learning. Try this one right now.

The calculator symbol means that your graphics calculator will be useful here.

1 A linear function is a function whose graph is a straight line. Use a graphics calculator

to draw each of the following functions. Which are linear functions?

a y = x + 1 b y = x2 c y = 4 ¡ x d y = 3

xe y = 3 f y = 1

2 x ¡ 6 g y =p

x + 1

2 Can you tell which functions are linear and which are not, just by looking at the

equations? How can you tell?

Look at this match-

stick pattern:



Have a look at the diagram. What do you see? Different people seedifferent things - some a patchwork quilt, some a climbing frame, whileothers a radioactive symbol. Lots of people start counting triangles. Did

you find yourself noticing the triangles or counting? Many situationsappear to have no mathematics involved at all and yet mathematics can be used to investigate and understand these situations better.


3 Test your theory on these functions:

a y = 5 ¡ 2x b y = x3 c y = 6 ¡ 1

x d y = ¡2

e y = ¡23 x ¡ 1 f y =

p 4x ¡ 1 g y =

2x ¡ 1

1 ¡ x

4 Write down a rule for deciding if a function is linear or not, by looking at the equations.

5 Once a pattern is noted, a mathematician then tries to determine why the pattern exists.

Can you see why your rule in question 4 always gives a linear function?

MODELLING

The process of using mathematics to help understand a situation is called mathematical modelling.

This is an important part of this course and you will find many opportunities to develop your skills.

Mathematical modelling can be very simple - in fact, we have looked at a model already. However,

many other situations can lead to very complex models which require a great deal of development.

The Consortium for Mathematics and its Applications (COMAP) in the United States has developed

the following structure which is useful for describing a mathematical modelling process. The steps

involved are general stages which may help you plan your activity. You may need to refine each of

the stages. It is important to note that mathematical modelling is a cyclic process - you will need

to revise your model (sometimes often) after you have evaluated it.

The following diagram shows the basic steps involved.

yes no

STEP 1 - IDENTIFY THE SITUATION

Notice something that you wish to understand.Pose a well defined question indicating exactlywhat you wish to know.

STEP 2 - SIMPLIFY THE SITUATION

Determine the assumptions on which your modelwill rest. List the key features and relationshipsamong those features you want to consider. Note features and relationships you choose toignore for now.Remember: keep it simple, you can make themodel more complex later.

STEP 4 - EVALUATE AND REVISETHE MODEL

Return to the original situation and see if the results of the mathematics make sense.Do they help to explain the situation better?

Reconsider theassumptions youmade in step 2 andrevise them to bemore realistic.

Use the model untilnew information

becomes available or assumptions change.

STEP 3 - BUILD THE MODEL

Interpret in mathematical terms the features andrelationships you have chosen.This might involve- defining variables and their type- writing equations- drawing shapes- gathering data- measuring objects- calculating probabilities. Add to this list as you

move through this course.



MODELLING ACTIVITY

VINTRODUCTION TO FUNCTIONS (Chapter 1) 21

Let us go back to see how these modelling steps could have been used in the

development of our model for the sports dinner.

Step 1: We wanted to know more about the cost of the dinner. The question could have

been “How does the number of guests affect the total cost of the dinner?” Other questions are also possible.

Step 2: We only considered one type of dinner guest - ignoring concessional entry, free

entry, etc. Make a list of key features of the situation. These might include size,

type of venue, seating arrangements, decorations, entertainment etc. Describe any

connections, for example, if you plan to have dancing, you need a dance floor, which

will affect the seating arrangement. Now go through your list and tick the ones we

have considered in our model. Mark the ones we have ignored up to this point.

Step 3: At the beginning of this chapter, we looked at interpreting our model in mathematical

terms. We defined variables (number of guests and cost) and their type (what were

they?). Go back over the information at the beginning of the chapter and make a

list of the procedures used.Step 4: Look back at the various ways we modelled the dinner. Which did you find most

helpful in understanding the situation?

Before you can effectively develop mathematical models, you need to have mastered the required

knowledge and procedures. This is covered in the next sections of this chapter, after which modelling

will be re-visited.

A linear function is a function whose graph is a straight line. One way to decide if a function is

linear or not is to draw the graph of the function. However drawing graphs is tedious. Is there a

better way? Think about the problem of the cost of the sports dinner. Could you have decided from

the statement of the problem that the graph of the function was a straight line?

If no-one comes to the dinner, you must still pay $200 for the

hire of the hall. On the graph, this corresponds to the point with

coordinates (0, 200). If one person attends, the corresponding

point on the graph is (1, 210). If two attend this is represented

by the point (2, 220), and so on. Note that for each additional

person, the total cost rises by a fixed amount, $10.

It is because of this fixed change that the graph of the function

is a straight line. We could plot each point as follows: first

plot the point on the y-axis, i.e., when x = 0. Then to plot

the rest of the points, just follow this pattern - across 1, up 10;

across 1, up 10; across 1 up 10; and so on.

The gradient of a straight line is a measure of the steepness of the line as you move from

left to right across the Cartesian plane.

GRADIENT OF A STRAIGHT LINE D

LINEAR FUNCTIONS

GRADIENT

200

205

210

215

220

225

230

0 1 2 3number of guests

cost ($)




You may recall that gradientAB = m = rise

run

where m is the variable commonly used to represent

gradient.

If the coordinates of point A are (x1, y1) and the coordinates of point B are (x2, y2), then

the gradient of the line joining A and B is given by

gradient = y2 ¡ y1

x2 ¡ x1

Mathematicians often write this as

¢y

¢x , where ¢ represents “the change in”.

So, ¢y is the change in y (the rise), and ¢x is the change in x (the run).

If a graph falls as you move from left to right, the gradient is negative, since the rise is negative.

For a straight line

gradient = m = rise

run =

y2 ¡ y1

x2 ¡ x1=

¢y

¢x

All of these different ways of representing the gradient are important, so you need to know them

all.

Find the gradient of the line joining the points A(¡3, 1) and B(2, ¡5).

It does not matter which point is called (x1, y1) and which is called (x2, y2).

We will do it both ways, just to show it.

Let A have coordinates (x1, y1)

and B have coordinates (x2, y2).

Then m = y2 ¡ y1

x2 ¡ x1

= ¡5 ¡ 1

2 ¡ ¡3

= ¡6

5

= ¡65

Let A have coordinates (x2, y2)

and B have coordinates (x1, y1).

Then m = y2 ¡ y1

x2 ¡ x1

= 1 ¡ ¡5

¡3 ¡ 2

= 6

¡5

= ¡65

(!z\\' \\@z)

(!x\' \\@x)

A

B

run

rise

y

x

EXAMPLE 1.4




1 From the diagram alongside, find the

gradient of the line joining

a A and B b A and C

c B and C d F and I

e E and H f B and F

g H and E h E and G

2 Find the gradient of the line joining

a (5, 1) and (3, 0) b (2, 3) and (7, 4)

c (¡3, 1) and (5, 0) d (¡3, 5) and (¡1, 7)

e (4, ¡8) and (¡2, 3) f (¡5, ¡5) and (3, ¡1)

g (0, 0) and (2, 8) h (2, 3) and (¡3, 3)

i (1, 4) and (1, 7) j (3:6, ¡1:9) and (¡2:4, 3:8)

k (3, 12 ) and (1, 2) l (¡3, 2

3 ) and (¡1, 12 )

m (a, b) and (c, d) n (m, 0) and (0, n)

3 a On grid paper, draw four lines, each with a gradient of ¡

2.

b On the same grid paper, draw four lines, each with a gradient of 12 .

c What do you notice about these lines?

4

a Plot these points on your graphics calculator: (¡6, 2), (¡3, 2), (¡1:5, 2), (0, 2), (4, 2)

What can you say about the gradient of a line that passes through points whose y-

coordinates are the same?

b Plot these points on your graphics calculator: (¡1, ¡5), (¡1, ¡2), (¡1, 0), (¡1, 1),

(

¡1, 3 1

2 )

What can you say about the gradient of a line that passes through points whose x-

coordinates are the same?

5 On grid paper, draw two oblique lines that are parallel to each other. Calculate the gradient

of each line. Make a conjecture about parallel lines. Can you explain why your conjecture is

true?

Note: Oblique lines are lines that are neither horizontal nor vertical.

6 On grid paper, draw two oblique lines that are perpendicular to each other (not horizontal or

vertical lines). Calculate the gradient of each line. Make a conjecture about perpendicular

lines. Can you explain why your conjecture is true?

EXERCISE 1D.1

y

xA

D

E

F

G

H

B

I

C

You can plot points using a graphics calculator by putting the -coordinates into List and the-coordinates into List , and then drawing a scatterplot of List vs List .

xy

12 2 1




PARALLEL LINES

If two lines are parallel, they have the same gradient.

From the figure, m p = mq.

The little p and little q are examples of subscript

notation. The notation ‘m p’ means the gradient of

line p. We say ‘m sub p’.

To show that line p is parallel to line q , show that

the gradient of line p = the gradient of line q .

If A is the point (¡3, 4), B is (4, ¡5), M is (¡1, 8) and N is (6, ¡1), show that AB is

parallel to MN.

In questions of this type, it is helpful to plot the points first.

mAB = y2 ¡ y1

x2 ¡ x1

= 5 ¡¡ 4

4 ¡¡3

= 9

7

= ¡

¡

97

mMN = y2 ¡ y1

x2 ¡ x1

= 1 ¡¡ 8

6 ¡ ¡1

= 9

7

= ¡

¡

97

Since mAB = mMN, then AB jj MN.

COLLINEAR POINTS

Collinear points are points which lie on the same

straight line. From the figure, if mAB = mBC, then

A, B and C are collinear.

To show that points A, B and C are collinear, show

that mAB = mBC.

run

run

rise

rise

p

q

y

x

EXAMPLE 1.5

y

x

M

A

B

N

Show that A(¡1, 4), B(1, 1) and C(5, ¡5) are collinear.

mAB = 4 ¡ 1

¡1 ¡ 1

= 3

¡2

= ¡32

mAC = 4 ¡ ¡5

¡1 ¡ 5

= 9

¡6

= ¡32

) Since mAB = mAC, then A, B and C are collinear.

EXAMPLE 1.6

A

B

C

y

x




PERPENDICULAR LINES

From the diagram, line p has a gradient of m since the

rise is m, and the run is 1, and hence the rise

run is

m

1 , or just m.

Now rotate line p by 90o in a clockwise direction, to

line q .

From the diagram, it is clear that the gradient of line q

is ¡ 1

m.

This quantity is called the negative reciprocal of m.

Note that, m p £ mq = m £ ¡ 1

m = ¡1, that is, the product of the gradients of perpendicular

lines is ¡1.

Prove that the points A(

¡3,

¡2), B(3, 1) and C(0, 7) are the vertices of a right-angled triangle.

From the diagram, it appears that the triangle is right-

angled at B. So we test the gradients of AB and CB.

mAB = y2 ¡ y1

x2 ¡ x1

= 1¡¡ 2

3¡¡ 3

= 3

6

= 12

mCB = y2 ¡ y1

x2 ¡ x1

= 7 ¡ 1

0 ¡ 3

= 6

3

= ¡

¡

2) mAB £ mCB = 1

2 £ ¡2 = ¡1 and so AB?CB

) A, B and C are the vertices of a right-angled triangle.

In these questions, remember to plot the points first.

1 Determine whether AB is parallel to CD, AB is perpendicular to CD, or neither.

a A(1, 4), B(5, 2), C(1, 2), D(3, 6) b A(2, 0), B(8, 2), C(4, 2), D(2, ¡

4)

If line q is perpendicular to line p, then the gradient of line q is the negative reciprocal

of the gradient of line p.

If line q is perpendicular to line p, and the gradient of line p is m and the gradient of line q is n, then mn = ¡1.

(1, )m

( )m

m

m

p

q

1

y

x

EXERCISE 1D.2

C

B

A

y

x

EXAMPLE 1.7




c A(¡3, 2), B(1, 6), C(¡3, 0), D(0, 3) d A(4, 8), B(7, 2), C(¡1, 1), D(2, ¡5)

e A(¡2, 5), B(3, 5), C(6, 3), D(¡1, 3) f A(2, 1), B(2, ¡3), C(5, 0), D(4, 0)

2 Given the points A(¡2, 0), B(4, 3), C(1, ¡1) and D(5, 1), show that AB is parallel to CD.

3 Given the points A(1, 2), B(3, ¡2), C(4, ¡1) and D(¡2, ¡4), show that AB is perpendicular

to CD.

4 Determine if the following points are collinear.

a (1, 1), (4, 5) and (¡2, ¡3) b (1, 1), (4, 0) and (7, ¡1)

c (2, ¡5), (6, 0) and (8, 4) d (¡4, 1), (¡1, 2) and (5, 4)

e (1, 2), (4, 6), (¡5, ¡6) f (¡6, ¡6), (¡1, 0), (9, 12)

g A(3, 4), B(¡6, 5), C(0, ¡4) h (a, b), (2a, 2b), (3a, 3b)

5 Show that the points H(1, ¡1), I(7, 3), J(3, 5) and K(¡3, 1) are the vertices of a parallelogram.

6 Show that the points A(1, 4), B(¡4, ¡1) and C(2, 1) are the vertices of a right-angled triangle.

7 Give the coordinates of another point R that is collinear with P(¡

1, 3) and Q(2, 0). Justifyyour answer.

8 Recall that the distance from A(x1, y1) to B(x2, y2) is given by

d =p

(x2 ¡ x1)2 + (y2 ¡ y1)2

a Show that A(1, 3), B(6, 3), C(3, ¡1) and D(¡2, ¡1) are the vertices of a rhombus.

b Show that AC is perpendicular to BD.

In every case this value is represented graphically as the point where the graph crosses the y-axis,

known as the y-intercept.

The y-intercept is the value of the dependent variable when the independent variable has a value

of 0. When we are working with mathematical models, the y-intercept represents a real value, and

should be written with its units.

For the above examples, the y-intercepts are 0 dollars, 15 mm of tread and $460.

The other important aspect of a linear function is how quickly the dependent variable changes. The

total rent increases by $120 per week. For each 8000 km driven, the tread wear is 1 mm, so the

tread wear is 1

8000 mm per kilometre driven. The variable cost of manufacturing ID cards is $0.25

per card.

The gradient of a linear function represents the rate of change of the dependent variable versus the

independent variable. In life-related problems, gradients have units.

LINEAR FUNCTIONS E

THE INTERPRETATION OF THE Y-INTERCEPT OF A LINEAR FUNCTION

THE INTERPRETATION OF THE GRADIENT OF A LINEAR FUNCTION

Consider question parts , and in Renting for zero weeks costs zero dollars,the amount of tread initially (with zero kilometres driven) is mm and the fixed cost of producingID cards (the cost if no cards are purchased by the students) is

Exercise 1B.3.

$ .15

460

1 a b c




For the rent, the unit is dollars/week, for the tread wear it is (mm of tread lost) / (kilometre driven),

and for the ID cards it is $/card. Note that the unit of a gradient is a rate.

1 For each of the situations below:

i What is the independent variable? ii What is the dependent variable?

iii Find the y -intercept. iv What are the units of the dependent variable?

v Find the gradient. vi What are the units of the gradient?

a For each millilitre of ink, a pen will draw lines and curves for a total length of 300 metres.

b Sam can type at a rate of 32 words per minute.

For each of the situations below:

i find the y-intercept ii find the unit of the dependent variable iii find the gradient. Include the units.

a

b

a i

ii

iii gradient = y2 ¡ y1

x2 ¡ x1

= 300 ¡ 30

3 ¡ 0

= 270

3= 90 km/h

EXAMPLE 1.8

100

200

300

1 2 3

y

x

d i s t a

n c e

( k m

)

hours

Travelling west

A a a

.

gardener is spraying his rose bushes with pesticide. He is spraying at constant rate.He begins the job with litres of pesticide in his spray pack and completes the task in

minutes. He notes that there are litres remaining in the spray pack.10

25 2 5

If we take noon to be our

starting time, that iswhen then the

-intercept is

t

y

= 0,

.30

The unit of the dependentvariable is kilometre.

b i y-intercept is 10.

ii The unit of the dependent

variable is litre.

iii gradient = 2:5 ¡ 10

25 ¡ 0

= ¡7:5

25

= ¡0:3 litres/min

2

4

6

8

10

5 10 15 20 25

l i t r e s

o f

p e s

t i c i d

e

minutes

x

y

Rate of spray

A a a TTT

person in car travelling west at constant speed from oowoomba observes that sheis km from oowoomba at noon. At pm she observes another sign showing that sheis km from oowoomba.

30300

3

EXERCISE 1E




c Telstra will connect your rural property with a phone line for $800 plus $1500 per kilo-

metre of telephone cable.

d A salesman receives a salary of $500 per week plus commission of 10% on his weekly

sales.

e A rule of thumb for converting degrees Celsius to degrees Fahrenheit is to double the

Celsius temperature and add 30 degrees.

One form of a linear equation is y = mx + c, where y is the dependent variable, x is the

independent variable, m is the gradient and c is the y-intercept.

To show this, consider a line passing through the point (0, c) with gradient m. Let point P with

coordinates (x, y) be any other point on this line.

mAP = m = y ¡ c

x ¡ 0

) m = y ¡ c

x

) mx = y ¡ c

y = mx + c

This is called the gradient-intercept form of a straight line.

If a line has gradient m and passes through the point with coordinates (x1, y1), its equation

is given by y ¡ y1 = m(x ¡ x1)

This is called the point-gradient form of a straight line.

You will be asked to derive this formula, as an exercise.

a Find the gradient and y-intercept of i y = 2x ¡ 3 ii 2x ¡ 3y + 6 = 0 b Find the equation of the straight line that passes through (¡2, 3) with gradient

m = ¡34 .

c Find the equation of the straight line that passes through the points (¡3, 2) and

(2, 12).

d Find the equation of the line that passes through (¡3, ¡2) and is perpendicular

to the line with equation 2x + y ¡ 6 = 0

a i The equation y = 2x ¡ 3 is in the form y = mx + c.

Therefore m = 2 and c =¡

3 by inspection.

EQUATIONS OF LINEAR FUNCTIONS F

(0' ^)

(!' @)

@= !+^m y

x

EXAMPLE 1.9

and rearranging gives

f A printer takes 15 seconds to warm up, and then prints 10 pages per minute after that.

g Marcus has a mobile phone. He pays $5 per month and then pays 80 cents a minute for

every minute for his calls.




ii Re-write this equation so it is in the form y = mx + c.

2x ¡ 3y + 6 = 0

) ¡3y = ¡2x ¡ 6 fsubtract 2x and 6 from both sidesg) y = 2

3 x + 2 fdivide both sides by ¡3g

) m = 2

3 and c = 2.

b Substitute y = 3, x = ¡2 and m = ¡34 into y = mx + c and solve for c.

y = mx + c

3 = ¡34 (¡2) + c fsubstituteg

3 = 32 + c fmultiply and simplifyg

c = 32 fsubtract 3

2 from both sidesgTherefore the equation is y = ¡3

4 x + 32 .

Note that some students prefer to use y¡

y1 = m(x¡

x1).

c As the gradient is not given, we must first calculate the gradient.

m = y2 ¡ y1

x2 ¡ x1

= 12 ¡ 2

2 ¡ (¡3)

= 105

= 2

Now use this gradient and either point to find the equation. Here we will use the point-

gradient form of a straight line equation. Students often ask which point to use. The

answer is that it does not matter! They both give the correct equation.

Can you explain why?

Using the point (¡3, 2)

y ¡ y1 = m(x ¡ x1)

y ¡ 2 = 2(x ¡ ¡3)

y ¡ 2 = 2x + 6

Rearranging: 2x ¡ y + 8 = 0

Using the point (2, 12)

y ¡ y1 = m(x ¡ x1)

y ¡ 12 = 2(x ¡ 2)

y ¡ 12 = 2x ¡ 4

Rearranging: 2x ¡ y + 8 = 0

d 2x + y ¡ 6 = 0 can be rearranged to y = ¡2x + 6. Therefore, m = ¡2,so the gradient of the line perpendicular to this one is the negative reciprocal of ¡2,

which is 12 .

Using the gradient-intercept form: y = mx + c

¡2 = 12 (¡3) + c

¡2 = ¡32 + c

c = ¡12

Therefore the equation is y = 12 x ¡ 1

2 :




1 For each equation below, write in the form y = mx + c. Give the gradient and y-intercept

of each.

a y = 3x + 4 b y = 2 ¡ 5x c y = 12 x

d 4y = 4x

¡5 e y =

¡3 f 2x

¡3y = 0

g 3x + y ¡ 4 = 0 h x ¡ 2y + 1 = 0 i 2x + y ¡ 8 = 0

j x = 0 k 3y ¡ 5x = 23 l y = 1200x ¡ 4500

m y ¡ 2

x = 3 n

3y + 3

x ¡ 4 = ¡3 o

2x ¡ y + 3

x + 3y ¡ 2 = ¡3

p the x-axis

2 If possible, graph the equations in question 1 using your graphics calculator. Choose an

appropriate viewing window by entering the domain and range. The best view is one which

shows the main features of the graph while using as much of the screen as possible.

3 Determine the equations of these straight lines.

gradient passing through gradient passing through

a 2 (¡1, 3) f undefined (¡2, 3)

b ¡1 (2, 4) g a (a, 0)

c 23 (¡2, ¡5) h ¡3 (¡3, 1)

d 12 the origin i 2 (2, 1

2 )

e 0 (¡2, 4) j 13 (¡1

2 , 23 )

4 Determine the equations of the straight lines passing through these points.

a (2, 3) and (5, 6) b (6, 7) and (4, 3) c (1, 3) and (¡1, ¡1)

d (¡2, 1) and (¡3, 4) e (¡6, 2) and (2, 3) f (4, ¡2) and (¡2, 5)

5 Find the equation of the straight line parallel to y = ¡3x + 1 that passes through (1, ¡3).

Graph both lines with your graphics calculator.

6 Find the equation of the straight line that passes through (0, 3) and is perpendicular to

y = 3 ¡ 13 x. Graph both lines.

SPECIAL CASES OF LINEAR EQUATIONS

If the straight line passes through the origin,

the y-intercept is zero, i.e., c = 0. So, the

equation of a line that passes through the origin

is y = mx.

Any linear function with an x-term, a y-term but

no constant term passes through the origin.

Now consider the horizontal line (parallel to the

x-axis) that passes through (0, 2). The gradient

is zero, i.e. m = 0 and the y-intercept is 2.

EXERCISE 1F.1

(0' ^)

( ' 0)k

( ' )k

!=k

@=c

@= !m

y

x




The equation then becomes:

y = 0x + 2, which simplifies to y = 2

As a wily old Frenchman once said, it is usually more instructive to do the same problem two

different ways than to do two different problems. So, let us find the same equation in a different

way.

Consider the graph of the horizontal line that passes through (¡3, 2), (¡2, 2), (¡1, 2), (0, 2),

(1, 2)and(2, 2). Note that every point on this line has a y-coordinate of 2. The best way to describe

this line is y = 2.

In general,

the equation of the horizontal line that passes through the point (k, c) has the equation y = c.

Finally consider the vertical line passing through the points (3, 4) and (3, 2). We will try to find

the gradient of this line by substituting the points (3, 4) and (3, 2) into the equation:

m = y2 ¡ y1

x2 ¡ x1=

4 ¡ 2

3 ¡ 3 =

2

0

Since dividing by 0 is not possible, the value of the gradient is undefined. All is not lost however.

Note that every point on this line has an x-coordinate of 3. The best way to describe the equation

of this line is x = 3: It follows that

the equation of the vertical line that passes through the point (k, c) has the equation x = k.

Note that the gradient of a vertical line is undefined. This means that it does not have a valuewhich is a real number.

All of the above equations can be expressed by a single equation, called the general form of a

linear equation ax + by + c = 0.

Find the equation of the line that passes through

a (3, ¡2) and (¡5, ¡2) b (2, ¡1) and (2, 3)

A quick plot of the points shows both equations can be quickly found.

a Since the y-coordinate of both points is ¡2, this is a horizontal line with equation

y = ¡2.

b Since the x-coordinate of both points is 2, this is a vertical line with equation x = 2.

EXAMPLE 1.10

GENERAL FORM OF A LINEAR EQUATION




a Write 2x ¡ 3y + 1 = 0 in the form y = mx + c.

b Write y = 23 (x ¡ 1) in the general form.

c Write ax + by + c = 0 in gradient-intercept form.

a 2x ¡ 3y + 1 = 0 fadd 3y to both sidesg) 2x + 1 = 3y fre-arrangeg

) 3y = 2x + 1 fdivide both sides by 3g) y = 2

3 x + 13

b y = 23 (x ¡ 1)

) y = 23 x ¡ 2

3 fexpandg) 3y = 2x ¡ 2 fmultiply both sides by 3g

2x ¡ 3y ¡ 2 = 0 fre-arrangeg c ax + by + c = 0

) by = ¡ax ¡ c

) y = ¡a

bx ¡ c

b

So, the gradient of the line with general equation is ¡a

b and the y-intercept is ¡c

b.

Summary:

You must know these different forms of a straight line equation.

² y = mx + c gradient-intercept form

² y ¡ y1 = m(x ¡ x1) point-gradient form

² y = mx line through the origin

² y = c horizontal line through (k, c)

² x = k vertical line through (k, c)

² ax + by + c = 0 general form of a linear equation

1 Write down the equations of the lines determined by the following information. Express your

answer in general form.

gradient y-intercept gradient y-intercept

a 2 3 e 34 1

b ¡1 2 f 12

34

c 3 0 g undefined 0

d 0 2 h a b

EXAMPLE 1.11

EXERCISE 1F.2



INVESTIGATION 1WHAT POINT SIZE ARE YOU?


2 Determine the equation of the straight line in general form passing through

a (3, 8) and (6, 8) b (¡1, ¡3) and (¡1, 11) c (2, ¡5) and (0, ¡8)

d ( 12 , 3

2 ) and ( 72 , ¡5

2 ) e (a, b) and (¡a, ¡b) f (2f , 2g) and (f , g )

3 Find the equations of the following lines.

a the x-axis

b the y-axis c the line parallel to the x-axis, passing through (¡2, 3)

d the line parallel to the y-axis, passing through (¡2, 3)

e the line parallel to the x-axis, passing through (c, d)

f the line parallel to the y-axis, passing through (c, d)

4 Find the equation of the line in general form passing through the point (6, 1) which is

a parallel to the line y = 2x + 3

b perpendicular to the line y = 2x + 3

5 Find the equation in general form of the line that passes through (1, 3) and is

a parallel to the line 2x + 3y ¡ 6 = 0 b perpendicular to the line 2x + 3y ¡ 6 = 0

6 A, B, and C are the points (3, 4), (¡5, 0) and (¡1, 12) respectively. Find the equations of AB

and AC and show that AB is perpendicular to AC.

7 Find the equations of the sides of the triangle with vertices A(3, 6), B(¡2, 4) and C(¡1, ¡3).

8 It costs the XYZ Company $3000 to produce 20 gadgets and $5000 to produce 60 gadgets.

a Determine the linear cost equation.

b Determine the fixed costs.

c Find the cost of producing 100 gadgets.

9 If a widget is priced at $5, there will be a demand for 75 widgets. If the widgets are priced at$10, there will be a decrease in demand to 50 widgets. Find

a the linear demand function

b the demand if the widgets are priced at $2.50 .

c At what price will the demand for the widgets decrease to zero?

10 On page 28, we derived the gradient-intercept form of the equation of a straight line. Using

this as a guide, derive the point-gradient form of the equation of a straight line.

1 Using a word processor, print out the letter “T”, in various

point sizes. Measure the heights of the letters. Record the

point size and the height both in a table and on a graph.

Which is the independent variable?

Point size is measure of the size of text. If the letter “T” was as tall as you, what point size would it be?

a

What to do:




2 You should find that your points almost lie in a straight line, but do not lie exactly in a

straight line. Give a reason why this might be.

3 Draw the straight line that best fits the data. A good method is to draw a line such that

as many points as possible fall on the line, and half of the remaining points are above the

line and half are below.

4 Determine the equation of this line. 5 Use the equation to determine the point size of a letter “T” that is as tall as you.

6 Compare your equation to that of your classmates. Are they the same? Explain.

7 On page 28, we derived the gradient-intercept form of a straight line.

Using this as a guide, derive the point-gradient form of a straight line.

Again it is time to take a step back, this time to take a look at the Communication and Justificationcriterion.

Communication in a mathematical context has two facets - knowing your audience and expressing

your mathematics in a way that is appropriate for that audience. In this course, good communication

includes:

² proper setting out of your solutions to exercises

The examples in this textbook and the worked solutions your teacher does in class

are good indicators of the setting out that is needed.

² logical, clear and concise solutions to problems, investigations and modelling activities

Once you have figured out the mathematics, you should expect to have to write a

number of drafts before you are satisfied with how you have communicated your

solution. Think of your audience as being a Year 11 Mathematics B student from

another school who has not previously seen the question you are trying to answer.

Would he understand your solution? Can you make it clearer by including a diagram

or a worked example? Find such a student and ask him to read what you have written.

Did he understand it?

² on assignments, a word-processed document with a title page containing the required information, headers and footers, and no spelling or grammatical errors

A high quality of presentation, while not a mathematical concept, is crucial to

communicating mathematics effectively.

1

COMMUNICATION AND JUSTIFICATIONG

COMMUNICATION

EXERCISE 1G.1

Re-read your solution to the Patterns problem posed on page Now assume that your audience is ear student who has just learned about linear functions. Re-write your solution so you have communicated your solution to this student as effectively as you are able.

19.a Y 10




JUSTIFICATION AND CONJECTURES

A conjecture is a statement that you believe to be true.

Justifying conjectures is an important part of communicating your mathematics to others. Once you

have developed a conjecture, you need to convince yourself that it is correct. After that, you should

try to convince a friend or others in your group. Often by explaining your ideas to others, you find

the parts where you are unsure or your thinking is unclear. Others can help with understanding and

finding just the right words. Sometimes friends will tell you places where they do not understand

your reasoning. In this part of the justification process, you may need to refine your conjecture or

your explanation.

When the “convince a friend” stage is complete, you will be ready to try to convince a sceptic. A

sceptic is someone who will not take what you are saying on trust. They will expect to be convinced

at every stage. Sceptics can be others in your class, your teacher or (if you are a mathematician)

the editorial panel of a mathematics journal.

Proofs are a specialised form of communication used by mathematicians. There are some formal

proofs in this text and more can be found on the accompanying CD. However, when you readthese, it is important to realise that before mathematicians were able to write the proofs in the

concise way they are presented, the same process of convincing themselves, a colleague and finally

a sceptic would have taken place with successive refinements of the language - both English and

mathematics.

1 In the Patterns problem, you made a conjecture about the rule for finding the number of

matchsticks in the nth pattern. Justify (or prove) that your rule is correct.

INVESTIGATION 2

EVEN FUNCTIONS

An even function is defined as one for which h(¡x) = h(x) for all x.

In this task, you will investigate even functions. Here are some ideas to help you start.

What to do:

1 Show that the following are even functions:

a f (x) = x

2 b g(x) = x

4 c h(x) = x

2

¡ 1

x2

2 Show that a + bx2 + cx4 is an even function. Does it matter what values you

choose for a, b and c?

3 What is special about even functions? What do they look like? Try graphing the

functions in 1 to look for patterns. Consider the symmetry of the graphs.

4 Make some conjectures about even functions.

5 Justify your conjectures.

Write a report on your investigation. In this report, pay special attention to how you

communicate your ideas.

EXERCISE 1G.2

Convincing sceptic is very challenging. As you work through this course, you will have the oppor-tunity to justify many conjectures. Justification is part of the development of

a.mathematical proof





Function name Equation Some facts

quadratic y = x2 ¡ 3x + 10 after linear functions, the most common

function

cubic y = x3 ¡ 2x2 ¡ x ¡ 1 a cubic is a special case of a class of

functions called polynomials

polynomial y = 0:1x(x ¡ 4)(x ¡ 1)(x + 3) this polynomial is called a 4th degree

polynomial

periodic y = 4sin(2x) note the repetitive nature of the graph

of this function

exponential y = 2x exponential functions are very common

in biology

logarithmic y = log(2x) look for the log button on your calculator

By hand, using pen and paper

Three methods are commonly used to graph a linear function by hand if you know the equation:

Method 1: Constructing a table, and plotting points

Method 2: Using the gradient and y-intercept

Method 3: Using the x-intercept and y-intercept

Consider the linear function 3x + y ¡ 6 = 0.

Method 1:

Rearranging gives y = ¡3x + 6.

Construct a table x 0 2 3¡3x + 6 6 0 ¡3

Plot points and draw the line passing through the points.

Note that only two points are needed to plot the graph of

a linear function, but we generally plot three points, as

this allows us to pick up any errors.

Here are examples of some of the functions that you will be studying in thiscourse. Use your graphics calculator to sketch the graph of each function. oumay need to change the viewing window so you can see the important features of each function.

For each type of function, make one or more conjectures about its graph.

Y

GRAPHING LINEAR FUNCTIONS H

y

x

3

3




Method 2:

Rearranging gives y = ¡3x + 6 ) m = ¡3, c = 6.

On a Cartesian plane first plot the y-intercept at (0, 6).

Now write the gradient as a fraction, i.e., rise

run =

¡3

1

For every 1 unit horizontally, the line must fall 3 units

vertically.

In this case the graph passes through (0, 6), (1, 3), (2, 0),

etc. Draw a line through these points.

Method 3:

Two points are enough information to draw the straight line.

The x and y intercepts can be easily determined for the graph

of 3x + y ¡ 6 = 0.

For the x-intercept, y = 0, so 3x ¡ 6 = 0x = 2

For the y-intercept, x = 0, so y ¡ 6 = 0y = 6

On a Cartesian plane, plot the x-intercept at (2, 0) and the

y-intercept at (6, 0) and draw a line passing through them.

Using a graphics calculator

Before a graphics calculator can be used, the equation must

be expressed in the form y = mx + c. In this case plot theequation y = ¡3x + 6:

1 Sketch these graphs by constructing a table, and plotting points. Check using a graphics

calculator.a y = 3x + 1 b y = 5 ¡ 3x c y = 100x + 60

d y = 5000000x ¡ 2000000 e y = 0:03x + 1 f y = 0

2 Sketch these graphs using the y-intercept and gradient.

a y = x ¡ 3 b y = 12 x ¡ 1 c y = 1000 ¡ 2x

d y = 40000 ¡ 5000x e y = 0:006x + 0:002 f y = 0:007x + 0:002

3 Sketch these graphs using the two-intercept method.

a 2x + 3y = 12 b x ¡ 3y = ¡6 c x ¡ 5y = 3

d 2x + y

¡8 = 0 e 1

2 x

¡2y =

¡1 f 1

3 x

¡ 14 y = 4

y

x

3

3

EXERCISE 1H

y

x

3

3




The constant of proportionality k = y

x

But, when x = 5, y = 2:75, ) k = 2:75

5 = 0:55

and so y = 0

:55

x

If x = 12, y= 0:55 £ 12

= 6:6 Therefore, 12 stamps cost $6:60.

If y = 9:9 then 9:9 = 0:55 £ x

) x = 9:9 ¥ 0:55

) x = 18 Therefore $9:90 buys 18 stamps.

The completed table is Number of stamps 5 12 18

Cost $2:75 $6:60 $9:90

1 Each of these situations below is an example of direct variation. For each:

i What is the independent variable?

ii What is the dependent variable?

iii Is the independent variable discrete or continuous?

iv Find the constant of proportionality.

v Write the relation between the variables in the form “/

”.

vi Write the equivalent linear equation.

a James sells roses at the market. Each rose costs $1:50. Find his earnings if he sells n

flowers.

b Sara uses railway sleepers to edge her garden. Each sleeper is 1:8 metres long. Find the

total length of garden edging if Sara has n sleepers.

c Each millilitre of brandy contains 0:17 millilitres of alcohol. Find the amount of alcohol

in a drink that contains n millilitres of brandy.

2 The circumference of a circle varies directly as the diameter.

a Write the relation between the variables in the form “

/”.

b What is the constant of proportionality? c Write the equation in the form y = kx:

3 In the following, does one variable vary directly as the other?

ax 2 4 6

y 4 16 36

bx 1 5 9

y 3 5 7

cx 2 4 8

y 6 12 24

d x 0:01 0:04 0:05

y 0:06 0:24 0:30

ex 1

434

32

y 1 3 5

f x 0:3 1:2 1:5

y 2¼ 8¼ 10¼

EXERCISE 1I.1




4 A spring is hung from a hook. The amount

the spring stretches is directly proportional

to the weight attached to the spring. Com-

plete the table.

Weight (g) 600 950

Stretch (mm) 108 126

5 The resistance R, of a copper wire, varies directly as its length L.

a Write this as an equation, using k as the constant of proportionality.

b If R = 516 when L = 3, find the value of k .

6 In defining direct variation, we wrote:

“If a variable y is proportional to x, we write y = kx, k 6= 0.”

Why did we add the restriction, k 6= 0?

7 As well as answering true or false, justify your answer for eacha True or False: the diagonal of a square is proportional to its side length.

b True or False: the area of a square is proportional to its side length.

c True or False: the volume of a cube is proportional to the length of its edge.

d True or False: the perimeter of any regular polygon varies directly as the length of the

side.

e True or False: the volume of a cylinder varies directly as its height.

f True or False: the volume of a cylinder varies directly as its radius.

8 Give a real life example of direct variation where k is negative.

What is the physical interpretation of k being negative?

1 The cost of hiring a car from Rent-a-Total-Wreck

is $20 a day plus 25c per kilometre. I decide to

rent a car for just one day, to drive around Central

Queensland and see the sights. I will drive for no

more than 6 hours at an average speed of 80 km

per hour. I am interested in the cost of renting the

car based upon the distance driven.a Are the variables ‘cost ’ and ‘distance driven’

discrete or continuous variables?

Explain your answer.

b What is the independent variable? The dependent variable?

c Draw up a table showing the cost of renting a car for a day, based on how far I drive.

d What is the domain?

e What is the range?

f Draw a graph from this table.

g What is the value of the y-intercept?

h What does the y-intercept represent in this application?

EXERCISE 1I.2

stretch

question below.




i Calculate the gradient. What are the units of the gradient?

j What does the gradient represent in this application?

k Find the equation that relates ‘distance driven’ and ‘cost ’.

2 The given graph indicates the total daily cost,

in dollars, to a factory manufacturing scien-

tific instruments. With the present facilitiesand staffing, the factory has a maximum ca-

pability of 120 instruments per day. Use the

graph to answer the following questions.

a Are the variables ‘total daily cost C ’ and

‘number of instruments manufactured I ’

discrete or continuous? Explain.

b Which is the independent variable?

The dependent variable? Explain your

choice.

c What are the domain and range of this

relation?d What is the value of the y-intercept and what does it represent in this application?

e By choosing points on the graph, calculate the gradient.

f What does the gradient represent in this application?

g Find the equation relating the total cost and number of instruments produced.

3 Assume that the relationship between ‘oven temperature’ and the ‘time it takes to heat an oven

to a given temperature’ is a linear relationship. It takes 7 minutes and 20 seconds to preheat

to 180 degrees Celsius. It takes 10 minutes and 45 seconds to preheat to 250 degrees Celsius.

a What is the independent variable, and what is the dependent variable?

b Find a linear equation that relates these two variables.

c Hence find the initial temperature of the oven before preheating.

d In your opinion, how realistic is the assumption that the relationship is linear?

4 Since the beginning of March, a local reservoir has been losing water at a constant rate. The

Water Resources Department estimate that on 12th March, the reservoir held 200 million kL

of water, and on the 21st March it held 164 million kL.

Determine a linear equation relating Q, the volume of water remaining (in millions of kL),

and n, the number of days since the beginning of March (when the reservoir was full). Hence

find the maximum capacity of the reservoir.

5 Copy and complete this table for the function y = 1

30x(39

¡10x2 + x4).

x ¡3 ¡1 0 1 3y

Plot the points on a number plane. What do you think the graph of this function looks like?

Use a graphing calculator to check your answer.

6 A recipe book suggests that the time required to cook a chicken is 20 minutes plus 25 minutes

for each kilogram of chicken.

a Represent this rule as a table, graph, and equation.

b How long does it take to cook a chicken that weighs 1.8 kg?

c The time required to cook a chicken is 1 hour. What does it weigh?

1000

2000

30 60 90 120

cost ($) C

Instrument manufacturing costs

number of instruments

I



MODELLING

V

Number of items 5 15 25Cost 875 1025 1175


Solution to matchstick pattern problem

Pattern number 1 2 3 4 5 ...

Number of matchsticks 1 3 5 7 9

a 10th pattern needs 19 matchsticks; 1000th pattern needs 1999 matchsticks.

b Number of matchsticks = 2 £ Pattern number ¡ 1. c

d The pattern in c is Number of matchsticks = 1 + 2 £ (Pattern Number ¡ 1)

which simplifies to Number of matchsticks = 1 + 2 £ (Pattern Number ¡ 2)

Therefore Number of matchsticks = 2 £ Pattern Number ¡ 1

This agrees with our rule in b.

1 Your friend is considering connection to the internet. He needs to decide

between two monthly plans offered by the Internet Service Provider and has

asked you for advice. Here are the two plans:

Easy User Plan: $10 per month, plus $1.50 per hour

Heavy User Plan: $20 per month, plus $0:80 per hour.

Use mathematics to gain a deeper understanding of this situation. Develop a mathematical

model by working through the mathematical modelling steps. Use your model to help

you advise your friend.

2 After driving for 12000 km, Jenny found the tread depth on her tyres to be 14 mm. After

driving a further 11000 km, she checked her tread depth again and found it to be 10 mm.

Tyres are illegal once the tread depth reaches 4 mm. Pose a specific question concerning

this information. Develop a mathematical model to explore this situation further.

3 The table shows the cost involved in the

production of an item. Use this table to

determine the cost of producing 28 items.

Take particular care to consider assumptions you are making. Evaluate the suitability of

your model.

4 In Australia we use the Celsius temperature scale, with the freezing point of water being

0oC and the boiling point being 100oC. In the USA the Fahrenheit scale is used, with

the freezing point of water being 32oF and the boiling point being 212oF.

Given that the formula for converting from one scale to the other is a linear equation,

find the formula for converting degrees Celsius to degrees Fahrenheit.




1 If h(x) = x3, show that h(a) =

¡h(

¡a). Is this true for all functions,

or is specialx3 in some way? Write a report of your findings and justify your

2 Write the equations of five lines that pass through (2, 1).

3 A student wishes to plot the image of an arrow on graph paper, and has arrived at the

following intervals to plot.

y = ¡x + 5 1 · x · 4

y = 1 3 · x · 4

x = 4 1 · y · 3

By plotting the intervals, show how she needs to correct the domain and/or range to

make the arrow symmetrical.

4 Find the three functions needed to draw

this arrow. The grid squares are one unit

on each side.

Warning: Do not forget that a function

has both a rule and a domain.

5 A factory worker earns $20 per hour for the first 8 hours in a day, and time and a

half after that (i.e., his overtime wage is $30 per hour). No worker is allowed to work

more than 12 hours in a day.

The function for total daily earnings for a worker who works between 0 and 8 hours a

day is given by E (t) = 20t, where t is the time in hours.

Find the equation for total daily earnings for a worker who works between 8 and 12hours in a day.

6 A builder is constructing a series of horizontal tunnels

on a building site. He is using an unmanned drilling

machine which is pre-programmed with a series of

grid co-ordinates and linear equations. The grid laid

on top of the site plan is shown below.

The machine must start at the point (0, 2) and travel

along y = 2x + 2 to the point (1, y). It then turns

clockwise 90o and travels to the point (x, 1). Finally

it travels vertically to point B on the boundary.

What are the coordinates for point B?

x

y

answer to the problem posed.

(0' 2)

(1' @)

(!' 1)

B

x

y

Note: This diagram is not to scale.O

Challenge!

7 Find the equation of a line which is parallel to the line y = 12 x ¡ 3, and where the

distance between the two parallel lines is 5 units. Note that the distance between two

parallel lines is measured perpendicular to the lines.

PROBLEM SOLVING J

EXERCISE 1J



CHAPTER 1 REVISION SET


Cartesian plane gradient problem solving

continuous variable independent variable range

dependent variable investigation relation

discrete variable linear relation variabledomain mapping whole numbers

function mathematical model y-intercept

ordered pairs

44 INTRODUCTION TO FUNCTIONS (Chapter 1)44 INTRODUCTION TO FUNCTIONS (Chapter 1)

1 a Which of the following variables are discrete and which are continuous?

i the batting average of Steve Waugh

ii the number of participants in the Sydney Olympics iii the mean annual rainfall for Brisbane in millimetres

iv the altitude of a communications satellite in kilometres

b Fill in the blanks below:

i The domain is the set of values of the ...... variable.

ii The range is the set of values of the ...... variable.

2 a Find the equation of the straight lines through each of the following pairs of points.

i (0, 2) and (4, 3) ii (¡1, 3) and (2, 3)

b Determine whether the following sets of points are collinear or not.

i (0, 2), (12, 5) and (4, 3) ii (¡

1, 3), (5, 7) and (2, 3).

3 a Find the gradient of the line whose equation is 2x + 3y + 4 = 0:

b Find the gradient of the line which is perpendicular to the line whose equation is

3y = 6x ¡ 5.

4

5 a Show that the three straight lines have the same y-intercept.

i

y + 2

x ¡ 3 = 2 ii y = 2x ¡ 8 iii 3y + 6x + 24 = 0

b By evaluating the gradients and intercepts of these lines show that they do not intersect

in a common point.

i 2y = 3x + 6 ii y = 1:5x + 4 iii y = x ¡ 1

6 a Which of these equations has a graph which is a straight line?

i y = 2x ¡ 7 ii 1

x =

1

y iii y = x2 iv y =

1

x + 3

b A family of straight lines has equation y = 3x + c. When the graphs of these straight

lines are drawn for different values of c, what geometric property is observed?

You have $ in your money box and decide to save by putting $ per fortnight into the

box. Find an expression for the amount in dollars in your account after saving for

weeks. Is the independent variable discrete? What is the range of the independent variable if

you save for exactly two years?

100 302s n

http://worksheets.pdf/



CHAPTER 1 TEST (KNOWLEDGE AND PROCEDURES)


7 a You have to pay the local council a service fee of $25 for the water supply and 80 cents

per kilolitre of water used. Does the amount of water you pay for vary directly with the

amount used? Justify your answer.

b If you travel d kilometres for t hours at a constant speed v kilometres per hour, what

equation relates d, t and v? Do any of these variables vary directly with any of the other

two variables.

1 Find the gradient of each of the following lines:

a a line that passes through the origin and the point (2, ¡2)

b a line perpendicular to the line y = 1

2x + 3

2 Find the equation of each of the following lines:

a the vertical line that passes through the point (2, 3)

b the line that passes through the points (¡2, 3) and (4, ¡1)

3 Show that the points (3, 5), (5, 1), (¡1, ¡2) and (¡3, 2) are the vertices of a rectangle.

4 A bamboo shoot is replanted when it reaches a height of 20 centimetres. For the next

7 days it grows 1.5 centimetres per day.

a The two variables are height, and days since re-planting. Which is the

independent variable and which is the dependent variable?

b Represent this function:

i as a table ii as a mapping iii as a graph

c State the domain. State the range.d Is this function discrete or continuous? Briefly explain your answer.

e What is the value of the y-intercept?

f What does the y-intercept represent in this situation?

g Calculate the gradient.

h What are the units of the gradient, and hence what does the gradient represent

in this situation?

i Find the equation of this function.

5 Bill has $600, is saving nothing, and is spending $20 a week on his new girl friend,

Amber. His sister Sue only has only $100, but saves an additional $30 a week.When will Sue have twice as much money as Bill?




EXTENDED MODELLING ACTIVITY

V46 INTRODUCTION TO FUNCTIONS (Chapter 1)

1 Linear Models

You have learned how to model life-related

situations with linear models. In the figurealongside are the graphs of six linear mod-

els associated with six different applications.

Match each graph with its appropriate applica-

tion. Note that specific values of gradients and

y-intercepts are not needed for this activity.

a This graph models the number of hours

of life left in a photocopier after a given

number of hours of use.

b This graph models the cost of manufacturing a hose as a function of its length.

Assume the cost of fittings on the ends of the hose is fixed, and does notdepend on the hose length.

c This graph models the cost of producing a hose similar to that in b, except

this hose is of a higher quality than that used in b .

d This graph models the cost of producing a given length of hose of the same

quality as that in b , but without the fittings.

e This graph models the cost of a quantity of potatoes as a function of the weight

of the potatoes.

f This graph models the cost of a quantity of a less expensive variety of potatoes

as a function of the weight of the potatoes.

2 Space Junk

A problem confronting the National Aeronautics and Space Administration (NASA)

is the accumulation of “space junk” in orbit. The space junk is man-made, and

consists of discarded stages of rockets, assorted bits of rockets and satellites (for

example, from collisions with other space objects) and many small particles.

In 1990, rocket scientists estimated that a total of 1:8 million kilograms of space

junk was in orbit. They also estimated that the amount of junk added in 1991 would

be 0:8 million kilograms, and that would rise to 1:2 million kilograms of space junk

being added in 2000.

Assume that the increase in the amount of space junk added per year follows alinear model, i.e., the weight of the space junk goes up by a constant amount each

year.

a Find the linear model to predict the weight of space junk in the year t, if t = 0in 1990.

b Using this model, what will be the total weight of space junk in 2005?

c In what year does this model predict that there will be 5 million kilograms of

space junk?

F

A

B

C

D

E



CHAPTER 2Quadraticfunctions

SUBJECT MATTER

practical applications of quadratic functions

relationships between the graph of and thegraphs of and for both positive and negative values of the constant

the absolute value function

f xf x a f x a af x

a

( )( )+ ( + ) ( ),



HISTORICAL NOTE +

Farmer Brown wishes to construct a rectangular chook pen, using her barn for one of the sides. The other three

sides will be constructed out of 80 m of chicken wire.

(See the diagram)

She can make the pen long and thin, or short and fat, or

any rectangular shape in between, as long as she uses

exactly 80 m of chicken wire. She is keen to know

which rectangular shape will give her the maximum area.

We can create a mathematical model for this problem. In Chapter 1, we looked at the steps

involved in developing the model.

Step 1: Identify the situation.

What would be a well-defined question which would indicate exactly what Farmer

Brown wants to know?

Step 2: Simplify the situation.

List the key features and relationships of this situation. Mark those we will ignore.

Step 3: Build the model.

Let x represent the width of the chook pen. As there are two sides, each x metres long, and the

total length of chicken wire is 80 metres, then the length of the pen, L, is given by L = 80

¡2x.

Over years ago, conducteda series of experiments on the paths of

projectiles, attempting to find a mathe-

matical description of falling bodies.Two of Galileo’s experiments consisted of rolling a balldown a grooved ramp that was placed at a fixed heightabove the floor and inclined at a fixed angle to the hori-zontal. In one experiment the ball left the end of theramp and descended to the floor.

In a related experiment a horizontal shelf was placed atthe end of the ramp, and the ball would travel alongthis shelf before descending to the floor.

In each experiment Galileo altered the release height( ) of the ball and measured the distance ( ) the ball

travelled before landing.The units of measurement were called ‘punti’.

400 Galileo

h d

A MATHEMATICAL MODEL OF A CHOOK PEN A

barn

chook pen

x

80 2 x

48 QUADRATIC FUNCTIONS (Chapter 2)



The area of a rectangle is given by A = width £ length. For this problem

A = x(80 ¡ 2x) ........ (1)

Expanding gives A = 80x ¡ 2x2 ........ (2)

This is an example of a quadratic function.

Equation (1) is in factorised form while equation (2) is in expanded form.From the equation, we

can make a table: width (x) 0 5 10 15 20 25 30 35 40area (A) 0 350 600 750 800 750 600 350 0

Draw a graph, by plotting the points and joining them

with a smooth curve.

Step 4: Evaluate the model.

From either the graph or the table, note the following:

² The maximum area is 800 m2, and is obtained

by making the width of the pen, x = 20 m.

² Both the table and the graph are symmetric

about this value.

Substituting into the formula for length: L = 80 ¡ 2x

= 80 ¡ 2 £ 20

= 40

The pen should be constructed with a width of 20 m and a length of 40 m, to give a maximum

area of 800 m2.

This makes sense in terms of the original situation. The model could be revised to include other

features, for example, adding a door, having a middle dividing fence, and so on. But the model issufficient for our purposes for now.

The equation of a quadratic function is given by

y = ax2 + bx + c, where a 6= 0.

The graph of a quadratic function is called a parabola.

The point where the graph ‘turns’ is called the vertex.

If the graph opens upward, the y-coordinate of the vertex

is the minimum, while if the graph opens downward, the

y-coordinate of the vertex is the maximum.

The vertical line that passes through the vertex is called

the axis of symmetry.

All parabolas are symmetrical about the axis of symmetry.

The point where the graph crosses the y-axis is the

y-intercept.

The points where the graph crosses the x-axis (if they exist) should be called the x-intercepts, but

more commonly are called the zeros of the function.

REVIEW OF TERMINOLOGY

y

x

x

y

vertex

a x

i s o

f s

y m m e t r y

zero

y-intercept

parabola

minimum

QUADRATIC FUNCTIONS (Chapter 2) 49



SURDS B

INVESTIGATION 1

SURD OPERATIONS

Before we continue our study of quadratic functions, we need to revise surds, as they will occur

often in this chapter.

Surds are expressions of the formp

a, where a is a rational number that is not a perfect square.

For example,p

2 andq

34 are examples of surds. Note that

p 2 and

q 34 are both irrational

numbers.

An irrational number written in decimal form will never terminate and never repeat. Here are the

first 12 digits of p

2: p 2 ¼ 1.41421356237

It is much simpler to label the place on the number line asp

2 than give its decimal approximation.

OPERATIONS ON SURDS

In order to work with surds we need to understand their properties. Can we add, subtract, multiply

and divide surds? will try these operations on some surds and examine the results.We

Use a calculator to test which of the following are true:p

12 +p

3 =p

12 + 3 = ?p

15

p 12 ¡ p 3 = p 12 ¡ 3 = ?p 9p 12 £ p

3 =p

12 £ 3 = ?p

36p 12p 3

=

r 12

3 =

p 4 = ?2

The rules that we have demonstrated are:

For any non-negative real numbers a and b,p

a £ p b =

p ab

p ap b

=

r a

b

We have also shown that in general, p

a +p

b 6= p a + b and

p a ¡ p

b 6= p a ¡ b.

Here we have pr oven that in general we cannot add or subtract surds, at least by merely adding or

subtracting the numbers under the radical sign. A single counter-example is sufficient to prove

that we cannot add or subtract surds using this method.

From the above investigation it appears that we can multiply and divide surds by simply multiply-ing and dividing the numbers under the , but we get a wrong answer when we add or sub-tract the numbers under the radical. We have not that we can multiply and divide surds. Wehave only demonstrated that these operations hold for a particular group of numbers.

radical proven




Carry out the given operations.

ap

2 £ p 18 b

p 3 £ p

6 cp

6 £ p 6 d

p 2 £ p

5 £ p 10

ep

36 ¥ p 2 f

p 24p 6

g

p 5p

20

EXAMPLE 2.1

ap

2 £ p 18

=p

36

= 6 fSince 36 is a square number, we can simplifyp

36g

bp

3 £ p 6

=p

3 £ 6

=p

18 fp 18 is an irrational number, and hence cannot be written

without the radical sign.g

c p 6 £ p 6=

p 6 £ 6

=p

36

= 6 fIn generalp

a £ p a = a. This is a very useful fact.g

d p

2 £ p 5 £ p

10

=p

2 £ 5 £ 10

=p

100

= 10

fWe can easily multiply three surds, or more.g

e p 36 ¥ p 2=

p 36 ¥ 2

=p

18

f

p 24p 6

=

r 24

6

=p

4

= 2

fDivision of surds is best written as a fraction.g

g

p 5p

20

=

r 5

20

=

r 1

4

= 12 fYou need to know the square roots of common fractions.g




Use your calculator to confirm thatp

72, 2p

18, 3p

8 and 6p

2 are all the same number,

written in different form. Of these, 6p

2 is said to be in simplest form, as it has the smallest value

under the radical.

We can use the fact that p

a £ p b =

p a £ b to simplify surds.

For example, p 72 can be expressed as 6p 2 as follows:p 72

=p

36 £ 2

=p

36 £ p 2

= 6p

2

To simplify a surd, write the surd as the product of two factors, where the first factor is the largest

perfect square that is a factor of the number.

In this example, 72 can be factorised as 72 £ 1, 36 £ 2, 24 £ 3, 18 £ 4, 12 £ 6 and 9 £ 8.

The largest factor of 72 that is a perfect square is 36, so we express 72 as 36£

2. Then follow the

above steps to complete the process.

Simplify: ap

120 bp

144 cp

42 d 5p

48 e

r 49

64

a The factors of 120 are 120 £ 1, 60 £ 2, 40 £ 3, 30 £ 4, 24 £ 5, 20 £ 6, 15 £ 8 and

12 £ 10. The largest factor that is a perfect square is 4.

So,p

120

= p 4 £ p 30

= 2p

30

bp

144 is a perfect square, sop

144 = 12.

c 42 has no factors that are perfect squares (other than 1), sop

42 is in simplest form.

d The factors of 48 are 48 £ 1, 24 £ 2, 16 £ 3. We do not need to find any

more factors, as we have found that 16 is the largest factor of 48 that is a perfect

square.So, 5

p 48

= 5£

p 16

£3

= 5 £ p 16 £ p 3= 5 £ 4 £ p

3

= 20p

3

e

r 49

64

=

p 49p 64

= 78

SIMPLIFYING SURDS

EXAMPLE 2.2




Find all solutions to the equations: a x2 = 9 b x =p

9 c x = ¡p 9

a x2 = 9 has two solutions, x = 3 and x = ¡3. Do not forget to include the

second solution.

b x =p

9 has only one solution, x = 3. By convention, the root symbol is taken to

mean the positive root.

c The negative root is symbolised as x = ¡p 9 which simplifies to x = ¡3.

Note: The correct setting out for solving x2 = 9 is x2 = 9

) x = §p 9

) x = §3

Or you could write x = 3 or x = ¡3 in the last line.

1 Simplify

ap

28 bp

8 cp

54 d p

27

ep

32 f p

40 gp

45 hp

700

i p

98 j p

64 k 2p

18 l 4p

9

2 Simplify

ap

147 bp

81 c 2p

8 d 3p

54

e ¡5p 50 f 2p a2 g p f 6

h

4p

8c3 i 4p

128 j ¡3p

4 k

¡3p x2

l

4p 10000000

3 Simplify

aq

19 b

q 1625 c

q 814 d

p 25

ep

0:01 f p

0:04 gq

1 79 h

p 2:25

4 Simplify

ap

3£

p 7 b

p 3

£p

2 cp

6£

p 6

d 2p 6 £ 2p 6 e p 2 £ p 32 f 2p 3 £ 3p 2 g 2

p 27 £ 4

p 3 h 5

p 8 £ 2

p 7 i

p 18 £ p

2

j p

2 £ 3p

3 £ 2p

6 k 8p

2 £ 3p

5 £ p 10 l ¡2

p 6 £ 3

p 2 £ ¡2

p 3

5 Simplify

a

p 32p 8

b

p 32

4 c

p 8p 8

d 3

p 6p

10

e

p 50p 8

f

p 12p 2

g

p 48p 24

h

p 4ap a

EXAMPLE 2.3

EXERCISE 2B.1




i

p 3 £ p

15p 2 £ p

10 j

p a2

p a3

k

p 6ap 8a

l

p 81p 16

6 Find the value(s) of x in simplest form if

a x2 = 100 b x =p

100 c x = ¡p

100 d x = §p

100

e x2

= 1 f x2

= 0 g x2

= ¡4 h x = ¡p 47 a Show that (

p ab)2 = ab.

b Show that (p

ap

b)2 = ab.

c Use these results to show thatp

ab =p

ap

b.

While we cannot add or subtractp

12 +p

3 to getp

15 (such surds are called unlike surds),

we can add and subtract like surds such as p 3 and 2p 3 as follows:p 3 + 2

p 3 =

p 3 + (

p 3 +

p 3) = 3

p 3.

In general,

The expressionp

12 can be simplified as 2p

3, so we can addp

12 +p

3 if we simplifyp

12 first.p

12 +p

3

= 2p

3 +p

3

= 3p

3

One of the reasons why we simplify surds is to be able to add them.

Add or subtract these surds, if possible.

ap

6 +p

24 bp

8 ¡ p 50 c

p 12 +

p 18 d 5

p 20 + 4

p 45

ap

6 +p

24

=

p 6 + 2

p 6

= 3p

6

bp

8 ¡ p 50

= 2

p 2 ¡ 5

p 2

= ¡3p

2

cp

12 +p

18 fThese are not like surds, therefore they cannot be added.g= 2

p 3 + 3

p 2 fThe answer in simplest form is 2

p 3 + 3

p 2.g

d 5p

20 + 4p

45

= 5 £ 2p

5 + 4 £ 3p

5

= 10p

5 + 12p

5

= 22p

5

ADDING AND SUBTRACTING LIKE SURDS

ap

n + bp

n = (a + b)p

n and ap

n ¡ bp

n = (a ¡ b)p

n

EXAMPLE 2.4




A common use of surds is to simplify algebraic expressions that contain surds.

If a > 0, simplifya

p a2 b

p 9a2 c 3

p 4a + 5

p 9a d

3p 2a

2p

3a

a Since squaring and taking the square root are inverse operations, the above

expression simplifies to a.So,

p a2 = a

bp

9a2

=p

9 £ a2

=p

9p

a2

= 3a

An alternative approach isp

9a2

=p

(3a)2

= 3a

c 3p

4a + 5p

9a

= 3p

4 £ a + 5p

9 £ a

= 3p

4p

a + 5p

9p

a

= 6p

a + 15p

a

= 21p

a

d 3

p 2a

2p

3a

= 3p 2p a2p

3 1

1

p a

fDivide out the common factor of p a.g

= 3

p 2

2p

3

1 Simplify mentally

a 2p

3 + 5p

3 b 5p

6 + 4p

6 +p

6 c 2p

4 ¡ 3p

4

d 3p 2 ¡ 5p 2 e ¡4p 2 + 7p 2 f ¡5p 3 ¡ 2p 3 2 Simplify

ap

6 ¡ 3p

6 ¡ 7p

6 b ¡(p

5 + 2p

5) c ¡(3p

2 + 2p

2)

d ¡(¡p 7 ¡ 2

p 7) e 3

p 2 ¡ 2(

p 2 + 5

p 2) f

p 6 ¡ 2

p 6 + 3

p 6 ¡ 4

p 6

3 Simplify

ap

12 +p

12 b 4p

3 + 5p

3 c 5p

2 + 2p

2 ¡ 9p

2

d 4p

17 + 3p

17 ep

3 +p

48 f p

8 +p

50

gp

12¡

p 27 h 3

p 2 +

p 8 i

p 1 +

p 2

¡p

3

SURDS AND ALGEBRA

EXAMPLE 2.5

EXERCISE 2B.2




j 3p

12 ¡ p 3 k 4

p 27 + 2

p 12 l 5

p 24 ¡ 5

p 6 ¡ 2

p 54

4 Simplify

ap

5 +p

20 b 2p

72 + 3p

32 c 3p

a ¡ 2p

a

d ap

2 + ap

3 e 5p

18 ¡ 2p

32 f p

a2 + 3p

a2

gp

c3 + 4p

c h dp

e¡

dp

e2

5 Simplify

a 3p

a + 2p

a bp

4a cp

4b2

d p

81a2 ep

a4 f ¡p 16b

g

r m2

25 h

1

2

r c2

16 i

p abc £ p

abc

j p

4xy £ p 9xy k (¡p

a) £ (¡p a) l

p bp

bp

bp

b

m

p 4a

p a n

p a2

a o

p ab

p a

RATIONALISING THE DENOMINATOR

When studying trigonometry in your Year 10 mathematics class, you may have learned that

sin45o = 1p

2or you may have learned that sin 45o =

p 2

2 .

Confirm using your calculator that these expressions are equivalent.

It is useful to be able to change from one form to the other. The expression p 22

is said to have a

rational denominator, while 1p

2has an irrational denominator.

Changing from 1p

2to

p 2

2 is called rationalising the denominator.

This was an important skill in the pre-calculator era, since dividing by 2 was much easier than

dividing byp

2. Rationalising the denominator is still useful when working with exact values.

For example, multiply 1

p 2 by 1 where 1 is written as

p 2

p 2.

1p 2

= 1p

2£

p 2p 2

fas 1 =

p 2p 2g

= 1 £ p

2p 2 £ p

2

=

p 2

2




In general, to rationalise ap

b, multiply by

p bp b

as follows:

ap b

= ap

b£

p bp b

fas 1 =

p bp bg

= ap bp b £ p

b

= a

p b

b

Rationalise the denominator.

a 2p

5 b

3p 3

c 2

3p

3d

3p

2

2p

3

a 2p

5

= 2p

5£

p 5p 5

fWe multiply by 1 in the form

p 5p 5

.g

= 2 £ p

5p 5 £ p

5

= 2

p 5

5 fas

p a £ p

a = ag

b 3p

3

= 3p

3£

p 3p 3


p 3p 3

.g

= 3 £ p

3p 3 £ p

3

= 3

p 3

3 fas

p a £ p

a = ag

= p 3

c 2

3p

3

= 2

3p

3£

p 3p 3

fWe only need multiply by

p 3p 3

, not by 3

p 3

3p

3:g

= 2

p 3

3 £ 3 fas

p a £ p

a = ag

= 2

p 3

9

EXAMPLE 2.6




d 3

p 2

2p

3

= 3

p 2

2p

3£

p 3p 3


p 3p 3

.g

= 3

p 6

2 £ 3 fas p a £ p a = ag

=

p 6

2

1 Rationalise the denominator. Simplify where possible.

a 3p

6 b

5p 2

c 1p

2d

p 12p 3

e

p 7p 3

f

p 3p 6

g 6p

6 h

3p 3

i 1p

3 j

p 3p 4

k ¡6p

3 l

¡3p 15

2

a 2

p 5p

3 b

4p

6p 5

c 3

p 2p

3d

2

3p

3

e 2

p 18

3p 2 f 6

5p 2 g 1 +

p 2

p 6 h 3 + 2

p 5

p 10

i 3

p 12p 3

j ¡5

p 3p

27 k

6p

2p 2

l

p 6p

2p 8


a 1p

a b

2p c

c ap

a d

ap

b

bp

a

e ¡2bp

b f

3cp 3c

g

p 2ap 3a

h ¡b

ap

b

Recall that (a + b)(c + d)

= a(c + d) + b(c + d)

= ac + ad + bc + bd

(a + b)(a ¡ b)

= a2 ¡ ab + ab ¡ b2

= a2 ¡ b2

(a + b)2

= (a + b)(a + b)

= a2 + ab + ab + b2

= a2 + 2ab + b2

EXPANDING BINOMIALS

EXERCISE 2B.3

REVIEW OF QUADRATIC EXPRESSIONS C


Express with a rational denominator Simplify where possible..



Note: ² a2 ¡ b2 is called the difference of two squares

² (a + b)2 is called a perfect square

Expand a (2x + 2)(x ¡ 3) b (4x + 3)(4x ¡ 3) c (4x ¡ 3)2

a (2x + 2)(x ¡ 3)

= 2x(x ¡ 3) + 2(x ¡ 3) fexpand the first factor g= 2x2 ¡ 6x + 2x ¡ 6 fexpand each termg= 2x2 ¡ 4x ¡ 6 fcollect like termsg

b We can expand as above, and get the answer 16x2 ¡ 9.

But (4x + 3)(4x ¡ 3)

= (4x)2 ¡ 32 fdifference of two squares with a = 4x and b = 3g= 16x2 ¡ 9

This is an example of an identity, a statement that is true for all values of a and b. We

suggest you expand the left-hand side (LHS) of this identity, and verify that it is true.

c We can expand by first writing the expression as (4x ¡ 3)(4x ¡ 3) and expanding

as usual. But we are better treating this as the expansion of a perfect square.

So, we have:

(4x ¡ 3)2

= (4x)2 + 2(4x)(¡3) + (¡3)2 f perfect square with a = 4x and b = ¡3g

= 16x2

¡ 24x + 9

1 Expand and simplify

a (x + 1)(x + 2) b (y + 3)(y + 8) c (t + 3)(t ¡ 2)

d (r ¡ 3)(r + 6) e (c ¡ 2)(c + 5) f (x ¡ 4)(x ¡ 2)

g (a ¡ 3)(a ¡ 3) h (z + 5)(z ¡ 5) i (k + 1)(k + 1)

2 Expand and simplifya (x + 3)(x + 2) b (x + 1)(x ¡ 6) c (x ¡ 9)(x ¡ 11)

d (2x + 3)(x ¡ 2) e (3x ¡ 9)(2x ¡ 5) f (x + 9)(2x ¡ 3)

g (9 ¡ x)(3 ¡ 2x) h (x ¡ a)(x + a) i (2x ¡ 2a)(3x ¡ 4a)

j 2(3x ¡ 1)(3x ¡ 1) k (x + 5)2 l (x + 1)2(x ¡ 3)

3 Expand using (a + b)2 = a2 + 2ab + b2.

a (x + 2)2 b (3x ¡ 2)2 c (9 ¡ x)2 d (2a ¡ 3x)2

e (5x ¡ 7)2 f (3 ¡ 2x)2 g (x + 1)2 h (1 ¡ 5x)2

i (3y

¡2x)2 j (6x

¡ 12 )2 k (100x + 1)2 l (

¡3

¡4x)2

EXAMPLE 2.7

EXERCISE 2C.1




4 Expand using (a + b)(a ¡ b) = a2 ¡ b2.

a (x + 9)(x ¡ 9) b (x + 6)(x ¡ 6) c (9 ¡ x)(9 + x)

d (2x + a)(2x ¡ a) e (2x + 2a)(2x ¡ 2a) f 2(3x ¡ 1)(3x + 1)

g ( 12 x + 1)( 1

2 x ¡ 1) h (x + 1)(1 ¡ x) i (10x ¡ 3)(10x + 3)

5 Expand using the most efficient method available.

a (3x + 1)(3x ¡ 1) b (3r + 5)2 c (2x ¡ 4)(2x + 3)

d (1 ¡ t)2 e (r ¡ 13 )2 f (5 ¡ r)(3 + r)

g 3(x ¡ 5)(x + 5) h ¡(r + 3)2 i ¡(t + 4)(t ¡ 2)

j d(d + 1)(d ¡ 1) k (3x ¡ 4)(4x ¡ 3) l (t + 1)(t ¡ 1)(t2 ¡ 1)

One of your goals in Mathematics B should be to become efficient at doing basic algebra. You

should practise this technique until you can factorise expressions such as this by inspection (i.e.,

mentally). You will save time and make fewer mistakes if you learn to do this.

Factorise 2x2 + 7x + 3.

As factorising is the reverse process of expanding, the steps for factorising should be

those for expanding, but in reverse order. The solution is:

2x2 + 7x + 3

= 2x2 + 6x + x + 3 frewrite 7x as 6x + xg= 2x(x + 3) + 1(x + 3) ffactorise the terms, in pairsg= (x + 3)(2x + 1)

fthe common factor is (x + 3)

g

FACTORISING QUADRATIC EXPRESSIONS OF THE FORM x2 + bx + c

Factorise the expression x2

¡x

¡12.

Factors of ¡12 + OK? Factors of ¡12 + OK?

12 £ ¡1 11 £ ¡6 £ 2 ¡4 £

¡12

£1

¡11

£ 4

£ ¡3 1

£6 £ ¡2 4 £ ¡4 £ 3 ¡1 X

So, the two numbers are ¡4 and 3. Hence, x2 ¡ x ¡ 12 = (x ¡ 4)(x + 3).

EXAMPLE 2.8

If it can be factorised easily, the answer is of the form . How do we find thetwo numbers and that give the correct factorisation? Here is a foolproof method.

Ask yourself, “What two numbers multiply to give (the constant term) and add to give(the coefficient of )?”. If the answer is not obvious, list the factors of in pairs, and

test each pair.

( + )( + )

121 12

x m x nm n

x¡

¡ ¡

FACTORISING QUADRATIC EXPRESSIONS WHERE a6= 1

EXAMPLE 2.9




The tricky step here is re-writing 7x as 6x + x in the second line of the setting out. If we

had chosen 4x + 3x, for example, we would not have been able to complete the question.

You may wish to confirm this for yourself.

Here is a method of factorising quadratic expressions of the form ax2 + bx + c, if the

expression can be factorised nicely:

² Multiply the coefficient of x2 by the constant term. In the example above: 2 £ 3 = 6.

² Ask yourself, “What two numbers multiply to give 6 and add to give 7?”.

In general, for the quadratic expression ax2 + bx + c, the question is,

“What two numbers multiply to give a £ c and add to give b?”.

² Try to do the calculations mentally. If you cannot, list the factors of 6 in pairs and test

each pair. The factors that multiply to give 6 and add to give 7 are 6 and 1.

Rewrite 2x2 + 7x + 3 as 2x2 + 6x + 1x + 3 and proceed as shown above.

Factorise 4x2 ¡ 10x ¡ 24:

First take out the common factor of 2. 4x2 ¡ 10x ¡ 24 = 2(2x2 ¡ 5x ¡ 12).

Now factorise the quadratic factor. What multiplies to give ¡24 and adds to give ¡5?

By inspection, or by making a table, the numbers are 3 and ¡8.

2x2 ¡ 5x ¡ 12

= 2x2 + 3x ¡ 8x ¡ 12 frewrite ¡5x as 3x ¡ 8xg= x(2x + 3) ¡ 4(2x + 3) ffactorise the terms, in pairsg= (2x + 3)(x

¡4)

ftake out the common factor of (2x + 3)

gThe complete factorisation is 4x2 ¡ 10x ¡ 24 = 2(2x + 3)(x ¡ 4):

Note that it would be possible to factorise 2x2 + 3x ¡ 8x ¡ 12 as x(2x + 3) + 4(¡2x ¡ 3).

This is not wrong, but it is not useful. As there is no common factor, we cannot proceed to the

final step. The third line of setting out needs to contain a common factor, in this case (2x + 3), in

order to factorise the expression fully.

1 Factorise completely

a x2 + 5x + 6 b x2 + 7x + 10 c x2 + 8x + 15

d x2 + 11x + 18 e x2 + 2x ¡ 15 f x2 ¡ 9x ¡ 22

g x2 + 2x ¡ 24 h x2 ¡ x ¡ 12 i x2 + 4x ¡ 21

j x2 ¡ 6x + 9 k x2 ¡ 7x + 6 l x2 + 9x + 18

m x2 ¡ 4x + 4 n x2 ¡ 15x + 14 o x2 ¡ x ¡ 42

p x2 + 5x ¡ 36 q t2 ¡ 8t + 15 r t2 ¡ 10t + 24

s r2 ¡ 14r ¡ 32 t h2 + 12h + 36 u s2 ¡ 11s ¡ 60


a x

2

+ 2ax + a

2

b x

2

¡ 2ax + a

2

c x

2

+ 0x ¡ 4 d x

2

¡ 25

EXAMPLE 2.10

EXERCISE 2C.2





a 2x2 ¡ 9x ¡ 5 b 3x2 + 8x ¡ 3 c 4x2 + 4x + 1

d 3x2 ¡ 4x ¡ 15 e 2x2 ¡ 11x + 12 f 6x2 + 5x ¡ 4

g 10x2 + 13x ¡ 3 h 12x2 + 2x ¡ 2 i 12x2 ¡ 18x + 6

j 16x2 + 10x + 1 k 2x2 ¡ 7x + 6 l 21x2 ¡ 4x ¡ 1

m 2x

3

¡ 3x

2

¡ 2x n 2x

2

+ 2x ¡ 24 o 18x

2

+ 15x ¡ 12 p 12x3 ¡ 26x2 ¡ 10x q 6r2 ¡ 21r ¡ 45 r 12r2 ¡ 2r ¡ 2

1 Factorise completelya a2 ¡ b2 b x2 ¡ 16 c x2 ¡ 1

d x2 ¡ 16 e c2 ¡ 25 f y2 ¡ 1600

g x2 ¡ 100 h x2 ¡ 14 i 4x2 ¡ 25

j 9x2 ¡ 16 k 4x2 ¡ 16 l [(2w)2 ¡ 9]

m x2 ¡ 19 n 16x2 ¡ 16 o 4x2 ¡ c2

p 5x2 ¡ 7 q 3x2 ¡ 16 r 14 t2 ¡ 1


a 2a2

¡32 b a4

¡16 c (x + 2)2

¡9

d x4 ¡ y4 e (x ¡ 1)2 + 2(x ¡ 1) + 1 f 12a2 ¡ 75b2

g 1 ¡ x4 h x4 + 5x2 + 6 i 4a2 ¡ 9a + 2

j d2 + 8d + 12 k 12t3 ¡ 25t2 ¡ 30t l 2x ¡ 4


a 5y ¡ 10 b 16 ¡ 4t c 4t + 2r ¡ 6s

d x2 ¡ 121 e x2 ¡ 11x + 28 f t2 + 6t ¡ 27

g 3 p2 ¡ 12 p + 9 h 2c2 ¡ 2 i 36 ¡ 4x2

j x3 ¡ 25x k 3n2 ¡ 6n + 3 l a4 ¡ b2

m 2d2

+ 13d ¡ 7 n 1 ¡ 100x2

o ¡x2

¡ 5x ¡ 6

FACTORISING EXPRESSIONS OF THE FORM a2¡b2

Factorise 16x2 ¡ 36.

16x2

¡36

= 4(4x2 ¡ 9) fAlways look for a common factor first.g= 4((2x)2 ¡ 32) fWe note that the factor 4x2 ¡ 9 is of the form a2 ¡ b2

where a = 2x and b = 3.g= 4(2x + 3)(2x ¡ 3) fdifference of two squaresg

You can check this by expanding the right-hand side (RHS).

EXAMPLE 2.11

EXERCISE 2C.3




INVESTIGATION 2

A FASCINATING DISCOVERY

4 How many ways can you demonstrate that f (x) = 2x2 ¡ 7x ¡ 15

x ¡ 5 where x 6= 5,

and g(x) = 2x + 3 where x 6= 5, are different forms of the same function?

5 a Expand and simplify i (a + b)(a2 ¡ ab + b2) ii (a ¡ b)(a2 + ab + b2)

b Use the results in a to factorise completely

i x3 + 27 ii x3 + 64 iii 8x3 + 1

iv 2x3 ¡ 16 v a6 ¡ b6

FINDING THE VALUES OF A QUADRATIC FUNCTION

As you learned in Chapter 1, the value of a function is found by substituting and evaluating.

Find the value of the function f (x) = 2x2

¡5x

¡3 for

a x = 0 b x = 3 c x = ¡12

a f (0) = 2(0)2 ¡ 5(0) ¡ 3

= ¡3

b f (3) = 2(3)2 ¡ 5(3) ¡ 3

= 0

c f (¡12 ) = 2(¡1

2 )2 ¡ 5(¡12 ) ¡ 3

= 0

Note: If x = 0, the value of the function is just the constant term, which is also the

y-intercept on the graph.

For both 3 and ¡12 , the value of the function is 0.

EXAMPLE 2.12

This email from a student was sent to a maths education mailing list:

Recently, my Maths teacher and I found an interesting thing. Take the equation

of , a standard, vertical parabola. Now, choose a point on the rightside, where the values are greater than zero. Then choose a point on the leftside. Now draw a line joining these two points and find its -intercept. The -intercept value is the negative of the product of the original two values youpicked previously! This works with any two points along the parabola. Myquestion is how exactly does this work?

f x x

y yx

( ) = 2

x


a

b

c

Could you please help the student by answering his question?

There is also a neat relationship to be discovered about the gradient of the line that joins

the two given points. See if you can find it.

For the above relationship, is it necessary for the two points to be in different quadrants?



1 Given

a f (x) = x2 + 2x ¡ 3 find: i f (1) ii f (¡3) iii f (0) iv f (a)

b f (x) = x2 ¡ 3x ¡ 10 find: i f (5) ii f (¡2) iii f (0) iv f (a)

c f (x) = 2x2

¡11x

¡6 find: i f (6) ii f (

¡1

2) iii f (0) iv f (2a)

2 Does the given point lie on the graph of the function?

a (¡2, 1); f (x) = x2 ¡ 2x + 1 b (2, ¡10); f (x) = 6 ¡ 2x ¡ 3x2

3 Does the given point lie above, below or on the graph of the function?

a (2, ¡6); g(x) = x2 ¡ 3x ¡ 6 b (1, ¡3); g(x) = (x ¡ 3)(2x ¡ 3)

4 For each function, give the coordinates of five points that lie on that function. Check with

your graphics calculator.

a y = 2x2 ¡ x + 3 b y = 5 ¡ x ¡ 4x2 c y = ¡(x + 1)(x ¡ 3)

A major part of the Mathematics B course is solving equations.

There are four common methods of solving quadratic equations of the form ax2 + bx + c = 0:

² graphically ² factorising

² completing the square ² quadratic formula

Recall that

if ab = 0, then either a = 0 or b = 0, or both a = 0 and b = 0.

EXERCISE 2C.4

SOLVING FOR THE ZEROS OF A QUADRATIC FUNCTIONBY GRAPHING

Solve 2x2 ¡ 5x ¡ 3 = 0 by graphing the function.

SOLVING QUADRATIC EQUATIONS D

EXAMPLE 2.13

Using a graphics calculator, draw the graph of

y = 2x2 ¡ 5x ¡ 3.

To find where this function has a value of 0, we need to find

where it crosses the x-axis, since y = 0 on the x-axis. From

the graph, it appears the solutions are x = 3 and x = ¡12 .

Your graphics calculator will also have a “zeros” menu

choice. Use this to confirm the above solutions.

SOLVING FOR THE ZEROS OF A QUADRATIC FUNCTIONBY FACTORISING




Solve 2x2 ¡ 5x ¡ 3 = 0 by factorising.

(x ¡ 3)(2x + 1) = 0 ffactorise the LHS of the equationg

) x ¡ 3 = 0 or 2x + 1 = 0 fsince ab = 0 implies a = 0 or b = 0g) x = 3 or x = ¡ 1

2 fsolve these two equations by inspectiong

As a check on the algebra, you should substitute these values back into the equation.

If they are the zeros of the function, the value of the function will be 0.

Solve 4x2 ¡ 16 = 0 by factorising.

4x2 ¡ 16 = 0

) 4(x2 ¡ 4) = 0 falways factorise out the common factor firstg) 4(x + 2)(x ¡ 2) = 0 fusing a2 ¡ b2 = (a + b)(a ¡ b)g

) x + 2 = 0 or x ¡ 2 = 0 fif 4ab = 0 then a = 0 or b = 0g) x = ¡2 or x = 2 fsolve the equations by inspectiong

1 Solve by graphing

a x2 ¡ x ¡ 6 = 0 b 2x2 ¡ 5x + 2 c x2 + x ¡ 5 = 0

2 Solve by factorising

a x2 + 2x ¡ 3 = 0 b x2 ¡ 11x + 24 =

=

0

0

c x2 ¡ 4x ¡ 12 = 0

d x2 + 6x + 9 = 0 e x2 ¡ 8x + 16 = 0 f 2x2 + 7x ¡ 4 = 0

g 3x2 ¡ 8x ¡ 3 = 0 h 4x2 + 8x + 3 = 0 i 9x2 ¡ 12x + 4 = 0


a x2 ¡ 1 = 0 b x2 ¡ 25 = 0 c x2 ¡ 4 = 0

d x2 ¡ 100 = 0 e 2x2 ¡ 50 = 0 f 2x2 ¡ 18 = 0

g 4x2 ¡ 25 = 0 h 4x2 ¡ 64 = 0 i 81x2 ¡ 1 = 0


a x2 ¡ 2x ¡ 15 = 0 b x2 + 7x + 10 = 0 c 3x2 ¡ 5x ¡ 2 = 0

d x2 + 2x ¡ 24 = 0 e 2x2 + 10x + 12 = 0 f x2 + 2x ¡ 15 = 0

g x2 ¡ 36 = 0 h x2 ¡ 16 = 0 i 16x2 ¡ 16 = 0

j x2 + 2ax + a2 = 0 k 12x2 + 2x¡

2 = 0 l 18x2 + 15x¡

12 = 0

EXAMPLE 2.14

EXAMPLE 2.15

EXERCISE 2D.1




COMPLETING THE SQUARE

If you try to solve x2 + 2x ¡ 2 = 0 by factorising, you will need to find two numbers that

multiply to give 2 and add to give ¡2. You will not be able to to do it! There are not any integer

or rational solutions. The solution, if it exists, must be irrational.

Using a graphics calculator, the graph of the function y = x2 + 2x ¡ 2 shows there are two

roots, and they are approximately equal to ¡2:7 and 0.73.

Completing the square allows us to find the exact values for the roots. The method is based on the

identity: (x + a)2 = x2 + 2ax + a2.

What has to be added to x2 + 2x to complete the square, i.e., to make an expression that

can then be written in the form (x + a)2?

A geometric solution:

Let x represent an unknown length. Then x2 + 2x

represents the area of the rectangle.

Now I want to make this into a square. Suppose we cut

the x £ 2 rectangle in half, and re-arrange the pieces as

shown.

We need to add a piece in the lower right corner to

complete the square.

An algebraic solution:

Recall that a perfect square of the form (x + a)2 can be expanded to x2 + 2ax + a2.

Now x2 + 2x is of the form x2 + 2ax, where a = 1.

To complete the square we need to add a2, which in this case is 12 (which is 1).

The completed square is x2 + 2x + 1, which can then be written in factorised form as (x + 1)2.

Study the above example until you can see the link between the geometric approach and the

algebraic approach.

EXAMPLE 2.16

x

x

2

x2 2 x

1

x

x

1

x2 x

x

1

x

x

1

x2

x 1

x

The piece that has to be added is a square with dimensionsunit . The area of the large square is .1 1 = 1 ( +1)£ 2 2x




Solve x2 + 2x ¡ 2 = 0 by completing the square.

x2 + 2x ¡ 2 = 0

) x2 + 2x = 2 fTranspose the constant to the RHS, so the LHS

is of the form a2 + 2a.g) x2 + 2x + 1 = 2 + 1 fAdd 1 to both sides, to complete the square,

since 1 = (half of the coefficient on x)2.g

EXAMPLE 2.17

² a is half of the coefficient of x.

² To complete the square we add a2.

² The resulting square term is (x + a)2.

FINDING THE ZEROS OF A QUADRATIC FUNCTION BY COMPLETING THE SQUARE

To complete the square follow this pattern:

If solutions exist to quadratic equation, then the technique of completing the square will find them.a

EXAMPLE 2.18


Find the number that must be added to complete the square, then complete the square for

a x2 + 4x b x2 + 13 x c 2x2 + 5x

a x2

+ 4x is of the form x2

+ 2ax so 2a = 4 and ) a = 2hence a2 = 22 = 4

We add a2 to complete the square: x2 + 2ax + a2

= x2 + 4x + 4

= (x + 2)2

b x2 + 13 x is of the form x2 + 2ax so 2a = 1

3 and ) a = 16

hence a2 =¡

16

¢2= 1

36

We add a2 to complete the square: x2 + 2ax + a2

= x2 + 13 x + 1

36

= (x + 16 )2

c 2x2 + 5x = 2(x2 + 52 x) so 2a = 5

2 and ) a = 54

hence a2 =¡

54

¢2= 25

16

We add a2 to complete the square: 2(x2 + 2ax + a2)

= 2(x2 + 52 x + 25

16 )

= 2(x + 54 )2



) (x + 1)2 = 3 fWrite the LHS as a square.g) x + 1 = §p

3 fIf X 2 = A, then X = §p A.g

) x = ¡1 § p 3 fSubtract 1 from both sides.g

By convention we write the square root term last.

These are the exact solutions. If you substitute either

x = ¡1 +p

3 or x = ¡1 ¡p 3 into y = x2 + 2x ¡ 2

and evaluate the RHS, the value of the function will be 0.

Using a calculator, ¡1 +p

3 is approximately 0.732while ¡1 ¡ p

3 is approximately ¡2:732, which are the

approximate solutions we find by graphing as shown along-

side.

Solve 2x2 + x ¡ 4 = 0 by completing the square.

2x2 + x ¡ 4 = 0) 2x2 + x = 4 fTranspose the constant to the RHS.g) x2 + 1

2 x = 2 fWe can only complete the square if the coeffic-

ient of x2 is 1, so we divide both sides by 2.g) x2 + 1

2 x + ( 14 )2 = 2 + ( 1

4 )2 fThe LHS was of the form x2 + 2ax where

a = 1

4 . So we add (1

4 )2

to both sides.g) (x + 1

4 )2 = 2 + 116 fComplete the square on the LHS.g

) (x + 14 )2 = 33

16 fWrite RHS as an improper fraction.g

) x + 14 = §

q 3316 fIf P 2 = Q, then P = §p

Q.g

) x + 14 = §

p 334 fSimplify the surd using

r a

b =

p ap b

.g

) x = ¡14 §

p 334 fSubtract 1

4 from both sides.g

) x = ¡1 § p 334

fWrite as a single fraction.

These are the exact solutions.gSo, x = 1:186 or ¡1:686 fThese are the approximate solutions.g

1 Write the following as perfect squares.

a x2 + 2x + 1 b g2 + 6g + 9 c x2 ¡ 14x + 49

d a2 + a + 1

4

e m2

¡10m + 25 f x2 + 200x + 1000

EXAMPLE 2.19

y

x

y x x 2 2 2

EXERCISE 2D.2




2

a (x + 2)2 = x2 + 4x + :::::: b (x ¡ 3)2 = x2 ¡ 6x + ::::::

c (x ¡ 4)2 = x2 ¡ 8x + :::::: d x2 + 10x + :::::: = (x + 5)2

e x2 + 10x + :::::: = (x + ::::::)2 f x2 ¡ 6x + :::::: = (x + ::::::)2

g x2 + ::::::x + 36 = (x + ::::::)2 h x2 + 3x + :::::: = (x + ::::::)2

i x2 ¡ 5x + :::::: = (x ¡ ::::::)2

3 Copy and complete the following to make a perfect square:

a a2 + 6a + ::::: = (a + :::::)2 b x2 ¡ 10x + ::::: = (x ¡ :::::)2

c y2 + ::::: + 36 = (y + :::::)2 d x2 ¡ 18x + ::::: = (x ¡ :::::)2

e y2 + 5y + ::::: = (y + :::::)2 f a2 ¡ 7a + ::::: = (a ¡ :::::)2

4 Find the number that must be added to complete the square.

a x2 ¡ 2x b x2 + 6x c x2 ¡ 5x d x2 + x

e 2x2 + 10x f 3x2

¡6x g 2x2

¡x h 3x2

¡4x

i 4x2 ¡ 16x

5 Solve by completing the square.

a x2 ¡ 2x ¡ 2 = 0 b x2 ¡ 6x + 9 = 0 c x2 ¡ 7x + 10 = 0

d x2 ¡ 5x + 4 = 0 e 2x2 ¡ 5x ¡ 3 = 0 f 6x2 ¡ 11x + 3 = 0


a x2 ¡ 4x + 1 = 0 b x2 + 2x ¡ 5 = 0 c x2 + 6x + 5 = 0

d x2 + 3x ¡ 3 = 0 e x2 ¡ 5x + 1 = 0 f x2 ¡ x ¡ 3 = 0

g 2x2

¡3x

¡3 = 0 h 2x2

¡6x

¡8 = 0 i 3x2 + 14x

¡5 = 0


a x2 ¡ 6x ¡ 4 = 0 b x2 ¡ 2x ¡ 1 = 0 c x2 ¡ 8x + 10 = 0

d x2 ¡ 3x + 1 = 0 e x2 + 5x ¡ 3 = 0 f t2 ¡ t ¡ 5 = 0

g 2x2 ¡ 6x + 3 = 0 h 3x2 ¡ x ¡ 1 = 0 i x2 ¡ 12 x ¡ 2 = 0

8 The graph of f (x) = x2 +2x +3 is along-

side. It shows that the function f (x) has no

zeros. What is the result when you complete

the square on x2 + 2x + 3 = 0?

The roots of the quadratic equation ax2 + bx + c = 0, a 6= 0 are

given by the quadratic formula: x = ¡b § p

b2 ¡ 4ac

2a

FINDING THE ROOTS OF A QUADRATIC EQUATION BY THE QUADRATIC FORMULA


Copy and complete these statements.

Then complete the square on



The derivation of this formula follows. It is not required that you know this derivation. It is included

to show that the quadratic formula is found by completing the square on ax2 + bx + c = 0:

Show that the roots of the quadratic equation ax2 + bx + c = 0, a 6= 0 are given by the

quadratic formula: x = ¡b § p

b2 ¡ 4ac

2a .

ax2 + bx + c = 0) ax2 + bx = ¡c fsubtract c from both sidesg

) a(x2 + b

ax) = ¡c ftake out a factor of ag

) (x2 + b

ax) = ¡ c

a fdivide both sides by ag

) (x2 + b

ax +

b2

4a2) = ¡ c

a +

b2

4a2 fcomplete the squareg

) (x + b

2a)2 =

b2

4a2 ¡ c

a fwrite LHS as a perfect square; reorder RHS

g) (x +

b

2a)2 =

b2

4a2 ¡ 4ac

4a2 fwrite RHS with a common denominator g

) (x + b

2a)2 =

b2 ¡ 4ac

4a2 fwrite RHS as a single fractiong

) x + b

2a = §

r b2 ¡ 4ac

4a2 ftake the square root of both sidesg

) x + b

2a =

§p b2 ¡ 4acp

4a2fsurd of a quotient equals the quotient of surdsg

) x + b

2a =

§p b2 ¡ 4ac

2a fsimplify denominator g

) x = ¡b

2a +

§p b2 ¡ 4ac

2a fsubtract

b

2a from both sides of the equationg

) x = ¡b § p

b2 ¡ 4ac

2a fwrite the RHS as a single fractiong

Find the solutions to 3x2 + x ¡ 1 = 0 using the quadratic formula.

a = 3, b = 1 and c = ¡1 flist the values of a, b and cg

Now x = ¡b § p

b2 ¡ 4ac

2a fthe quadratic formulag

) x = ¡1 §

p (¡1)2 ¡ 4(3)(¡1)

2(3) fsubstitutingg

) x = ¡1 § p

13

6

EXAMPLE 2.20




So, the exact solutions are ¡1 +

p 13

6 and

¡1 ¡ p 13

6 .

The approximate solutions are x = 0:434 and x = ¡0:768.

We now have four methods of finding the roots of a quadratic equation. Which one should you

use?

If a = 1 and the roots are integers, the method that is easiest, fastest and least prone to error is

factorising.

If integer factors do not exist, and an exact solution is desired, then either completing the square

or the quadratic formula can be used.

The graphics calculator can be used for finding the zeros of almost all of the functions studied in

this course, and not just quadratic functions. It has the drawback of not providing exact answers,

and not providing much understanding, either.

1 Solve using the quadratic formula.

a x2 ¡ x ¡ 6 = 0 b x2 + 4x ¡ 5 = 0 c 2x2 ¡ 5x ¡ 3 = 0

d x2 ¡ 4x + 1 = 0 e x2 + 2x ¡ 5 = 0 f x2 + 6x + 5 = 0

g x2 + 3x ¡ 3 = 0 h x2 ¡ 5x + 1 = 0 i x2 ¡ x ¡ 3 = 0

j 2x2 ¡ 3x ¡ 3 = 0 k 2x2 ¡ 6x ¡ 8 = 0 l 3x2 + 14x ¡ 5 = 0

m 9x2 ¡ 6x + 1 = 0 n 15x2 + 2x ¡ 8 = 0 o 3x2 + x + 1 = 0

2 Write in the form ax2 + bx + c = 0, and then solve using any method you wish.

a x(x + 9) = ¡8 b x2 = 3(2x ¡ 3) c 34 x2 ¡ 1

4 x = 54

d x2 + x = ¡14 e 18 = x(9 ¡ x) f x + 2 =

5

x + 3

3 Solve each quadratic equation by

i graphing

ii factorising

iii completing the square

iv quadratic formula

a x2 + 4x + 3 = 0 b x2 + x

¡6 = 0 c 2x2 + 7x + 3 = 0

The quadratic formula for ax2 + bx + c = 0 gives two solutions,

x = ¡b +

p b2 ¡ 4ac

2a and x =

¡b ¡ p b2 ¡ 4ac

2a .

These values are the zeros of the quadratic function, i.e., they are the x-coordinates of the points

where the graph crosses the x-axis.

EXERCISE 2D.3

THE DISCRIMINANT




Now consider the following functions and their graphs:

The first graph has two zeros, the second has only one, and the third one has no zeros. How can

we reconcile this with the fact that the quadratic formula always appears to give two solutions? Let

us investigate. Finding the zeros of the above functions using the quadratic formula gives:

a x = ¡2 § p

16

2

) x = 1

or x = ¡3

b x = ¡4 § p

0

2

) x = ¡2

c x = ¡1 § p ¡8

2

The first result shows two real solutions.

The second has only one real solution, ¡2. Note that this solution is given by ¡ b

2a, as the square

root term is zero.

The last result has no real solutions, as the expression includes the square root of a negative number.

(As Mathematics C students will learn, it does have two solutions. These are not real numbers, but

what are called complex numbers.)

So, the algebra and the graphs are in complete agreement!

From the examples, it should be clear that the key to determining the number of solutions to thequadratic equation ax2 + bx + c = 0 is the nature of the expression under the square root, which

is b2 ¡ 4ac.

This is called the discriminant, and is often symbolised by the Greek letter ¢ (pronounced ‘delta’).

The above examples should make it clear that,

if ax2 + bx + c = 0, and we write ¢ = b2 ¡ 4ac then

¢ < 0 indicates that there are no real roots

¢ = 0 indicates that there is one real root, which is ¡ b

2a

¢ > 0 indicates that there are two real roots. If both a and b are rational numbers and

¢ is a perfect square not equal to 0, then the square root term simplifies to

a positive integer, and the two roots are both rational numbers.

a 3x2 ¡ x ¡ 4 = 0 b 3x2 ¡ 2x + 4 = 0

a b c

y

x

944 2 x x y

y

x

442 x x y

y

x

322 x x y

Decide if each of the following quadratic equations has zero, one or two real roots, by examin-ing the discriminant. If there are two roots, decide if they are rational or irrational.

EXAMPLE 2.21




a ¢ = (¡1)2 ¡ 4(3)(¡4) = 49

Since ¢ is positive and a perfect square, there are two rational roots.

b ¢ = (¡2)2 ¡ 4(3)(4) = ¡44Since ¢ is negative, there are no roots of this equation.

1 Find the discriminant of each of these quadratic equations.

a x2 + 8x + 1 = 0 b x2 ¡ x + 7 = 0 c t2 + t ¡ 45 = 0

d 2x2 + 5x ¡ 4 = 0 e 12 ¡ x ¡ 3x2 = 0 f 4c2 ¡ 12c + 9 = 0

g x2 ¡ 10x + 21 = 0 h 9x2 ¡ 18x + 9 = 0 i 3x2 + 5x + 3 = 0

2 Determine if each of the following quadratic equations has zero, one or two real roots, by

examining the discriminant. If there are two real roots, determine if they are rational or

irrational.a x2 ¡ 6x + 9 = 0 b 2x2 ¡ 2x + 4 = 0 c 3 ¡ x ¡ x2 = 0

d x2 + x + 7 = 0 e 9x2 + 6x + 1 = 0 f 5 ¡ 2x ¡ x2 = 0

g 16 ¡ 5t2 = 0 h 2x2 ¡ 6 = 0 i 8x ¡ 3x2 = 0

Just as there are a variety of methods of solving quadratic equations, there are a variety of methods

of graphing quadratic functions. One of these is using a graphics calculator.

Sketch the graph of f (x) = x2 + 2x + 15 on your graphics calculator.

If you are using the standard window on your graphics calculator, you will probably not

see the graph.

Here is why: f (x) = x2 + 2x + 15

= (x2 + 2x + 1) + 14

= (x + 1)2

+ 14The smallest value that (x + 1)2 can be is zero

(when x = ¡1) so the smallest value of f (x) is 14.

For most graphics calculators that use the standard win-

dow, this is larger than Y max. The graph is ‘above’ the

viewing window.

If you change Y min to 13 and Y max to 16, you should

see the graph. Adjust the values of X min and X max so

the graph nicely fills the screen. One possible graph is

shown alongside.

EXERCISE 2D.4

GRAPHING QUADRATIC FUNCTIONS E

EXAMPLE 2.22




The other methods that you should know are

² graphing quadratic functions by making a table of values

² graphing quadratic functions by finding the zeros, y-intercept and the coordinates

of the vertex

² graphing quadratic functions by transformations of the function y = x2.

The last two methods in particular are important, as they help us to understand the relationship between a function and its graph. The zeros and the coordinates of the turning points of any func-

tion give us much information about the function. Learning to sketch a graph using transformations

helps develop a deeper understanding of the relation between a equation and its graph.

1 Sketch the graph of each of these quadratic equations over the domain [¡3, 3] by first making

a table.

a y = x2 ¡ x ¡ 12 b y = 2x2 ¡ 3x ¡ 2 c y = x2 ¡ x ¡ 6

2 Draw a table of values for

a y = x2 b y = (x + 2)2 c y = x2 + 2 d y = 2x2

What do you notice? Predict what would happen if you changed the 2 to ¡

3.

GRAPHING QUADRATIC FUNCTIONS BY MAKING A TABLE OF VALUES

Sketch the graph of y = 2x2 ¡ 5x ¡ 3 by first making a table.

x ¡3 ¡2 ¡1 0 1 2 3 4

2x2 18 8 2 0 2 8 18 32

¡5x 15 10 5 0 ¡5 ¡10 ¡15 ¡20

¡3 ¡3 ¡3 ¡3 ¡3 ¡3 ¡3 ¡3 ¡3

y 30 15 4 ¡3 ¡6 ¡5 0 9

(x, y) (¡3, 30) (¡2, 15) (¡1, 4) (0, ¡3) (1, ¡6) (2, ¡5) (3, 0) (4, 9)

In general this method has a few drawbacks.

Often we do not know the domain over which

we should be graphing. The interesting parts

of the graph may occur for values of x be-

tween 15 and 17, so plotting points between

¡3 and 4 is a waste of time. Also, because the

method is time-consuming, we tend to only

plot a few points, and guess that the function

is behaving itself in between those points.

EXAMPLE 2.23

EXERCISE 2E.1

y

x

y x x 2 5 32




The “interesting” points of a graph are the zeros, the points where the graph reaches a maximum

or a minimum value, and the point where the graph crosses the y-axis (the y-intercept). So it is

common to determine and plot only these points, and then join them up with a smooth curve.

You already know how to find the zeros of a quadratic function. The y-axis has the equation x = 0,so to find the y-intercept, we find the value of f (0), which is the constant term, c. So all that needs

to be found are the coordinates of the vertex. And this is easy, since the x-coordinate of the vertex

is midway between the two zeros. We can calculate this by adding the roots, and dividing by two.

We then find the y-coordinate of the vertex by substitution.

Sketch the graph of y = x2 + 4x ¡ 5:

We find the zeros to be x = 1 and x = ¡5 by factorising.

The y-intercept is the constant term, ¡5.

The x-coordinate of the vertex is the average of the zeros,

or (1 + ¡5) ¥ 2 = ¡2.

The y-coordinate of the vertex is found by substituting ¡2for x: f (¡2) = (¡2)2 + 4(¡2) ¡ 5

= 4 ¡ 8 ¡ 5

= ¡9

GRAPHING QUADRATIC FUNCTIONS BY FINDING ZEROS,THE COORDINATES OF THE VERTEX AND THE -INTERCEPT Y

EXAMPLE 2.24

y

x

y x x 2 4 5

INVESTIGATION 3

FINDING FORMULAS FOR THE COORDINATES OF THE VERTEX

1 The x-coordinate of the vertex of a quadratic function equals the arithmetic mean of

the two roots,

x = ¡b +

p b2 ¡ 4ac

2a and x =

¡b ¡ p b2 ¡ 4ac

2a .

Show that the x-coordinate of the vertex is given by x = ¡b

2a.

What to do:


2 Find the expression for the y-coordinate of the vertex, simplified if possible.

The formula x = ¡b

2a for the x-coordinate of the vertex is particularly useful if the

function has no roots.



1 For each of the functions below, find the zeros, y-intercept and the coordinates of the vertex.

Then sketch the graph of the function using this information.

a y = x2 + 2x ¡ 8 b y = x2 ¡ 12x + 35 c y = x2 + 3x ¡ 10

d y = x2 ¡ 25 e y = 4x2 ¡ 1 f y = 2x2 + 3x + 1

2 Find the y-intercept and the coordinates of the vertex of these functions, and where possible,

find the zeros. Check your answers using a graphics calculator.

a y = x2 + 2x + 5 b y = 3x2 ¡ 5x ¡ 2 c y = 2x2 ¡ 3x + 6

3 What are the zeros, y-intercept and the coordinates of the vertex of these quadratic functions?

a y = (x ¡ 4)(x + 2) b y = 3(x + 1)(x ¡ 3) c y = (x ¡ 1)(x + 4)

d y = ¡2(x ¡ 2)(2x ¡ 5) e y = x2 ¡ 2dx + 4 f y = ax2 + bx + c

EXAMPLE 2.25

EXERCISE 2E.2


y = x2

Consider the equation y = Ax2 where A can be any real number.

For each value of A, we get a different graph. The equation y = Ax2 is said to define a

family of functions.

A variable such as A that gives rise to a family of functions is called a parameter.

TRANSFORMATIONS OF THE QUADRATIC FUNCTION F

TRANSFORMATIONS OF


Find the coordinates of the vertex of the function y = x2 ¡ 4x + 5.

This function has no zeros, as ¢ = (¡4)2 ¡ 4 £ 1 £ 5 = ¡4 which is < 0.

We can use x = ¡b2a

to find the x-coordinate of the vertex.

x = ¡b

2a

= 4

2

= 2

and y = 22 ¡ 4(2) + 5

= 1

fsubstitutingg

The coordinates of the vertex are (2, 1).

Note that the graph opens upward and the vertex is

above the -axis. It follows that there will be no zeros.x

642

8

6

4

2

y

x



A parameter transforms the graph of a function in a fundamental way.

1 Investigate graphs of the family of functions y = Ax2, by drawing the graphs of these

functions on a graphics calculator.

y = x2 y = 2x2 y = 12 x2 y = ¡x2 y = ¡2x2

What effect does the magnitude of the parameter A have on the graph of the quadratic

function? What is the effect of the sign of A? Tables of values may help you spot

patterns.

2 Investigate graphs of the family of functions y = x2 + C , by drawing the graphs of

these functions on a graphics calculator.

y = x2 y = x2 + 1 y = x2 ¡ 2

What are the effects of the constant term C on the graph of the quadratic function?

3 Investigate graphs of the family of functions y = (x

¡B)2, by drawing the graphs of

these functions on a graphics calculator.

y = x2 y = (x ¡ 3)2 y = (x + 2)2

What are the effects of the parameter B on the graph of the quadratic function?

4 Investigate functions of the type y = A(x ¡ B)2 + C . Do your conclusions still hold

when these transformations are combined?

5 Write a paragraph to summarise your findings.

Sketch the graph of y = x2 and the graph of the given function on the same set of axes.

a y = 3x2 ¡ 2 b y = ¡2(x ¡ 1)2 + 3

a The graph of y = 3x2 ¡ 2 is the graph of y = x2

transformed in two ways:

² the vertex is shifted two units down, to (0, ¡2)

² the graph is stretched vertically (or squeezed

horizontally, which has the same effect) by a

factor of 3.

b The graph of y = ¡2(x ¡ 1)2 + 3 is the graph of

y = x2 transformed as follows:

² the vertex is shifted three units up, and 1 unit

to the right, to (1, 3)

² the graph opens downward

² the graph is stretched vertically (or squeezed

horizontally, which is the same thing) by a

factor of 2.

What to do:

EXAMPLE 2.26




1 Sketch the graph of y = x2 and the graph of the given function on the same set of axes.

a y = 3x2 b y = ¡2x2 c y = 12 x2

d y = x2 ¡ 5 e y = x2 + 1 f y = (x ¡ 2)2


a y = 2x2 + 1 b y = 2(x + 1)2 c y = 3 ¡ x2

d y = ¡(x + 2)2 e y = 3(x ¡ 4)2 f y = (x ¡ 4)2 + 1


a y = ¡2(x ¡ 3)2 b y = 12 x2 ¡ 4 c y = 2(x + 3)2 ¡ 1

d y = ¡12 (x + 2)2 ¡ 4 e y = 2(x + 2)2 + 2 f y = ¡3

4 (x ¡ 1)2 ¡ 3

Transform the function y = 3x2 ¡ 4x + 1 into the form y = A(x ¡ B)2 + C .

Then sketch the function.

You will need to study this example carefully, as the algebra is more difficult than

examples.y = 3x2 ¡ 4x + 1

= 3[x2 ¡ 43 x + 1

3 ] fTake out a factor of 3.g= 3[x2 ¡ 2( 2

3 )x + ( 23 )2 ¡ ( 2

3 )2 + 13 ] fComplete the square.g

= 3[(x ¡ 23 )2 ¡ 4

9 + 13 ] fWrite as a perfect square.g

= 3[(x ¡ 23 )

2

¡ 49 +

39 ] fGet common denominator.g

= 3[(x ¡ 23 )2 ¡ 1

9 ] fAdd fractions.g= 3(x ¡ 2

3 )2 ¡ 13 fExpand to put into desired form.g

The vertex of the graph is at ( 23 , ¡ 1

3 ),

and the shape of the graph is that of

y = x2 stretched vertically by a factor of

3. The y-intercept is +1 (found from the

original equation).

Find the zeros of the above function, by setting y = 0, and solving for x.

3(x ¡ 23 )2 ¡ 1

3 = 0 fSet y = 0.g) 3(x ¡ 2

3 )2 = 13 fAdd 1

3 to both sides of the equation.g) (x

¡ 2

3)2 = 1

9 fDivide both sides by 3.

g

EXERCISE 2F.1

EXAMPLE 2.27

EXAMPLE 2.28

in previous




) x ¡ 23 = §1

3 fIf P 2 = Q then P = §p Q.g

) x = §13 + 2

3 fAdd 23 to both sides.g

) x = 1 or 13 fSimplifying the fractions.g

The roots are x = 1 and x = 13 .

1 For each quadratic function below, use completing the square to re-write the function in the

form y = A(x ¡ B)2 + C: Then sketch the graph of the function.

a y = x2 ¡ 2x ¡ 2 b y = x2 + 6x + 8 c

d e f

2 a Use completing the square to re-write the function y = 2x2 ¡ 13x ¡ 7 in the form

y = A(x ¡ B)2 + C:

b What are the coordinates of the turning point of the graph?

c What is the y-intercept of the graph of this function?

d Find the zeros of the equation.

e Sketch the graph, showing all relevant information.

3 Sketch the graphs of the following functions using your method of choice.

Find the coordinates of i the zeros

ii the y-intercept

iii the vertex

a y = 2x2 ¡ 12x + 18 b y = x2 ¡ 2x + 6 c y = x2 + 5x ¡ 1

d y = 4x

2 e y = 3x

2

¡ 4 f

y = 12 ¡ 3x

2

INVESTIGATION 4

A NOVEL METHOD OF SOLVING QUADRATIC EQUATIONS

We will solve the quadratic equation x2 ¡ 2x ¡ 3 = 0 using yet another method.

1 Rewrite the equation x2 ¡ 2x ¡ 3 = 0 as x2 = 2x + 3.

2 Sketch on the same set of axes y = x2 and y = 2x + 3.

3 Find the coordinates of the points where these two graphs intersect. The x-coordinates

are the solutions to the original quadratic equation.

4 Explain why this method works.

5 Use this method with a graphing calculator to solve the equationp 1 ¡ x2 ¡ 2x2 + 1 = 0

What to do:

EXERCISE 2F.2

Note that since these roots are rational numbers, we could also have found the zeros byfactorising. could also have used the quadratic formula with the original equation.We


y = 2x2 ¡ 4x + 6

y = 2x2 + 4x + 7 y = 2x2 ¡ 6x + 8 y = ¡3x2 ¡ 12x ¡ 8



4 For each of the following quadratic functions, transform the function into the form

y = A(x ¡ B)2 + C . Sketch the function. Include the coordinates of the vertex and the

y-intercept in your sketch.

a y = x2 ¡ 4x + 4 b y = x2 + x + 1

c y = 3x2 ¡ x ¡ 3 d y = 4 ¡ 2x ¡ 12 x2

5 Each set of coordinate axes below contains the graph of f (x) = x2

, and the graph of g(x).Determine the equation of g(x) from its graph.

a b c

d e f

6 Use a graphics calculator to sketch the following functions on the same set of axes. Describewhat you see. How could you have predicted this without having sketched the graphs of these

functions?

a y = x2 + x ¡ 2 b y = 2x2 + 2x ¡ 4 c y = 3x2 + 3x ¡ 6

d y = ¡x2 ¡ x + 2

7 Give three quadratic functions whose zeros are¡1 and 3. Check your answer using a graphics

calculator.

8 Give three quadratic functions whose vertex is at (3, ¡4).

9 Give three quadratic functions whose y-intercept is

¡2.

10 Give three quadratic functions whose graphs are concave down.

The concept of transformations is central in the study of functions.

For any function f (x), the graph of y = Af (x ¡ B) + C is the graph of y = f (x) shifted B

units to the right, C units up, and scaled by a factor of A in the vertical direction.

As an example, let us examine transformations of a function called the absolute value function.

y

x

y

x

y

x

y

x

y

x

y

x

TRANSFORMATIONS OF OTHER FUNCTIONS G




1 Write a paragraph explaining how the graph of y = A jx ¡ Bj + C is transformed as the

parameters A, B and C are changed. You may wish to use the absolute value function on

your graphics calculator to explore this family of functions.

2 Sketch the graph of y = jxj and the graph of the given function on the same set of axes.

a y = jxj + 2 b y = ¡2jx ¡ 3j c y = 12 jx + 1j ¡ 3

3 Sketch the graph of y = jxj and the graph of the given function on the same set of axes.

a y =¡ j

xj

b y = 1

3 jx + 1

j c y = 2

¡ 1

2 jx + 1

j

THE ABSOLUTE VALUE FUNCTION

The absolute value function is an example of a piecewise function, which consists of two different

rules, each defined over a given domain.

The symbol used for the absolute value of x is jxj.

The function is defined as follows:

y = jxj =

( x if x ¸ 0

¡x if x < 0

For example, j3j = 3, j2:57j = 2:57 and j0j = 0.

What is j ¡ 3j? By the definition, it is ¡(¡3) which is 3.

Similarly, j ¡ 2:57j is 2.57.

We can sketch the graph of y =

jx

j by first

x ¡3 ¡2 ¡1 0 1 2 3y 3 2 1 0 1 2 3

For each question, sketch the graph of y = jxj and the graph of the given function on the

a y = 2jxj b y = ¡jxj + 1 c y = 12 jx + 2j ¡ 1

a b c

making table:a

set of same axes.

EXAMPLE 2.29

y

x

y

x

y

x

EXERCISE 2G

y

x

x y





4 For each exercise, the graphs of f (x) = jxj and g(x) are given. State the equation for g(x).

a b c

d e f

5 The function f (x) = 1

x is called the reciprocal function.

Make a table of values from x = ¡6 to x = 6, and use the table to sketch its graph.

6 For each question, sketch the graph of f (x) = 1

x and the graph of the given function on

the same set of axes.

a f (x) = 3

x b f (x) =

1

x + 4 c f (x) =

¡2

x

d f (x) = 1

x + 2 ¡ 3

7 The graph of a function y = f (x) is given

alongside. Sketch the graph of each of the

functions given below. Sketch each graph on

its own number plane.

a y = 2f (x) + 3

b y = ¡12 f (x ¡ 2)

c y = 3f (x + 1) ¡ 5

y

x

y

x

y

x

y

x

y

x

y

x

y

x



Quadratic functions arise in many real-life applications.

Area, the motion of a thrown object and revenue are just some of the situations where quadratic

equations are used.

1 A cricket ball is hit directly upwards at a speed

of 20 m/s. The height, h, of the cricket ball

above the ground after t seconds is given by the

equation: h = 20t ¡ 5t2.

a List some of the assumptions that have been

made in choosing this model.

b The cricket ball will strike the ground when

the value of h is 0. Solve a quadratic equa-

tion to determine when the cricket ball will

strike the ground.

c Make a table and draw a graph that shows

the height of the cricket ball after t seconds.

d What is the maximum height that the cricket ball will reach?

e Estimate from your graph the height of the ball after 2.5 seconds.

f Calculate the height of the ball after 2.5 seconds, using the equation.

g When is the cricket ball exactly 15 metres above the ground? Answer this in two ways:

i estimate the answer from your graph

ii solve a quadratic equation by factorising h When is the cricket ball exactly 10 metres above the ground? Answer this in two ways:

i estimate the answer from your graph

ii using the quadratic formula (to 2 decimal places)

2 The total cost of producing x pottery bowls per week is given

by: C (x) = 200 + 4x + 0:1x2.

a What assumptions are made in using this model?

b What is the total cost of producing 100 pottery bowls?

c If the total cost is $496:40, how many bowls were

produced?

d What would be a sensible domain for this model?

3 The efficiency, E percent, of a particular brand of spark plug

is given by E = 400(x ¡ x2) where x represents the size

of the gap in millimetres.

a What size gap gives 100% efficiency?

b What gap sizes will give an efficiency of 90% or greater?


MODELLING USING QUADRATIC FUNCTIONS H

EXERCISE 2H



6

a

b

c

d a b

e

f

A roll of fence wire me-

tres long is to be used to en-

close and separate two identical

rectangular paddocks as shown.

Let the length of each

paddock be , and the

width of the paddock be .

Find an expression for the

total area of the two pad-

docks, in terms of and .

Find an expression for the total length of fence wire needed to enclose the two paddocks,

in terms of and .

Set this expression equal to , and solve it for .

Use your answers to and to get an expression for the area in terms of only.

Use a graphics calculator to draw the graph of the function and to locate the coordinates

of the maximum value.

Find the dimensions of each enclosure, so that the total area enclosed will be maximised.

1200

1200

x

y

x y

x y

y

x


4 The height h, of an object that is thrown straight up at time t, is given by the following

function: h(t) = d + ut ¡ 12 at2, where

h(t) = the height in metres

t = time in seconds

d = the initial height above the ground from which the object is thrown

u = the initial velocity in metres per second, and

a = the acceleration of the object due to the force of gravity.

Assume that the acceleration due to the force of gravity is 10 m/s2.

An object is projected straight up from a platform situated 20 metres above the ground at a

speed of 50 m/s. Assuming the object reaches the ground (i.e., it does not strike the platform

on the way down), how many seconds after it is projected does it reach the ground?

5

a

b

c

d

e

f

g

A manufacturer of tricycles finds that the number of

tricycles sold, , is related to the price per tricycle ,

by the equation .

What price must be set to sell tricycles?

How many tricycles will sell if the price per tri-cycle is $ ?

Total revenue, , from sales is calculated by:

(price per tricycle) (number of tricyclessold).

What is the total revenue if the price per tricy-cle is $ ?

Determine the formula for calculating total reve-nue in terms of , the number of tricyclessold.

Graph the revenue function.

Determine from the graph the maximum salesrevenue.

How many tricycles must be sold in order tomaximise sales?

n p

p n

R

R

R n

= 2 + 200

30

150

=

80

¡

£

happy cows contented sheep




1 a Write down five rational expressions that simplify top

2.

b Solve without a calculator: Which is larger, 5p

6 or 6p

5?

(Hint: square both sides)

c What relation must hold between m and n if mp

n > np

m?

2 Find three numbers, x, y , z, none of which are perfect squares, such thatp x +

p y =

p z.

How many sets of such numbers can you find?

3 Give the equation of a quadratic function that passes through (1, 2).Show your reasoning.

4 There are an infinite number of functions that have zeros at (¡3, 0) and (2, 0).

What are two of them? Explain your answer.

5 Solve for x:

q p p x = 2

6 p

x +p

x +p

x =p

xp

xp

x. Find all values of x that make this equation true.

7 Find the number that exceeds its square by the greatest amount.

x mm

EXERCISE 2I

PROBLEM SOLVING I

7

a

b

Australia Post is conducting research intothe optimum distance between perforationson a book of stamps. Too close and the

books of stamps may fall apart; too far apart and the stamps will be hard to tear out. Let represent the distance between

perforations.

Based on their research the Post Office hasassigned a “convenience value” , to vari-ous values of . In particular, the value of is when (too close) and when

(too far apart). The conveniencevalue is symmetric for values of betweenthese two values, and rises to a maximumvalue of .

Find a quadratic model for the “conven-ience value” of the perforation distanceof books of stamps. Check that your

model gives for and. and a maximum value of

between these two values.

Are there models other than the quadratic model that meet these assumptions? Explainyour reasoning.

x

C x C

x :x :

x

C

C x :x C

0 = 0 4= 1 8

= 1

= 0 = 0 4= 1 8 = 1




8 Green Globs was a popular maths computer game in

the early 1990s and is still available commercially.

The game board consisted of a number plane with

spots on it, as in the diagram. The object of the

game was to find the equation of a graph that passed

through as many spots as possible, and then to type

in its equation. The player received a point for every

spot that the graph passed through.

Find a quadratic equation whose graph passes through the three points in the diagram.

The lines are one unit apart.

9 You are an architect who has been commissioned to design a bridge. A simple scale

drawing of the bridge is shown. For aesthetic purposes, you make the arch in the shape

of a parabola. Find an equation of the parabolic arch. Use metres for the units.

10 Solve the following equation, setting out clearly and describing each significant step

in your solution:

p (x2 ¡ x) = 2

p x ¡ x (Hint: to start, square both sides)

11 The diagram shows the graphs of f (x) = 12 x2

and g(x) = x + 4. A and B are points that can

slide along f (x) and g(x) respectively, between

x = ¡2 and x = 4, such that AB is always

parallel to the y-axis. Find the greatest length of

AB.

12 A farmer wants to put some fish into his dam in

March, and then sell them at the end of the grow-

ing season, in November. Due to competition for

available food, the amount that each fish gains

in weight w , is a function of the number of fish

per cubic metre of water n, and is given by the

equation w = 60 ¡ 25n where w is measured

in grams.

Determine the number of fish per cubic metre of

water that will maximise the total weight gain.

scale 1:1000

x

y

4224

12

10

8

6

4

2

y

x

ƒ( ) = 0.5 x x2

g x x( ) = + 4A

B




13 A sculptor commisioned to design a monument

for the Longreach City Council has chosen a

parabolic shape that stands 15 m high with sup-

ports at 45o as shown in the diagram. For aes-

thetic reasons the sculptor chose the shape given

by the function y = 15 ¡ x2

¼ .

Find the length of the support beam marked A.

14 A rectangular area of land bordering onto a

straight river bank is to be divided up into 6identical rectangular paddocks as shown, us-

ing a total of 3600 m of fencing. The river

bank side is not to be fenced, and the total

area enclosed is to be the maximum possible.

Find

a the dimensions of the small paddocks

b the total area enclosed.

15 An engineer designing a gutter for cyclonic con-

ditions determines it requires a cross-sectional

area of 60 cm2. His job is to find the minimum

width of metal that can be bent into the shape

that will meet the design specifications given in

the diagram.

Write an equation that expresses the width of the metal in terms of the variable x, and

then find the minimum width of metal required using a graphics calculator.

16 Given below is a fallacy, which is a mathematical argument that leads to an impossible

statement or a contradiction because of a subtle flaw in the mathematics reasoning. Give

the algebraic reason for each step of this argument, and then discuss where the flaw exists.

a = b

a2 = ab

a2 ¡ b2 = ab ¡ b2

(a + b)(a ¡ b) = b(a ¡ b)

a + b = b

2b = b

2 = 1

A

45°

river

60 cm2

2 cm3 cm

x cm x cm

y cm






axis of symmetry maximum radical sign

coefficient minimum rational number

constant term parabola revenue

counter-example parameter root of an equationdifference of squares perfect square surd

discriminant piecewise function turning point

family of functions quadratic equation vertex

identity quadratic formula y-intercept

irrational number quadratic function zero of a function

1 a Simplify

i p

98 ii p

8 £ p 32 iii 2

p 50 ¡ 7

p 32 iv

p 16b4

b Rationalise the denominator. Simplify where possible.

i 2

p 2p

6 ii

3ap a

2 Simplify

ap

75 ¡ p 48 +

p 108 b 5

p 7 ¡ 8

p 28 + 4

p 63

c 2p 12 £ p 96 d (7 + 2p 3)2

e (6p

7 + 3p

5)(6p

7 ¡ 3p

5) f (1 +p

2)3


a ¡2(2x ¡ 1)(5x + 1) b (3x + 2)(3x ¡ 2)

c (2x + 1)(x ¡ 3) ¡ (x + 5)2 d x(x ¡ 3) ¡ (3 ¡ 2x)2

4 Factorise, then simplify: (x ¡ 1)2 ¡ 4

5 Solve

a 4x2

¡4x

¡3 = 0 by factorising

b 10x2 ¡ x ¡ 3 = 0 using the quadratic formula

c 2x2 + 5x ¡ 12 = 0 by completing the square

6 By examining the discriminant, determine whether each of the following have two, one or

a x2 + 4x + 9 = 0 b 2x2 ¡ 5x ¡ 3 = 0 c 4x2 ¡ 12x + 9 = 0

7 Transform the function y = 6x2 + x ¡ 2 into the form y = A(x ¡ B)2 + C . Hence

sketch the graph of y = 6x2 + x ¡ 2 by considering the transformation.

zero real roots.






8 The number plane alongside contains the

graph of f (x) = x2, and the graph of

g(x). Determine the equation of g(x)from its graph.

9 Give the coordinates of two points that lie on the function y = 2x2 ¡ x + 6.

10 For what value of a would the function f (x) = a(x ¡ 1)2 + 2 pass through the point

(3, ¡10)?

11 A roll of fence wire 1800 metres long is to

be used to enclose and separate three identical

rectangular paddocks as shown.

Find the dimensions of each enclosure, so thatthe total area enclosed will be maximised.

12 By considering the transformation of y = jxj, sketch the graphs of

a y = jxj ¡ 3 b y = jx ¡ 3j c y = jxj + 2

d y = jx + 2j e y = 2 jx + 1j ¡ 3 f y = ¡0:5 jx ¡ 1j + 2

1 Simplify

a 2p

32 £ 4p

2 b 2p

a ¡ 3p

4a

2 Rationalise the denominator, and simplify 3

2p

6


a (2x ¡ 2)(2x + 2) b (3a ¡ 2x)2


a x2 ¡ 7x + 12 b 6x2 ¡ 7x ¡ 2 c 9x2 ¡ 4c2 d a4 ¡ 1

5 f (x) = x2 + 3x ¡ 3. Find

a f (¡3) b f (0)

6 Solve x2 ¡ 6x ¡ 8 = 0 by

a graphing b factorising

c completing the square d quadratic formula

7 For the function y = x2 ¡ 2x ¡ 3, find the zeros, y-intercept and the coordinates of

the vertex. Then sketch the graph of the function using this information.

8 Sketch the graph of f (x) = x2 and g(x) = 3(x + 1)2 on the same set of axes.

@= (!) g @=ƒ(!)

b

l




EXTENDED INVESTIGATION

ABSOLUTE VALUE FUNCTIONS


V


9 The total monthly cost of maintaining an inventory of x items in a warehouse is given

by: C (x) = 350 + 3:4x + 0:004x2

a What is the total monthly cost of maintaining an inventory of 400 items?

b If the total monthly cost is $1450, how large is the inventory?

When a keen student of mathematics “plays around” with an interesting mathematical situation,

questions arise naturally. Often a first step in answering such questions is gathering information,

and looking for patterns in the information. From the patterns, the student forms a conjecture.

The student then tests the conjecture to see if it appears to be true. If so, the next step is for

the student to justify or prove the conjecture - to satisfy herself that the conjecture must be

true. Finally, once she understands the solution to the problem well, she should communicate

her problem and its solution to others.

The absolute value function can give rise to some fascinating graphs. Draw the graphs of

functions such as those given below.

Y 1 = jxjY 2 = jxj + jx + 1jY 3 = jxj + jx + 1j ¡ jx ¡ 2j

Pose some questions, form some conjectures, and test to see if your conjectures appear to be

true. If so, try to justify your conjectures.

Here are a few hints, to get you started.

² Start with a simple function. Make a small change to the function, and see how the

corresponding graph changes. Before moving on to a more complicated function, tryto fully understand the relationship between the equation and graph of this simple

function.

² A good way to test a conjecture is to make predictions (“if I change the 1 to a 2, I

expect the graph to ....”) and check to see if your predictions are correct.

ERASER TOSS Materials: one metre stick, one eraser, seven white board markers and eight

matics B students.

Rule a line 3 metres long across the bottom of the white board and mark off every 50 centime-

tres, starting from one end. Rule a vertical line up from each of these marks.

Now one student will attempt to throw the eraser in an arc from one end of the line to the other.

Repeated throws might be necessary. The other students are each in charge of one “vertical

line”. They are to mark the position where the eraser crossed their line.

If this is done accurately, the points marked should lie on a parabola.

Find the equation of this parabola.

Check the location of the marked points against that predicted by the quadratic model.

Year 11 Mathe



CHAPTER 3Exploring data

SUBJECT MATTER

identification of variables and types of variables

and data (continuous and discrete); practical appli-cations of collection and entry of data

choice and use in context of appropriate graphicaland tabular displays for different types of data in-cluding pie charts, barcharts, tables, histograms,stem-and-leaf plots and boxplots

use of summary statistics including mean, median,standard deviation and interquartile distance asappropriate descriptors of features of data incontext

use of graphical displays and summary statistics indescribing key features of data, particularly incomparing data sets and exploring possiblerelationships



HISTORICAL NOTE +

92 EXPLORING DATA (Chapter 3)

Data are numbers that have meaning.

For example, 54, 67, and 98 are numbers, but if they represent the marks on your last three

There is a story in data, and your job as a student of statistics is to find that story. If the situation

from which the data are gathered is interesting, then the story the data tell may be interesting too.

The first graphical displays of information were charts and maps of the Earth’surface, which date back thousands of years. However it was not until the th centurythat numerical information or patterns were displayed graphically An early example was the

map of central London, created by in that showed the location of recentdeaths from cholera, as well as the locations of eleven public water pumps. The graphic clearlyshowed that the deaths were centred around the Broad Street water pump. Dr Snow had the

pump handle removed. This simple action ended an epidemic that took over lives.

s,

.

,

19

1854

500

Dr John Snow

O X F O R D

S T R E E T

G T. M

A R L B

O R O U

G H S T

.

R I N

G

S T R E E T

S A V I L L E

R O W

N E W

B O N D

S T R E E T

C O N D U I T

S T R E E T

B R O A D

S T R

E E T

D E A N

S T R E E T

W

A R D O U R

S T R E E T

B R E W

E R

S T R E E T

P I C A D I L

L Y

R E G

E N

T S Q U A D R A N T

R

E

G

E

N

T

S

T

R

E E T

G O L E N

S Q U R E

G O L D E NS Q U A R E

yards

d e a t h s f r o m c h o l e r a

p u m p

5 0 0 5 0 1 0 0 1 5 0 2 0 0

DATA A

Mathematics examinations, they are data.B



EXPLORING DATA (Chapter 3) 93

The tools that are available to help us learn what data can tell us can be broadly classified:

² Graphical displays show the data pictorially.

² Summary statistics describe a large set of data using just a few numbers, such as the

mean and the standard deviation.

² Probability is the formal study of chance. It is an essential tool in making good

decisions based on the data.

² Inferential statistics is a set of analytical tools that helps us makes decisions based on

data, and gives us a measure of the reliability of our decision.

In this chapter we will study graphical displays and summary statistics. Probability and an intro-

duction to inferential statistics are studied in Year 12.

PATTERNS IN DATA

The first thing a statistician does with a dataset is look at the data - looking for patterns, departures

from patterns and unusual values. The data may be summarised in a table that shows counts and/or

percentages, or it may be displayed graphically, for example as a bar chart, pie chart or histogram.

1 The tables below contain a summary of data collected about a particular incident involving a

large number of people.

a Look at the data, and list all unusual patterns that you find. For example, one unusual

pattern is the difference in the overall female death rate (27 deaths per 100) and the overall

male death rate (80 deaths per 100).

b Use the patterns you have found to help you decide what this unusual incident might be.

Justify your answer using the data, and the patterns that you have noticed.

EXERCISE 3A

Economic P opulation exposed Deaths per 100

statusto occurrence

Number of deathsexposed to occurrence

Adult Child Both Adult Child Both Adult Child Both

I (high) 319 6 325 122 0 122 38 0 37II 261 24 285 167 0 167 64 0 59III (low) 627 79 706 476 52 528 76 66 73

Other 885 0 885 673 0 673 76 - 76Total 2092 109 2201 1438 52 1490 69 48 67

Data cross-tabulated by economic status and age

Economic

statusto occurrence

Number of deathsexposed to occurrence

Male Female Both Male Female Both Male Female Both

I (high) 180 145 325 118 4 122 65 3 37

II 179 106 285 154 13 167 87 12 59

III (low) 510 196 706 422 106 528 83 54 73

Other 862 23 885 670 3 673 78 13 76

Total 1731 470 2201 1364 126 1490 80 27 67

Data cross-tabulated by economic status and gender

Deaths per 100 Population exposed




2 How well do you know Queensland? The graphs below are climate charts for eight locations

in Queensland. Match each location with its chart.

The locations are ² Cairns ² Coolangatta ² Lady Elliot Island

² Longreach ² Rockhampton ² Stanthorpe

² Thursday Island ² Toowoomba

Some information about climate and geography

a b

c d

0

5

10

15

20

25

30

J F M A M J J A S O N D

month

t e m p e r a

t u r e

( d e g r e e s

C )

0

20

40

60

80

100

120

140

r a i n f a l l

( m m

) median rainfall

daily mean max. temp

daily mean min. temp

J F M A M J J A S O N D0

5

10

15

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25

30

0

20

40

60

80

100

0

5

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30


0

20

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35


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140

0

5

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35


0

50

100

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350

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450

The further north town is located, the higher the temperature tends to be.

The further town is from the sea, the more extreme are the differences in daytime and nighttime

temperatures.

owns in northern Queensland have monsoonal climate, with very pronounced wet and dry

seasons. owns in western Queensland tend to receive less rain than those on the coast.

The vertical scales on the charts are not all the same.

The chart of Brisbane given below will help you interpret the charts.

a

a

T a

T

Note:



SOME TERMINOLOGY B


e f

g h

Source: The original data is from the Bureau of Meteorology website.

A population is the entire group of people or things in which a statistician is interested.

A unit or a case is one member of the population.

The particular characteristics of a population about which data are collected are called variables.

Here are some examples.

Population Variables

wattle trees height, growth rate, colour of bark

home computers system clock speed, amount of RAM

Australian rivers length, level of salinity

refrigerators capacity, hours of operation before needing repair

Mathematics B students gender, favourite movie, enjoyment of problem solving

The collection of data from every member of a population is called a census.

If data is only collected from some members of the population, it is called a sample.

Variables have values. For example, in my Mathematics B class, Sam is a female who rates her

enjoyment of problem solving as 9 out of 10, while Peter is a male who rates his enjoyment of

problem solving as 3 out of 10.

0

5

10

15

20

25

30

35

40


10

20

30

40

50

60

0

5

10

15

20

25

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35


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100

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200

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400

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200

For example, if we are interested in the attitude of the students in a Mathematics class towards

problem solving, and we survey them to determine this, we have census. On the other hand, if we

are interested in the weight of young koalas as they grow to maturity we would work with

sample, as we cannot capture and weigh every young koala in Australia.

B

a

, a




A single value is called a data value. A collection of data values is a called a dataset.

The collection of responses of my Mathematics B students about their attitude towards problem

solving forms a dataset.

A parameter is a number associated with a variable of a population.

The mean weight of 3 month old koalas is a parameter.

A statistic is the corresponding number associated with the sample.

For example, from our data we can say that the mean weight of three

month old koalas in our sample is 5.3 kilograms. The number 5.3 is

a statistic.

In inferential statistics we use a sample statistic, which is calculated from the data that have been

collected, to estimate a population parameter. The population parameter is what we really want

to know, but usually do not know, because we usually do not have data on all members of the

population.

TYPES OF VARIABLES

SAMPLING

The height of trees, length of rivers and capacity of refrigerators are all numerical data for which

calculations such as differences and averages make sense. Data collected on such variables are

called quantitative data.

Quantitative data are either discrete or continuous. Variables associated with discrete data are

sometimes referred to as discrete variables and those associated with continuous data are often

referred to as continuous variables. You have already met these terms in Chapter 1. The number

of children in a family is discrete, as it can only take on whole number values. You cannot have

0.3 of a child. The height of these children is continuous. A child that grows from 1.2 m to 1.3metres in a year must have been 1.23 m in height at some time during that year.

An effective way to distinguish between discrete and continuous data is:

² if you count, the data is discrete ² but if you measure, the data is continuous.

We count children, but we measure their heights.

On the other hand the gender of students is an example of categorical data, since the responses

can be categorized, as male or female.

The arithmetic operations on categorical data are counts, proportions and percentages. It makes

sense to say that there are 12 males in the class, or 12 out of 30 students are male or that 40% of

the class is male.

The purpose of statistics is to be able to make a statement about

one or more variables of a population.

For example, a scientist working for the Great Barrier Reef

Marine Parks Authority wants to be able to make a statement

about the number of each species of fish at a certain reef. In

this example, the population is all of the fish on the reef, and

variable is the species.

Since it is impossible, practically speaking, to catch and record

the species of every fish on the reef, we must take a sample.




As the point of gathering data is to learn something about a population, a census is obviously

superior to a sample. But for most populations it is not practical, or even possible, to take a census.

So we usually have to sample, and then use that sample to draw conclusions about the population

from which the sample is drawn.

As another example, consider pineapples. One variable

is its sugar content, which is a major factor in determin-

ing its market value. The only way to measure the sugar content of a pineapple is to either lop off the top and

measure the sugar content directly, or to paint it with

a special dye that changes colour, where the colour re-

flects its sugar content. Either way the pineapple can no

longer be sold. If you took a census of a pineapple crop,

you would know a great deal about its sugar content, but

there would be no pineapples left to sell. A census is

not practical.

These are extreme examples, but even when a census is possible, there are two reasons why it

usually is not done.

Collecting, recording and analysing data is a expensive, and b time consuming.

If sampling will give a result almost as accurate as taking a census, and for far less cost, then

sampling is the preferred choice. But, if sampling is used, the size of the sample and deciding how

to select the sample are important issues, and issues that can only be addressed using knowledge

about statistics.

To keep this in perspective, if a census can be done, it should be done.

If the question to a Year 11 Mathematics B class is,

“Over the weekend, would this class prefer to a have homework or b not have homework?”,

a census, by a show of hands, is the best way of knowing the class decision.

1 A scientist is studying the pattern and frequency of

car accidents involving teenagers in the local com-

munity.

a For this study, identify the

i population ii variables.

b State one parameter of the population.

c Should the scientist take a census or a sample?Justify your decision.

2 The Student Council at your school wants to know

the students’ opinions about increasing the number

of school dances next year from four to six.

a For this study, identify the

i population ii variables.


c Should the Student Council take a census or a

sample? Justify your decision.

EXERCISE 3B




3 For each of the following:

i state a quantitative variable that may be of interest to a researcher

ii state a categorical variable that may be of interest to a researcher

a voters in the seat of Mundingburra

b planets in the solar system

c fish on the reef at Lady Musgrave Island

d students in your Mathematics B class

4 Determine if each of these variables is discrete (D) or continuous (C).

a the number of cancer patients who survived for at least five years after treatment

b the height of basketball players in the National Basketball League

c the number of medals won by each country at the 2000 Olympics

d the number of computers in each Queensland high school

e the total cost of the computers in each Queensland high school

f the rainfall in Ingham over the past twelve months

g shoe sizes of Queensland Year 11 students

h the number of pieces of luggage on the Monday to Friday 6:45 a.m. flight from Rock-

hampton to Brisbane

i the weight of the luggage on the Monday to Friday 6:45 a.m. flight from Rockhampton

to Brisbane

j the colour of the luggage on the Monday to Friday 6:45 a.m. flight from Rockhampton

to Brisbane

5 For these variables, decide if data are best collected on a population or a sample. Briefly

justify your decision.

a favourite food sold at your school’s tuckshop b length of holes on golf courses in Australia

c species of fish on the Fitzroy Reef, north of Cairns

d batting average of cricketers who played in the last Test match

e voting intentions of voters in the seat of Mundingburra at the next election

6 For which of these questions is a census practical? Briefly explain your decision.

a A scientist is trying to determine the lifespan of the giraffe.

b A scientist is trying to determine if giraffes kept in wild animal parks live longer than

those kept in zoos.

c The P&C of your high school is interested in the students’ opinions of the school uniform.

d The P&C of your high school is interested in

the staff’s opinions of the school uniform.

e A restaurant owner wants to know his cus-

tomer’s opinions about the quality of the ser-

vice and food.

f The Australian government needs to gather

data on a range of issues, including household

income, the number (and age and gender) of

people living in each house, and how people

travel to work.




There are four common methods of collecting data.

These are ² a survey

² an observation

² an experiment

² using available data.

A survey is a series of questions asked of a group of people, either orally, or on paper.

For example, pollsters ask voters their opinion on whether Australia should be a republic.

An observation is made when the researcher observes members of the population, and records data

on the variables of interest.

An experiment is performed when the researcher actively controls the data gathering process.

For example, an environmental scientist establishes three aquariums with different oxygen levels,

to determine the effect of differing levels of oxygen on the reproduction of aquatic snails.

Available data are data that were previously collected, and can be used to help answer the present

questions.

As leukemia is a serious disease that requires the services of a hospital, the local hospital

would most likely have the information required.

The most appropriate method would be to use this available data.

GOOD DATA AND BAD DATA

EXAMPLE 3.1

COLLECTING DATAC

For example, a researcher observes and records the time of day and the species of wild animals

drinking from a waterhole in an African game park.

The residents of Emerald believe that the number of cases of leukemia in the region has risen,and that the local aerial spraying of pesticides on cotton crops has caused this. One step in

determining if aerial spraying is causing leukemia is to collect data on the number of leukemia

cases in the region. Which method of collecting data: survey, observation, experiment or

using available data, is most appropriate? Justify.

The final outcome of most statistical studies is a decision, based on the data. To make that decision,

the data needs to be analysed, after they are collected. Before data are collected, the researcher

must think very carefully about exactly what questions he or she wants answered, what information

needs to be gathered to answer those questions, and how that information can best be collected.

If the wrong data are collected, or if the method of data collection is flawed, then no amount of

sophisticated statistical analysis of the flawed data can help to find good answers to the questions

being asked.




A news show host asked viewers, “Do you think Australia should have the death penalty?”.

Viewers were asked to dial one number for ‘Yes’ and another for ‘No’. Assume that the news

show was trying to determine the percentage of the Australian population that was in favour

of the death penalty. Comment on this method of collecting data.

The results of this survey reveal nothing about the views of the Australian population.

This survey has two flaws:

² Only people with a very strong view on the death penalty (which generally means

those in favour of the death penalty) will spend the time and the cost of the phone

call to register their views.

² People can vote as many times as they wish, so a special interest group could skew

the results by voting multiple times.

Note: The population parameter in this example is the percentage of the Australian population in

favour of the death penalty. Other studies have shown that this is about 50%. The samplestatistic is the corresponding percentage in the sample, which in a properly designed

survey should have been close to 50%. For this study, it was over 90%, which indicates

that this survey was not properly designed to give valid data about the population.

THE SIMPLE RANDOM SAMPLE

The method of sampling in the above example is called a self-selecting sample. To obtain a sample

with a statistic that is a good estimator of the corresponding population parameter, a simple random

sample (SRS) is the most common choice.

In an SRS, every possible sample has an equal chance of being selected. It is equivalent to puttingslips of paper with the names of all members of the population into a hat, shuffling them up

thoroughly, and then picking the sample from the hat.

Such a sample is said to be unbiased.

A self-selecting sample is biased because only those with strong opinions are in the sample.

How would you obtain a SRS of size twenty-five from each of these populations?

a all students in your school b all voters in the seat of Keppel

a Here is one method that is reasonably efficient. In secondary schools, every student

has a unique student number. Say the student numbers range from 12307 to 13 512.

² Use your graphics calculator to generate forty random numbers between 12307and 13512 inclusive and store them in List1. You need more than 25 numbers

because some numbers may be generated more than once, and some students

with numbers in this range may have left. Forty numbers should be sufficient.

² Starting at the top of your list, use your school’s student database to identify the

students with those student numbers. Ignore any duplicates. Do this until you

have selected 25

students currently enrolled.

EXAMPLE 3.2

EXAMPLE 3.3




b The electoral roll for Keppel contains the list of all voters in the electorate. To

generate a SRS, for each case, generate a random page number and then a random

number to select a person on that page. This process will need to be repeated until

you have a sample of size 25 voters.

Often it is impossible, practically speaking, to get a true SRS.

BIASED SAMPLES

A sample is said to be biased if it systematically favours certain outcomes.

Explain why each of the following samples is biased. Then make a practical suggestion to

the researcher as to how they could obtain an unbiased (or less biased) sample.

a A researcher wants to know the opinion of Mackay shoppers about whether local

shops should be allowed to trade on Sunday if they wish. He surveys shoppers at a

Mackay shopping centre between 9:00 a.m. and 11:00 a.m. on a Monday morning.

b A researcher is trying to ascertain voting intentions in the next election, so he is

knocking on doors in a particular street, asking the residents about their voting

intentions in the next election.

a The survey excludes people who are at work on a Monday morning, and who may be

strongly in favour of Sunday trading, as that is a time when they are free to shop. To

obtain a less biased sample, the researcher could also survey shoppers on Saturday

morning, so people who work typical Monday to Friday hours are included in the

survey.

b People living in a particular suburb, because they tend to have common beliefs and

common concerns, tend to have common voting patterns. A preferable method of

gathering data about voting intentions is to conduct a phone survey of the residents

of that electoral division (though see the section above).

EXAMPLE 3.4

For example, a supermarket is conducting a customer survey to find out how they can improve

their service. Since they do not have a list of every customer, they do not have a list from which

they can select an SRS. The best they can do is to ask all customers who enter the store over, say,

a two-week period, to complete the survey. If they are successful in getting a significant response,

they should have collected some good data that will help them to improve their service.

The self-selecting sample discussed earlier is an example of a biased sample. A biased samplerarely gives useful data about the population it is supposed to represent, and hence is usually of

little value.

A researcher has to be aware of possible sources of bias, and design a method of data collection

that minimises bias (while staying within budget!).

When gathering data on people, removing all bias when choosing a sample is almost impossible.

For example, phone surveys are common, but exclude people who cannot afford a phone.

Mailing questionnaires to a randomly selected sample often results in a poor return rate, thus biasing

the sample towards those who had enough interest in the topic to respond. Many people refuse to

participate in such surveys, so they are excluded as well.




1 These questions require data to be collected before they can be answered. For each question,

which method of collecting data (survey, observation, experiment or using available data) is

most appropriate? Be prepared to defend your decision in class.

a There is a crossing near your school that some parents think is dangerous. How many

students use the crossing at the end of the school day?

b How many students at your school think that wearing the school uniform should be

optional?

c A friend of yours claims that he can tell, by taste alone, whether he is drinking Coke or

Pepsi. Can he really do this?

d The media claim that the percentage of female teenagers who smoke is increasing. Is this

true of female students in your school?

e A cricket fan has a theory that left-handed batsmen have on average a higher run rate

than right-handed batsmen. Is he correct?

f Mathematics B students are expected to become capable users of technology (graphics

calculator and/or computer) in their study of mathematics. Are students in this classcapable users of technology?

2 Explain how a SRS could be obtained in each case.

a An Internet Service Provider wants to survey fifty customers about the quality of the

service.

b

c A biologist is studying the movement patterns of koalas in an area where it is proposed

a freeway be built.

3 Each of the following methods of collecting data will give a biased sample. State how the bias

will occur, and give an alternative method that will give an unbiased (or less biased) sample.

a In the campaign period prior to a state election, the Rockhampton Morning Bulletin

newspaper advertises each day a phone number for readers to ring to record who they

will vote for.

b A computer magazine conducts an annual survey on the reliability of the various brands

of computers, printers, etc. It includes a survey form in one issue of the magazine, and

asks readers to complete the survey and then mail it to the magazine.

c A Department of Transport researcher is gathering data on the use of boat ramps throughout

Queensland. Each day, from Monday to Friday, he observes and records the usage of oneramp in the morning, and a different ramp in the afternoon.

4 In 1992, the youth radio station Triple J conducted a survey of its listeners about their views

on the decriminalisation of marijuana use. The results of the phone-in poll were:

² 9924 out of more than 10000 callers (96%) favoured decriminalisation.

² Only 389 believed that possession of the drug should be a criminal offence.

² Many callers said that they were in favour of decriminalisation, even though they did

not use marijuana themselves.

Just prior to the poll, the Australian Capital Territory had decriminalised the use of marijuana.

Discuss possible sources of bias in this survey.

EXERCISE 3C

A fruit and vegetable wholesaler is buying oranges, which come in large crates. She

needs to determine the percentage of damaged fruit, as this affects the price she pays for

the oranges.




5

² What exactly is litter? Does it include organic matter that will decompose, such as a banana peel? How large must a piece of paper or plastic wrap be before it is classified

as litter?

² For which areas of the school grounds does data need to be gathered? Is the SC only

concerned about the grounds near the classrooms, or should the school oval be included,

or should data be gathered about every square metre of the school grounds, even those

areas where students rarely go? The time and cost of gathering the data needs to be

considered.

² Does the SC want to know the amount of litter in each area of the school, before and

after the campaign, or is it sufficient to just know an overall figure?

² Which method, or methods, of collecting data (survey, observation or experiment) willgive the most accurate data?

Let us now assume that the data-gathering process was carefully designed, we have gathered our

data and we feel confident that we have obtained a simple random sample, or near enough. Thenext step is to put the data into a spreadsheet file or a statistics program, or enter the data into lists

on a graphics calculator.

This should be a reasonably straightforward process, but there are some issues to consider.

Missing data

Recording errors and transcription errors

RECORDING DATA D

If you conduct a survey, it is highly likely that some people will leave questions unanswered,

or one of the plants in your biological experiment has died, so measuring its nitrogen uptake is

rather meaningless. You need to know how the statistics program you are using handles missing

data. Some programs, for example, treat a blank cell as a 0. The average of 3, 4, - , 5, where

the ‘-’ represents a missing datum (the singular of data), is 4, but such programs would give the

average as 3.

Humans doing repetitive tasks make errors. It is easy to write down an incorrect number while

recording data, and easy to type in an incorrect number when putting the data into a computer

or graphics calculator. Some students find that working in pairs reduces such errors. After the

data has been entered, checking that the data has been recorded and entered correctly is a critical

step in statistical analysis.

In an email, a researcher wrote that she often uses graduate students to enter the data from her

research. As there is a large amount of numerical data to be entered, she has the same data

entered multiple times, by different graduate students, as there are always errors in data entry.

She then averages the values entered, and uses those averages in her analysis.

Mathematical modelling A number of students have complained to the Student Council (SC)

about the amount of litter on the school grounds, so the Council has decided to implement an

anti-littering campaign. To determine the effectiveness of the campaign, they need data on the

amount of litter both before and after the campaign. Since you are currently studying the col-

lection and recording of data, the Student Council has asked your class to advise on how the

data should be collected. In advising the council, some questions that need to be considered are:

Having considered the above, write a report with recommendations for the SC on how the litter

data should be collected. Be prepared to put forward your views in your next Mathematics B

lesson. Think about this as a mathematical model (see Chapter 1).




Inconsistent symbolsIt is important that a consistent symbolism is used for data entry. Some years ago, one of the

authors collected data from a large sample of students, and had his Maths A class enter the data

into a database.

In the ‘Gender’ field for males, the students entered ‘M’, ‘m’, ‘male’, ‘Male’, ‘ M’ and ‘ m’ (the

last two have a space before the letter). The database treated each of these as being different.

1owned by the girls in her class into an Excel spreadsheet.

Here are her results for the number of CDs

10 25 30 15 0 50 5 20 20 15

she owned, so Yanna wrote down a zero for her.

a If Yanna uses the spreadsheet to calculate the average

number of CDs, what will the result be?

b What is the correct average?

2 In 2000, a state Education Department gave a Mathematics test to a group of secondary students

to determine their mastery of basic mathematics. The funding for each school in the state was

partly based on how well the students of that school did on this test. If a student was absent

on the day of the test, the student’s result was recorded as a zero.

a Comment on this method of recording test results for absent students.

b Why might the Education Department have done this?

3 Enter these numbers into the first three lists of your

graphics calculator. Now calculate the total of each list

(your calculator can do this for you). Check your totals

against that given in the answers, and correct any errors.

Comment on how accurate you were in entering these

numbers.

0:8245 0:7644 0:1784

0:7723 0:8639 0:1095

0:0775 0:7285 0:1968

0:7301 0:4231 0:1729

0:2054 0:6964 0:1573

0:1898 0:5592 0:0329

0:7693 0:1817 0:4380

0:4311 0:1287 0:6198

0:8084 0:7704 0:6964

0:8330 0:5124 0:89540:2722 0:1975 0:3187

0:6047 0:8615 0:6510

0:2599 0:0281 0:9480

0:4948 0:2117 0:2931

0:2310 0:8008 0:9410

0:9102 0:1652 0:8349

0:1580 0:4762 0:6756

0:6802 0:0279 0:3510

0:7587 0:2854 0:4600

0:7571 0

:0481 0

:5607

EXERCISE 3D

Y aanna recorded the results of survey on the number of CDs

The fifth student anna asked did not know how many CDsY




4 You have just completed a large survey about the buying habits of teenagers, and have employed

the students in your Mathematics B class to enter the data into a spreadsheet. One of the

questions was “How knowledgable are you about Internet shopping?”.

The choices were: ² very knowledgable

² knowledgable

² somewhat knowledgable

² what is the Internet?

² what is shopping?

Decide how you want this data to be entered into a spreadsheet, and then write a clear set of

instructions for the students doing the data entry.

“A picture is worth a thousand words.” This is true in statistics as well, though it might be phrased

as, “A picture is worth a thousand data values.” Visual displays show patterns that exist in a massof data. The historical note at the start of this chapter is one striking example. Here is another.

THE CHALLENGER DISASTER

On January 28, 1986 the space shuttle Challenger ex-

ploded. Seven astronauts died because two large rubber

O-rings leaked during takeoff. These rings had lost their

resiliency because of the low temperature at the time of

the flight. The air temperature was about 00 Celsius,

and the temperature of the O-rings about 6 degrees be-

low that.

The link between O-ring damage and ambient temper-

ature had been established prior to the flight. The en-

gineers at Morton Thiokol Inc had recommended that

the flight be delayed. Unfortunately their argument was

not accepted by their management, and the launch pro-

ceeded with disastrous consequences. Might this scat-

terplot have helped?

Temperature (oC) Damage index

12 11

14 4

14 4

17 2

19 0

19 0

19 0

19 0

19 0

20 0

21 4

21 0

21 4

21 0

21 0

22 0

23 0

24 4

24 0

24 0

26 0

26 0

27 0

Data from previous flights

VISUALISING DATA E

0

2

4

6

8

10

12

0 5 10 15 20 25 30

temperature (°C)

d a m a g e

i n d

e x

O- ring damage index v temperature




Most graphs of data in newspapers, textbooks and magazines, and on television, tend to be one of

four types:

Column graphs (which include horizontal bar graphs and picture graphs)

These graphs show (and compare) the frequency or relative frequency of data values in each

category.

Pie graphs and 100% bar graphs

Pie graphs and 100% bar graphs show the percentage or relative size of each category.

COLUMN GRAPHS, PIE GRAPHS, TIME SERIES GRAPHS

AND SCATTER PLOTS

0

50

100

150

200

car

bus

bike walk train

method of travel

n u m b e r

o f

p e o p l e

Travelling to work

m e

t h o

d o

f t r a

v e

l

number of people

Travelling to work

0 50 100 150 200

car

bus

bike

walk

train

car bus bike walk train

method of travel

n u m b e r

o f

p e o p l e

Travelling to work

50

100

150

Travelling to work

car

bus

bike

walk

train

method number

car 154

bus 68

bike 35

walk 22

train 10

Travelling to work

0% 20% 40% 60% 80% 100%

trainwalkbikebuscar




Time series graph

Scatterplot

Scatterplots show the relationship between two variables.

For each of the following, state whether a column graph, pie graph, time series graph

or scatterplot would be most appropriate.

a The Student Council has collected data on the percentage of students in each year

level who smoke.

b A scientist has collected data on the number of road deaths in Queensland over the

last ten Easter long weekends.

c The Student Council has collected data on the sources of its income, and wishes to

know which fund-raising activities have been most successful.

a

b The scientist is interested in the trend in road deaths, and a time series graph will

show this.

c This data is best displayed as a pie graph, which will show what percentage of total

income comes from each source.

EXAMPLE 3.5

650

600

550

500

450

45

40

35

'000

moving annual total car sales

national

our state

'000 NEW MOTOR VEHICLE SALES

MAY'94 MAY'95 MAY'96 MAY'97 MAY'98 MAY'99 MAY'00 MAY'01

Time series graphsshow trends over time. Showing morethan one line graphon the same axes al-

lows for two trendsto be compared.

length (m)1.5

2

2.5

3

0 5 10 15 20 25 30 35

a g e

( y e a r s

)

Length of dugongs

Given some data and a question to be answered, your job is to decide how to best display the story

that the data have to tell. Often one of the above graph types will be useful. If so, use software (for example, the graphing program in Microsoft Excel) to draw the graph.

This data is best shown as a column graph. Column graphs can show the frequencies

of different categories, which is what is wanted here.




The column, pie and time series graphs above are very useful. They clearly display important

patterns in the data, but they are all one-dimensional. They only show a count, a percentage,

a trend or a relationship. It is instructive to view some graphs that are able to clearly show

relationships between multiple variables.

The graphs in the example and the exercises are from The Visual Display of Quantitative Information

by Edward R. Tufte, and the website Gallery of Data Visualisation

This graph is considered by many to be the best graph ever drawn. It was drawn by Charles Joseph

Minard, a French engineer, in 1885 and shows the terrible fate of Napolean’s army in Russia. It

combines a time series graph with a data map.

The grey band represents the march to Moscow. The width of the band shows the size of the army(initially 422 000 men) at various stages on the march. Numbers are given at intervals, which

show the size of the army at that point. The black band represents the return journey. It is linked

to a temperature scale near the bottom of the map, as the severe cold of this Russian winter was

the cause of death of many of the soldiers. The map shows six variables: the size of the army at

different stages of the campaign, the location (considered as two variables), the direction the army

was travelling, the temperature during the retreat, and dates during the retreat.

Besides showing the movement of the main army, the graph also shows the movement of soldiers

sent to protect the rear and the flank of the army. It shows the horrible human cost of crossing the

Berezina River, and how the army finally struggled back to Poland, with only 4 000 men surviving

the march to Moscow.

( .http://www.math.yorku.ca/SCS/Gallery/)

GRAPHICAL EXCELLENCE




1 For each of the following, state which graph or graphs (column graph, pie graph, time series

graph or scatterplot) would best display the data.

a The student council has data on the number of pieces of litter on the oval at the end of

each lunch hour for the past fortnight.

b A trade union wants to inform its members about how its budget is allocated for the next

financial year.

c Johann Kepler carefully gathered data on the orbital period of a planet (i.e., the number

of years it takes to revolve around the Sun) and the average distance that the planet is

from the Sun. He was hoping to find a relationship between the two.

d Prior to a new dam being built, an ecologist has gathered data on the number of each

species of animal currently living in the affected area.

e A cricket statistician has data on which part of the boundary Steve Waugh has hit each

of his fours in his Test career.

f A researcher has collected data on the monthly unemployment rate and inflation rate over

the past twenty years. He wants to know if there is a relationship between the two.

2 This graph, showing the life cycle of the Japanese beetle, is very rich in the information it

provides. Study this graph carefully. What does this graph tell you about the life cycle of the

Japanese beetle?

3

EXERCISE 3E.1

Causes of mortality in the army in the East

April 1854 to March 1855

non-battle

battle June

July August

May April 1854

March

February

September

October

November

December

January 1855

Jan Feb Mar Apr May Jun Jul Aug Sept Oct Nov Dec

Florence Nightingale is best known as the mother

of modern nursing. In statistical circles she is

also known as one of the early proponents, along

with Playfair, of the visual display of data. Shewas the originator of the Coxcomb graph. This

graph was used to vividly show that the major-

ity of deaths during the Crimean War were at-

tributable to non-battle causes (largely poor san-

itation) and hence were preventable.

a

b Comment on the effectiveness of this graph.

In this graph, the number of deaths is proportional to the area. What proportionof deaths in January was caused in battle?

1855




GRAPHICAL INTEGRITY

This graph shows road deaths in Tasmania from 1970 to 1994. It clearly shows how the number

of road deaths is decreasing, and the effects of three major policy changes: compulsory seatbelts,

random breath testing and speed cameras. This graph has graphical integrity. It tells the story of

road deaths in Tasmania clearly and honestly.

Graphical integrity means that the visual display must be honest in how it shows the data.

1972 1983 19930

20

40

60

80

100

120

140

after compulsoryseatbelts

after compulsoryseatbelts

after 0.05randombreathtesting

after 0.05randombreathtesting

after speedcameras

after speedcameras

Tasmanian road deaths

1981 8.8

2001 13.7

1999 13.4

1997 12.6

1995 11.9

1993 11.41991 10.9

1989 10.41985 9.6

Fuel economy standardsfour cylinder cars (1981 – 2001)

(in kilometres per litre of fuelhighway driving)

Average distance travelled

Lie Factor = ratio depicted by the graphic

ratio in data

The graph shows the fuel economy standards to be met by manufacturers of four cylinder cars from

1981 to 2001. The numbers on the right represent the standard for each year, in kilometres driven

per litre.

From the data, the ratio of distances travelled per litre of fuel is 13:7

8:8 + 1:56: Now the graph

implies that the length of each horizontal line represents the standard for that year. Based on the

lengths of the lines representing 1981 and 2001, the ratio of line lengths is 787 + 11:1. The Lie

Factor quantifies the visual effect of misleading graphs:

For this graphic Lie factor = 11:1

1:56 + 7:1

The ratio of line lengths is 7:1 times as great as the ratio of their corresponding data values.




Study this graph carefully. Discuss why it is dishonest in its state-by-state comparison.

Date Index

Jan 17 3240:7Jan 20 3207:5Jan 21 3189:4Jan 22 3216:5Jan 23 3163:9Jan 24 3254:2(y-scale from 3120 to 3280)

EXAMPLE 3.6

Basic parliamentary salariesInterstate comparison – October 1993

40000

45000

65230

50000

55000

60000

65000

70000

$

Fed. Qld NSW Vic. SA WA NT ACT Tas.

n e w

s a

l a r y

3120

3140

3160

3180

3200

3220

3240

3260

3280

1 7 / 0

1 / 0 1

1 8 / 0

1 / 0 1

1 9 / 0

1 / 0 1

2 0 / 0

1 / 0 1

2 1 / 0

1 / 0 1

2 2 / 0

1 / 0 1

2 3 / 0

1 / 0 1

2 4 / 0

1 / 0 1

All ordinaries graph

i n d e x

date

If you turn to the financial section of any issue of newspaper and look for theyou will see time-series graph similar to the first graph below

The Courier-Mail ,, a .

all

ordinaries graph

Compare the Federal and Tasmanian salaries.

From the data, the ratio of Federal salaries to Tasmanian salaries is given by (67 500)

(47000) = 1:44.

By measuring the bars in the bar graph, the ratio of the height of the Federal salaries bar

to the height of the Tasmanian salaries bar is given by 287 = 4. The Lie Factor is given by:

Lie Factor = ratio depicted by the graphic

ratio in data

= 4

1:44

= 2:78

The Lie Factor equals 2:78. An honest graph would have a Lie Factor equal to 1.

All Ords graph with this data




Even though the vertical scale starts at 3120 and not 0, this graph has graphical integrity. The

differences between this graph and the previous one are:

Line vs column graphWhen viewing a column graph, the reader first sees and mentally compares the heights and areas

of the columns. This mental picture is not formed so readily with line graphs.

Seeing changeThe primary purpose of the all ords graph is to show the change in the all ords index, as that is

what people who view the graph want to know. By restricting the range on the vertical scale,

the change is more apparent.

The audiencePeople who use all ords graphs know what the graph shows. If this graph was shown to a class

of Year 8 students, much care would have to be taken to ensure that the students realised that

the changes that the graph shows are in reality not as large as they appear.

1 Calculate the Lie Factor for The

Shrinking Family Doctor graphic.

Note that this graphic uses area to

show one-dimensional data. The size

of the effect is given by the compara-

tive areas of the doctors, and not their

heights. If the shapes are similar, as

they are here, the ratio of areas is thesquare of the ratio of heights.

Area ratio =

µHeight 1

Height 2

¶2

2 a

b

EXERCISE 3E.2

4·

%55 4

· %60 4

· %90

5·

%00

first3-month period

second3-month period

third3-month

period

fourth3-month period

Note that the fontsize increases as thesize of the doll in-creases. Why do youthink the advertise-ment was designedthis way?

Calculate the Lie Factor for this advertisement which claimsto show how the interest rate increases every

three months if you keep your moneyin this account for up to year a .

1:2 247 RATIO TO POPULATION

1:3167

1 : 4 2 3 2

THE SHRINKING FAMILY DOCTOR

Percentage of Doctors Devoted Solely to Family Practice

196427%

197516.0%

199012.0%

8 023

6 694

6 212

Use the largest and smallest graphics

to determine the area ratio.



CLASS ACTIVITY

GATHERING DATA FROM YOUR CLASS


3 Date $US

Jan 2 55:45Jan 3 55:63Jan 4 56:48Jan 5 55:59Jan 8 57:03

Jan 9 56:81

This table shows the value of the Australian dollar from 2 January,

2001 to 9 January, 2001.

a Enter this data into your graphics calculator, and display it

using a y-scale from 53.5 to 57.5.

b Now change the y-scale so it is from 0 to 57.5:

c Comment on the effectiveness of the second graph.

4

Some statisticians think it has earned the title

The Worst Graph in the World, partly because

it gives a clear picture that is exactly opposite

to what was intended.

a What message do you think the graph is

trying to convey? b What message does it convey, at least at first glance?

c List the features (or lack of features) that put it in the running as The Worst Graph in the

World.

d Give some suggestions to the graphic artist who created this graph about how the graph

could be improved.

5 a By drawing a connected scatterplot on your graphics calculator, redraw the graphic for

Fuel Economy

b What additional information does this graph show that the original graph did not?

The remainder of this chapter is about the display and summarising of quantitative data.

1 Guess the width of the room in metres.

2 Estimate the number of hours you spend watching TV in a typical week.

3 Estimate the number of hours you spend using a computer in a typical week.

4 Estimate the number of hours you spend on the telephone during a typical week.

5

6 Estimate the amount of money you earn (or are given) in a typical week.

7 Estimate the number of hours you spend playing sport in a typical week.

8 Estimate the number of CDs that you own.

QUANTITATIVE DATA F

fees

year

University’sranking

University’sannualfees

1999

1999

1989

1965

Estimate the number of hours you spend doing homework and assignments in typical week.a

For the student of statistics, there are two main sources of datasets - interestingdata collected by others, and data collected in class. Both sources should be

used. The datasets from outside the classroom highlight how statistics is usedin the real world, while data collected in the classroom are interesting becausethey are about The survey for the class is as follows: you.

This is replica of time series graph in-tended to compare the change in the annualfees of prestigious university with its rank-ing amongst all universities.

a a

a

Standards on page as simple time series.110 a




Instructions for carrying out surveys and the survey form (which can be photocopied for class use)

While the density of the earth is not uniform (the centre of the Earth is heavier than the crust), the

value of the mean density is important in determining the Earth’s composition.

The units are grams/cm3.

The data are: 5:50 5:57 5:42 5:61 5:53 5:47 4:88 5:62 5:63 4:07 5:295:34 5:26 5:44 5:46 5:55 5:34 5:30 5:36 5:79 5:75 5:295:10 5:86 5:58 5:27 5:85 5:65 5:39

For this dataset the variable is

A histogram is a graphical display that shows the shape of the data.

To construct a histogram, we first need to make a frequency table. The most important decision is

that of class interval, since a poor choice of class interval can give a false impression of the shape

of the data. (Class interval is sometimes referred to as ‘bin width’.)

For the above dataset, one choice is to group the data into these intervals - 4.00 to 4.09, 4.10 to

4.19, up to 5.70 to 5.79. Here the class interval is 0.1 unit.

The frequency table and the histogram are shown below.

x tally freq.

4:0 - 4:09 j 14:1 - 4:19 0

4:2 - 4:29 0

4:3 - 4:39 0

4:4 - 4:49 0

4:5 - 4:59 0

4:6 - 4:69 0

4:7 - 4:79 0

4:8 - 4:89 j 1

4:9 - 4:99 0

x tally freq.

5:0 - 5:09 05:1 - 5:19 j 1

5:2 - 5:29 jjjj 4

5:3 - 5:39 jjjj © © 5

5:4 - 5:49 jjjj 4

5:5 - 5:59 jjjj © © 5

5:6 - 5:69 jjjj 4

5:7 - 5:79 jj 2

5:8 - 5:89 jj 2

A histogram shows the shape of the dataset and displays unusual data values. This histogram shows

a dataset that is reasonably symmetric, with the exception of a single value a long way from the

rest of the data values and another that is less far from the main group of observations.

Values significantly different from the majority of the data values are called outliers. Outliers are

of particular interest. They may be errors and hence can be discarded, or they may be the seed for

a momentous scientific discovery.

DISPLAYING QUANTITATIVE DATAG

In Henry Cavendish measured the mean density of the earth using an instrument calledtorsion balance.

1798, a

the mean density measurement.

For example, the satellite that was monitoring the ozone level in the atmosphere was programmed todiscard outliers - readings that were “too low”. It took ten years for scientists to find out that the read-ings being discarded were in fact accurate and that there was an ozone hole

above the South Pole.

are in Appendix 2.




The metric dataset

Guesses: 8 9 10 10 10 10 10 10 11 11 11 11 12 12 1313 13 14 14 14 15 15 15 15 15 15 15 15 16 1616 17 17 17 17 18 18 20 22 25 27 35 38 40

a What is the variable?

b Construct two histograms, each with a different class interval.

c Choose the class interval that gives the clearest picture of the dataset. Justify your

choice.

a The variable is the estimated width of the lecture hall.

b The first display uses the default settings of a particular

model of graphics calculator. In the second display the

class interval has been decreased to one unit.

c

Datasets are classified by shape as follows:

This is a symmetric dis-

tribution, in which the left

half looks roughly like the

right half.

This dataset is skewed, as

it has a “tail” on one side.

This histogram is called

right-skewed as the tail is

towards the right.

A multimodal dataset has more

than one “hump”. Such a dataset

may contain data from two groups

with different characteristics, so

each hump represents the centre

of one of the groups. Note that

a multimodal dataset may also be

symmetric or skewed.

EXAMPLE 3.7

One of the main purposes in constructing histogram is to show the shape of the dataset. The

choice of class interval can have an effect on the picture that the histogram reveals. Hence it iscommon to construct number of histograms for the same set of data, and choose the one that gives

the ‘best’ picture of the dataset. As histograms are very tedious to construct by hand, statisticians

use statistics program to construct histograms. Once you have constructed few histograms by

hand yourself, to get feel for the process, we recommend that you use graphics calculator or

computer for this task.

a

a

a a

a a a

Shortly after metric units were introduced in Australia, forty-four students were asked to guess,

to the nearest metre, the width of the lecture theatre in which they were sitting. The true width

of the theatre was 13.1 metres.

The second histogram is more informative. It shows

that the values of 10 and 15 appear more frequently than

neighbouring values. This pattern is common when the

data values are estimates - people often round to the

nearest multiple of 5 or 10. The pattern is not evident

in the first histogram.



CLASS ACTIVITY

GREED


1 Here is a set of 27 numbers:

1 0 6 1 6 8 5 1 3 1 3 5 1 5 7 1 0 8 5 8 1 1 8 5 8 5 1 8 3 4 5 9 1 2 2 1 6

a Construct a histogram from the numbers, using these class intervals: 1-2, 3-4, .... 17-18:

b Construct a histogram from the numbers, using these class intervals: 1-3, 4-6, .... 16-18.

c Construct a histogram from the numbers, using these class intervals: 1-4, 5-8, ..., 17-20.

d Construct a histogram from the numbers, using these class intervals: 1-5, 6-10, ..., 16-20.

e Describe the shape of each histogram. How similar are they?

f Which of the above histograms gives a “truest” picture of the data? Justify your response.

2 The Rainfall dataset gives the average annual rainfall for 20 cities around the world, measured

in millimetres.

Algiers 762Athens 406Beirut 889

Berlin 583Bogota 1067Bombay 1803Cairo 25

Dublin 762Geneva 864Havana 1219

Lagos 1829La Paz 584Lima 51London 586

Madrid 432Moscow 635Oslo 686

Paris 559Rome 765Vienna 660

a Enter this data into your graphics calculator. Now draw at least three different histograms,

choosing a different class interval each time.

b Choose the histogram that best displays the data, and comment on what the data tell

about

A stemplot (also called a stem-and-leaf plot) is a display similar to a histogram. It alerts us to

outliers, and gives a picture of the overall shape. Its main advantages over the histogram are that

it is easier to construct by hand, and each data value is retained.

Because the original data is available, it is possible to make some statements about the spread and

the centre of the data. It is most effective for displaying small datasets with only positive values.

The entire class stands up. The teacher throws die twice, and adds the numbers.

This is the starting score of every student. Students who are happy with that score sit down, andrecord their score for that round. The others remain standing.

The teacher throws the die again. If the number is the students still standing add thenumber to their total. Students may now elect to sit down, and record their total, or remainstanding for another throw If number then the students still standing lose all of their

points, sit down and record score of for that round. The round continues until all studentshave sat down, or is thrown.

The game consists of rounds, and the object of the game is to maximise your score after rounds.

a

a ,

. a a ,a

a

not

is

2

20

2

5 5

STEMPLOTS H

EXERCISE 3G

rainfall in these cities around the world.




A class of year 11 Maths B students played a game called Greed as detailed above.

Each value below represents the score for one student in the class.

5 0 1 7 2 7 0 5 6 6 6 8 0 3 5 3 7 2 4 1 3 1 7 1 2 2 3 3 2 5 2 2 0 2 8 3 1 2 6 2 4

Display the data with a stemplot. Discuss what information the stemplot gives us about the

data.

We will divide each number into a stem and a leaf . The tens digits form the stem, and they

are written in a column, as shown in A. The units digit of each number is a leaf. For the first

number, 50, 5 is the stem, and 0 is the leaf. The number is added to the stemplot as shown in

B. The next number, 17, has a stem of 1 and a leaf of 7, while 27 has a stem of 2 and a leaf

of 7. The number 0 is represented with a stem of 0 and a leaf of 0. These numbers have been

added to the stemplot in C. The completed stemplot is shown is D.

Stem

0

123456789

A

Stem Leaves

0

12345 06789

B

Stem Leaves

0 0

1 72 7345 06789

C

Stem Leaves

0 0 1 7 2 0

1 7 3 22 7 4 3 8 6 43 5 7 2 145 0 6 26 678 09

D

For a stemplot to give an accurate indication of the shape of a dataset, it is important to write theleaves neatly in columns.

THE STORY THAT THE STEMPLOT TELLS US

This stemplot shows us that the majority of the students had scores under 40.

The range of the data is found by subtracting the minimum score from the maximum score.

So, the range is 80 ¡ 0 = 80.

The score of 80 is an outlier. As it is a real value, it needs to be retained in this dataset.The data appear to be bunched towards the lower scores, with a smaller number of larger scores, so

the shape of the dataset is said to be right-skewed (Note: for a stemplot, “right-skewed” means

the skew is towards the bottom of the stemplot).

The row with the greatest number of scores is the row with the stem of 2, so this group of scores,

from 20-29, is called the modal class.

There are 23 scores altogether, so the middle score, called the median, can be found by pairing

off the smallest value with the largest, then the second smallest with the second largest, and so on.

The number remaining unpaired at the end is the 12th score, which is 26.

Sometimes it may be useful to draw a back-to-back stemplot. A second set of data is added as

leaves to the left of the stem.

EXAMPLE 3.8




The students then played a second game of Greed. We can add this data to the stemplot by

adding the new set of leaves to the left of the stem rather than the right. What story does the

data tell us?Second game Stem First game

0 0 0 1 7 2 02 4 1 7 3 28 2 2 7 4 3 8 6 4

9 0 8 2 3 5 7 2 19 5 5 0 0 1 4

7 4 6 5 0 6 20 1 6 6

4 7 4 78 09

For the second game, the minimum is still 0, while the maximum is 77, so the range is

77 ¡ 0 = 77. This is not significantly different from the first game.

The modal class is 40-49. There are three scores in the 70s, but there are no outliers in this

dataset.

The dataset appears to be roughly symmetric, with maybe a slight skew towards the smaller

values. The median is 41.

ADVANCED STEMPLOTS

Advanced stemplots really is a contradictory phrase as stemplots, by their nature, should be simple

to construct. However, there may be times when a stemplot is desired and constructing it involves

a little bit more effort than usual. An advanced stemplot includes one or more of these features:

Split stemsOne purpose of a stemplot is to display the shape of the distribution. To achieve a satisfactory

display of some datasets, the stem is best split into two parts, for example, with one part

containing values from 0 to 4 and the other part from 5 to 9. Other datasets may benefit if the

stem is split into five parts: 0-1, 2-3, 4-5, 6-7, 8-9. Very occasionally it is best to split the stem

into ten parts.

Truncated data valuesFor data with a large number of significant digits, it is common to truncate the data so the leaf

consists of a single digit. Very little information is lost by doing this, as the essence of the

original data is retained. Data are truncated rather than rounded as it is easier to do.

OutliersImagine a dataset that contains an extremely large outlier. It is not sensible to extend the stem to

include the outlier, if it means including row after row of empty stems. Most computer-generated

stemplots display the outlier as a data value outside of the stemplot proper, at the top or bottom

of the stemplot as appropriate, and labelled as HIGH or LOW. It is a matter of judgement when

to adopt this approach.

EXAMPLE 3.9

Overall the scores in the dataset have increased in the second round, which suggests that the

students may have used more effective strategies.




Scaling

As with histograms, different choices on how the stem is split, if at all, may result in significantly

different “pictures” of the data. If you are not sure how the stem should be split, do it in a number

of ways and choose the display that you feel reveals the most information about the dataset.

The density dataset revisited (page 114)

Construct a stemplot from the Density dataset.

This stemplot was constructed using a freeware statistics program available from the Internet,

called NCSS 6:0 Jr. This software is on the accompanying CD.

Low 40748 8495051 052 679953 0446954 246755 0357856 123557 5958 56

Unit = 0:01

Comments:

The metric dataset revisited (page 115 )

Display the Metric dataset using a stemplot.

0 89

1 00000011111 223331 444555555551 66677771 882 02 22 5

High 27, 35, 38, 40

Comments:

To achieve the best display, NCSS Jr has split the stems intofive parts. The first row that has a stem of 1 contains the

values 10 and 11, the next row with a stem of 1 contains the

values 12 and 13 and so on.

The two peaks are visible in this display. Since the original

data are retained we can see that the reason for these peaks is

that 10 and 15 were common guesses.

There are four high outliers that are given in the ‘High’ row

at the bottom of the stemplot.

There is no scaling necessary for this stemplot.

EXAMPLE 3.10

EXAMPLE 3.11

If the values to be plotted are extremely large or extremely small, it may be necessary to scale the

data by multiplying or dividing by power of For example if the data contain decimal points,

the statistics program NCSS Jr scales the data to remove them.

a .

.

10

NCSS has chosen a two-digit stem with single digit leaves.

For example, the entry 5.50 is displayed with a stem of 55 and a

leaf of 0. NCSS arranges the leaves on each row in order, from

smallest to biggest.

The outlier is labelled as ‘Low’ and the entire value (407, repre-

senting 4.07) is given in its row.

The scale is given at the bottom. For this dataset NCSS multiplies

each value by 100 to remove the decimal point.

For example, 54

j 2 represents a value of 542. Multiply this by

the unit (0.01) to obtain the original value: 542 £ 0:01 = 5:42




Surnames dataset

I collected data from my students on the length of their first names and surnames.

Here is the dataset:

4 4 5 5 5 5 6 6 6 6 6 6 6 7 7 7 7 7 8 8 8 9 10 10 11

Display this data with a stemplot.

Stem Leaves

4 005 00006 000 000 07 000 008 0009 0

10 0011 0

Comments:

For this dataset I wish to graph one value per row. Since most

numbers contain only one digit, I need to use the digit 0 as a

placeholder for the leaves.

The Unit legend at the bottom shows a slightly different

method of giving the scale.

Unit = 4 j 0 represents 4.0

The Bradmanesque dataset

Name NameCountry Country A Average verage

D G Bradman Aus 105:72E de C Weekes WI 82:46

A D Nourse RSA 68:60A R Morris Aus 67:00S G Barnes Aus 66:53

D C S Compton Eng 61:95B Mitchell RSA 59:55

A L Hassett Aus 58:10

A Melville RSA 55:27

W J Edrich Eng 53:09L Hutton Eng 51:97

J D B Robertson Eng 51:90R S Modi Ind 49:78

V S Hazare Ind 49:77E A B Rowan RSA 48:10C Washbrook Eng 47:85

J Hardstaff jnr Eng 47:16

D G Phadkar Ind 46:16C L Walcott WI 45:00

4 5 6 7 7 8 995 1 1 3 5 8 96 1 6 7 878 29

10 5

EXAMPLE 3.12

EXAMPLE 3.13

The table displays the est cricket batting averages in the period for all batsmenwho averaged over Construct stemplot for this data.Comment on the shape, and the presence of outliers.

Bradmanesque, the oMIC Project, Geiger et al,

T -.

a

AT

1946 4940

1997Source:

As there are too many digits, we will truncate the data. All we

have lost are a few insignificant decimal places.

The data is right-skewed since the stemplot drawn on its side would

show a skew in that direction. The average scores of Bradman and

Weekes are both outliers as their Test averages are considerably

higher.




OUTLIERS

If a dataset contains one or more outliers, you have to make a decision - do you toss them out or

keep them in? As is usual with questions about statistics, the answer is, “It depends.”. There is no

“rule” you can apply to answer this question, so you have to use your judgement. Is the data value

a valid member of the dataset?

In Example 3.13, Bradman’s test average is an outlier, but it is certainly a valid member of the

dataset. There is no justification for removing it.

Similarly the high guesses in the Metric dataset are genuine guesses, so they probably should be

retained. But the low value in the Density dataset is different. While it is a genuine measurement,

the instruments that were used were very sensitive, and it was easy to make an error. The density

of the Earth is a fixed (but unknown) value, and it is likely that the other measurements are much

closer to this value. We would choose to remove this value from the dataset, on the basis that it is

probably in error due to the difficulty of obtaining accurate measurements.

There is no correct decision about retaining or removing an outlier, but you should be able to justify

the decision that you have made.

1 a Using the data in Exercise 3G, Question 1, construct a stemplot:

i without splitting the stems

ii splitting the stems into 0-4, 5-9, 10-14, 15-19 iii splitting the stems into 0-1, 2-3, ..., 16-17, 18-19 iv splitting the stems into 0, 1, 2, ... , 16, 17, 18

b Use the stemplot in part iii above to determine the median value.

c Which stemplot gives the “truest” picture of the data? Justify your response.

2 This question refers to the class survey that was discussed on page 113.

a Construct, by hand, a back-to-back stemplot of the number of hours doing homework or

assignments, comparing the male responses to the female responses. Describe the dataset

in a few sentences.

b Construct a histogram of the number of hours of paid work, using the combined data.

Describe the dataset in a few sentences.

3

0 0 0 0 30 50 90 90 120 150 155 170 210 230 270 270 340 560 570

a Construct a stemplot from this data.

b Combine this with the data from your class survey to construct a back-to-back stemplot.

c Comment on the shapes of the distributions.

d Compare the two datasets, noting any unusual or interesting differences.

e Coins commonly used in the US are the penny (1 cent), nickel (5 cents), dime (10 cents)

and quarter (25 cents). We are interested in determining how this different currency affects

the amount of change in the pockets of students. The data below is from Prospect High

School, Illinois. Draw a back-to-back stemplot comparing the data from the US high

school with data from your class. Write a few sentences about what the stemplot tells

you.97 97 15 53 139 31 38 0 50 96 89 111 85 125 98 8 46 150

83 45 84 420 74 5 133 0 155 256 0 15 7 65 98 201 59 111

EXERCISE 3H

This is the amount of change in the pockets of class of students in Rockhampton, measuredin cents. Only coins were counted, not notes.

a




4 Here are the lengths of surnames of students from a class of secondary students. Construct

histograms from this data and the data from your class survey, using the same scale, to compare

the lengths of names of students in your class. Write a few brief sentences on what the

histograms tell you. 7 7 7 5 7 6 7 5 4 5 6 5 5 5 5 6 9 67 5 6 4 8 7 11 6 4 8 7 4 6 6 8 7 8 5

5 This is the Rowing dataset. It contains the weights in kilograms of the crews participating in

an Oxford and Cambridge rowing race.

Cambridge 85:7 83:2 88:4 84:1 97:3 92:5 84:6 81:1 49:5Oxford 84:6 83:9 92:7 83:9 88:9 92:0 79:1 83:2 49:8

a Construct a back-to-back stemplot. b Comment on the dataset.

6 This is the Hitchcock dataset. The

stemplot gives the distribution of the

times (in minutes) of the videotape

versions of 22 movies directed by Al-

fred Hitchcock. What is the median

length of the movies?

8 19

10 133588811 13667912 00068

13 026 Units: 8 j 1 means 81 minutes

7

a

b Find the median area.

0 001111112222222333333444556689991 000001112233456792 112834 295 167 46

8 19 3

Unit = 1000 Example: 1 j 2 represents 1200

8 A dotplot is another graphical display. It is similar to a stemplot or a histogram, where each

data value is represented by a single dot. Two companies, The Tool Company and The Machine

Company, have made prototype devices to automatically throw softballs a fixed distance. The

dotplot below shows the results of 100 throws for each device. Each device was set to throw

each ball a distance of 55 metres.

Each company argued that its prototype is better. In a sentence or two write an argument to

support each company’s prototype.

12 machine

24 36 48 60 72

tool12 24 36 48 60 72

R ,eference: The Independent, March 31 1992

This is a stemplot showing the sizes of wildlife management areas measured in hectares.

Three of the areas

are 53, 506 and 5157hectares respectively.

Identify these val-

ues in the stemplot

alongside.




One useful statistic about a dataset is its centre.

There are two common measures of the centre of a dataset, ²

the mean (or average)

² the median.

As computers and graphics calculators are so fast and accurate at calculating these statistics, we

will confine our paper-and-pencil calculations to small datasets only. You should know how to

calculate these summary statistics, but more important than calculating these statistics is being able

to choose the measure of the centre that is most appropriate for a particular dataset.

MEAN

If we give a class two mathematics exams and we want to know in which exam the students scored

higher, we can compare the centres of the two datasets, by finding the mean (or average).

The formula for the mean is x =

Px

n

The symbolP

is the capital Greek letter sigma, and is an instruction to add the values.

The x represents each data value, while n is the number of data values.

What this formula tells us is, “add the data values, and divide by the number of values”.

4.07 4.88 5.10 5.26 5.27 5.29 5:29 5:30 5.34 5:345:36 5.39 5:42 5:44 5:46 5.47 5.50 5.53 5:55 5.575:58 5:61 5:62 5:63 5:65 5:75 5:79 5:85 5:86

There are 29 values with a sum of 157.17.

So, the mean x =

Px

n

= 157:17

29= 5.42

MEDIAN

The median is the “middle” value of a dataset. There are as many values in the dataset less than

the median as there are values greater than the median.

To find the median, simply list the data values in order, from smallest to biggest, and then

² if there are an odd number of values, the median is the middle value

² if there are an number of values, the median is halfway between the two middle values.even

We will use the Density of the Earth dataset again. What is the mean value of these densities?

EXAMPLE 3.14

SUMMARISING DATA – MEASURES OF THE CENTRE

I




a Find the median of the Density dataset.

b Find i the mean ii the median of the data in the Metric dataset.

a The 29 values in the Density dataset above are already in order. The middle value

is the 15th value, as there are 14 values smaller than it and 14 values larger. Themedian therefore is 5.46.

b i The mean value of the Metric dataset is found by x = 705

44

= 16.0 metres

ii There are 44 ordered values in the Metric dataset, so the median is halfway

between the 22nd and the 23rd data values. Since they are both 15, the median

value is 15.

CHOOSING BETWEEN THE MEAN AND THE MEDIANWhich of the two measures of the centre, the mean or the median, should we use? The answer

depends on the shape of the dataset as well as the presence of outliers, which is why you always

draw a histogram or stemplot before calculating any summary statistics.

Consider the dataset containing the weights in kilograms of

the Cambridge University rowing eight from the previous

exercise, reproduced here.

85:7 83:2 88:4 84:1 97:3 92:5 84:6 81:1 49:5

Shall we choose the mean or the median as our measure of

the centre? Calculating both statistics gives:

mean = 82.9 and median = 84.6

The mean has changed substantially, while the median has only changed slightly. The median is

said to be a robust measure, which means that it is not greatly affected by outliers. This is because

the median does not use the actual values of the data in its calculation. If the largest value of a

dataset were 10 and it changed to 10 000, the value of the median would not change. The meandoes use all of the values in its calculation, so it can be greatly affected by an outlier.

Generally the median gives a better measure if

² the data are skewed, or

² if there are one or more outliers.

Because of the outlier, the median is a better choice as a measure of the centre for this dataset. If

a set of data is symmetric, you can use either the mean or the median, as both measures would be

about the same value.

In the Density of the Earth dataset, the value 4.07 is so far from the other data values, the researcher

may decide that the value is in error and delete it from the dataset.

EXAMPLE 3.15

Looking at the dataset, we notice that all of the weights are greater than 80 kg, except one, which

is 49.5 kg. This is the weight of the cox. If we are only interested in the weights of the rowers

themselves, we can eliminate this value, and recalculate the statistics. We now have:

mean = 87.1 and median = 85.2:




If it is deleted, the median only changes slightly, from 5.46 to 5.465, while the value of the mean

rises from 5.42 to 5.47. This is a good indication that the median is preferable in describing the

centre of this dataset, especially if the outlier is retained.

MODE

The mode is the value that has the greatest frequency. Some textbooks call the mode a measure of

the centre of a dataset, but this is not very sensible, since the mode could be a long way from the

centre. (See Exercise 3H, question 3.)

The Density of the Earth dataset is multi-modal, as it has two modes, 5.29 and 5.34, both of which

occur twice. Note that neither of these is close to the mean or the median.

1 Why is the median considered to be a more robust statistic than the mean?

2 For each set of numbers below, calculate the i mean ii median.a 2 2 3 3 3 4 5 6 7 8 9 1 0 1 1 1 2 1 3 1 3 1 3 1 3 1 3 1 4 1 4

b 23 20 18 25 34 22 22 500

c 0.9 0.9 0.9 0.9 0.9 0.9 2.0 20.0

3 Using data from the class survey on the number of hours students spend on the computer each

week:

a Draw a histogram for the Male data, and one for the Female data.

b Calculate the mean and the median for each gender.

c Decide which of the two measures of centre is more appropriate. Justify your decision.

d Summarise what the data says about gender differences in the class regarding computer usage.

4 a Using data from the class survey, construct a histogram of the estimated width of the

room.

b Calculate the mean and median measurements.

c Which measure of the centre would you use, and why?

5 Give an example of a dataset with ten data values for which the median is a better measure of

the centre than the mean.

6 Of nine numbers in a dataset, eight are given: 5 5 6 6 6 7 7 8

What must be the missing value if the mean is equal to the median?

7 a What is the mode of the pocket money dataset for the Rockhampton students, in Exercise

3H, Question 3 ?

b For this dataset, is the mode a good measure of the centre of the data?

8 Calculate the mean and the median of the Hitchcock dataset, in Exercise 3H , Question 6.

Which measure of the centre should be used? Why?

9 Calculate the mean and median of the dataset (Exercise 3G, Question 2 ). Decide

which is the best measure of the centre. Justify your choice.

Rainfall

EXERCISE 3I




10 The mean height of a team of 6 basketball players was 180 cm. A new player was recruited,

and because of this, the mean height rose to 183 cm. How tall was the new player?

The other important measure of a dataset is the amount of spread or variation in the data. In other

words, are the data generally close to the centre, or are they widely spread?

Manufacturers are particularly interested in measures of spread, as controlling the amount of vari-

ation in a manufacturing process is important in quality control.

Suppose that you sell computers, 30% of which fail within a week, but the rest last for 30 years.

Only 3% of your competitor’s computers fail within a week, but the rest last for 5 years. On average

your computers last longer, but whose customers will be most satisfied with their purchase?

Often the measure of spread is more important than the measure of the centre. In Mathematics B

we study three measures of spread.

These are: ² the range

² the standard deviation and

² the interquartile range.

RANGE

The range is found using the formula:

range = the maximum value ¡ the minimum value

For example, the range of weights of the members of the Cambridge rowing team is

range = 97.3 kg ¡ 49.5 kg

= 47.8 kg

The range as a measure of spread is not very robust.

For example, if we drop the cox from the Cambridge dataset, the range changes from 47.8 kg to

16.2 kg.

STANDARD DEVIATION

A logical method of calculating a measure of spread is to find out how far, on average, the datavalues are from the centre. A large answer would indicate that on average the data are very spread

out, while a small answer would indicate that the data is generally close to the centre of the dataset.

A measure of spread based on this principle is called the standard deviation.

To illustrate how the standard deviation is calculated, consider the following dataset of the ages of

8 children:2 2 5 6 7 7 13 14

First we find the mean of the dataset: x = 56

8 = 7

Next we calculate how far each value is from the mean, by subtracting the mean from each.

SUMMARISING DATA – MEASURES

OF SPREAD

J




These are the values in the second column of the table below, labelled x ¡ x.

These values are called the deviations from the mean.

To find the average deviations, we start by adding up these deviations. Unfortunately when we add

the deviations, the sum is 0, as the positive and negative values cancel out.

There are two ways around this.

We can ignore the sign of the numbers, by taking their absolute values, and then adding these

absolute values. This results in a measure of spread called the mean deviation. The mean deviation

is not commonly used, as mathematical formulas containing the absolute value are difficult to use.

The alternative is to square the deviations, as squaring either a positive or a negative number will

return a positive answer. This is what we have done in the column headed (x ¡ x)2. The sum of

these squared deviations is 140.

x x x ¡ x (x ¡ x)2

2 7 ¡5 252 7 ¡5 25

5 7 ¡2 46 7 ¡1 17 7 0 07 7 0 0

13 7 6 3614 7 7 49

sum 0 140

We now divide this sum by the number of scores, 8, to get the average of the squared deviations:

This value is called the variance, symbolised by ¾2, which is read as “sigma squared”.

Now the units on the ages are years, while the units on the variance are years2, since we squared the

deviations. In order to get our units back to years, we need to take the square root. This gives us

a measure of spread called the standard deviation, which is shown by the symbol ¾ (pronounced

“sigma”).

We can summarise this whole process with a single, neat formula for the standard deviation:

¾ =

r P(x ¡ x)2

n

This formula says² “Find the mean, x.

² Subtract it from each of the scores, x.

² Square these values, and then add up the answers.

² Divide this total by the number of scores.

² Finally take the square root to counteract the effect of the earlier squaring”.

While it is important to understand how this formula gives us a measure of spread, you do not

need to do extensive calculations with it. Any scientific calculator, graphics calculator or statistics

software package will do the calculation for you. It is more important that you understand what

information the standard deviation is giving you about the data.

P(x ¡ x)2

n =

140

8 = 17.5




Note that we used the mean to calculate the standard deviation. If we use the mean as the measure

of centre, then we should use the standard deviation as the measure of spread.

Calculate the standard deviation for the following set of numbers: 2, 5, 4, 6, 7, 5, 6

x = 2 + 5 + 4 + 6 + 7 + 5 + 6

7

= 35

7

= 5

¾ =

r P(x ¡ x)2

n

= r 16

5

+ 1:79

Score ( x ) x x ¡ x (x ¡ x)2

2 5 ¡3 94 5 ¡1 15 5 0 05 5 0 06 5 1 16 5 1 17 5 2 4

35 16

1 Calculate by hand the standard deviation of the following sets of numbers. Check your answers

using your graphics calculator.

a 1 1 2 2 3 4 5 6

b 1 1 2 2 3 4 5 54

c 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2d 48 49 50 51 52

e 0 25 50 75 100

INTERQUARTILE RANGE

The interquartile range is the measure of spread that is normally used with the median.

The median gives us the middle of the dataset, the value that divides the dataset into two halves.

The interquartile range gives us the range of the middle 50 percent of the data values.

Here is how the interquartile range is calculated:

² Find the median. Now consider two subsets of the dataset: the set of those values

smaller than the median, and those values larger than the median. Note that neither

set contains the median itself.

² Find the median of the subset of values smaller than the median.

This gives us the first quartile, or Q1.

² Find the median of the subset of values larger than the median.

This gives us the third quartile, or Q3.

² The interquartile range, (IQR ), is found by subtracting: IQR = Q3

¡ Q1.

EXAMPLE 3.16

EXERCISE 3J.1




Find the interquartile range of the set of ages: 2 2 5 6 7 7 13 14

The median is 6.5, halfway between 6 and 7. The two sets are:

The median of the first set, Q1, is halfway between 2 and 5, so it is 3.5.

The median of the second set, Q3, is halfway between 7 and 13, so it is 10.

The interquartile range is IQR = Q3 ¡ Q1

= 10 ¡ 3:5

= 6:5

Find the interquartile range of this set of ages: 1 2 2 5 6 7 7 13 14:

The median is 6, which is the middle value. The two sets are:

Note that we do not include the median in either set.

The median of the first set, Q1, is halfway between 2 and 2, so it is 2.

The median of the second set, Q3, is halfway between 7 and 13, so it is 10.

The interquartile range is IQR = Q3 ¡ Q1

= 10 ¡ 2

= 8

If we use the median as the measure of centre, then we must use the interquartile range as the

measure of spread.

The median, Q1 and Q3 can all be calculated by all graphics calculators and statistics software.

You should understand how these statistics are calculated, but you will only be asked to calculate

them by hand if the datasets are very small. For large datasets you are expected to use either a

calculator or computer software.

FIVE-NUMBER SUMMARY

We are now able to summarise a dataset, using the five-number summary.

EXAMPLE 3.17

EXAMPLE 3.18

2 2 5 6 7 7 13 14

1 2 2 5 6 7 7 13 14

The five-number summary consists of (in order) the thethe the and the

minimum value first quartile

median third quartile maximum value

, ,, , .

Q

Q1

3

Note that the interquartile range, like the median, is robust measure. If the largest age is rather than the interquartile range would not change.

a,

9914

We will add one more data value to our set of ages, to illustrate how we find the IQR if we have

an odd number of data values.





85:7 83:2 88:4 84:1 97:3 92:5 84:6 81:1

What to do:

1 Assume that every member of the team gained 5 kilograms over the Christmas break.

What effect on the data set would this have on the

a mean b median c standard deviation d interquartile range?

2 You wish to publish this dataset in an English magazine. They have asked you to convert

all of the weights to pounds, which you can do by multiplying the number of kilograms

by 2:2. What effect on the data set does this have on the


The first question looks at the effect of adding a constant to all members of a dataset, whilethe second question looks at the effect of multiplying all members of a dataset by a constant.

The following question combines the two operations. This is called an affine transformation

of the dataset.

3 The following data is the body temperature, measured in degrees Fahrenheit, of a group of

34 people. You wish to convert these temperatures to degrees Celsius, using the formula

C = 59 (F ¡ 32):

96:3 97:0 97:1 97:1 97:3 97:4 97:4 97:5 97:6 97:6 97:8 97:897:9 98:0 98:0 98:2 98:2 98:3 98:4 98:5 98:6 98:6 98:6 98:798:7 98:8 98:8 98:8 99:0 99:1 99:2 99:3 99:4 99:5

What effect on the data set will this have on


1 Calculate by hand the interquartile range of the following sets of numbers:

a 1 1 2 2 3 4 5 6 b 1 1 2 2 3 4 5 54

c 12 12 12 12 12 12 12 12 12 12 d 48 49 50 51 52

e 0 25 50 75 100

EXERCISE 3J.2

Recall the dataset of the weights in kilograms of the members of the CambridgeUniversity rowing eight, not including the cox.

Using our original set of ages (Example 3:17), the five-number summary is:

minimum = 2

Q1 = 3:5

median = 6:5

Q3 = 10

maximum = 14

Graphics calculators and statistics software will calculate the five-number summary for any dataset.

Note that there are a variety of methods for calculating Q1 and Q3, so your calculator may not give

exactly the answers you were expecting. For large datasets they should be close, though.

f 12 3 4 2 11 8




2 For each of the sets of numbers in question 1, determine the five-number summary.

3 Consider the Rowing dataset (Exercise 3H, Question 5 ).

a Calculate, by hand, the standard deviation of each set of data with the outlier included.

b Calculate, by hand, the standard deviation of each set of data with the outlier not included.

c How much effect do the outliers have on the standard deviations?

d Calculate, by hand, the IQR of each set of data with the outlier included.

e Calculate, by hand, the IQR of each set of data with the outlier not included.

f How much effect did the outlier have on the IQR?

g

4 Use the data in the class survey on the number of hours spent playing sport.

a Draw a histogram or stemplot of the data.

b Calculate the mean and the median.

c Calculate the standard deviation and the interquartile range.

d Give the five-number summary.e Decide which of the two sets of summary statistics is more appropriate. Justify your

answer.

5 Use the data in the class survey on the amount of money earned in a given week, for both

Males and Females.

a Display the data using appropriate graphical displays.

b Calculate the mean and the median for each set of data.

c Calculate the standard deviation and the interquartile range of each set of data.

d Give the five-number summary.

e Decide which of the two sets of summary statistics is more appropriate. f Make a statement comparing and contrasting these two datasets.

6 Consider the Rainfall dataset (Exercise 3G, Question 2).

a Calculate, using any method you wish, the standard deviation and the interquartile range.

b Give the five-number summary.

7 Give an example of a dataset with five data values that has a standard deviation of 0.

8 A dataset consists of 1000 zeros, 1000 ones, 1000 twos and 1000 threes, that is, there are 4000

data values altogether.a Find the median. b Find the first quartile, Q1.

9 Match each histogram with a set of summary statistics. Note the horizontal scale.

a Amean 10.5standard deviation 1.4median 10.7IQR 2.0

0 5 10 15 20

Give the five-number summary for both datasets, with and without the outlier.




b B mean 10.1standard deviation 2.7median 10.1IQR 4:2

c C mean 10.2standard deviation 2:1median 10.5IQR 2.5

d D mean 10.2standard deviation 4.1median 11.9IQR 6.8

e E mean 8:8standard deviation 2:8

median 8:0IQR 1:9

A simple boxplot (also called a box-and-whisker plot) shows the five-number summary graphi-cally. Here is how it is constructed:

² Draw a number line from the minimum value to the maximum value. The number line can

be drawn either vertically or horizontally.

² Mark the minimum, Q1, the median, Q3 and the maximum on the number line.

The interquartile range is shown by drawing a rectangular box between Q1 and Q3.

The median is displayed as a line across the box. Lines, called whiskers, are drawn from

the sides of the box out to the minimum and maximum values.

The following example will clarify this.

BOXPLOTS K

6 8 10 12 14

4 7 10 13 16

5 11 18 24 30

2 6 9 13 16




Draw a simple boxplot from the five-number summary in Example 3.16.

The 5 number summary is:

minimum = 2, Q1 = 3.5, median = 6.5, Q3 = 10, maximum = 14.

The simple boxplot can occasionally give a misleading picture of the dataset. A long whisker

implies that the dataset is skewed in that direction, but the reality may be that there is a single large

outlier to which the whisker is drawn. With a simple boxplot there is no way of interpreting a long

whisker.

A standard boxplot (usually just called a boxplot) is a bit more difficult to construct, but is a much

more useful display, as outliers are specifically identified. A long whisker on a standard boxplot isa strong indicator that the data is skewed in that direction. An example will clarify this.

Construct both a simple and a standard boxplot using the Density of the Earth dataset,

using a graphics calculator. Compare the two displays.

From the simple boxplot alone we can not tell if the long

whisker to the left is because of a dataset that is skewed to

the left, or due to one or more outliers.

A standard boxplot better shows the information, and indicates

that this dataset is symmetric (since the box is symmetric), with

a possible slight skew to the left (since the left whisker is longer

than the right whisker), and has

CONSTRUCTING A STANDARD BOXPLOT BY HAND

The box of a standard boxplot is identical to that of a simple boxplot. It is only the whiskers that

may be drawn differently. If there are no outliers, the simple boxplot and the standard boxplot are

identical.With a standard boxplot, the left whisker extends to the smallest data value that is within 1.5 IQRs

of Q1. All data values smaller than that are marked separately, as outliers.

The right whisker extends to the largest data value that is within 1.5 IQRs of Q3. All data values

larger than that are marked separately, as outliers.

Here is an example to clarify this process, using the Density of the Earth dataset:

Draw the box

Find the five-number summary.

Draw the number line, the box and the median.

min = 4.07 Q1 = 5.295 median = 5.46

Q3 = 5.615 max = 5.86

EXAMPLE 3.19

EXAMPLE 3.20

2 3 4 5 6 7 8 9 10 11 12 13 14

one outlier.




Draw the left whisker

Calculate the interquartile range.

Multiply the interquartile range by 1.5:

Find Q1 ¡ 1.5 £ IQR.

Mark any data value smaller than 4

.815as an outlier.

Extend the whisker to the smallest data

value that is not marked as an outlier.

IQR = Q3 ¡ Q1 = 5:615 ¡ 5:295 = 0:32

0:32 £ 1:5 = 0:48

5:295 ¡ 0:48 = 4:815

In this case there is one outlier at 4

.07

.

Draw the left whisker to 4.88.

Draw the right whisker

Find Q3 + 1.5 £ IQR.

Mark any data value larger than 6.095 as

an outlier.

Extend the whisker to the maximum data

value in the dataset.

5:615 + 0:48 = 6:095

In this case there are no outliers.

Extend the right whisker to 5.86.

Comparing two or more datasets

One of the uses of boxplots is to compare two or more (sometimes many more) datasets graphically.

Lord Raleigh was one of the earliest scientists to study the den-

sity of nitrogen. In his studies, he noticed something peculiar.

The density of nitrogen produced from chemical compounds

tended to be smaller than the density of nitrogen produced from

the air. However, he was working with fairly small samples,

and the question is, was his conjecture correct?

Lord Raleigh’s measurements which first appeared in Proceed-

ings, Royal Society (London, 55, 1894 pp. 340-344) are shown.

The units are the mass of nitrogen filling a certain flask under

specified pressure and temperature.

Chemical Atmospheric

2:301 43 2:310172:298 90 2:309862:298 16 2:310102:301 82 2:310012:298 69 2:310242:299 40 2:310102:298 49 2:310282:298 89 2:311632:300 74 2:309562:30054

We will use a graphics calculator to draw side-by-side boxplots

of the two datasets. The boxplots make it clear that the two

samples have significantly different masses, and that the vari-

ation of the chemical sample is much larger than that of the

atmospheric sample.

Lord Raleigh thought that the difference in mass may have been

due to an undetected gas. He was correct, and this eventually

led to the discovery of the gas argon.

EXAMPLE 3.21

4.0 4.2 4.4 4.6 4.8 5.0 5.2 5.4 5.6 5.8




In 1970, the U.S. Congress instituted a random selection process for drafting young men to

serve in the Vietnam war. All 366 possible birth dates were placed in plastic capsules in a

rotating drum and were selected one by one. The first date drawn from the drum received

draft number one and eligible men born on that date were drafted first. The next date drawn

received draft number two, and eligible men born on that date were drafted second, and so on.What does the boxplot below reveal about the selection process?

Note that month number 1 = January, month number 2 = February, and so on.

The most striking feature is that months towards the end of the year tended to have lower draft

numbers. We find that the capsules were placed in the drum in order, starting with January

dates, so December dates were added in last, and the capsules were not fully mixed! Males

born towards the end of the year had a greater chance of getting a small draft number and

hence being sent to fight in the Vietnam war.

1 Draw, by hand, a simple boxplot for each of the following sets of numbers. Check with your

graphics calculator.

a 2 3 3 4 4 5 6 7 9 10 10 b 0 0 1 2 3 3 4 5 5 6 6 13

c 0 6 6 6 6 7 7 8 15 25

2 Draw, by hand, a standard boxplot for each set of numbers in question 1 . If possible, check

your answers with your graphics calculator.

3 Draw, by hand, side-by-side boxplots for the three sets of numbers below.

A: 11 20 19 2 1 16 12 13 9 13 B: 6 10 13 8 10 9 0 1 10 13 10 8

C: 4 7 3 2 16 15 4 4 3 4

4 Use the data from the class survey on the number of hours of TV watched each week.

a Calculate the five-number summary for each gender.

b Draw side-by-side boxplots of the data.

c Calculate the mean and the median of each set of data.

d Decide which of the two measures of centre is more appropriate.

e Depending upon which measure you chose in d , calculate the standard deviation or the

IQR.

f Write a brief report on what these datasets tell you.

EXERCISE 3K

EXAMPLE 3.22

100

200

300

400

1 2 3 4 5 6 7 8 9 10 11 12

1970 Draft Lottery

month number

d r a f t

n u m b e r




5 Using the data from the class survey on the number of hours spent on the telephone in a typical

week, draw side-by-side boxplots of the Male data and the Female data. Comment on what

the boxplots reveal.

6 While baseball is not a major sport in Australia, it is a rich source of interesting statistics.

The Home Run dataset contains data about the number of home runs hit in a single season is

US baseball. From 1927 to 1960, the holder of the record was Babe Ruth. In 1961, Roger Maris eclipsed the record with 61 home runs. In 1998, two players broke this 27 year old

record, with Mark McGuire hitting 70 home runs and Sammy Sosa hitting 66. Hank Aaron

holds the record for the greatest number of career home runs.

Using a graphics calculator, put this data into five lists. Using a variety of graphical displays or

summary statistics, determine who was the greatest home run hitter of all time. Write a report

supporting your choice. Include in your report the summary statistics and graphical displays

you used to support your decision.

7

Note:

23/25 means she scored on 23 outof 25 attempts.

Sharelle had less time on the court

during the World Championships

than she had playing for Melbourne

Phoenix.

Game

World Championships Commonwealth

1999 Season 20001 23 / 25 34 / 40

2 18 / 22 27 / 32

3 22 / 28 28 / 29

4 12 / 14 37 / 41

5 7 / 9 40 / 49

6 32 / 40 25 / 31

7 20 / 26 36 / 44

8 4 / 6 19 / 19

9 6 / 8 27 / 31

10 35 / 4311 50 / 55

12 42 / 50

13 41 / 48

14 41 / 49

15 32 / 35

Put this data into lists in your graphics calculator. Using a variety of graphical displays and/or

summary statistics, compare her shooting statistics in the 1999 World Championships and the

2000 Commonwealth Bank season. Write a report on your comparison, using the graphical

displays and summary statistics you have generated.

Yr 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35

No 0 4 3 2 11 29 54 59 35 41 46 25 47 60 54 46 49 46 41 34 22 6

Yr 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76

No 13 27 26 44 30 39 40 34 45 44 24 32 44 39 29 44 38 47 34 40 20 12 10

Yr 57 58 59 60 61 62 63 64 65 66 67 68

No 14 28 16 39 61 33 23 26 8 13 9 5

Yr 86 87 88 89 90 91 92 93 94 95 96 97 98

No 3 49 32 33 39 22 42 9 9 39 52 58 70

Yr 89 90 91 92 93 94 95 96 97 98

No 4 15 10 8 33 25 36 40 36 66

Sammy Sosa

Babe Ruth

Hank Aaron

Roger Maris

Mark McGuire

Sharelle McMahon was a goal at-

tack for the Australian netball team

that won the World Championship

in 1999, and for the Melbourne

Phoenix in the 2000 Common-

wealth Bank season.

Here are her shooting statistics

for the 1999 World Championships

and the 2000 Commonwealth Bank

season.




8 Refer to the table in Example 3.13 which displays

the Test batting averages in the period 1946-49for all batsmen who averaged over 40.

a Draw a boxplot.

b Is the Don Bradman batting average an outlier,

over the period 1946-49?

9 Following are the mean minimum and maximum monthly temperatures for Perth and Adelaide,

arranged chronologically (i.e., Jan, Feb, Mar, etc). Using any method or methods of displaying

and/or summarising the data that you feel are useful, compare both the maximum temperatures

and the minimum temperatures in Adelaide and Perth.

Perth

max. 31.4 31.7 29:5 25:2 21.4 18.8 17.7 18.3 20.0 22.3 25.4 28.5min. 16.7 17:3 15:7 12:6 10:2 9.1 8:0 7:9 8:7 10.1 12.3 14.6

Adelaide

max. 27.8 27.9 25.4 22.1 18.4 16.0 14.9 15.9 17:9 20.8 23.5 25:5min. 15:5 15.7 14.2 11.6 9.4 7.4 6:8 7.5 8.6 10.4 12.3 14.2

10 A marketing consultant observed 50 consecutive shoppers at a small independent supermarket,

and recorded how much money each shopper spent in the store.

2.32 6.61 6.90 8.04 9:45 10:26 11.34 11.63 12.66 12:9513.67 13:72 14:35 14:52 14:55 15:01 15:33 16:55 17:15 18:2218:30 18:71 19:54 19:55 20:58 20:89 20:91 21:13 23:85 26:04

27:07 28:76 29:15 30:54 31:99 32:82 33:26 33:80 34:76 36:2237:52 39:28 40:80 43:97 45:58 52:36 61:57 63:85 64:30 69:49

Using any method or methods of displaying and/or summarising the data that you feel are

useful, write a report to the supermarket manager about the spending pattern of shoppers.

11 Immediate release medications quickly liberate their drug con-

tent into the body, with the maximum concentration reached in

a short time, followed by a rapid decline in concentration. Sus-

tained release medications, on the other hand, take longer to reach

maximum concentration in the body and stay active for longer pe-

riods of time.

Age Circ Csrc

33 181:8 195:740 466:9 167:041 136:0 217:343 221:3 375:725 195:1 285:730 112:7 177:224 84:2 220:344 78:5 243:542 85:9 141:633 85:3 127:238 217:2 345:239 49:7 112:143 190:0 223:4

a Construct histograms for both irc and src.

b Determine which summary statistics are most appropriate, and justify your decision.

c Calculate the summary statistics.

A study of two such pain relief medications compared the maxi-

mum concentration in mg/mL of immediate release codeine (irc)

with sustained release codeine (src). Thirteen healthy patients

were randomly assigned to one of the two types of codeine and

treated for 2.5 days. After a 7-day wash-out period, the same

patients were given the other type of codeine. Thus, each patient

received both treatments.




d Construct side-by-side boxplots of the concentrations of irc and src.

e Write a report comparing the maximum concentration of codeine in the blood for irc and

src.

12 For each of five datasets, a histogram and a boxplot have been constructed. Match each

histogram with its boxplot.

a A

b B

c C

d D

e E



MODELLING

VEXPLORING DATA (Chapter 3) 139

1

a

² What is the proposed population in this study?

² What is the population which would be represented by a sample of 200 people at Metro bus stops?

² How similar are these populations?

b If you were planning this survey, how would you design the collection of the data?

Some questions you may need to consider are:

² How many bus stops would you use?

² What times of day would you conduct the survey?

² What training would you provide the volunteers conducting the survey?

² How would you handle a situation with someone who refused to answer

the survey?

² What specific questions would you ask?

2 The design of controls

Subject Right-hand Left-hand

Subject Right-hand Left-hand

thread thread thread thread

1 113 137 7 103 148

2 130 133 8 107 87

3 138 115 9 103 146

4 87 103 10 104 135

5 116 145 11 89 93

6 96 107 12 100 116

Is the design of this study satisfactory? Some questions you may need to consider are:

The design of controls has large effect on how easily people can use them. welveright-handed people were asked to turn knob that moved an indicator There were twoidentical instruments, one with right-hand thread, and one with left-hand thread. Thetable below gives the times required (in seconds) to move the indicator fixed distance.

The project co-ordinator hoped to show that right-handed people find right-hand threadseasier to use. Analyse the data by using what ever graphical displays or summary statis-tics that you feel are suitable. rite report to the project director on your findings. In-

clude in your report any graphs or calculations needed to support your argument.

a Ta .

a aa

W a

Bus stop view of breastfeeding mothers

Attitudes on breastfeeding in public are the subject of a study in southern Tasmania.

The study, by the Department of Community and Health Services and the Nursing Mothers

Association (now known as the Australian Breastfeeding Association), will involve volun-

teers surveying at least people at Metro bus stops.

Department nutritionist Roger Hughes said bus stops would be used as survey sites because

they provided a broad cross-section of society and because Metro buses would later be used

in a campaign to promote breastfeeding in public as a normal activity.

200

Mr Hughes said that while research showed a large number of developmental, health

and economic benefits were associated with breastfeeding, mothers were often reluctant

to breastfeed in public.



PROBLEM SOLVING L

EXERCISE 3L


1 A set of eleven data values has a mean of 8 and a standard deviation of 2. At least one of the

data items is the value 8. If this single value is removed, without replacement, from the dataset, what will be the effect on the standard deviation of the dataset?

A The standard deviation will decrease. B The standard deviation will increase.C The standard deviation will not change. D There is insufficient information to answer this question.

Justify your answer.

2 The Australian Bureau of Meteorology collects

data on rainfall across Australia. The table

shows the mean monthly rainfall in Broken

Hill as well as the median monthly rainfall.

(in millimetres) 1900 to 1990

Month Mean Median

Jan 23 9Feb 24 10Mar 18 9Apr 19 9May 22 13Jun 22 15Jul 17 15

Aug 19 17Sep 20 12Oct 25 15

Nov 19 10

Dec 20 7

a Note that the median monthly rainfall in

January is much smaller than the mean

monthly rainfall. What does this imply

about the shape of the distribution of the

rainfall data for the month of January?

b Which measure of central tendency, the

mean or the median, is more appropri-

ate for describing rainfall in Broken Hill?

Justify your answer using knowledge of

mean and median.

c Use the above table to estimate the total

yearly rainfall for Broken Hill.

d

3 In an exam, seven students, A to G, obtained

the following marks: A B C D E F G

18 26 34 42 50 58 66a Calculate the mean and standard deviation

of this dataset. b Re-scale the marks so the new dataset now has a mean of 60 and a standard deviation

of 15. You would like to retain the relative differences in the exam results.

4 Prove that any dataset that consists of seven consecutive integers has a mean and a standard

deviation that are also both integers.

5 Dataset A is the amount of change in the pockets of 100 people.

Dataset B is the amount of change in the pockets of these people after I gave them each 50c.

Dataset C is the amount of change in the pockets of these people after I doubled their money.

Assume that I draw side-by-side boxplots of these 3 datasets. Describe what you would

expect to see.

In the north of Australia, the wet season occurs from November to April. Broken Hill,

in central Australia, is occasionally drenched by a northern storm during these months.

These storms tend to drop a large amount of rain in a comparatively short time. How

does the table reflect this fact?

Average monthly rainfall in Broken Hill





average

bivariate data

boxplot

centre

class interval data

experiment

five-number summary

frequency table

histogram

interquartile range

leaf

maximum

mean

mean deviation

median

minimummodal class

mode

multi-modal

outlier

quartile

random variation

range

1

a For this study, identify the i population ii variables.


c Should the scientist take a census or a sample? Justify your decision.

d Is the thickness of the eggshell a discrete or a continuous variable?

e

f

2

a Explain why this may result in a biased sample.

b Explain how she can obtain an unbiased (or less biased) sample.

Aa

.

scientist is interested in determining the number of days that different varieties of potatoes will last before they go rotten. She goes to her local supermarket and buys dozen potatoes of each variety

Which of the methods of gathering data discussed in the chapter does the biologist use?

Outline cost-effective and time-effective method of obtaining simple random sample.a a

A biologist is varying the amount of calcium in the diet of battery hensto determine whether the amount of calcium in the diet affects the thicknessof the shells of their eggs.


6 The two most common measures of the centre of a dataset are the mean and the median. The

median is called a robust measure of location as a very large or very small value in the dataset

does not have undue influence on the value of the median.

Two less commonly used measures of location are the ‘mid-hinge’ and the ‘mid-range’. The

mid-hinge is the value that lies exactly halfway between the first quartile and the third quartile.

The mid-range is the value that lies exactly halfway between the largest and smallest values.

Which of these measures of the centre, the mid-hinge or the mid-range, is the more robust

measure of the centre? Justify your answer.

c Which type of graph (column graph, pie graph, time series graph or scatterplot) would best display the data?

robust

sample

scatterplot

simulation

skewed spread

standard deviation

stem

stemplot

symmetric

trial

variance





3 Compare and contrast the terms

a graphical integrity b graphical excellence

4 Wind roses display the direction, strength and frequency of winds at a

location, for a given season and time of day. The percentage of calms

is represented by the size of the centre circle. Each branch represents

wind coming from that direction.

The varying thicknesses of the branches represent wind speeds from

that direction. The length of each branch segment is proportional to

the percentage of winds in that speed range. See the legend for more

detail.

5

Refinery 45 30 38 42 63 43 102 86 99 63 58 34 3755 58 153 75 58 36 59 43 102 52 58 30 2140 141 85 161 86 71

DoE 12:5 20 4 20 25 170 15 20 15

a i Draw a stemplot of the refinery data.

ii Discuss the shape of the data, and the presence of any outliers.

iii Determine the median from the stemplot.

b Enter the data into two lists of your graphics calculator.

c Draw a number of histograms of the refinery data. Which class interval best displays

the data?

d

A aa W awind rose is rich source of information about the wind strength, direction and frequency

at location. rite report comparing the wind in Brisbane at a.m. in summer and winter.9

A.

refinery is planning to increase its production, which will increase the overall emissionlevel of carbon monoxide by the refinery In order to establish the level of carbon monoxideemissions by the refinery prior to the increase, both the refinery and the Department of theEnvironment have measured the existing carbon monoxide levels.

Here is the data that each gathered.

Based on the histogram, which measure of the centre is best used: the mean or themedian?

NNE

E

SESSW

W

NW

calm

calm

20%

1-10

75%

calms11-20 21-30 >30

km/h

Brisbane, summer 9 am Brisbane, winter 9 am





e For each set of data, calculate:

i the mean and standard deviation

ii the median and interquartile range

f Draw side-by-side boxplots of the two datasets.

g Comment on the measurements made by the two bodies.

h Can you explain why the measurements are so different?

1

a a 40 degree day in Rockhampton

b a 35 degree day in Brisbane.

2 Ms. Sweetwater’s biology class had a standard deviation of 2.4 on a standardized test,

while Ms. Quincy’s biology class had a standard deviation of 1.2 on the same test.

What can be said about these two classes?

A Ms. Sweetwater’s class is more homogeneous than Ms. Quincy’s.

B Ms. Quincy’s class is less heterogeneous than Ms. Sweetwater’s.

C Ms. Quincy’s class did not do as well on the test as Ms. Sweetwater’s.

D Ms. Sweetwater’s class performed twice as well on the test as Ms. Quincy’s.


3 A distribution of 6 scores has a median of 21. If the highest score increases 3 points,

the median will become

A 21 B 21:5 C 24 D Cannot be determined without additional information.

E None of these.


4 If you are told a population has a mean of 25 and a standard deviation of 0, what must

you conclude?

A Someone has made a mistake.

B There is only one element in the population.

C There are no elements in the population.

D All the elements in the population are 25.

E None of the above.

5 If the mean and median of a distribution are 6 and 5 respectively, then the distribution is

A skewed left B not skewed C skewed right

D symmetric E multimodal

6 A manufacturer of shock absorbers claims that its product lasts longer than the shock

absorbers supplied with the car originally. To test this claim, eight cars each had one

shock absorber from each manufacturer installed on the rear of the car, and were driven

until the shock absorbers were no longer effective. The number of kilometres driven is

given in the table which follows.

Explain briefly how you would decide which of the following two eventsis the more unusual:





a Draw a back-to-back stemplot of this data.

b Calculate the mean and the standard deviation.

c Calculate the median and the IQR.

d Draw side-by-side boxplots.

e Comment on the manufacturer’s claim.

Car Original shock absorber Replacement shock absorber

1 42:6 43:82 37:2 41:33 50:0 49:7

4 43:9 45:75 53:6 52:56 32:5 36:87 46:5 47:08 39:3 40:7

Thousands of kilometres



CHAPTER 4Modellingdata withfunctions

SUBJECT MATTER

This chapter is optional, as the content is in thelearning experiences in the syllabus, but not the subjectmatter.

However, the topic does allow students to apply their knowledge of various functions to real-life datasets.Gathering and analysing bivariate data is part of thestudy of many disciplines, including all sciences, business and economics.



HISTORICAL NOTE +In , gave his Presidential address before the anthropol-ogy section of the British Association for the advancement of Science. He talkedabout a study in which he compared the heights of tall

children with the heights of their parents. Galton madetwo interesting observations about parents who wereeither tall or short. If a parent was very tall, the childrentended to be tall, but shorter than the parent. Similarly, avery short parent tended to have short children, but taller than the parent. He called this “regression to the mean,”with the word regression meaning “to come back to”.Today the term “regression” is attached to a line that isused to represent a set of data: the .

1885 Sir Francis Galton

regression line

A FIRST MODEL (THE REGRESSION LINE) A

146 MODELLING DATA WITH FUNCTIONS (Chapter 4)

You have ten different secondary mathematics textbooks. You look on the last page of each to see

how many pages are in each book, and then put each text onto your digital kitchen scales to find

its weight.

The results are given in this table:

Number of pages 568 584 656 576 506 430 534 594 422 582

weight (g) 1178 1197 1350 1056 916 867 998 1151 779 1079

When studying univariate data, the first thing we did was look at the data using graphical displays.

We do the same with bivariate data, using a graphical display called a scatterplot.

If we assume that the weight depends to some extent on the number of pages, then the number of

pages is the independent variable, weight is the dependent variable. The data can be plotted as a

set of 10 ordered pairs, as shown.

data.

Such a line is called a line of best fit or a regression line. This line will be a mathematical model

for this dataset.

Now you could calculate some statistics on each of the variablesseparately, say, find the average weight of your sample of mathe-matics texts, and draw a boxplot to show the distribution of weights. But what you are interested in is the betweenthe number of pages and the weight.

Since there are two variables and you are studying them together,you are working with , which is also known as

paired data. A single variable by itself is called .

relationship

bivariate data

univariate data

The data appear to be roughly linear, so we might wish to find the straight line that best fits these

Now what can we say about these data? While the data do not lie perfectly in straight line, thereis general pattern that shows that books with more pages tend to weigh more.

aa



INVESTIGATION 1

FINDING A LINE OF BEST FIT

One method of finding a line of best fit would be to

choose a line that passes through as many points as pos-

sible, with half the remaining points above the line and

half below the line. But this may not always be possible.

For example, in the diagram alongside, if a line is drawn that passes through as many points as

possible, the remaining points are all above the line. Or a line could be drawn so half of the points

are above the line and half below. But I could draw such a line, and you could draw such a line,

and they might be quite different. What is needed is an objective method of finding the line of best

fit.

Not all data fit a linear pattern. In this diagram, the data

values tend to follow a curved pattern. Fitting a linear

function to this data is not appropriate.

Later in the course you will learn how to fit non-linear

functions to such data.

1

a x 0:5 0:7 1 1:1 1:8 1:9 2:8 3 3 3:2 3:9y 3:2 4:5 5 5:1 7:2 7:1 10:5 11:6 12 11:4 14:3

b x 1:2 1:7 2:5 2:8 4:5 4:8 7 7:5 7:5 8 8:8y 5 20 30 22 34 41 35 34 52 41 70

Materials needed:

For the class:

For each group, a photocopy of these scatterplots (they are also on thewebsite in a suitable size and format) on an overhead transparency and a soluble overheadtransparency pen.

one overhead projector.

For each of the scatterplots below, draw in the line that you think best fits the data. Now compare your lines with those of your classmates by overlaying the scatterplots. How closely didyour lines agree?

y

x

y

x

y

x

EXERCISE 4A

MODELLING DATA WITH FUNCTIONS (Chapter 4) 147

Enter each of the sets of data into a graphics calculator, and draw the scatterplot. For each

dataset, decide whether the data are better modelled by a straight or a curved line.



CLASS DISCUSSION

c x 0 1 1:5 1:6 2:2 2:4 2:6 3:7 4:1 5 5:1y 0 0:8 2:4 2:9 4:1 5 7 12 20 28:1 30:2

2

Area (cm2 ) 489 489 489 434 438 439 428 429 432 424

weight (g) 1178 1197 1350 1056 916 867 998 1151 779 1079

a Draw a scatterplot of this data.

b Is a linear model appropriate for this dataset?

c Discuss any interesting features of this dataset.

The diagram on the right is the scatterplot

of the weight vs page data with the least

squares regression line included. Its equation

is W = 2:2P ¡ 146 where W = the weight

and P = the number of pages.

1 What are the largest and smallest values of the

a independent variable b dependent variable?

2 What are the interpretations of the gradient and y-intercept for this situation?

3 What are the units of the gradient, and on the y-axis?

4 Use the regression equation to predict the weight of a Maths text with 450 pages.

5 Use the regression equation to predict the weight of a text with 850 pages.

6 Use the regression equation to predict the number of pages if a book weighs 1 kilogram.

When we predict the weight of a book with 450 pages, we are using interpolation, since 450 pages

is in the domain. Predicting the weight of a book with 850 pages is called extrapolation, since

this value is larger than the largest book in the dataset.

LEAST SQUARES REGRESSION LINE B

What to discuss:


An alternative to counting the number of pages of text is measuring the area of the cover The data for the same set of textbooks are given below comparing the area of the cover withthe weight.

a .,

There are a number of methods of finding a line of best fit. The most common is the least squares

method, which results in the least squares regression line (LSR line). This method is available

on all graphics calculators. In Maths B we do not need to be able to derive the equation of the

LSR line, but we need to understand that a regression line is a line that represents paired data, that

the LSR line is the most commonly used regression line, and what is meant by “least squares”. A

fuller explanation of the least squares method is given later in the chapter.

For now, we must be able to find the equation of the least squares regression line using a graphics

calculator.



A statistic called r2 which is pronounced ‘r-squared’, and called the coefficient of determi-

nation, is a measure of the proportion of the variation of the dependent variable (distance, in this

case) that is explained by the independent variable (year since 1900).

If all of the data values lay exactly on the line, then all of the variation in the dependent variable

could be explained by the independent variable, and the value of r2 equals 1 (i.e., 100% of the

variation can be explained by the independent variable).

If there is no relationship whatever between the independent and dependent variables, and the points

of the scatterplot are randomly placed on the graph, then none of the variation in the dependent

variable can be explained by the independent variable, and the value of r 2 is 0.

Generally, the closer the data values are, on average, to lying on a straight line, the closer r2 is to

1. The more the data values are scattered randomly about the number plane, the closer r 2 is to 0.

The value of r2 for the discus data is 0.920, while for the long jump data, r2 = 0.648. This indicates

that over 92% of the variation in the length of the discus throw can be explained by the year inwhich the throw took place, while only about 65% of the variation in the long jump distances can

be explained by the year.

MEASURING THE ‘FIT’ OF A LINEAR MODELC

6.0

6.5

7.0

7.5

8.08.5

9.0

-20 0 20 40 60 80 100

Olympic Long Jump Gold Medal Performances

Year since 1900

D i s t a n c e

( m )

25.0

32.5

40.0

47.5

55.0

62.5

70.0

-20 0 20 40 60 80 100

Olympic Discus Gold Medal Performances

Year since 1900

L e n g

t h o

f t h

r o w

( m ) The scatterplots show Olympic gold medal perform-

ances for discus and long jump, with their least squares

regression lines.For each graph, the data lie roughly within an ovalshape, the general trend is for the distance to increaseover the years and the data are approximately linear.

However, on average the long jump data, compared tothe discus data, lie further from the least squares line.Can this difference be quantified in some way?

For the long jump data, the dependent variable varies between . and . . This variation is not entirely ran-dom. Much of the variation appears to be explained bythe fact that the length of the winning long jump in-creases over time.

6 3 8 9


For the Mathematics B course, we do not need to know how to calculate r2, but we must understand

what it measures, and have a feel for the shape of a distribution of data values for different values

of r 2. We must be able to find r 2 using a graphics calculator or statistics software.

The diagrams following are of five scatterplots with the value of r2 for each.

Note: When we find the linear regression equation with a graphics calculator, we may be

given the value of r2 as well. If not, we will be given the value of r, which is called

the correlation coefficient. Just square this value to find the value of r

2

.



1 While writing this textbook, one of the authors conducted a little experiment.

Day No. 0 1 4 5 6 7 8 9 11 12 17 19 21 22

Weight (g) 124 121 103 96 90 84 78 71 58 50 27 12 8 6

a Enter the given data into two lists in a graphics calculator.

b Draw a scatterplot of the data.

c Using the regression function of your graphics calculator, find the least squares regression

line.

d What is the value of r 2?

e Comment on how well this linear function fits the data.

2 The data below show the number of deaths caused by firearms in Australia from 1983 to

1997, expressed as a rate per 100,000 of population.

a Enter the data given into two lists in a graphics calculator.

b Draw a scatterplot of the data.

r2 = 0:998 r2 = 0:600 r2 = 0:001

r2 = 0:250 r2 = 0:81

I had a hypothesis that the daily weight of the bar of soap in my shower was not a linear func-tion, the reason being that the tiny little bar of soap at the end of its life seemed to hang around just about forever. I wanted to throw it out, but I felt I should not do so until it becameunusable. And that seemed to take weeks.

Also I had recently bought some digital kitchen scales and felt I needed to use them to justifythe cost. I hypothesised that the daily weight of a bar of soap might be dependent upon surface area, and hence would be a quadratic function.

Year 83 84 85 86 87 88 89 90

Rate 4.31 4.42 4:52 4.35 4.39 4.21 3.4 3:61

Year 91 92 93 94 95 96 97

Rate 3.67 3.61 2.98 2:95 2:72 2:96 2:3

EXERCISE 4C


Let 1983 be year 0.



c Using the regression function of your graphics calculator, find the least squares regression

line.

d What is the value of r2?

e Comment on how well this linear function fits the data, and on any unusual features of

the dataset.

3 Match the scatterplot with the coefficient of determination.a b c

d e f

A r2 = 0:77 B r2 = 0:94 C r2 = 0:84 D r2 = 0:03 E r2 = 0:88

F r2 = 0:26

4 Estimate the value of r 2.

a b c d

e f g h

5 a Measure the diameter and the circumference of at least 10 tin cans of different sizes.

b Put the data into two lists in a graphics calculator.

c Construct a scatterplot of the data.

d Find, and then plot, the least squares regression line.

e What is the value of r2?

f Find the gradient of this line.

g If the tin cans were perfectly round, and your measurements were perfect, what should

the value of the gradient be?


10

15

20

0

10

20

30

40

50

0 5 10 15 200

20

40

60

80

100

120

0 5 10 15 200

10

20

30

40

50

60

0 5 10 15 20

0

5

10

15

20

0 5 10 15 200

10

2030

40

50

60

0 5 10 15 200

10

20

30

40

50

60

70

0 5 10 15 20

0

20

40

60

80

0 5 10 15 200

10

20

30

40

0 5 10 15 200

10

20

3040

50

60

0 5 10 15 200

5

0 5 10 15 20

0

20

40

60

80

100

0 5 10 15 20

0

20

40

60

80

0 5 10 15 20

0

10

20

30

40

50

0 5 10 15 20

0

10

20

30

40

0 5 10 15 20




On the scatterplot given the residuals are shown as vertical line segments between the predicted

values and actual values in the dataset. For the year 1912 our predicted value was 0:32 metres too

small.

Note the large residual in 1968. This was the year in which Bob Beaman made his astounding long

jump of 8.90 metres, an Olympic record that still stands.

Values that have unusually large residuals are called outliers.

Note that another outlier is the long jump in 1896, in which the actual distance is much less than

the predicted distance.

A residual plot is a scatterplot of the values of the residuals versus the values of the

independent variable.

We will come back to the long

jump data after we have lookedat how to construct a residual plot

with a simpler dataset. The scatter-

plot in the diagram is the residual

plot for the long jump data.

RESIDUALS AND THE RESIDUAL PLOT D

Returning to the dataset for thelong jump event, let us now look

at a particular data value in somedetail.

The actual winning jump in

was . metres while the pre-

dicted value is found by substitut-

ing for in the regression equa-

tion,

which equals . metres.

The difference between theactual value and the predictedvalue is called the .

In this instance the

residual .

19127 60

12(12) = 0 0180 12 + 7 06

7 28

= 7 60 7 28 = 0 32

x

f : :

: : :

£

¡

residual

– 0.80

– 0.40

0

0.40

0.80

-20 0 20 40 60 80 100

Residual Plot of the Long JumpData

Year since 1900

R e s

i d u a

l s

6.0

6.5

7.0

7.5

8.0

8.5

9.0

-20 0 20 40 60 80 100

Olympic Long Jump Gold Medal Performances

Year since 1900

D i s t a n c e ( m )

1912

Just as outliers in univariate data are unusual, so are outliers in bivariate data. An outlier in bivari-ate data may represent significant event, as in the above example, or it may represent an error inmeasurement or an error in recording measurement.

aa



Many graphics calculators have the residual plot available from a menu, however a knowledgeable

user can produce a residual plot using any graphics calculator (see Example 4.1). All commercial

statistics software programs will produce a residual plot.

A residual plot shows the distance each datapoint is from its predicted value.

If all of the data values were on the least squares line, all of the residuals would be 0 and hence

all of the points on the residual plot would be on the horizontal line passing through 0. For points

not on the regression line, the distance of the point above or below this line is its residual.

Construct the residual plot for this dataset.

x 1 2 3 4 5 6 7

y 1 4 4 9 10 15 16

1 Use the linear regression facility of your graphics cal-

culator to find the least squares regression line. For this

dataset, it is y = 2:6x ¡ 2. Draw the scatterplot with

its regression line.

2 Using the regression equation, calculate the predicted

value for each value of x in the table. These are called

the ‘fitted’ values or the fit. This is the 3rd row of the

table following.

3 The residual is the difference between the actual data value (in row 2 of the table) and

the fitted value (row 3). This is row 4 of the table.

The rule is residual = data ¡ fit. (This can also be written as residual = yactual ¡ ypredicted:)

x 1 2 3 4 5 6 7

data 1 4 4 9 10 15 16

fit 0:6 3:2 5:8 8:4 11:0 13:6 16:3

residual 0:4 0:8 ¡

1:8 0:6 ¡

1:0 1:45 ¡

0:3

4 To draw a residual plot, draw a scatterplot of thex values

vs the residuals. Set your window so Y min is about ¡2and Y max is about 1.5, so the residual plot nicely fills

the screen.

EXAMPLE 4.1

Note: The above calculations can be done using listsin your graphics calculator Consult your calcu-lator manual or teacher if you are not sure howto do this.

.


If we compare the residual plot of the long jump data with the scatterplot of the data itself, we

should understand how a residual plot is constructed.



WHAT DOES THE RESIDUAL PLOT SHOW?

x 2 2:1 2:2 2:3 2:4 2:5 2:6

y 0:4 0:441 0:484 0:529 0:576 0:625 0:676

The first diagram shows a scatterplot of the data along

with the least squares regression line, which has equation

y = 0:46x ¡ 0:525.

The data appears to be linear, and r2 = 0:999, whichimplies that the model is an excellent predictor for the data.

The second diagram, which is the residual plot, tells a dif-

ferent story. There is a strongly curved pattern to the data,

which indicates that the data are not linear.

The residual plot exaggerates any patterns in the data, mak-

ing the patterns much easier to see.

Note that despite its apparent excellent fit, the linear model

is not useful for extrapolation.

For x = 10, the actual value is y = 0:1 £ 102, or y = 10, while the linear model predicts the value to be

I = 0:46 £ 10 ¡ 0:525, or y = 4:075.


Residual plots help decide if the mathematical model we have chosen is appropriate. Residual plots

are useful because they emphasise patterns in the data.

Patterns in a residual plot may indicate that there are additional features of the dataset that have to

be considered. Ideally, the residuals are random, with no pattern, which implies that the residuals

are the natural variation of the dependent variable.

Lack of a pattern indicates that the function chosen to model the data is appropriate.

The residual plot for the long jump data shows no strong pattern. This implies that a linear function

is an appropriate choice to model the data.

There are a variety of patterns in a residual plot that would alert a statistician to potential problems

with the data or with the function being used to model it. In Mathematics B, we will focus our

attention on the pattern that occurs when an incorrect function is used to model the data.

Here is an example.

The following table contains values of the function y = 0:1x2, which is a quadratic function

rather than a linear function. What does the scatterplot and the residual plot show when we try tomodel this data with a linear equation?

Most graphics calculators can also calculate the least

squares quadratic regression line. This is the quadratic

function that best fits the data. Residual plots can be con-

structed for the quadratic regression line as well. Since the

data fit a quadratic equation perfectly, the residuals of the

least squares quadratic regression line are all 0. The residual

plot on the right shows that the quadratic function models

the data perfectly.



MODELLING DATA WITH A FUNCTION

In Chapter 1, we looked at the steps involved in mathematical modelling. Step 3 (Build the

model) involves interpreting the situation in mathematical terms.

Stride Rate (strides/sec) 3.05 3.12 3:17 3:25 3:36 3:46 3:55

Speed (feet/sec) 15:86 16:88 17:50 18:62 19:97 21:06 22:11

a Plot the data on a scattergram. Decide, by eye, if the data appear to be linear.

b Find the equation of the least squares regression line.

c What is the gradient of this equation? What is the physical interpretation of the gradient

of this function? In other words, what are the ‘units’ of the gradient?

d Plot the least squares regression line on the scatterplot. Comment on the fit.

e Using the equation of the least squares regression line, make a prediction of the speed if

the stride rate is 3.5 feet per second.

f Using the equation of the least squares regression line, make a prediction of the stride

rate if the speed is 19 feet per second.

g What is the speed if the stride rate is 0? Interpret your result. Do the results make sense?

What should be done about this?

h Draw a residual plot of the data. Do you notice any patterns? Can you find a better

mathematical model for this data?

EXAMPLE 4.2


One measure of form for runner is stride rate, defined as the number of steps per second.runner is considered to be efficient if the stride rate is close to optimum. The stride rate isrelated to speed; the greater the stride rate, the greater the speed (measured in feet/second).

In study of top female runners, researchers measured the stride rate for different speeds.

The following table gives the average stride rate of these women versus the speed.

a A

a 21

Here is a process for finding a mathematical model for paired data.

Step 1:

Step 2: Look at the data, by graphing a scatterplot. The human eye is fantastic at spotting

patterns - far better than, say, a mere supercomputer.

Step 3:

Step 4: Calculate the regression equation, using your graphics calculator.

Step 5:

Step 6: Decide how closely the function models the data, by calculating r2 (using your calcu-

lator). The closer r 2 is to 1, the better the function will be at prediction.

Think about the context. For example, if we think the dependent variable changes at aconstant rate, then we would expect a linear function to be appropriate.

Choose the type of function that we think will best model the data, based on Steps 1

and 2. At the moment you only know of two types of functions, linear and quadratic,

but you will learn about others over the next year.

Graph the residual plot. If there are no obvious patterns in the residual plot, then the

function is probably an appropriate choice. If there are patterns, then we need to go

back to Step 1.



Enter the following

data into two lists. Stride Rate 3:05 3:12 3:17 3:25 3:36 3:46 3:55

Speed 15:86 16:88 17:50 18:62 19:97 21:06 22:11

a Construct a scatterplot from this data. The data appear

to be linear.

b The regression equation is y = 12:4x ¡ 21:9

c The gradient of the graph is 12.4. Since the units on the

y-axis are feet per second, and the units on the x-axis

are strides per second, the units of the gradient are

feet per second

strides per second, which simplifies to feet per stride.

d The least squares regression line with its scatterplot is

shown. The regression line appears to fit the data ex-

tremely well.

e Substitute x = 3:5 into the equation of the least squares

regression line.

y = 12:4 £ 3:5 ¡ 21:9 = 21:5 feet per second

f Substitute y = 19 into the equation of the least squares regression line, and solve.

19 = 12:4x ¡ 21:9

) 12:4x = 40:9

) x = 3.30 strides per second

g The speed for a stride rate of 0 is the y-intercept,

¡21:9 . This means that if the runner

is not striding at all, she is moving backwards at a speed of 21.9 feet per second! Our linear model is not valid when we extrapolate back to the value x = 0. The data for

the independent variable has a low value of 3.05 and a high value of 3.55. If we restrict

ourselves to values between 3 and 3.6, we should be fairly safe. We are not confident

that we can make predictions outside of this range.

h A scatterplot of the residuals helps to determine the va-

lidity of the linear model. Ideally, the residuals should

show no strong patterns.

However, the scatterplot shows the data has a strongly

curved pattern. Another model, say a quadratic model,

may give a better fit. The results of applying quadraticregression are shown below. The residual plot shows no

pattern, which indicates that a quadratic model is more

accurate than a linear model.




1 Enter the following data into two lists in your graphics calculator.

x 0:6 0:65 0:7 0:75 0:8 0:85 0:9

y 0:565 0:605 0:644 0:682 0:717 0:751 0:783

a Construct a scatterplot and least squares regression line.

b Find the equation of the least squares regression line.

c Find the value of r2.

d Based on the above, how well does the linear model fit the data?

e Now construct a residual plot of the data.

f Answer the question in part d again, using the information from the residual plot.

g Find the equation of the quadratic regression line.

h Construct a residual plot based on the quadratic regression line.

i How well does the quadratic model fit the data?

2 Repeat the above questions

for this dataset.x 97 105 121 147 151 169 180 201y 158 130 168 291 311 328 278 343

3 Consider these four bi-variate datasets:

x1 y1

10 8:048 6:95

13 7:589 8:81

11 8:3314 9:966 7:244 4:26

12 10:847 4:825 5:68

x2 y2

10 9:148 8:14

13 8:749 8:77

11 9:2614 8:16 6:134 3:1

12 9:137 7:265 4:74

x3 y3

10 7:468 6:77

13 12:749 7:11

11 7:8114 8:846 6:084 5:39

12 8:157 6:425 5:73

x4 y4

8 6:588 5:768 7:718 8:84

8 8:478 7:048 5:25

19 12:58 5:568 7:918 6:89

a Confirm that the summary statistics of each of these datasets are identical:

n 11Mean of x 9:0

Mean of y 7:5

Equation of regression line y = 3 + 0:5x

r2 0:67

b Draw the scatterplot of each dataset.

c Frank Anscombe created these datasets to emphasise the importance of looking at the

data, and not just relying on the summary statistics (and r2 in particular) to decide on

the appropriateness of modelling the data with the linear function y = 3 + 0:5x. For

which of these datasets is this linear model appropriate?

EXERCISE 4D




In the premiere ice hockeyleague in the world, the NationalHockey League, was in trouble. Scor-ing was at its lowest level in yearsand the percentage of shots on goalsaved by the goalie was at its highest

level since that statistic was first re-corded in The general manag-ers of the teams in the NationalHockey League were meeting to dis-cuss this. They needed to predict themean number of goals per game infuture years, based on past data.

There are three variables:

ear (the season is coded as and so on)

Save Percent (the percentage of shots on goal saved by the goalie)

Goals per Game (mean goals scored per game, for an entire season)

1997

42

198226

1984 85 84

,

.

Y - ,

4

Carats 0:17 0:16 0:17 0:18 0:25 0:16 0:15 0:19

Cost( $ ) 355 328 350 325 642 342 322 485

Carats 0:21 0:15 0:18 0:28 0:17 0:18 0:17 0:18

Cost( $ ) 483 323 462 823 353 438 318 419

Carats 0:17 0:15 0:17 0:32 0:32 0:15 0:16 0:16

Cost( $ ) 346 315 350 918 919 298 339 338

a Find a linear model for this data.

b Interpret the gradient and y-intercept.

c Construct a scatterplot with a least squares regression line.

d Construct a residual plot.

e Are you satisfied using a linear model for this data? Justify your answer.

5


On February, a full page ad-vertisement for diamond rings was

placed in thenewspaper.

The advertisement contained picturesof diamond rings and listed their

prices, the size of the diamond (meas-ured in carats) and the gold purity(also measured in carats, with caratrepresenting pure gold).

There were rings of varying de-signs, with the diamonds weighingfrom . to . carats (one carat . gram) and priced between $ and $ .

The data below comes from of these rings. Each ring is a carat gold ladies’ ring,mounted with a single diamond. Consider only the size of the diamond in the ring and its costand determine if a linear equation is a suitable model for this data.

29 1992

24

48

0 12 0 35 = 0 2 223 1086

24 20

Singapore Straits Times



Year 82 83 84 85 86 87 88 89

Save Percent 0:875 0:873 0:874 0:874 0:88 0:881 0:879 0:881

Goals per Game 7:73 7:8 7:68 7:86 7:25 7:34 7:39 7:28

Year 90 91 92 93 94 95 96 97

Save Percent 0:886 0:882 0:885 0:895 0:901 0:898 0:905 0:906

Goals per Game 6:82 6:87 7:18 6:4 5:9 6:2 5:75 5:19

a Draw scatterplots of

i Save Percent vs Year ii Goals per Game vs Year

b Find the least squares regression line for each of the above.

c Draw a residual plot for each, to decide if a linear function is appropriate for these data.

d Estimate the mean number of goals per game in the year 2005.

e Estimate the mean number of saves per game in the year 2005.

6

Now that you understand residuals, the meaning of

‘least squares’ should make some sense.

The least squares regression line is the line that

gives the smallest sum of the squares of the resid-

uals.

To clarify what this means, consider the follow-

ing paired data (1,2), (2,4), (3,6), (4,8), (5,15)

which are plotted on the scatterplot along with two

lines, L1 and L2.

Line L1 has equation y = 3x ¡ 2

Line L2 has equation y = 2x

LEAST SQUARES REGRESSION LINE REVISITED E

L1

L2

3

6

9

12

15

1 2 3 4 5

y

x


Over the years, goalie equipment in the

National Hockey League is becoming pro-gressively lighter. According to an articlefrom , “Since ...goalie equipment is not only lighter, it isalso ridiculously bigger. Leg pads were al-lowed to become percent wider, frominches to , under a - rule revision,and along with them, every other piece of goalie gear from catching gloves to toe capsto shoulder pads has mutated into bizarresizes and shapes.”

Split the data for Years and Goals per Game

into two datasets - one for the data fromto , and the second for the data

from to .

For each, calculate the equation of the regression line. How do the gradients of these linescompare? Does your analysis confirm that the rule revision has made it more difficult to score?

The New York Times 1990

20 1012 1989 90

1982 19891990 1997



For each line, we will calculate the sum of the squares of the residuals.

Recall that:

yactual is the y-coordinate of the paired data

ypredicted is the corresponding value of the linear function

residual = yactual

¡ ypredicted

Line L1 y = 3x ¡ 2x 1 2 3 4 5

yactual 2 4 6 8 15

y predicted 1 4 7 10 13

residual 1 0 ¡1 ¡2 2

(residual)2 1 0 1 4 4

The sum of the squares of the residuals is 1 + 0 + 1 + 4 + 4 = 10

Line L2 y = 2xx 1 2 3 4 5

yactual 2 4 6 8 15

y predicted 2 4 6 8 10

residual 0 0 0 0 5

(residual)2 0 0 0 0 25

The sum of the squares of the residuals is 0 + 0 + 0 + 0 + 25 = 25

Since the sum of the squares of the residuals is smaller for line L1 than L2, line L1 ‘fits’ thedata better, even though line L2 passes through four of the five points. If you followed this, the

definition at the start of this section should now make some sense:

The least squares regression line is the line that gives the smallest sum of the squares

of the residuals.

The least squares regression line has the equation y = ax + b where:

a = Pxy ¡ 1n

(Pn)(P y)Px2 ¡ 1n (Px)2

and b = y ¡ ax

Recall thatP

is a symbol that tells you to sum what follows, and x and y are the means of the x

and y variables respectively.

We will illustrate these formulas by using them to find the equation of the least squares regression

line for the above data.

Note that your graphics calculator is very good at summing lists, so these calculations can be done

using your calculator.




To calculate a:

We construct a table to find the sum of x, y, xy and x2.x y xy x2

1 2 2 1

2 4 8 4

3 6 18 9

4 8 32 16

5 15 75 25P 15 35 135 55

Substitute these values into the formula for a.

a =

Pxy ¡ 1

n(P

x)(P

y)Px2 ¡ 1

n(P

x)2

= 135 ¡ 1

5 (15)(35)

55¡

15 (15)2

= 135 ¡ 105

55 ¡ 45

= 30

10

= 3

To calculate b:

x =

Px

n =

15

5 = 3 and y =

Py

n =

35

5 = 7

and b = y ¡ ax) b = 7 ¡ 3(3)

) b = 7 ¡ 9

) b = ¡2

Therefore, the line y = 3x ¡ 2 is the least squares regression line.

You do not need to know how to derive the formulas for a and b; this is a topic for a mathematical

statistics course at University.

You do not even need to use the above formulas, as they are built into your graphics calculator.

But, you need to understand that a regression line is a line that represents paired data, and the least

squares regression line is the most commonly used regression line. You should have knowledge of what ‘least squares’ represents.

EXERCISE 4E.1


Doing one or two exercises “by hand”, even though they can be done using a calculator, is often

instructive.

1 Use the above method to find the least squares regression equation for this set of data.

(1, 1), (2, 7), (3, 10), (4, 7), (5, 13), (6, 12)

Confirm your answer using a graphics calculator.



2 For the data below, find the least squares regression equation “by hand”. Confirm your answer

with a graphics calculator.

(2, 11), (4, 12), (5, 5), (6, 8), (7, 4)

LEAST SQUARES REGRESSION LINE THROUGH THE ORIGIN

Danielle was confirming Hooke’s Law. She attached weights to a

spring, and measured the extension of the spring due to the attached

weights. She obtained the following data.

Weight (g) 50 100 150 200 250 300

Extension (mm) 21 49 64 94 111 137

She wishes to find the least squares line that passes through the

origin, as the extension is zero when no weight is attached.

spreadsheet).

Then List 3 = List 1 * List 2, and List 4 = (List )1 2 . To find a, calculate: sum(List 3)

sum(List 4)

x y xy x2

50 21 1050 2500

100 49 4900 10 000

150 64 9600 22 500

200 94 18 800 40 000

250 111 27 750 62 500

300 137 41 100 90 000P 103 200 227 500

a =

PxyPx2

= 103 200

227 500

= 0:454


Our least squares regression equation is y = 0:454x. We now need to look at the residual plot,to see if this function y = ax is a good model for this data.

To draw the residual plot:

² Set Y 1 = 0:454x. Turn off the graph of this equation.

² The formula for residuals is: residual = yactual ¡ y predicted.

We will put the residuals into List 5, as follows:

List 5 = List 2 ¡ Y 1(List 1)

² Draw the residual plot of List versus List The resid-ual plot shows no strong patterns, so this is good model

for the data.

5 1.a

Enter the first two columns of this table into List 1 and List 2 of your graphics calculator (or use a

In other words, the equation that models this data has the form y = ax, a > 0 and she wants

to find the value of a that minimises the sum of the squares of the residuals.

This particular regression equation is not available in Danielle’s calculator, so she will have to use

the formula to find a: a =

PxyPx2



Notes: The derivation of this formula for a will be done as an example in the Year 12 textbook,

after you have studied the topic Optimisation using derivatives.

All graphics calculators are programmable. Writing a program that finds the least

squares regression equation of the form y = ax for data in List1 and List2 would be

useful, and not too difficult.

1 For each dataset i find the least squares regression model of the form y = ax

ii construct a residual plot

iii discuss the appropriateness of the equation as a model for the data

a x 1 2 3 4 5y 5 17 16 27 39

b x 13:4 21:6 28:9 31:0 56:2

y 47:5 83:2 126:7 118 198:3

2 Justin was confirming Hooke’s Law. He attached weights to a spring, and measured the

extension of the spring due to the attached weights.

He obtained the following data.

Weight (g) 50 100 150 200 250 300

Extension (mm) 17 27 37 53 57 68

a Find the least squares regression line of the form y = ax.

b Draw the residual plot.

c The above analysis implies there is a problem with Justin’s data. Briefly describe what it

is, and how he might fix it.

3 Three 11 Mathematics B students from Glenmore High School repeatedly rolled a steel ball

down an inclined plane, and recorded displacement (s) and time (t) for each trial.

Their results are:

time (t) 0 0:5 1:0 1:5 2:0 2:5 3:0 3:5 4:0 4:5 5:0

displacement (s) 0 4:5 7:7 13:4 36:6 52:2 88:5 110:5 144:5 161:5 188:8

From their knowledge of Physics, they expect the least squares regression equation to be thenon-linear function s = at2 where s = displacement, t = time, and a depends upon the

steepness of the inclined plane.

To find this regression line, the students need to determine the value of a.

Here is the least squares equation for finding this value: a =

Px2

iyiP

x4

i

a Use your graphics calculator and the above formula to find the least squares function of

the form s = at2 that models this data.

b Draw the residual plot.

c Decide if the function is appropriate for this data set. Justify your decision.

EXERCISE 4E.2


, for a > 0:



MODELLING

1 The tables below contain the world record times and the year they were set for the 100metre sprint for males and females. Use this data to find a model to predict when the

female world record holder will have a faster time than the male world record holder.

Discuss the assumptions that underlie your prediction.

Female

Year ( 19 ) 34 37 48 52 55 61 65 68 70

Time 11:7 11:6 11:5 11:4 11:3 11:2 11:1 11:08 11:0

Year ( 19 ) 73 73 83 84 88

Time 10:9 10:8 10:79 10:76 10:49

Male

Year ( 19 ) 12 20 30 36 56 60 68 83 88

Time 10:6 10:4 10:3 10:2 10:1 10:0 9:95 9:93 9:92

Year ( 19 ) 91 94 96

Time 9:90 9:85 9:84

2 Six students were asked to form a line at the front of the classroom, and to hold hands. A

seventh student was asked to act as the Timer. When the teacher said “Start!”, the timer

started his watch, and the person at the head of the line squeezed the hand of the next

person. When that person felt a squeeze, they squeezed the hand of the next person, and

so on. When the last person felt his hand being squeezed, he shouted “Stop!”. The Timer

recorded the time taken between ‘Start’ and ‘Stop’.

Six more students joined the group and the same process was carried out. The process

was repeated until all students in the class were in the line.Here are the data:

Number of students 6 12 20 28 38

Time (sec) 2:22 3:03 4:63 6:44 9:25

a Find a mathematical model for this dataset. Justify your choice.

b What do you predict would have been the time if 24 students had been in line?

c What would you predict if 50 students had been in line?

You may wish to carry out this experiment with your class, and compare the predicted

time with the actual time.

3 In 1981, two new varieties of a tiny biting insect called a midge were discovered by

biologists W. L. Grogan and W. W. Wirth in the jungles of Brazil. They named one kind

of midge an Apf midge and the other an Af midge. The biologists found out that the Apf

midge is a carrier of a debilitating disease that causes swelling of the brain when a human

is bitten by an infected midge. Although the disease is rarely fatal, the disability caused

by the swelling may be permanent. The other form of the midge, the Af, is quite harmless

and a valuable pollinator. In an effort to distinguish the two varieties, the biologist took

measurements on the midges they caught. The two measurements taken were of wing

length and antennae length, both measured in centimetres.




Af Midges

Wing Length (cm) 1:72 1:64 1:74 1:70 1:82 1:82 1:90 1:82 2:08

Antenna Length (cm) 1:24 1:38 1:36 1:40 1:38 1:48 1:38 1:54 1:56

Apf Midges

Wing Length (cm) 1:78 1:86 1:96 2:00 2:00 1:96

Antenna Length (cm) 1:14 1:20 1:30 1:26 1:28 1:18

Is it possible to distinguish an Af midge from an Apf midge on the basis of wing and

antenna length? Use your knowledge of regression to find a method of distinguishing the

two types of midges.

1

Year 75 76 77 78 79 80 81 82 83

Lean 642 644 656 667 673 688 696 698 713

Year 84 85 86 87

Lean 717 725 742 757

The TI-83 screen shots above show: A the scatterplot, with the least squares regression line

B the linear regression output screen

C the residual plot

a Is a linear model an appropriate choice for this set of data? Justify your decision.

b In 1918, the lean was 71. What does the linear model predict for 1918?

c Discuss briefly how well the model fits the data when we extrapolate back to 1918.

* The data represent the amount of lean greater than 2.9 metres, written without a

decimal point. For example, the lean of the Tower in 1975 was 2.9642 metres, and is

given in the table as 642.

2 a Draw a scatterplot with 15 data values that you think has an r 2 value of about 0.6 .

b Now enter the coordinates of these 15 points into two lists in your calculator, and use

them to determine the actual r2 value.

c If necessary, modify your original data values, and repeat b until your r2 value is close

to 0.6 .

The Leaning Tower of Pisa, an ancient tower in Italy, is falling over! The

table below gives the amount of lean in each of the years from

to .

*

1975 1987

A B C


PROBLEM SOLVING F

EXERCISE 4F





bivariate data interpolation residual

coefficient of determination least squares regression line residual plot

correlation coefficient line of best fit scatterplot extrapolation outliers univariate data

fitted values (fit) regression line

Skid length (m) 5 8 11 15 25 31 37 44 52 60 80 100

Speed (kph) 32 40 48 56 64 72 80 88 96 104 120 144


1 Explain what a least squares regression line is, in your own words.

2 When investigating a car accident, the police need to determine how fast each car was

travelling when the accident occurred. One indicator of the speed is the length of the skid

mark left by the car. If a car leaves a skid mark, how accurately can the speed of the car

be determined from its length?

The following data was gathered by the Department of Transport under controlled conditions

(i.e., for a particular surface type and particular weather conditions).

a Enter this data into two lists of your graphics calculator.

b Draw the scatterplot. Describe the shape of the data.

c i Fit a least squares linear regression model to this data.

ii Find the value of r2.

d Construct the residual plot for the linear model and discuss any patterns in the residual

plot.

e i Fit a least squares quadratic regression model to this data.

ii Find the value of r2.

f i Construct the residual plot for the quadratic model

ii Discuss any patterns in the residual plot.

g Now use the information you have gathered to decide which of these three models best

fits the data. Determine the domain over which you feel the model is valid. Justify

each of these decisions.

h Use your chosen model to estimate the speed a vehicle was travelling if it left a skid

mark 110 metres long.

i Use your chosen model to estimate the length of the skid mark left by a vehicle

travelling at 155 kilometres per hour.






1 a When undertaking exploratory data analysis, an initial step is to

plot the data. Why is this important?

b What is a regression line? Is it always a linear function?

c What symbol is given to the coefficient of determination?

d Give two reasons why outliers might occur in a dataset.

e Write a paragraph explaining what information the coefficient of determination can

give about a dataset.

f Explain why knowing the value of r2 is not, by itself, sufficient to determine how

well a model fits a dataset.

2 Enter the following data into two lists in your graphics calculator.

x 10 15 26 29 36 41 45

y 81 187 251 197 266 304 297

a Graph the scatterplot.

b Find the equation of the least squares regression line, and graph it along with the

data.

c Find the value of r 2.

d Based on the above, how well does the linear model fit the data?

e Now construct a residual plot of the data.

f Answer the question in part d again, using the information from the residual plot.

g Find the equation of the quadratic regression line.

h Construct a residual plot based on the quadratic regression line. i How well does the quadratic model fit the data?





V168 MODELLING DATA WITH FUNCTIONS (Chapter 4)

Kochel numbers¨

W. .A Mozart (January 27, 1756 - December 5, 1791) is one of the most famous composers of

all times. Because much of Mozart’s work is undated, and he did not have a publisher, DrK ochel, an Austrian botanist and mineralogist, composed a chronological thematic catalogue of

Mozart’s works. The catalogue gives each work a K ochel number by which the work is now

universally identified. K ochel’s catalogue was first published in 1862, and has been periodically

revised.

Colin Fox is a professional mathematician who also hosts programs on the Australian Broad-

casting Corporation’s FM Classical Music radio network. Colin Fox claims that there is a linear

relationship between the K ochel numbers and Mozart’s age when the particular work was writ-

ten.

The catalogue of Mozart’s works is available from a number of websites.

One such is www.classical.net/music/composer/works/mozart/

It is also available from the website for this text.

From the complete collection of Mozart’s compositions, select a random sample of 20 works.

Describe briefly how you selected your sample to ensure it was random.

Using your sample, find a linear relationship between the K ochel number and Mozart’s age.

Use a residual plot to determine if a linear relationship is a good model for your data.

From your linear relationship, devise a “rule of thumb” (a simple rule) that allows someone to

quickly estimate Mozart’s age from the K ochel number. Test your rule of thumb on a random

selection of ten of Mozart’s compositions.

The website also contains the rule of thumb used by Colin Fox. Compare your rule of thumb

with that of Colin Fox for:

² ease of use

² accuracy.

Let us speculate on what might have been. If Mozart had lived to the age of 70, and continued

to produce compositions at the same rate until his death, what would have been the K ochel

number of his last composition?

http://www.classical.net/music/composer/works/mozart/

http://www.classical.net/music/composer/works/mozart/



CHAPTER 5Trigonometry

SUBJECT MATTER

simple definitions and practical applications of thesine, cosine and tangent ratios

simple practical applications of the sine and cosinerules

definition of a radian and its relationship withdegrees

definition of the trigonometric ratios sin, cos andtan of any angle in degrees and radians



INVESTIGATION 1

PROPERTIES OF SIMILAR TRIANGLES

HISTORICAL NOTE +Trigonometry

Hipparchus

Ptolemy

sine

was first developed to help people understand the heavens. Thestars were thought to be fixed on a large crystal sphere, on which the planets, theSun and the Moon moved.

To aid in this understanding, ( - BC) produced the first trigonometric table.This was a table of chords for angles in a large circle. The angles were measured in “steps”,where one step equalled .

( - BC) used a different large circle and extended the table to include anglesfrom to in steps of and even provided a means of interpolating results to getgreater accuracy. Although trigonometry was originally based on the chord of a circle, todaywe base our trigonometry on the “half-chord”, which we call the .

190 120

15

178 100180

o

o o o12 12

We will start with a brief review of the trigonometry that you learned in the Junior school.

Similar triangles are triangles with the same shape.

Materials: protractor, ruler, calculator

1 In the above triangles, by measuring confirm that angles A, D and G are congruent.

2 Confirm that these ratios are equal: BC

AB,

EF

DE, .

HI

GH

12

12

SIMILAR TRIANGLES A

What to do:

A

B

C

D

E

F

G

H

I

While historically trigonometry was based on circles, it is more useful to the beginstudent ning thestudy of trigonometry to start with triangles, and in particular with similar right-angled triangles.

The basis of trigonometry of the right-angled triangle is that the ratios of the lengths of the sides

of similar triangles is equal, irrespective of the size of the triangle.

This fact allows us to find inaccessible lengths, distances and angles.

170 TRGONOMETRY (Chapter 5)



Name of Ratio Abbreviation Ratio of Sides

sine sin oppositehypotenuse

cosine cos adjacent

hypotenuse

tangent tan opposite

adjacent

cosecant cosec hypotenuse

opposite

secant sec

hypotenuse

adjacent

cotangent cot adjacent

opposite

The last three ratios are referred to as the reciprocal ratios as they are the reciprocals of the first

three ratios respectively.

In particular,

cosec µ = 1

sin µ sec µ =

1

cos µ cot µ =

1

tan µ

As in our earlier work on trigonometry we will concentrate mainly on the sin, cos and tan ratios

but in some situations a knowledge of the reciprocal ratios will be very useful.

As it is very important to remember the ratios of sides, some students learn a mnemonic to help

them remember these ratios.

For example, some old hippies cannot always hide their old age helps us to remember:

TRIGONOMETRIC RATIOS B

As the ratios of the sides of right-angled triangles areso critical in the study of trigonometry, we havenames for the sides of a right triangle, as well as for

the ratios of these sides.If is the angle of interest, we will name the sides asshown in the diagram.

The is opposite the right angle, theand the hypotenuse form the angle, and

the side is the side furthest from the angle.

Taking any two sides at a time it is possible to makeup different ratios.

µ

hypotenuse

adjacent side

opposite

six

B

A

C

o p p o s

i t e h y p o t e n u s e

adjacent

sin µ = opp

hyp cos µ =

adj

hyptan µ =

opp

adj

TRGONOMETRY (Chapter 5) 171



sin µ = opposite

hypotenuse = 4

5

cos µ = adjacent

hypotenuse = 3

5

tan µ = opposite

adjacent = 4

3

cosec µ = hypotenuse

opposite = 5

4

sec µ = hypotenuse

adjacent = 5

3

cot µ = adjacent

opposite = 3

4

Calculating the unknown sides and angles in a triangle is referred to as solving the triangle.

Solve ¢PQR, giving length answers to 2 decimal places.

Given: P = 90o, Q = 35o, PQ = 8 cm. We are required to find ]R, PR and QR.

If we know two angles, we can find the third angle because the sum of the angles of atriangle is 180o. .So, R = 180o ¡ 90o ¡ 35o = 55o

Once we have found the unknown angle, we can now use either 35o or 55o as our reference

angle for our trigonometric ratios. Sometimes a calculation may be made a little easier by

selecting a particular reference angle over the other.

To find PR: tan35o = opp

adj falways write the trig ratio firstg

) 0:7002 = PR

8 fsubstituteg

) PR = 5.60 fsolve for PR, and round to two decimal placesg

To find RQ: cos35o = 8

RQ

) 0:8192 = 8

RQ

) RQ = 8

0:8192 fmultiply by RQ, then divide by 0.8192g

) RQ = 9.77 fround to two decimal placesg

Therefore: P = 90o, Q = 35o, R = 55o; PQ = 8 cm, PR = 5.60 cm, RQ = 9.77 cm.

EXAMPLE 5.1

EXAMPLE 5.2

8 cm

R

QP 35°

4 5

3

A

BC


Find sin µ, cos µ, tan µ, sec µ, cosec µ

and cot µ ratios from the diagram.



INVESTIGATION 2

TRIGONOMETRIC RATIOS AND GRADIENT

If sin A = 0.7071, find angle A.

We want to find the angle whose sine is 0.7071. We write: sin A = 0.7071

There are two common notations for finding an unknown angle:

A = sin¡1 0.7071 We read this as “A equals the inverse sine of 0.7071”.

or A = arcsin 0.7071 We read this as “A equals the arcsin of 0.7071”.

Using the inverse sine key, we find that A + 45o.

Use the diagram to find a relationship between a

trigonometric ratio and the gradient of the line OA.

FINDING AN UNKNOWN ANGLE

EXAMPLE 5.3

EXAMPLE 5.4

Given: B = 90o, AB = 6 cm, BC = 5 cm. We are required to find ]A, ]C and AC.

To find A: tan A = opp

adj = 5

6

) A = tan¡1( 5

6 ) + 39:8o

To find C : C + 180o ¡ 90o ¡ 39:8o + 50:2o

To find AC: Use Pythagoras’ Theorem (although we could also use the sine ratio).

AC2 = 52 + 62

) AC =p

52 + 62 + 7:8

Therefore: A = 39.8o, B = 90o, C = 50.2o

AB = 6 cm, BC = 5 cm, AC = 7.8 cm.

A

C

B6 cm

5 cm

y

x

A

BO


Solve ¢ABC giving angles and sides correct to one decimal place.



1 Using your calculator determine the following (correct to 4 decimal places).

a sin56o b cos18o c tan72o d sin30o

e cos60o f tan45o g cosec 56o h cot17o

i sec39o j cot45o k sec60o l cosec 89o

2 Find A (give your answer to the nearest

a sin A = 0:6428 b cos A = 0:2250 c tan A = 5:6713

d sin A = 0:5 e cos A = 0:7071 f tan A = 1:7321

g sec A = 1:5 h cot A = 0:702 i cosec A = 10

3 Solve the following triangles.

a b c d

4 a For y = sin x, if the domain of x is 0o 6 x 6 90o, then determine the range of y.

b For y = cos x, if 0o 6 x 6 90o, then determine the range of y .

c For y = tan x, if 0o 6 x < 90o, then determine the range of y .

5 By selecting any three angles between 0o to 90o, demonstrate that sin A

cos A = tan A.

6 A clinometer is used to measure angles.If the Greek geometer Thales had access

to a clinometer, what angle would he

have measured to the top of the Great

Pyramid?

7 Use the following set of angles: A = f10o, 20o, 30o, .... , 70o, 80og to verify the following

relationships:

a sin A = cos(90o ¡ A) b cos A = sin(90o ¡ A)

c What do you think the co in cosine is referring to?

d What is the relationship between tan A and cot A?

8 For a circular track of radius r metres, banked at

µ degrees to the horizontal, the ideal velocity (the

velocity that gives no tendency to sideslip) in metres

per second is given by the formula:

v =p

gr tan µ, where g = 9:8 m/s2.

a What is the ideal velocity for a vehicle travelling

on a circular track of radius 100 m, banked at

an angle of 15o?

b At what angle should a track of radius 200 m be banked, if it is designed for a vehicle

travelling at 20 m/s?

EXERCISE 5B.1

H

G

F

4 cm

5 cm

A

B

10 cm

20° C

D

F

E

6 cm

50°

A

BC

12 cm5 cm

185 m

148 m


degree).



9 A regular hexagon has a side length of 1 metre. Determine the area of this hexagon in m2.

1 degree = 60 minutes 1o = 600

1 minute = 60 seconds 10 = 6000

so, 1 degree = 3600 seconds 1o = 360000

Therefore an angle of 28o 150 3600 (which we say as “28 degrees, 15 minutes, 36 seconds”)

equals 28 + 1560 + 36

3600 degrees, which in decimal form is 28.26o.

The decimal form is becoming more common, but both methods are still in use. Expressing angles

in degrees, minutes and seconds is referred to as sexagesimal form.

a Change 25.63o from decimal form to sexagesimal form.

b Change 36 o 400 3000 to decimal form.

a 0.63o = 37.80 fmultiply 0:63 £ 60g0:80 = 4800 fmultiply 0:8 £ 60g

So, 25.63o = 25o 370 4800

b 36o 400 3000

= 36 + 4060 + 30

3600 degrees

= 36.675o

DIVIDING UP DEGREES

EXAMPLE 5.5

EXERCISE 5B.2


Although one degree is a very small angle there are many situations where scientists require a more

accurate angle measurement. Again thanks to the Babylonians, the traditional fractional system for

degrees is based on number 60. Here are the conversions, and the symbolism:

If you use a scientific calculator, you may have a key on your calculator that does these conversions

automatically. Many graphics calculators have decimal to sexagesimal conversions as a menu choice.

Alternatively, you can (if you know your calculator’s programming language) write a short program

to do these conversions for you.

1 Convert into sexagesimal form.

a 25.6o b 105.38o c ¡63:125o d 235:366o

e 14:04o f ¡546:2o g 0:1o h 12

o

2 Convert into decimal form.

a 15o 240 3600 b 125o 80 4800 c ¡44o 150 2400 d 83o 420 5500

e 123o 450 600 f ¡90o 120 1200 g 0o 500 h 0o 00 500

3 Using your calculator find the following trig values.

a sin18o 320 2400 b cos84o 320 5400 c tan46o 450 1500

d sin12o 340

5600

e cos67o 120

16:700

f tan89o 560

4300



ANGLES OF ELEVATION AND ANGLES OF DEPRESSION

EXAMPLE 5.6

EXERCISE 5B.3

205 m

84°

6°

d

176 TRIGONOMETRY (Chapter 5)

a b c d

Commonly angles are measured up or down from

the horizontal. Such angles are called angles of

elevation and angles of depression respectively.

Note that the angle of elevation equals the angle

of depression as they are measured from parallel

lines.

Always draw a diagram first. The angle in

the triangle is the complement of 6o, or 84o.

tan84o = oppadj

) tan84o = d

205

) d = 205 £ tan84o

) d = 1950

The fire is 1950 metres from the base of the hill.

1 Find the height of a vertical cliff if the angle of elevation is 35o to the top, from a point which

is 170 metres from the base of the cliff.

2 Lindsay is standing on level ground, 100 metres from a statue. Using a clinometer, he measures

the angle of elevation to the top of the statue to be 12o.

a Ignoring Lindsay’s height, what is the height of the statue?

b A more realistic model takes Lindsay’s height into account. What is the height of the

statue if Lindsay’s eye is 1:6 metres above the ground?

3 The angle of elevation to the top of a lighthouse 175 metres above sea level from a trawler is

4 Solve these triangles expressing all angles in sexagesimal form.

H

G

F

6 cm

10 cm P

Q

R

7 m

24 m

angle of depression

angle of elevation

angle of elevation

A a a.

fire-spotter atop metre hill sees wisp of smoke in the forest at an angle of depres-sion of Assuming the smoke is at the same altitude as the base of the hill, how far away isthe smoke?

2056o

7o

. Calculate the horizontal distance of the boat from the lighthouse.

17 8

A B

C

41

40

Z Y

X



EXACT TRIGONOMETRIC RATIOS

P R

Q

T

TRIGONOMETRY (Chapter 5) 177

Angle Sin Cos Tan

30o 12

p 3

21p

3

45o 1p 2

1p 2

1

60op

32

12

p 3

A

B

C

1

1

From questions and in the previous exercise, youshould have discovered these trigonometric ratios:exact

9 10

4 Kylie measures the angle of elevation from a point on level ground to the top of a building

120 metres high to be 32o. She walks towards the building until the angle of elevation is 45o.

How far did she walk?

5 From a point A, 40 metres from the base of a building B, the angle of elevation to the top of

the building C is 51o, and to the top of the flagpole D is 56o. Find the length of the flagpole.

6 A vertical tree is growing on the side of a hill with a slope of 10o400 to the horizontal. Froma point 50 m downhill from the tree, the angle of elevation to the top of the tree is 18o200.Find the height of the tree.

7 A balloon travels horizontally at a distance h kilometres above the ground between two points

A and B, which are two kilometres apart. From a point C on the ground, the angle of elevation

to the balloon at A is 40o and at B is 25o. Assume that A, B and C are in the same plane and

that A and B are on the same side of the observation point. Find the height h of the balloon.

8 The hypotenuse of a right-angled triangle is five times the length of the shortest side. Find

the tangent of the angle that is opposite the remaining side, i.e., the angle adjacent to the short

side.

9 Triangle ABC is isosceles, with legs AC and BC = 1 unit.

a Use the Theorem of Pythagoras to find the length AB.

b Hence, find exact values of sin 45o, cos45o and tan45o.

10 a Triangle PQR is equilateral with each side of length

2 units.

If QT ? PR, then determine ]P, ]PQT and QT.

b Hence, find exact values of sin30o, cos30o

and tan30o.

c Find the exact values of sin60o, cos60o and

tan60o.



The other justifications for using exact trigonometric ratios are and Often it ismuch faster to work without calculator and with exact trigonometric ratios, and the chance of making silly error is much reduced. never make transcribing error when writing whilewe could easily make transcribing error when writing

efficiency accuracy.a

a We aa . .1 732060808

90°

60°120°

30°150°

0°

360°180°

330°210°

300°240°

270°

starting position

anticlockwise= positive

clockwise

= negative

ANGLES OF ANY MAGNITUDE C


It is very satisfying at the end of a tricky problem to find, for example, that an unknown length can

be expressed very simply asp

3. It is not at all satisfying to find that the length is 1.732050808.

p 3,

To the ancient Greeks, all angles had a magnitude between 0oand 180o; in fact 0o and 180o were

not considered to be angles. But once we think about angles in settings other than triangles, for

instance the angle through which a golf club turns during a swing, it becomes necessary to studyangles outside of this range.

An angle can be defined very simply as an amount of rotation.

In the previous section, our discussion was based on the right-angled triangle, which limited the

magnitude of any angle from 0o to 90o.

However, from our calculator we can see that sin150o = 0:5 and tan(¡135o) = 1.

It appears that trigonometric ratios exist for angles smaller than 0o and larger than 90o.

To represent an angle we need a starting position. By convention this is the horizontal position

pointing to the right.

Consider the circle diagram above to be a Cartesian Plane with the centre of our circle at the origin

of the Cartesian Plane.

We define the starting position for measuring angles as the positive direction of the x-axis.

A positive angle is an anticlockwise rotation from the starting position while a negative angle is

a clockwise rotation from the starting position.

Instead of limiting our discussion byreferring only to right-angled triangles,

let us use circle diagram to displayangles. Recall from the historical notethat trigonometry was originally devel-oped using circles and not triangles.

ou have learned previously that onecomplete revolution has It isthought that this number of origi-nated from the Babylonians who used

base number system. Becausecan be divided by so many divisors,later cultures probably found it veryconvenient to retain the degrees in

circle rather than change it to sayunits or even units.

a

Y.

a

a

360360

60 360

360100

10

o

As these are used extensively in Mathematics B, you must either commit them to memory (it is nothard, as there are many patterns in this table) or be able to quickly reproduce them from the dia-

grams given in questions and in Exercise 5B.3.9 10



3

a

b

4 a

b

c

d c

A (n mile) is the distance on the Earth’s sur-face that subtends an angle of minute of the Great Circlearc measured from the centre of the Earth. A is aspeed of nautical mile per hour.

Given that the radius of the Earth is km, show thatn mile is approximately equal to . km.

Calculate how long it will take a plane to fly from Rock-hampton to Brisbane (a distance of km) if the planecan fly at knots.

A figure skater does a triple axle (i.e., three complete rotations). How many degrees doesshe turn through?

An engine is running at rpm (revolutions per minute). How many degrees does it turnin one second?

Some children are playing ‘pin the tail on the donkey’. A child is facing the donkey whenthe blindfold is put on. She is spun around until she is facing away from the donkey.Through what angles might she have been rotated?

Can you find an elegant way to answer question that includes possible solutions?

nautical mile

knot

1

1

64001 1 852

650400

5000

all

EXERCISE 5C

210°

120°

480°

II

III

I

IV

For example, 210o is shown in the first

diagram alongside.

The angle ¡120o is shown in the second

diagram alongside.

We do not have to stop after one revolu-

tion. An angle of 480o is shown in the

first diagram on the right.

Any angle of any magnitude (positive or

negative) can be represented by a rotation

measured from the positive x-axis.

The circle diagram, like the Cartesian Plane, can be divided into four quadrants with the quadrants

being numbered from the starting position in order in the positive direction. Often Roman numerals

are used to number the quadrants. An angle from 0o

to 90o

is sometimes referred to as a firstquadrant angle, an angle from 90o to 180o is referred to as a second quadrant angle and so on.

1 Represent the following angles with a circle diagram.

a 240o b 330o c ¡60o d ¡180o

e 450o f 720o g ¡240o h ¡540o

i 0o j 90o k 180o l 270o

2 Classify the following as 1st , 2nd, 3rd or 4th quadrant angles.

a 45o b ¡120o c 300o d ¡215o

e 495o f ¡95o g 2000o h ¡1079o

i 90o j 180o k 1000000o l ¡1000000000o


N

C

S

P

Q

1'

1 n mile



CIRCULAR FUNCTIONS D


The previous section discusses angles of any magnitude. Your calculator confirms that trigonometric

ratios are defined for any angle. So far we have defined the various trigonometric ratios using the

sides of a right-angled triangle.

But definitions such as sin µ = opp

hyp have a major drawback

in that the range of µ is restricted to 0o < µ < 90o.

The unit circle is a circle of radius 1 unit with centre the origin.

The unit circle has intercepts of (1, 0) and (¡1, 0) on the x-axis,

and intercepts of (0, 1) and (0, ¡1) on the y-axis. Consider

the point P with general coordinates (x, y) situated on the cir-

cumference of the unit circle. Let µ be the angle formed as P

rotates from the x-axis (the starting position for measurement

of angles) and initially let µ be a first quadrant angle.

We will now extend our definitions of sin µ, cos µ and tan µ so they can be applied to angles of

any magnitude.

Given a circle of radius 1 unit centred at the origin O and any point P(x, y) on the circumference,

let the angle formed by the positive x-axis and ray OP be µ. Then we define cos µ to be thex-coordinate of P and sin µ to be the y-coordinate of P. The tangent ratio will be defined as the

ratio the y-coordinate : the x-coordinate.

These definitions are consistent with the original defini-

tions of sin µ and cos µ for angles less than 90 degrees.

To show this, we will construct a right-angled triangle

in the first quadrant of the unit circle. See the diagram.

In ¢OPN: sin µ = opp

hyp

= PNOP

= y

1

) sin µ = y

Similarly cos µ = adj

hyp = x

and tan µ = opp

adj =

y

x

1 unit

y

x

(0, 1)

(1, 0)( 1, 0)

(0, 1)

y

x

P( , ) x y

y

x

P( , ) x y

1 y

x

O N

Hence, we must rethink our definition of the trigonometric

ratios. In the previous section we introduced circle to explain

angles of any magnitude. Similarly we will use circle to

re define our trigonometric ratios. In fact to make things as

simple as possible we will use unit circle.

a

a

a



HISTORICAL NOTE

+The origin of the term “sine” is quite fascinating. , a Hindu mathe-matician who studied trigonometry in the th century AD called the sine-leg of thecircle diagram “ardha-jya” which means “half-chord”. This was eventually shortened to“jya”. Arab scholars later translated Aryabhata’s work into Arabic and initially phoneti-cally translated “jya” as “jiba” but since this meant nothing in Arabic they very shortly

began writing the word as “jaib” which has the same letters but means “cove” or “bay”.

Arbyabhata

5

Finally in an Italian,, translated this work into Latin and re-

placed “jaib” with “sinus” which means“bend” or “curve” but is commonly used inLatin to refer to a bay or gulf on a coastline.

The term “sine” that we use today comes fromthis Latin word “sinus”. The term “cosine”comes from the fact that the sine of an angle isequal to the cosine of its complement. In ,

introduced the abbreviated“co sinus” for “complementary sine”.

1150

1620

Gerardo of Cre-

mona

Edmund Gunter

sinleg

chordof circle


Therefore, based on a unit circle, the definitions of the trigonometric ratios are:

If point P is any point on the unit circle, then

cos µ = x-coordinate of P

sin µ = y-coordinate of P

tan µ = y-coordinate of P

x-coordinate of P

Rather than say that P has co-ordinates (x, y), we can

now say that the co-ordinates of P are (cos µ, sin µ).

Let us see how the definition works in a quadrant other

than quadrant I. If the angle is in the third quadrant,

i.e. 180o < µ < 270o, then sin µ and cos µ are both

negative while tan µ will be positive.

To remind us of the meaning of the x and y-coordinates

of a point on the unit circle, we can also call the x-axis

the cosine axis and the y-axis the sine axis.

As the line segment OP moves around the unit circle mapping out angles of any magnitude, the

values of sin µ and cos µ change from a maximum value of 1 to a minimum value of ¡1. Therefore

for angles of any magnitude, the range of sin µ and cos µ are:

¡1 6 sin µ 6 1 and ¡1 6 cos µ 6 1 respectively.

In geometry you have learned that a tangent is a line that intersects a circle at just one point.

However, when we refer to the tangent ratio in trigonometry we mean either adj

hyp or

sin µ

cos µ or

y

x.

Are these two seemingly different ideas, the geometrical and the trigonometrical, related in some

y

x

P(cos , sin )

cos

sin

1

sine axis

cosine axis




way? Let us return to our original unit circle diagram which we used to define the sin, cos and tan

ratios.

Draw a line perpendicular to the x-axis (cosine

axis) which touches the unit circle at A(1, 0).

Extend OP so that it intersects this tangent at B.

¢ONP is similar to ¢OAB

) NP

ON =

AB

OA

But NP

ON = tan µ

) AB

OA = tan µ

) AB = tan µ since OA = 1 for the unit circle.

A tangent line is drawn to the circle at A (and hence is perpendicular to the x-axis). If OP isextended to intersect this tangent line at B, then the length of AB is equal to the tangent of the

angle. Hence the tangent ratio does have a strong connection with the tangent to a circle.

We have already noted that the range for sine and cosine is from ¡1 to 1. The tangent ratio is

more interesting. For an angle of 89o, the tangent ratio is very large. For an angle of 89:9999o,

the tangent ratio is very large. Try this on your calculator. Can you see why this is so?

For an angle of 90o, the value of tangent does not exist since the ray OP will never intersect the tan

axis. We say that the value of tan90o (and similarly 270o) is undefined. While sine and cosine

only take on values in [¡1, 1], the value of the tangent ratio can be any real number.

The values of sin, cos and tan ratios can be either positive or negative now that we are considering

angles of any magnitude.

We need to develop a simple way of quickly determining whether a trigonometric ratio for a given

angle is positive or negative.

cosine axis

sine axis

P

y

xO N

A

B

INVESTIGATION 3

TRIGONOMETRIC RATIOS IN OTHER QUADRANTS

What to do:only in

valuesare positive

S All trig

valuesare positive

only osvaluesare positive

Conly anvalues

are positive

T

1

2

A S T C

Use your calculator to find sin, cos and tan values of variousangles in the range , and satisfy yourself thatthis diagram is correct for angles in that particular quadrant.

Based on the unit circle definitions, develop an argument as towhy only sine is positive in quadrant II, only tangent is posi-tive in quadrant III and only cosine is positive in quadrant IV.

This can be summarised using the diagram on the right. There area variety of ways of remembering which letter goes in which quad-

rant. is one common mnemonic.

Perhaps you would like to develop your own.

0 360o o< µ <

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ l l t a t i o n s o e n t r a l S A

T C




Because definitions of the trig ratios can be developed from a unit circle, mathematicians quite oftenrefer to sin, cos and tan as circular functions. Consider the sine ratio. It is a function because for

each value in the domain (i.e. for each angle) there is exactly one value in the range (i.e. one value

of sin). The study of trigonometry is included in our study of functions.

1 Refer to the third

diagram on page 180.

Using this, copy and

complete the table.

Angle sin µ cos µ tan µ

0o

90o

180o

270o

360o

2 Draw a unit circle and label the sin axis, the cos axis and the tan axis. A useful diagram is

available from the website. Use the diagram to estimate the value of

a sin60o b cos60o c tan60o d sin315o

e cos315o f tan315o g sin150o h cos150o

i tan150o j sin( )¡30o k cos( )¡30o l tan( )¡30o

m sin570

o n cos 570

o o tan570

o

Now check all of your estimates using your calculator.

3 Without using your calculator determine whether the ratio is positive, negative or zero.

a sin210o b cos300o c tan( )¡60o d sin130o

e cos( )¡150o f tan480o g sin750o h cos( )¡450o

4 If 0o 6 µ 6 360o, determine the range of µ values for which

a sin µ is positive b cos µ is negative

c tan µ is positive d sin µ is negative

e sin µ is positive and tan µ is negative f cos µ is positive and tan µ is positive

g tan µ is negative and cos µ is negative.

INVESTIGATION 4

OTHER TRIGONOMETRIC RELATIONSHIPS

As mentioned previously there are many relationships between the trig ratios. A few that we

tan µ = sin µ

cos µ, sin µ = cos(90o ¡ µ), cos µ = sin(90o ¡ µ)

A set of relationships that will be required in later sections of this chapter are

sin(180o ¡ µ) = sin µ

cos(180o ¡ µ) = ¡ cos µ

tan(180o ¡ µ) = ¡ tan µ

By using a variety of values for angle µ, satisfy yourself that these relationships are true.

have encountered already are

EXERCISE 5D



CLASS INVESTIGATION 5


5 If 0o 6 µ 6 90o, find µ in each of the following. Drawing a diagram of a unit circle may help.

a sin160o = sin µ b sin135o = sin µ

c cos100o = ¡ cos µ d cos152:8o = ¡ cos µ

e tan140o = ¡ tan µ f tan115o3204500 = ¡ tan µ

6 Angles are said to be coterminal if they have the same position on the unit circle diagram,

i.e. if the angles terminate at the same point on the unit circle. For example, 30o

and 390o

would be coterminal angles.

a Give six angles of different magnitude that are coterminal with 60o.

b What can be said about the sine values of all angles coterminal with 60o?

c What can be said about the cosine values of all angles coterminal with 30o?

d What can be said about the tangent values of all angles coterminal with 90o?

7 If sin µ = ¡0:89 determine two possible values for µ if 0o < µ < 360o. Drawing a diagram

of a unit circle may help.

8 If sin µ = a, then determine the following in terms of a, where 0o < µ < 90o. Drawing a

diagram of a unit circle or a right-angled triangle may help.a cos µ b tan µ

9 If 0o < µ < 360o, then determine for what values of µ

a sin µ = cos µ b sin µ = tan µ c tan µ = cos µ

In our study of geometry and trigonometry work so far we have used degrees as the unit for the mag-nitude of an angle. As already mentioned the Babylonians introduced this unit and mathematicians

over the ages have seen little need to change this basic unit of 1o = 1360 of a revolution.

RADIAN MEASURE E

Materials: a A , aa a

a

A , aa

f

sheet of paper pair of compasses, length of garden twist’em or piece of whipper-snipper chord or piece of string(something flexible but not too flexible).

On an sheet of paper draw large circlewith pair of compasses. Do not change the dis-tance between the points and step this distanceof around the circumference of the circle.

4

4

The division of circle into equal parts is quite arbitrary Why not divide the circle into equal parts or into parts? The division of circle into units has some appeal as this would give usright-angle of parts. This has been tried from time to time but it is not very convenient.

more natural way to measure angles is to use measure known as radian is simply an angle subtended at the centre of circle by an arc equal in length to the radius. The term initiallyused with this measure was radial angle but in James Thompson introduced the term radian.

a .a a

A a . Aa

360 100400 400

100

1873

radian measure

Q

A

P




Note the distance AP is a chord length equal in length to

the radius and in fact goes around the circle exactly six

times. This is the well known construction of a regular

hexagon. The triangle AOP is equilateral and ]AOP is

exactly 60o.

Now measure the radius of the circle with your piece of flexible material. Then place one end of the material at

A and carefully, with your fingers, lay it along the arc

of the circle.

If you do this carefully you will find that the end does

not quite reach P. Call this point B. Join OB, and the

angle ]AOB is one radian.

Note that one radian is slightly less than 60o. (1 radian + 57o)

How many times would the radius fit around the curvature of the circle to reach A again?Obviously it is six plus a “little bit”.

We can work out how many times the arc equal to the radius will divide into the circumference as

circumference

arc =

2¼r

r = 2¼ so 360o = 2¼ radians.

60°O A

P

A radian is the angle at the centre of a

circle subtended by an arc whose length

is equal to the radius.

O A

B

r

r r

Establish the magnitude of the angles in these diagrams, in radians.

a b c

a µ = 85 = 1.6 (do not write anything after the number 1.6)

b µ = 66 = 1

c No lengths are given. But, in any circle C = 2¼r so, half the circumference will be ¼r.

Hence, µ = ¼r

r = ¼.

From we see thatExample 5.7 180o = ¼ radians

EXAMPLE 5.7

8 cm

5 cmO

6 cm

6 cmO

O

O A

Pflexiblematerial

B




Symbol for radians

In this text, we use the radian as a unit of angle measure. Different notations are used in different

texts. For example, we could write sin2 (radians), sin2R or sin2c (c for circular measure;

this is not used very much in print as c could be confused with o for degrees).

In more advanced mathematics you will learn that radians are simply real numbers and the conven-

tion is to write sin 2, the implication being that without being specific, the angle is in radians.

CONVERTING FROM DEGREES TO RADIANS, AND VICE VERSA

Simple conversions Because 360o = 2¼R

180o = ¼ R

90o = ¼2

R

45o = ¼4

R

60o = ¼3 R

30o = ¼6

R

You should be able to convert multiples of 90o, 45o, 60o, 30o to radians without a calculator.

Convert to radians a 240o b 135o c 330o

a 240

o

= 4 £ 60o

= 4 £ ¼3

= 4¼3

b 135

o

= 3 £ 45o

= 3 £ ¼4

= 3¼4

c 330

o

= 11 £ 30o

= 11 £ ¼6

= 11¼6

Harder conversions

Convert to radians a

180

o b 300

o c 500

o

EXAMPLE 5.8

EXAMPLE 5.9

To do any conversion, multiply by desired unit

existing unit. Since ¼ radians = 180o, we multiply by ¼

and divide by 180:

a 180o

= 180 £ ¼180 radians

= ¼R

= 3:14R

b 300o

= 300 £ ¼180 radians

= 5¼3 radians

= 5:236R by calculator

c 500o

= 500 £ ¼180 radians

= 8:727R




Convert these angles from radians to degrees: a ¼4 b 11¼

6 c 1

To convert from radians to degrees, we multiply by 180o

¼

,

i.e., we multiply by 180 and divide by ¼ .

a ¼4

= ¼4 £ 1800

¼

= 45o

b 11¼6

= 11¼6 £ 1800

¼

= 11£180o

6

= 330o

c 1

= 1 £ 1800

¼

= 1800

¼

+ 57.3o

An angle of magnitude 1 radian is slightly less than 60o. So a rough approximation for a radian is

a bit less than 60o.

EXAMPLE 5.10

1 Determine the magnitude of each of the following angles in radian measure.

a b c

d e f

2 Express the following in radian measure, leaving your answer in terms of ¼.

a 30o b 90o c 45o d 120o

e 15o f 60o g 22:5o h 240o

i 300o j 225o k 180o l 270o

3 Express the following in radian measure (correct to 4 decimal places).

a 47:8o b 153:47o c 221o d 418o

e 64o180 f 257o450 g 98o1503000 h 338o2704500

i ¼o

WARNING: Now that you have two methods of naming the magnitude of angles, you must be careful that your calculator is in the correct degrees or radians.mode,

EXERCISE 5E

4

2

O6

9

O5

15

O

O x

x2

O

4r

r O



4 Copy and complete this table.

degrees 45o 60o 120o 180o

radians ¼6

¼2

5¼4

3¼2 2¼

5 Convert the following into degrees.

a ¼4 b 3¼ c ¼

3 d 4¼3 e ¼

6

f 7¼6 g 11¼

4 h 7¼5 i 20¼

9

6 Convert the following into degrees, giving your answer in sexagesimal form.

a 1 b 0:76 c 1:39 d 2 e 3:64

f 0:27 g 4:85 h 2:71 i 5

7 Find the value of each of the following. Give the exact value where possible.

a sin( ¼3 ) b cos( 3¼

4 ) c tan( ¼4 ) d sin2¼ e cos(¡¼

6 )

f tan(¡

2¼

3 ) g

sin1:49

h cos(¡3

:6)

i tan1

8 Find the angles µ in each case, where 0 6 µ 6 2¼. Give your answer in radian measure.

a sin µ = 0:5 b cos µ = 0:5 c tan µ = 1

d sin µ = 0:84 e cos µ = ¡0:563 f tan µ = ¡2:8

A formula for the area of a triangle is Area = 12 base £ perpendicular height.

In both triangles a perpendicular is constructed from A to D on BC (extended if necessary).

sin C = h

b

) h = b sin C

sin(180o ¡ C ) = h

b

) h = b sin(180o ¡ C )

but sin(180o ¡ C ) = sin C

) h = b sin C

AREA OF TRIANGLES F

B

A

CD a

bc

h

acute

B

A

C Da

b

ch

obtuse

180°C

This formula is sufficient if we have right-angled triangle or we know the perpendicular height inthe triangle that is not right-angled. real life there are many triangles that are not right-angled or where it is often very difficult to measure the perpendicular height accurately

aIn

.

Using trigonometry we can develop an alternative formula that does not depend on perpendicular height. Any triangle that is not right-angled must be either acute or obtuse. will consider both cases.

, aWe




From our old formula for the area of a triangle,

Area = 12 base £ perpendicular height

= 12 ah

we have been able to develop another formula for the area of a triangle

Area = 12 ab sin C

This formula will work for all angles between 0o and 180o.

In particular, if we let C = 90o,

Area = 12 ab sin90o

= 12 ab since sin 90o = 1

which shows that our formula for the area of a triangle that we learned in Junior school is just a

special case of this formula. This is a common occurrence in Mathematics. A more general formulais developed that includes an existing formula as a special case.

In general, to use this formula we need to know any two sides of the triangle and the included

angle. There are three variations of the rule depending on how the sides and angles are named:

Area of ¢ABC = 12 ab sin C

= 12 ac sin B

= 12 bc sin A

In words,

Find the area of ¢ABC shown.

We know B = 110o, a = 15 cm, c = 12 cm

(i.e., 2 sides and the included angle)

Area = 12 ac sin B

= 12 £ 15 £ 12 £ sin110o

= 84:6 cm2

B

A

Ca

bc

B

A

Ca

bc

B

A

C

110° 15 cm12 cm

EXAMPLE 5.11


The area of triangle equals half the product of any two sides and the sine of the included angle.

a



1 Calculate the area of the following triangles. All lengths are in cm.

a b c

d e f

2 Calculate the area of the following shapes.

a b

3 Find the angle A in this triangle if the area is 15 cm.

4 Find the side x in each case.

a b

5 Find the area of the parallelogram ABCD in which AB = 6.2 cm, BC = 8.9 cm and B = 40o.

6 A triangle has an area of 5 cm2. Two sides are 5 cm and 4 cm and the angle between these

sides is µ . Find µ , giving your answer in sexagesimal form.

7 A rhombus has an area of 60 cm2 with an internal angle of magnitude 75o. Find the length of

the side of this rhombus.

8 Prove

Hint: Draw a diagram.

that the formula for the area of a regular hexagon is as follows:

Area of hexagon = 3

p 3s2

2 where s = length of one side.

EXERCISE 5F

70°

8 8

100°6

4

52°

7

10

107°40'8

5

50° 45°

85°

8

6

36°52'12"

6

8

10

68°

12 cm

6 cm

105°

73°

5 m

4 m

2.8 m3.6 m

A

B C

5 cm 4 cm

15 cm

55°

Area 100 cm 2

x

Area 3 cm 2 ~3 cm

x

60°




9 The Australian 50 cent coin has the

shape of a dodecagon (12 sides).

Eight of these 50 cent coins will fit

exactly on an Australian $10 note

as shown below. What fraction of

the $10 note is not covered?

Previously we saw that the area of ¢ABC can be written in three ways.

Area of ¢ABC = 12 bc sin A or 1

2 ac sin B or 12 ab sin C .

Since each represents the same area, we can write

This may also be written in inverted form as

12 bc sin A = 1

2 ac sin B = 12 ab sin C

) bc sin A = ac sin B = ab sin C fmultiply by 2g

) bc sin A

abc =

ac sin B

abc =

ab sin C

abc f¥ by abcg

) sin A

a =

sin B

b =

sin C

C fsimplifyingg

a

sinA

= b

sinB

= c

sinC

The sine rule states:

In any triangle ¢ABC sin A

a =

sin B

b =

sin C

c

or a

sin A =

b

sin B =

c

sin C

The sine rule allows us to solve any triangle provided that two angles and one side are known.

THE SINE RULE G

A

CB a

bc

This formula which relates the three sides and the sin ratios of the three angles is called the sine rule.


We

We a Wea a

have seen earlier that in any right-angled triangle if we know two sides, or one side and an an-gle other than the right angle then we can use Pythagoras’ Theorem or trigonometric ratios to solve

the triangle. now wish to extend this to triangles that do not have right angle. will use our newly found formula for the area of triangle to develop formula relating the sides and angles of any triangle.



A = 105o a = ? cm B = 25o b = 14 cm C =? c = ? cm

Finding C : C = 180o ¡ 105o ¡ 25o = 50o

Finding a: a

sin A =

b

sin B

) a = b sin A

sin B

) a = 14 £ sin105o

sin25o

) a = 32

Finding c: c

sin C =

b

sin B

) c = b sin C

sin B

) c = 14 £ sin50o

sin 25o

) c = 25:4

Solution is: C = 50o, a = 32 cm, c = 25:4 cm

Note: If you are given two angles it is a very simple procedure to find the third angle. As a

check against having made calculation errors, in any triangle the longest side is always opposite

the largest angle and the shortest side is always opposite the smallest angle.

1 In ¢ABC, find

a a if A = 72o, B = 23o and b = 35 cm

b c if A = 66o, B = 51o and a = 5:1 m

c a if A = 47o, B = 65o and b = 6:8 cm

d b if B = 59o, C = 73o and a = 12 m

e a if B = 69o, C = 41o and b = 3 cm

2 Solve the following triangles (i.e. find all unknown sides and angles).

a b c d

3 Find

a ]ADB

b length of BD

c length of CD

EXERCISE 5G.1

AC

B

50°

65° 20 cm

R

Q P52°

48°8 m

X

Y

Z

105°

200 km 36°

25° 40°

A B C

D

100 m


E

D

F

80°

45°

0.16 m

Solve ¢ABC.

EXAMPLE 5.12A

B

C25°

105° 14 cm



INVESTIGATION 6

THE AMBIGUOUS CASE

4 The angle of elevation of the top of a tall building,

T, observed from a point A on ground level is

18o. The angle of elevation from a point B, 300metres closer to the base of the building in the

same plane as A and T, is 30o. Find the height

of the building using the sine rule.

Task 1: Draw AB = 10 cm. At A construct an angle of 30o. Using B as centre, draw an

arc of a circle of radius 6 cm. Let the arc intersect the ray from A at C. How many

different positions may C have and therefore how many different triangles ABC may

be constructed?

Task 2: As before, draw AB = 10 cm and construct a 30o angle at A. This time draw an

arc of radius 5 cm based on B. How many different triangles are possible?

Task 3: Repeat, but this time draw an arc of radius 3 cm on B. How many different trianglesare possible?

Task 4: Repeat with an arc of radius 12 cm from B. How many possible triangles?

In this investigation you should have discovered that when you are given two sides and a non-

included angle there are a number of different possibilities. You could get two triangles, one

triangle or it may be impossible to draw any triangles from the given data.

Let us consider the calculations involved in each of the cases of the investigation.

Task 1: Given: c = 10 cm, a = 6 cm, A = 30o

Finding C : sin C

c =

sin A

a

) sin C = c sin A

a

) sin C = 10 £ sin 30o

6 = 0:8333

Because sin µ = sin(180o ¡ µ) there are two possible angles:

C = 56:44o or 180o ¡ 56:44o = 123.56o

On your calculator check that the sin ratio of both of these angles is 0.8333.

THE AMBIGUOUS CASE OF THE SINE RULE (OPTIONAL)

Y a , a , a a.

ou will need blank sheet of paper ruler protractor and compass for the tasksthat follow In each task you will be required to construct triangles from given infor mation.

Ya

a

ou may have noticed in the previous exercise that all calculations where the sine rule was usedinvolved finding side (having been given two angles but only one side). The sine rule may also beused to solve triangles where and are given. However you have to

be very careful because different triangles are sometimes possible from the same set of given data.tw sides non-included angleo

10 cmA B

C

C

6 cm

6 cm

30°

18° 30°A B C

T

300 m




Carrying on to solve the triangle in both cases:

If C = 56:44o

) B = 180o ¡ 30o ¡ 56:44o

) B = 93:56o

b

sin B =

a

sin A

) b = a sin B

sin A

) b = 6 £ sin93:56o

sin30o

) b = 11:98 cm

If C = 123:56o

) B = 180o ¡ 30o ¡ 123:56o

) B = 26:44o

b

sin B

= a

sin A

) b = a sin B

sin A

) b = 6 £ sin 26:44o

sin30o

) b = 5.34 cm

Therefore two solutions are possible:

c = 10 cm C = 56:44o C = 123:56o

a = 6 cm with B = 93:56o or B = 26:44o

A = 30o b = 11:98 cm b = 5:34 cm


Finding C : sin C

c =

sin A

a

) sin C = c sin A

a

) sin C = 10 £ sin30o

5) sin C = 1

There is only one possible solution for C in the range from 0o to 180o and that is C = 90o.

So only one triangle (i.e. one set of solutions) is possible. Complete the solution of the triangle

yourself.


Finding C : sin C c

= sin Aa

) sin C = c sin A

a

) sin C = 10 £ sin30o

3

) sin C = 1:6667

There is no angle that has a sin ratio > 1. Therefore there is no solution for this given data, i.e.,

no possible triangle can be drawn.

10 cmA B

3 cm

30°

10 cmA B

5 cm

30°

C





Finding C : sin C = 10 £ sin 30o

12

) sin C = 0:4167

Two angles have a sin ratio of 0.4167

C = 24:62o or 180o ¡ 24:62o

C = 24:62o or 155.38o

However, in this case only one of these two angles is valid. If A = 30o then C cannot possibly

equal 155.38o because 30o +155:38o > 180o. Therefore, there is only one solution,C = 24:62o.

Once again, you may wish to carry on and complete the solution.

1 Construct two different triangles for each of the following and then proceed to solve both

triangles.

a ¢ABC with a = 15 cm, c = 13 cm and C = 45o

b ¢PQR with r = 6 cm, q = 4:5 cm and Q = 30o

2 ABC has B = 40o , b = 8 cm and c = 11 cm . Show that C can have two possible values

and solve the triangle in each case.

3 Solve ¢ABC in each case, given:

a a = 75 cm, c = 80 cm and A = 60o

b a = 24 cm, c = 19 cm and A = 36o

c a = 3 cm, c = 8 cm and A = 100o

d a = 30 cm, b = 15 cm and A = 120o

e a = 60 cm, b = 76 cm and B = 35o

4

The cosine rule, like the sine rule, involves the

a2 = b2 + c2 ¡ 2bc cos A

or b2 = a2 + c2 ¡ 2ac cos B

or c2 = a2 + b2 ¡ 2ab cos C

We will develop the first formula for both an acute and an obtuse triangle.

EXERCISE 5G.2

10 cmA B30°

12 cm

C

68°

85°9.8 cm

11.4 cmIs it possible to have triangle with measurements as shown?Explain!

a

A

CBa

bc

sides and angles of a triangle.

THE COSINE RULE H

In any ABC¢


Conclusion: Each situation using the sine rule with two sides and non included angle must be

examined very carefully

a

.



In both triangles drop a perpendicular from C to meet AB (extended if necessary) at D.

Let AD = x and let CD = h.

Apply the Theorem of Pythagoras in ¢BCD

a2 = h2 + (c ¡ x)2 a2 = h2 + (c + x)2

) a2 = h2 + c2

¡2cx + x2 ) a2 = h2 + c2 + 2cx + x2

In both cases, applying Pythagoras and substitute for to ¢ADC: h h2 2+ x2 = b2

) a2 = b2 + c2 ¡ 2cx ) a2 = b2 + c2 + 2cx

In ADC: cos A = x

b Now cos(180 ¡ A) =

x

b

) b cos A = x ) b cos(180 ¡ A) = x

) a2 = b2 + c2 ¡ 2bc cos A But, cos(180 ¡ A) = ¡ cos A

) ¡b cos A = x

) a2 = b2 + c2

¡2bc cos A

The other variations of the cosine rule could be developed by rearranging the vertices of ¢ABC.

Note that if A = 90o, cos A = 0 and a2 = b2 + c2 ¡ 2bc cos A reduces to a2 = b2 + c2.

This shows that Pythagoras’ Theorem is just a special case of the cosine rule.

In ¢ABC find:

a b

b A

Ta o find b we need the variation of the formula

b2 = a2 + c2 ¡ 2ac cos B

) b2 = 62 + 82 ¡ 2 £ 6 £ 8 £ cos114o

) b2 = 139:05

) b = 11:8

EXAMPLE 5.13B

A

C

114°

8

6

b

B D A

C C

B A

h ha ab b

c x c x x

180°A

D




T b o find A it is best to rearrange the formula to make cos A the subject.

a2 = b2 + c2 ¡ 2bc cos A

) 2bc cos A + a2 = b2 + c2

) 2bc cos A = b2 + c2 ¡ a2

) cos A = b2 + c2

¡a2

2bc

) cos A = 11:82 + 82 ¡ 62

2 £ 11:8 £ 8

) cos A = 0:8858

) A = 27:7o

Find the angle opposite the longest side in this triangle.

Given: a = 20, b = 15, c = 8, A =?

From before: cos A = b2 + c2 ¡ a2

2bc

) cos A = 152 + 82 ¡ 202

2 £ 15 £ 8

) cos A =¡

0:4625

) A = cos¡1(¡0:4625)

) A = 117.55o or 117o3205400

Note: If cos A is negative this means that A must be a second quadrant angle.

The cosine rule can be used to solve triangles given:

² two sides and an included angle

² three sides.

There is no ambiguity possible using the cosine rule.

1 Rearrange the formulas for the cosine rule:

a to make cos A the subject b to make cos B the subject

c to make cos C the subject

2 In ¢ABC, given

a b = 4 cm, c = 5 cm and A = 68o, find a

b a = 11 cm, b = 7 cm and C = 60o, find c

EXAMPLE 5.14

A

B

C

8 cm

20 cm

15 cm

EXERCISE 5H




c a = 135 m, b = 215 m and C = 115o, find c

d b = 20 m, c = 15 m and A = 95o, find a

e a = 22 cm, c = 18 cm and B = 55o, find b

3 In ¢PQR, given

a p = 8 m, q = 5 m and r = 4 m, find P

b p = 30 cm, q = 20 cm and r = 41 cm, find Q c p = 12 m, q = 15 m and r = 7 m, find R

4 Find the value of x in each of the following.

a b

5 Find the smallest angle in a triangle with sides of 4, 5 and 6 cm.

6 Solve these triangles:

a b c

d e f

7 Find the largest angle in the triangle with sides of 2.1 m, 3 m and 1.8 m.

8 The lengths of two adjacent sides of a parallelogram are 15 cm and 9 cm with the angle

between the sides being 40o. Find the length of the longer diagonal.

9 Use the cosine rule followed by the sine rule to solve ¢ABC where b = 5, c = 8 and A = 56o.

10 If two sides of a triangle are of length 26 cm and 22 cm and include an angle of 69o3004500,find the length of the third side and the other two angles.

11 Two cyclists depart from the same point. One travels due East at 18 km/hr while the other

travels North-West at 20 km/h. How long (to the nearest minute) will it be before they are 80km apart?

x

5

6

60°

A C

B

10 m

5 m 6 m

X

Y

Z

56°

11.2 km

14.6 km

8 cm

16 cm12.5 cm

15 m

40°25'

18 m

11 cm9 cm

5 cm

12 cm

36°42'

9 cm

x12

120°7




Surveyors and navigators use trigonometry to make calculations in many real-life practical situations.

These calculations usually involve directions or bearings.

The direction of a point is stated as a number of degrees East or West of either North or South. In

the diagram, the direction of A from O is N40oE. The direction of B from O is S60oW.

A party of bushwalkers start out at Base Camp and walk for 12 km on a bearing of 231.

They then walk for 19 km on a bearing of 117. Find

a how far they are now from Base Camp

b the bearing they must travel along to return to Base Camp.

First draw a labelled diagram showing all data. Other angles can be calculated and added on

to the diagram.

We need to find BC and the bearing

of B from C (i.e. angle µ).

In ¢ABC: A = 117o ¡ 51o o= 66 ; b = 19 km, c = 12 km

As we know two sides and an included angle we will use the cosine rule.

Finding BC:a a2 = b2 + c2 ¡ 2bc cos A

) a2 = 192 + 122 ¡ 2 £ 19 £ 12 £ cos66o

) a2 = 319:53

) a = 17.88 km

Therefore the distance from Base Camp is 17.88 km.

EXAMPLE 5.16

51°

66°

117°

A

19 km

12 km

231°

51°

B

C

A

N

S

EW

B

30°

40°


Another way to measure a direction is by measuring

it from true North in a clockwise direction, called abearing. A bearing is given as a 3 digit number from

001 to 359, without the degree symbol. From the

diagram given the bearing of A from O is 040 while

the bearing of B from O would be 240. The reverse

bearings of O from A would be 220 (add on 180) and

of O from B would be 060 (either add on 180 and

then subtract 360, or just subtract 180).



Finding \ b ABC: cos B = a2 + c2 ¡ b2

2ac

) B = 76o (to nearest degree)

) O = 76o ¡ 51o

= 25

o

Therefore bearing of Base Camp from C is 335.

1

2 From the foot of a building I have to look upwards at an angle of 10o to sight the top of a

tree. From the fourteenth floor of the building, 50 metres above ground level, I have to look

down at an angle of 60o below the horizontal to sight the tree top.a How high is the tree?

b How far from the building is this tree?

3 A Communications Tower is constructed

on top of a building as shown. Find the

height of the tower.

4 A tower, 50 metres high, stands on top of a hill. From a point some distance from the base

of the hill, the angle of elevation to the top of the tower is 12.5o. From the same point the

angle of elevation to the bottom of the tower is 9o. What is the height of the hill?

5 A man walks a distance of 8 km in a direction S56oW . He then walks due East for 20 km.

What is the distance and bearing of the man from his starting point?

6 A hiker walks 5.5 km in a direction S20oE . He then turns and walks due North for a distance

of 6 km. What is the distance of the hiker from his starting point?

7 A point A is 12 km from B along a bearing of 132. C is 17 km from B along a bearing of

063. Find the bearing of A from C.

EXERCISE 5I

A aa

B a

.

student wishes to determine the height of flag pole. Shetakes sighting of the top of the flagpole from point A.She then moves further away from the flagpole bymetres to point and takes second sighting.The information is shown in the diagram

below How high is the flagpole?

30

B A

30 m

25° 55°

tower

building

20°

20°

100 m




8 A yacht sails on a bearing of 065 for 4 km and then sails 8 km along a bearing of 195. Find

the distance and bearing of the yacht from its starting position.

9 A football goal (soccer) is 8 metres wide. When a player is 24 metres from one goal post and

18 metres from the other, he shoots for goal. Assuming that the goal keeper is out of position

and will not interfere, within what angle must the shot be made if he is to score a goal?

10 An aircraft takes off from an airport and flies 400 km in a direction of N40o

E. It then changescourse and flies in a direction S20oE until it is 500 km from the airport.

a What is the direction of the airport from the aircraft’s final position?

b How far did it fly along the second leg of the journey?

11 The solid figure shown is a rectangular prism.

Find the magnitude of angle CAB.

12 The triangle ABC has a = 5:8 cm, b = 7:5 cm and A = 45o.

Find

13 Three circles with radii of 3 cm, 4 cm and 5 cm touch each other externally. Using the centres

of the circles as vertices, a triangle is drawn. Find the area of this triangle.

14 Two points A and B on the same bank of a river are 60 metres apart. C is a point on the other

side of the river. If ]BAC = 82o420 and ]ABC = 68o260, find the width of the river.

15 Jack and Jill are standing on level ground 100 metres apart. A large tree is due North of Jack

and on a bearing of 065 from Jill. The top of the tree appears at an angle of elevation of 25o

to Jill and 15o to Jack . Find the height of the tree.

16

A ! B: due North

A ! C: 028A ! D: 060B ! C: 070B

! D: 140

Find the area of the property in hectares.

A

B

C

5 cm

3 cm

6 cm

450 m

A

B

C

D


A property owner wishes to know the area

of his property, the shape of which is

shown alongside. He measured the length

of side AB to be metres and then took

the bearings as shown.

450

the area of the triangle.



1 A triangle has an area of 85 cm2. If two of the sides are 18 cm and 11 cm,

2 ABCD is a square. M is the mid-point of side AB and E divides DC in the ratio of 2 : 1,

i.e, DE : EC = 2 : 1. Find ]AEM, giving your answer in sexagesimal form.

3 A pilot takes off from the RAAF base at Amberley for a destination 400 km due West of

the base. After covering 250 km, the pilot checks his instruments and finds that he has

been flying on a bearing of 280o since leaving the base. What course should he now fly

to arrive at his destination?

4 At 12 noon a ship leaves port and travels at 8 km/hr along a straight course in the direction

N25oE. At 2 pm a second ship leaves the same port and travels at 9 km/hr in the direction

S45o

E. The radio communication between the two ships has a range of 75 km. Will thetwo ships be able to communicate by radio at 6 pm?

5 You now know two formulas for the area of a triangle:

Area = 12 bh and Area = 1

2 ab sin C

a Show that for a right-angled triangle, these formulas are equivalent.

b Show that for any triangle, these formulas are equivalent.

6 In ¢ABC, prove that c = a cos B + b cos A:

8 A painting 2 metres by 2 metres is hung on a wall such that the lower border is 1 metre

above eye level of a normal person. An art expert gives advice that the viewer should stand

far enough away so that the viewing angle is about 20o. How far from the wall should the

viewer stand so that this angle of 20o occurs?

9 Recall the alternative formula for the areaof a triangle using trigonometry is:

Area = 12 ab sin C or

Area = 12 ac sin B or

Area = 12 bc sin A

By applying this formula to the triangles in the diagram given:

a Show that: sin(µ + Á) = h

a sin µ +

h

c sin Á

b Hence show that: sin(µ + Á) = sin µ cos Á + cos µ sin Á

determine the length of the third side.

A C

B

c

b

ah

N


7 The diagram alongside shows a circular

entertainment area. It has a paved hexag-onal area with plants growing in the gar-

den (shown as the shaded sectors).

If the radius of the circle is 7 metres, find

the area of the garden.

PROBLEM SOLVING J

EXERCISE 5J




10 Three equal circles with radius r are drawn as

shown, each with its centre on the circumfer-

ence of the other two circles. A, B and C are

the centres of the three circles. Prove that an

expression for the area of the shaded region

is:

A = r2

2 (¼ ¡ p 3)

11 Vee belts are often used to form the drive link between pulleys. The length of the vee belt

is very important. To save time and reduce the chance of error, maintenance engineers use

a formula to determine the correct belt length given the radii of the pulleys and the distance

between their centres.

a = radius of larger pulley

b = radius of smaller pulley

c = distance between pulley centres.

Establish that:

belt length = (a + b)¼ + 2(a ¡ b) + 2c cos µ

where µ = sin¡1

µa ¡ b

c

¶ in radians.

12 Suppose you are standing an unknown distance

d away from a cliff of height h. You need to

know the height t of a tower located on top of

the cliff. You know that the angle of elevation

of the bottom of the tower is B and the angle of

elevation of the top of the tower is A. Show that

the formula for the height of the tower, t, is given

by

arc cosine rule segment

bearing cotangent sexagesimal

central angle coterminal similar triangles

chord direction of a point sine

circular functions knot sine axis

corresponding sides nautical mile sine rule

cosecant radian tangent

cosine secant

cosine axis sector

A B

C

ab

c

t = h

µtan A

tan B ¡ 1

¶AB

d

h

t






1 a For A = 29o, 58o and 85o use your calculator to verify that

cos(270o¡A) = ¡ sin A:

b Solve the triangles i ii iii

c At a distance of 160 metres from Nelson’s Column in Trafalgar Square in London, the

base of the statue is at an angle of elevation of 29:25o while the top of the statue is at an

angle of elevation of 31:80o. What is the height of the statue of Nelson?

2 The Great Pyramid in Egypt is 146:6 metres high

and has a square base of side 203:3 metres. Find

the angle which the edge at the intersection of

two faces makes with the horizontal.

3 a Convert from sexagesimal to decimal the following angles:

i 28o3303600 ii 135o1001200

b Convert from decimal to sexagesimal the following angles:

i 36:7o ii 137:2o

c Convert the angles in b to radians.

4

5 Solve the following triangles:

a b c

6 cm

5 cm

A

B C

13.9 cm

11.3 cm

P

Q

K V W

U

5 5

4

203.3 m 203.3 m

146.6 m

12 cm

58°

39°

11 m

25°

7 m42°

11 km

13 km

Euclid, a racing pigeon, averages 90 km per hour in peak condition. He is taken by rail to

Plains 200 km north west of Mourne and then by road to Arid which is 50 km beyond Plains

on a bearing of 280 degrees. How long will he take to fly back to Mourne?





6 Find the areas of the following figures:

a b

7 Use your calculator to find µ given 0 < µ < 180o and

a cos µ = ¡0:6791 b sin µ = 0:4672

c cos µ = 0:6154 d sin µ = 0:7777

8 a A regular octagon is inscribed in a circle of radius r . Show that the area of the octagon

is 2p

2r2:


1 Solve the triangle alongside.

2 Convert 35o1402600 into decimal form.

3 If 0o 6 µ 6 360o, determine the range of µ values for which cos µ is positive and

tan µ is negative.

4 Express 342.8o in radian measure (correct to 4 decimal places).

5 Convert 11¼4 into degrees.

6 A triangle has an area of 9 cm2. Two sides are 6 cm and 4 cm and the angle betweenthese sides is µ . Find µ, giving your answer in sexagesimal form.

7 Solve ¢ABC, given a = 62 cm, c = 73 cm and A = 62o.

8 Two cyclists leave town at midday. One travels along a bearing of 037 at 25 km/hr

while the other travels on a bearing of 153 at 30 km/hr. How far apart are they at

3 pm?

9 From a point X, a mountain peak which is due North of X has an angle of elevation

of 25o. From point Y, 3 km West but on the same level as X, the mountain peak

is on a bearing of 050. Find the height of the mountain peak.

8 cm 9 cm

A

BC

11 cm

28° 15 cm

70°

12 m

8 m

30°

10 m

b In the given diagram, angle BPC = 2(angle BAC).

Find a formula for r in terms of a and A.

A

BC

r

r

r

P

a




CHAPTER 6Revisionexercises forchapters

1 to 5

Review exercises are repro-

duced as student worksheets

on the CD and are accessible by clicking on this icon



REVIEW OF CHAPTER 1 A

208 REVISION EXERCISES FOR CHAPTERS 1 TO 5 (Chapter 6)

1 A salesman earns a fortnightly salary of $800, plus 15% commission on his fortnightly sales.

a What is the i independent variable ii dependent variable?

b Show the relation as a

i table ii set of ordered pairs iii mapping iv graph v equation

2 For the function f (x) = x2 + 3x, find the value of

a f (¡1) b f (3) c f (0)

3 Which of the following variables are discrete and which are continuous?

a hat size

b capacity of the glasses in your kitchen cupboard

c number of colours that your computer monitor displays

4 As a casual worker, Jenna earns $12 per hour for the first eight hours she works in a day, and$18 per hour for every hour after that. Union regulations state that she must be employed for

at least 4 hours per day, and no more than 12 hours per day.

a Which is the i dependent variable ii independent variable?

b Are the variables discrete or continuous?

c Represent the function as

i a table ii a graph

iii the rule for the domain to0 8 iv the rule for the domain to8 12

d State the domain and the range of the function.

5 Find the gradient of the line joining

a (¡3, 1) and (3, 2) b (1, ¡5) and (2, ¡5)

6 Show that the points H(¡1, 2), I(2, 4,), J(0, ¡3) and K(3, ¡1) are the vertices of a parallelo-

gram.

7 The cost of curtains is $16 per metre for the material, plus a labour charge of $125.

a What is the independent variable? b What is the dependent variable?

c Find the y-intercept. d What are the units of the dependent variable?

e Find the gradient. f What are the units of the gradient?

8 Find the gradient and the y-intercept of

a y ¡ 3 = ¡2(x ¡ 4) b x + 1

y ¡ 3 = 2

9 Determine the equations of these straight lines.

a gradient 5 and passing through (2, ¡5) b gradient ¡2 and passing through (1, ¡3)

c a vertical line through (1, 2) d a horizontal line through (1, 2)

10 a Find the equation of the straight line parallel to y = ¡2x + 3 that passes through

(¡1, 2). Graph both lines with your graphics calculator.

b Find the equation of the straight line perpendicular to y = ¡2x+3 that passes through

(¡

1, 2). Graph both lines with your graphics calculator.




REVISION EXERCISES FOR CHAPTERS 1 TO 5 (Chapter 6) 209

11 a Sketch the graph of y = ¡3x + 7 by constructing a table, and plotting points.

b Sketch the graph of y = ¡13 x ¡ 1 using the y-intercept and gradient.

c Sketch the graph of 4y ¡ x = 8 by finding the x and y-intercepts.

12 Broccoli costs $3.29 per kilogram. What is the cost of

n kilograms?

a What is the independent variable?

b What is the dependent variable?

c Is the independent variable discrete or continuous?

d Find the constant of proportionality.

e Write the relation between the variables in the form “/”.

f Write the equivalent linear equation.

13 A laser printer takes 15 seconds to print the first page and

prints 12 pages per minute after that.

a Are the variables ‘time’ and ‘number of pages printed’

discrete or continuous variables? Explain your answer.

b What is the

i independent variable ii dependent variable?

c Draw up a table showing the relationship between time

and number of pages.

d What is the domain?

e What is the range?

f Draw a graph from the table.

g Calculate the gradient. What are the units of the gradient?

h What does the gradient represent in this application? i Find the equation for this situation.

14 Which of these relations is a function? Explain your answers.

a f(1, 3), (2, 3), (2, 5), (3, 6), (4, 7)g b

15 Explain the sentence, “The time taken to dig the trench is proportional to the length of the

trench”. You may wish to include a graph as part of your explanation.

16 The size of a sheet of A4 paper is 297 mm by 210 mm. A sheet of A3 paper has twice the

area of a sheet of A4 paper, with dimensions 420 mm by 297 mm. Similarly, a sheet of A2 paper is twice the size of a sheet of A3, and so on. Now if you cut a sheet of A4 paper in

half, you make two sheets of A5 paper, each 210 mm by 148:5 mm. Smaller paper sizes can

be made similarly.

a Copy and complete the following table.

1

2

3




Size A0 A1 A2 A3 A4 A5 A6 A7 A8

Width (mm) 297 210 148:5

Length (mm) 420 297 210

b On your graphics calculator, graph length vs width.

c Find an expression for the length, given the width.

d Explain why this is an example of direct proportion.

e What is the value of the constant of proportionality? Do you recognise this value?

1 Simplify

ap

56 bp

121 cp

x2

d 5

p 12

p 27

e 2

p c3 f

p 6 £

p 12

g 3p

8 £ 2p

2 h 3p

2 £ ¡3p

2 i ¡3p

6 £ 2p

2 £ p 3

j

p 27p 3

k

p 34p 17

l

p 18gp 2g

2 Solve for x.

a x2 = 36 b x =p

36 c x = ¡p 36 d x = §p

36

3 Simplify

a 6p

13

¡9p

13 b 2p

5 + 3p

20 cp

18 +p

50

d ¡3p 24 ¡ 2p 54 e 6p x ¡ 9p x f 6p a + 5p a ¡ 12p a g 5

p d2 + 3

p d2 h

p f 4 ¡ 3f 2 i

p 36n2

j

r a2

16 k

p w

p w

p w l

p a2bc3

abc2


a 5p

10 b

7p 7

c 6

p 2p

6d

p 8 + 6p

2

e

3

2p 3 f

1

p b g

7

p 2 h

2c

p c

5 Expand and simplify.

a (x + 6)(x ¡ 5) b (2x ¡ 1)(3x + 2) c (x ¡ 9)2

6 Expand using (a + b)2 = a2 + 2ab + b2:

a (x ¡ 8)2 b (2x + 5)2 c (5 ¡ 2x)2

7 Expand using (a + b)(a ¡ b) = a2 ¡ b2:

a (x + 7)(x ¡ 7) b (3x + 1)(3x ¡ 1) c (4 ¡ 2x)(4 + 2x)

REVIEW OF CHAPTER 2 B





8 Factorise completely.

a x2 ¡ 10x + 16 b 6x2 ¡ 7x ¡ 3 c x2 ¡ 36

d 3x2 ¡ 27 e 2x2 ¡ 12x ¡ 32 f 5x2 + 20x + 15

g 2x4 ¡ 2y4 h (x + 2)2 ¡ y2 i 2x2 + x ¡ 6

j 3x2 ¡ 11x ¡ 4 k 4x2 + 2x ¡ 6 l 6x2 ¡ 21x + 15

9 Given f (x) = 2x2 ¡ 17x + 8, find

a f (2) b f (8) c f ( 12 ) d f (0)

10 Factorise completely.

a x3 + 1 b x3 ¡ 125

11 Solve by graphing.

a 3x2 ¡ 5x ¡ 2 = 0 b 2x2 ¡ 8 = 0

12 Solve by factorising.

a x2

¡9x

¡22 = 0 b x2

¡6x + 9 = 0 c 6x2 + x

¡12 = 0

13 Solve by factorising.

a x2 ¡ 144 = 0 b 3x2 ¡ 75 = 0 cp

5x2 ¡ 4p

5 = 0

14 Write the following as perfect squares.

a x2 + 12x + 36 b 4g2 ¡ 12g + 9 c 25 ¡ 20x + 4x2


a x2 ¡ x ¡ 1 = 0 b 2x2 + 8x ¡ 1 = 0 c x2 + 5x + 4 = 0

16 Solve using the quadratic formula.

a x2 ¡ x ¡ 1 = 0 b 2x2 + 8x ¡ 1 = 0 c x2 + 5x + 4 = 0

17 Find the discriminant of each of these quadratic functions.

a y = 3x2 ¡ x + 6 b y = x2 ¡ 9x ¡ 14 c y = 3t2 + 6t + 3

18 Sketch the graph of y = x2 + x ¡ 6 over the domain [¡3, 3], by first making a table.

19 For the function y = 24 ¡ 2x ¡ x2, find

a the zeros b the y-intercept c the coordinates of the vertex.

Then sketch the graph of the function using this information.

20 For y = 3(x

¡1)(x

¡2) find the

a the roots b the y-intercept c the coordinates of the vertex.

21 a Sketch the graph of y = x2 and the graph of y = ¡3x2 on the same set of axes.

b Sketch the graph of y = x2 and the graph of y = 2x2 + 2 on the same set of axes.

c Sketch the graph of y = x2 and the graph of y = 6 ¡ (x + 3)2 on the same set of

axes.

22 Use completing the square to re-write the function y = x2 ¡ 4x + 10 in the form

y = A(x ¡ B)2 + C . Then sketch the graph of the function.

23 a Use completing the square to re-write the function y = 3x2 ¡ 8x + 1 in the form

y = A(x ¡ B)

2

+ C .




b What are the coordinates of the turning point of the graph?

c What is the y-intercept of the graph of this function?

d Find the zeros of the equation.

e Sketch the graph, showing all relevant information.

24 Sketch the graph of y =

jx

j and the graph of the given function on the same set of axes.

a y = jxj ¡ 4 b y = ¡jx + 5j 25 For each exercise, the graphs of f (x) = jxj and g(x) are given. State the equation for g(x).

a b

26 For each question, sketch the graph of f (x) = 1

x and the graph of the given function on the

same set of axes.

a f (x) = ¡ 1

x b f (x) =

1

2(x + 1)

27 One important feature of the overhead projector (OHP) in the mathematics room is the area

of projection A, for a given distance from the screen d. The OHP has been designed to have

the following relation between A and d: A = 0:15d2 + 0:4 where A is measured in square

metres and d is measured in metres.

a List some of the assumptions that have been made in the design of this mathematicalmodel.

b Make a table and draw a graph that shows that area for various distances d.

c What is the area of projection when the OHP is 2.1 metres from the screen?

d If a projection area of 2 square metres is needed, how far from the screen should the

projector be?

28 Give the coordinates of two points that lie

a on the function y = x2 ¡ 4x ¡ 6 b below the function y = x2 ¡ 4x ¡ 6

c above the function y = x2

¡4x

¡6

29 Consider the rectangle alongside, called the

golden rectangle. It has width of 1 unit and

a length of x units. It has the property that

if a 1 £ 1 square is cut from the rectangle,

the remaining rectangle has the same shape

as the original rectangle. In other words, the

ratio BC : AB is equal to the ratio EF : FC.

a Use this information to find the exact value of x.

This value is called the golden ratio.

b What is the value of the golden ratio to four decimal places?

ƒ(!)

g (!) y

x

ƒ(!)

g (!)

y

x

B

A

CF

DE

1

1 x




x 3 5 7 9 11 13 15 17

y

distance from origin

REVIEW OF CHAPTER 3C

30 Here is a “proof” that 4 = 5, which uses completing the square:

16 ¡ 36 = 25 ¡ 45 fobviously true, since ¡20 = ¡20g) 42 ¡ 9 £ 4 = 52 ¡ 9 £ 5 fwrite each term in an equivalent formg

) 42 ¡ 9 £ 4 + 814 = 52 ¡ 9 £ 5 + 81

4 fcompleting the squareg) (4

¡ 92 )2 =

=

(5

¡ 92 )2

) 4 ¡ 92 5 ¡ 9

2 ftaking the square root of both sidesg) 4 = 5 fadding 9

2 to each sidegWhere is the error in this proof?

31 One root of the equation x2 ¡ 2x + k = 0 is 8 more than the other. Find the value of k .

32 Consider the function y = x2 ¡ 1

2 , for odd values of x starting at x = 3.

Copy and complete

the table.What do you notice?

33 Solve the equation 2x

x ¡ 4

x ¡ 2

= x ¡ 3, rounded off to two decimal places.

1 A medical researcher needs to determine for how many days a person who has the Melbourne

flu is infectious.

a For this study, identify the i population ii variables.


c Should the scientist take a census or a sample? Justify your decision.

2 There is a large population of possums on Great Keppel Island.

a State a quantitative variable that may be of interest to a researcher.

b State a categorical variable that may be of interest to a researcher.

3 Determine if each of these variables is discrete (D) or continuous (C).

a the length of time a person with the Melbourne flu is infectious

b the highest sound, measured in hertz, that a person can hear

c the amount of memory in a computer

4 For these variables, decide if data are best collected on a population or a sample. Briefly

justify your decision.

a the official handicaps of the members of the Royal Brisbane Golf Club

b the number of spelling mistakes in an edition of The Courier Mail

5 These questions require data to be collected before they can be answered. For each question,

which method of collecting data (survey, observation, experiment or using available data) is






6 Explain how a SRS could be obtained in each case.

a the number of spelling mistakes in an edition of The Courier Mail

b the number of each species of bird in the Brisbane Botanical Gardens

7 To find out the views of her electorate about gun control, a candidate for mayor talks to every

registered voter in her street. This method of collecting data will give a biased sample. State

how the bias will occur, and give an alternative method that will give an unbiased (or less

biased) sample.

8 For each of the following, state which graph or graphs (column graph, pie graph, time series

graph or scatterplot) would best display the data.

a A school is bringing in a program to reduce absenteeism. To measure the effectiveness

of the program, the school needs to know current levels of absenteeism.

b The number of each species of bird in the Brisbane Botanical Gardens.

9 The graph shows the profile of

Australia’s population by age

and sex in both 1901 and 1999.

It is a graph rich in information.

Comment on what information

the graph conveys.

Profile of Australia’s population

by age –1901 and 1999

1901

1999

150 100 50 0 1501005000

5

10

15

20

25

30

35

40

45

50

55

60

65

70

75

80

85

90

95

male female

age

thousands of people

appropriate?most

a the official handicaps of the members of the Royal Brisbane Golf Club

b the number of spelling mistakes in an edition of The Courier Mail

c the number of hours of paid work done by 11 Mathematics B students at your school




10 Two Year 11 Mathematics B students collected the following data on the number of students

in their school in each year level who claimed that they smoked at least one cigarette a week.

They drew the following graph to illustrate how the number of smokers increased with year

level.Year level 8 9 10 11 12

Number of smokers 19 23 28 30 35

Calculate the lie factor of the graph.

11 A random dot stereogram is the name given to a picture in which a three dimensional shape

can be seen if the viewer stares at the picture while crossing his or her eyes. Usually it takes

some time before a viewer can see the three-dimensional shape. The data given here are the

times it took 43 subjects to see the three dimensional shape, in seconds. The group was given

no information about the nature of the three dimensional object.

47.2 22.0 20.4 19.7 17.4 14.7 13:4 13.0 12.3 12.2 10:3 9.7

9:7 9:5 9:1 8:9 8:9 8:4 8:1 7:9 7:8 6:9 6.3 6.1

5:6 4:7 4:7 4:3 4:2 3:9 3:4 3:1 3:1 2:7 2:4 2:3

2:3 2:1 2:1 2:0 1:9 1:7 1:7

a Construct three different histograms of these data, using a different class interval for each.

b Comment on which histogram best shows the shape of the dataset.

c Construct a stemplot of these data.

d Using your graphics calculator, calculate for these data the i mean ii median iii standard deviation

iv interquartile range

e Which set of summary statistics, mean and standard deviation, or median and IQR, is

more appropriate for this set of data. Justify your decision.

f Calculate the five number summary.

g Draw the standard boxplot of this data.

h

Number of smokers, by year level

10

20

30

40

Yr 8 Yr 9 Yr 10 Yr 11 Yr 12

W aa

rite paragraph about what these data tell you about the time it takes for subjects tovisualise stereogram if they have no prior knowledge of the shape of the three dimen-sional object it contains.




12 Your class has gathered some data on the number of CDs

owned by class members.

a Construct a histogram using appropriate class intervals.

b Construct a stemplot.

c Using your graphics calculator, calculate for these data the

i mean ii median

iii standard deviation iv interquartile range

d Which set of summary statistics, mean and standard deviation, or median and IQR, is

more appropriate for this set of data. Justify your decision.

e Calculate the five number summary.

f Draw the standard boxplot of these data.

g Write a paragraph about what these data tell you about the number of CDs owned by

members of your class.

13 When is the median more appropriate than the mean as a measure of the centre of a dataset?

14 Give an example of a dataset for which the mode is not an appropriate measure of the centre.

15 Give an example of a dataset with ten data values for which the median is a better measure of

the centre than the mean.

16 Give an example of a dataset that contains an outlier.

17 Does the following dataset contain an outlier? Justify your answer.

0 0 0 1 1 1 1 2 2 2 3 12 20 20 21 21 21 21 22 22 23

18 When would a modified boxplot and a standard boxplot be identical?

19 The diagram given shows the standard boxplot of this set of data:

1 1 2 2 3 3 3 3 4 4 5 5 6 6 7 7 8 9

a If you add 3 to each of these numbers, how will this

change

i the mean

ii the standard deviation

iii the median

iv the interquartile range

v the boxplot

b If you multiply each of the original numbers in the dataset by 10, how will this change

i the mean ii the standard deviation iii the median

iv the interquartile range v the boxplot

20 A group of eleven rugby league players has a

mean weight of 100 kg. The heaviest member

who weighs 121 kg, is replaced by a player

who only weighs 89 kg. What is the mean

weight of this new group?




21 Give an example of a dataset with five data values that has an interquartile range of 10.

22 A dataset consists of 999 zeros, 999 ones, 999 twos and 999 threes, i.e., there are 3996 data

values altogether.

a Find the median. b Find the first quartile, Q1.

1 a Enter these data into two lists in a graphics calculator.

x 1 2 3 4 5 6 7 8

y 68 66 58 59 49 32 39 26

i Draw a scatterplot of the data.

ii Using the regression function of your graphics calculator, find the least squares

regression line.

iii What is the value of r2?

iv Comment on how well this linear function fits the data.

b Repeat the above steps for these data.

x 0 1 2 3 4 5 6

y 0 0:5 5 10 15 23 30

2 In a 1948 book called The Song of Insects, George W. Pierce, a Harvard physics professor,

presented data relating the number of chirps per second for striped ground crickets to the

temperature.

a Find the least squares regression line. b What is the gradient and y-intercept?

c What is their physical interpretation?

d Construct the residual plot.

e Decide if a linear model is useful for these data, using the value of r2, and the residual

plot. Justify your decision.

Temp oC 31:4 22:0 34.1 29:1 27.0 24:0 20.9 27:8

Chirps/sec 20 16 19.8 18:4 17:1 15:5 14:7 17:1

Temp oC 20.8 28:5 26:4 28:1 27:0 28:6 24.6

Chirps/sec 15:4 16:2 15 17:2 16 17 14:4

3

Speed (km/h) 50 60 70 80 90 100 110

Total stopping distance (m) 25 34 43 54 66 80 ?

Data is from the Queensland Department of Transport website.

a Find a mathematical model that fits the above data. Justify your model.

REVIEW OF CHAPTER 4 D

When talking about cars, speed kills. One reason speed kills is that the faster car is travel-ling, the greater the distance it takes to stop. quantify this, here are some data from theQueensland Department of ransport.

aTo

T





b Use your model to predict the total stopping distance of a car travelling at 110 km/h.

c If the speed limit on residential streets was lowered from 50 km/h to 40 km/h, what would

be the difference in the stopping distance?

4 Draw ten squares, of different sizes. Measure the side length and the length of the diagonal of

each.

a Put the data into two lists in a graphics calculator. b Construct a scatterplot of the data.

c Find, and then plot, the least squares regression line.

d What is the value of r 2?

e Find the gradient of this line.

f Construct the residual plot. Comment on what it tells you.

g If the drawings and the measurements were perfectly accurate, what should the value of

the gradient be?

5 For the dataset below,

a find the least squares regression model of the form y = ax b construct a residual plot

c based on the residual plot, discuss the appropriateness of the equation as a model for the

data

x 1 3 5 7 9 11

y 4 23 31 34 58 78

1 Using your calculator, determine the following (correct to 4 decimal places):

a sin23o b cos14o c tan211o d sec30o

2 Find A to the nearest minute, if 0 6 A 6 90o and

a sin A = 0:6 b cos A = 0:501 c sec A = 3:6 d cot A = 0:87654

3 Solve the following triangles.

4 A lighthouse is located on the edge of a

cliff. From a small boat on the water, the

angle of elevation to the top of the light-

house is 24o and to the base of the light-

house is 20o. The top of the lighthouse is

100 metres above sea level. How tall is

the lighthouse?

REVIEW OF CHAPTER 5 E

24°

14 cm

A

CB

12 cm16 cm

A

BC

24°20°

100 m

h





5 Using four different angles, one from each quadrant, verify that cos A = sin(90 ¡ A):

6 Convert into decimal form:

a 15o2403600 b 125o804800 c ¡44o1502400 d 83o4205500

7 Using your calculator find tan16o1205400:

8 Represent the following angles with a circle diagram.a 150o b 330o c ¡120o d ¡840o

9 State the value of

a tan180o b sin270o c cos0o d sec90o

10 If 00 6 µ 6 360o, determine the range of µ values for which

a sin µ is negative b cos µ is positive c cot µ is positive d sin µ is zero

11 Give six angles of different magnitude that are coterminal with 150o.

12 If cos µ = 0:3 determine two possible values for µ if 0o < µ < 360o.

13 Determine the magnitude of each of the following angles in radian measure.

a b c

14 Express the following in radian measure, leaving your answer in terms of ¼:

a 60o b 210o c 315o d 180o

15 Express the following in radian measure (correct to 4 decimal places):

a 25.7o b 98o1503000 c 257o450 d 1000o

16 Convert the following into degrees:

a ¼3 b 5¼

3 c 7¼6 d 11¼

9

17 Convert the following into degrees, giving your answer in sexagesimal form:

a 1:5 b 0:23

18 Find the value of each of the following. Give the exact value where possible.

a sin 2¼3 b sin ¼ c tan1

19 Find the angle µ in each case, where 0 6 µ 6 2¼. Give your answers in radian measure.

a cos µ = 0:5 b tan µ = ¡1 c cos µ = ¡0:165

20 Calculate the area of the following shapes. All lengths are in cm.

a b

4

1

5r

r

16

1835° 130°

32

25




21 Find the angle A in this triangle if the

area is 15 cm2.

22 In ¢ABC, find a if A = 32o, B = 71o and b = 35 cm.

23 Solve the following triangle (i.e., find all

unknown sides and angles).

24 Construct two different triangles ABC and then proceed to solve both triangles.

¢ABC has a = 9 cm, c = 8 cm and C = 40o

:

25

26 Find the length of each of the diagonals of this parallelogram.

27 A regular polygon of nine sides is inscribedin a circle of radius 10 cm. Find its perimeter.

28 From a point on level ground, the angle of elevation of the top of a radio tower is 26o. From

another point 60 m closer to the base, the angle of elevation of the top of the radio tower is

63o. What is the height of the tower?

29 A rule of thumb useful in trigonometry is that, for small angles, the tangent of an angle is

approximately equal to the angle divided by 60.

For what range of angles is this approximation suitably accurate?

30 Given that ]ABC = 90o, ]BCP = 124o

and BC = 50 mm,

a find the measure of ]OCM.

b Express MC in terms of r .

c Hence find the value of r.

23 cm

32 cm A

130°

32

25

A

Q

B M C D

P

r

O

Two cyclists depart from the same point. One travels Northwest at km/hr while the other travels South at

2214 80km/hr How long to the nearest minute) will it be before they are km

apart?. (

8 cm

42°

85°

A

B

C



Indices andlogarithms

SUBJECT MATTER

index laws and definitions

solutions of equations involving indices

definitions of and , for > 1

logarithmic laws and definitions

use of logarithms to solve equations involvingindices

applications of indices and logarithms

a x axalog

CHAPTER 7



HISTORICAL NOTE +The development of modern notation for indices is enlightening.

In 1572 Bombelli used to represent the unknown, for

our modern x2

, for x3

and so on.

Vieta in 1591 used a, a quad, a cub etc. This was improved

upon by Herigonus in 1634 who wrote a, a2, a3 etc. for a, a2,

a3.

Finally in 1637 Descartes

Descartes

(whom you met in Chapter 1) wrote

a1, a2, a3 but for positive whole number indices only. Work

continued with Wallis who, in 1659, found meanings for a¡1,

a1

2 , etc. Newton finally used the idea of xn where n is any real

number in his development of the binomial theorem.

Integers can be written as a product of numbers called factors.

For example, the positive factors of 15 are 1, 3, 5 and 15 since 15 can be written as 3 £ 5, or as

1 £ 15.

In a similar way, 32 can be written as 2 £2£2£2£2 and this is written in mathematical shorthand

as 25

. The 5 is called an index or exponent and indicates the number of factors while the 2 iscalled the base. 32 is called the 5th power of 2. Since 32 = 25, we often refer to the expression

25 as a power, as well.

Some further examples:

² 10 000 = 10 £ 10 £ 10 £ 10 = 104

² ppppppp = p7

² x21 would be too time consuming to write out, so we write x21 = xxx ::::: x to 21 factors.

This enables us to define am as:

am = aaa ::::: a to m factors

In this definition a can be any real number. However, (for the moment) m is restricted to positive

whole numbers, i.e., m = 1, 2, 3, 4 .......

Express (3ab2)3

a in factor form

b in the form pasbt where p, s and t are integers.

1 2

3

A REVIEW OF INDEX NOTATION – A MATHEMATICAL SHORTHAND

A

EXAMPLE 7.1

222 INDICES AND LOGARITHMS (Chapter 7)



a (3ab2)3

= 3ab2 £ 3ab2 £ 3ab2

= 3abb £ 3abb £ 3abb

= 3 £ 3 £ 3 £ a £ a £ a £ b £ b £ b £ b £ b £ b

b

1 Express in index notation.

a 10 £ 10 £ 10 £ 10 £ 10 b 6 £ 6 £ 6 £ 6 £ 6 £ 6 £ 6 £ 6 £ 6

c 2

xxxxxxxxx d pqpqpqpq

e aaaabbaabbab f y

2 Evaluate

a 2n, where n = 1, 2, 3, .... 10

b 23, 33, 23 £ 33, and 63. What does this suggest might be true?

c¡

13

¢n, where n = 1, 2 , ... 5.

3 Express in factor form.

a 72 as prime factors b 1296 as prime factors c 5040 as prime factors

d 2310 as prime factors e a3

b5

f 2y6

g (2y)6 h (abc)3 i c3d3

j (cd)3 k (2d)3 l (5a2b)3

m a n 5 p3q 2 o (¡3a)3

p (3a2b3)2 q 3(2y2)2 r 3(¡2a2b2)3

Once mathematicians define a new mathematical object (such as a power), they often want to

determine if there is a logical way to define mathematical operations (such as multiplication and

division) on that object. The word “logical” here means that the mathematical operations are

consistent with existing mathematics and do not lead to any contradictions.

MULTIPLICATION OF POWERS

OPERATIONS WITH POWERS B

EXERCISE 7A

3 £ 3 £ 3 £ a £ a £ a £ b £ b £ b £ b £ b = 33a3b6

Hence, (3ab2)3 = 33a3b6

= 27a3b6 so p = 27, s = 3, t = 6:

Consider b3 £ b4 b3 = bbb and b4 = bbbb

) b3 £ b4 = bbb £ bbbb

= bbbbbbb

= b

7

INDICES AND LOGARITHMS (Chapter 7) 223



In general,

m factors n factors

am £ an =

z }| { (aaa::::aa) £

z }| { (aaa::::aa) | {z }

m + n factors

) am £ an = am+n

Consider k7

k3.

Now k7

k3 =

kkkkkkk

kkk = k4

In general, and assuming that m > n,

am

an =

aaa:::::::a

aaa:::::::a

m factors of a

n factors of a

n factors of a in the numerator divide out n factors of a in the denominator

) am

an = aaa:::::::a (m ¡ n) factors

)

am

an = am¡n

RAISING A POWER TO A POWER

Consider (a2)4

Now (a2)4 = a2 £ a2 £ a2 £ a2

= aa £ aa £ aa £ aa

= a8

In general,

(am)n

amn(am)n

= am £ am £ £am ...... xm n factors }{

= (aaa::::a) | {z } £ (aaa:::::a) | {z } £ ::::: (aaa:::::a) | {z } n factors }{

m factors m factors m factors

= aaaaaa ::::: a n £ m factors{ }

= amn

) =

DIVISION OF POWERS




THE POWER OF A PRODUCT

Consider (ab)4

(ab)4 = ab £ ab £ ab £ ab

= aaaabbbb

= a4b4

In general, (ab)n = anbn

THE INDEX LAWS FOR POSITIVE INTEGER EXPONENTS AND BASES

Index Law 1 aman = am+n To multiply powers with the same base, add the indices.

Index Law 2 am

an = am¡n To divide powers with the same base, subtract indices.

Index Law 3 (am)n = am£n To raise a power to a power, multiply the indices.

Index Law 4 (ab)n = anbn The power of a product equals the product of the powers.

EXERCISE 7B


1 Simplify

a y2 £ y5 b r7 £ r9 c 2 £ y4 £ y3

d 2x4 £ 3x9 e (5d2)(¡3d4) f h3 £ h2 £ h7

g bb2b3b4 h a2 £ b £ a3 £ b5 i x2y3 £ x5y2

j (6ab3

)(4a3

b3

) k 3g2

n2

£ 2n3

g5

£ (¡5gn4

) l 3a2

b £ 4c2

d3

2 Simplify

a ¡3x2 £ ¡2x3 b (6x)(¡2x)(4x) c ¡5d13d2d4

d (¡6y2)(¡8y3)(¡y) e rr2r3r4r5 f 6 £ (¡4t) £ (¡5t) £ (¡6t)

g xy £ xy £ xy £ (¡xy) h (¡3f 2g3) £ (¡2f g3) i (¡3kt2)(5k4t3)

3 Simplify

a a7

a2 b

b4

b c

4c5

c3 d

6a6

3a2

e 6y4

9y2 f a2

a2 g 5a8

a2 h 4x5

x5

i a5b4

a3b2 j

6a4b3

2a2b3 k

5a8b9c10

10abc l

16 p3q 2r

12q

4 Simplify

a (y2)3 b (x5)4 c 3(c2)5 d 4 £ (a2)4

e 5c2(c2)3 f (d2)3 £ (d3)2 g (c3)4 £ (c2)5 h 4(a3)2 £ 2(a2)6

i 2(c5)3[

¡5(c2)2] j

(c3)3

(c2

)3

k 4( p5)2

6( p4

)2

l (a4)3(b4)2

(a3

)2

(b2

)2




5 Write without brackets, simplifying where possible.

a (ab)2 b (3c)3 c (2xy)5 d (xyz)4

e (2 pqr)6 f (ab2)3 g (4d2c3)4 h (a2b)2(a3b)4

i 3st5(2st)2 j (a2b3)2

a2b3 k

(4a2b2)2

(2ab)2 l

4a6b8

(2ab2)2

6 Simplify

a d2 £ d5 b b6

b2 c 2 £ a4 £ a2

d 3a5 £ 2a3 e 6f 6

3f 2 f

x3

2x2

g 3x4

x4 h 5a3 £ ¡2a2 i 4 p6 £ 3 p

j (5m3n2)(¡4mn2) k y3 £ y2 £ y5 l 3a £ 2b £ c

m a3b5

ab3 n 4g7h3

g6h3 o 4 p2s £ 2 ps2 £ ps

p ¡x3y2 £ xy2 £ 2y q ¡b2 £ 4b £ ¡5b r (2k)2 £ (¡k)3

s (2y)2 £ (3y)3 t (¡q 3) £ (2q )3 u (¡2t)2 £ (t4)3

v (¡4a) £ (2a2)3 w (3r)2 £ (¡2r)3 x (2ab)2 £ (3a2b)

y (c2d)2

cd z

(2ab2)5

4a4b2

7 Simplify

a 6a2b £ 4a2b3 ¡ (¡2ab)4 b (4 pq )2 ¡ (¡2 p) £ (3q ) £ (2 pq )

c 5a2 2£¡2b £3b + 7a2 £ 2b2 + (¡3a)(¡5a)(4b)

Clearly a¡1 cannot be defined using the definition of an given at the start of this chapter. Under

this definition, the expression a¡1 implies ¡1 factors of a, which is meaningless.

In a similar way b1

2 cannot mean half a factor of b.

However, it is possible to give a logical meaning to these expressions. Extending the set of indices

to include all real numbers, in particular zero, fractions and negative numbers, is very worthwhile.

Throughout this chapter, we will assume that the base is positive. While (¡2)5 has a value of ¡32,

the expression (¡2)1:5 is not a real number (try it on your calculator) and hence is outside the scope

of Mathematics B.

INDICES OTHER THAN NATURAL NUMBERS C

The four index laws given at this point are consistent with our knowledge and understanding of arithmetic, given that the indices are positive whole numbers. Is it possible to define an index to be

any real number in such way that the definition is consistent with both the index laws for positivewhole number indices and with arithmetic?a




Most students, when asked to guess the value of a0 will say 0, reasoning that if there are no factors

of a, then you have nothing at all, or 0. But defining a0 to be 0 is not consistent with existing

arithmetic.

Consider: b4

b4 = 1

But from Index Law 2 b4

b4 = b4¡4 = b0

Therefore, b0 = 1

In general, am

am = am¡m = a0

But also, am

am = 1

Therefore,

THE MEANING OF , WHEREa a0 0

INVESTIGATION 1

A NUMBER TO A ZERO POWER

CLASS DISCUSSION

ZERO TO THE ZERO

To be consistent with existing mathematics, we say that any number raised to the index 0 equals 1.

Use your calculator to evaluate:

a 100 b 40 c ¼0 d 17.460 e ¡20 f

Ã p 5 ¡ 1:7622

1:543 + 3p

6:5

!0

THE MEANING OF an

The next task is to find a logical definition of a negative index.

Consider d 4

d 6 =

dddd

dddddd =

1

d 2

From Index Law 2 d 4

d 6 = d 4¡6 = d ¡2

Therefore, d¡2 = 1

d 2

In general, a¡n = 1

an

In words, a¡n is defined to be the reciprocal of an.

A class was discussing how to interpret .

James says, “Zero raised to any power is , so equals ”. Sommer says, “But any number raised to the power of equals . So, equals .” What do you think?

0

0 0 00 1 0 1

0

0

0

a0 = 1, a > 0



GRAPHICS CALCULATOR EXERCISE

FRACTIONAL INDICES


This definition of a negative index is used because it is consistent with our index laws for positive

whole numbers.

Sketch the graph of y = x1

2 .

Use your trace button to find the value of

a 41

2

b 91

2

c 21

2

Comment.

Recall that for a > 0,p

a2 = a

From Index Law 3 (a2)1

2 = a2£ 1

2 = a1 = a

Hence,p

a2 = (a2)1

2

This suggests that taking the square root is equivalent to raising an expression to the index 12 .

Therefore, a1

2 = p a

Similar arguments show that a1

3 = 3p

a,

a1

4 = 4p

a, etc.

In general, a

1n = n

p a

The meaning of a

mn can be derived from a

1n = n

p a and

Index Law 3: a

mn = am£ 1

n = (am)1n = n

p am

In general, a

mn = n

p am

Note: p 3,

3

p 5 and

n

p am

are examples of radical notation.

THE MEANING OF a1n

THE MEANING OF amn




a Evaluate 4¡3 b Write 3

p y4 without the radical

a 4¡3

= 143

= 164

b 3

p y4 = y

4

3

a Evaluate 163

4 b Simplify 2n £ 4n

a 1634

= 4p 163

= 4p

16 £ 16 £ 16

= 4p

16 £ 4p

16 £ 4p

16

= 2 £ 2 £ 2

= 8

b 2n

£4n

= 2n £ (22)n fwrite as a common baseg= 2n £ 22n

= 23n

1 Simplify without using a calculator.a 60 b 123:60 c c0 d (4y)0

e

µ2x

3

¶0

f ¼0

2 Evaluate without using a calculator, simplifying where possible.

a 4¡2 b 2¡6 c 3 £ 6¡2 d 1

2¡2

e 2¡2 £ 3¡3 f (2¡2)(3¡3) g 6(3¡2) h 5¡1

i 2

5¡2

j 3¡1

£2¡2 k (3

£2)¡2 l

¡2

(3 £ 2)¡2

SUMMARY OF FURTHER INDEX LAWS

Index Law 5 a0 = 1

Index Law 6 a¡n = 1

an

Index Law 7 a

1

n = np a

Index Law 8 a

mn = n

p am

EXAMPLE 7.3

EXAMPLE 7.2

EXERCISE 7C




m 2¡2 £2¡2 £2¡2 n 6¡2 £ 2¡2 o (1 £ 2 £ 3)¡2 p 1

3¡3

q 1

4¡2 r

5

2¡2 s

2¡2

22 t

22

2¡2

3 Express with positive indices.

a a¡2 b b¡

1 c 5 £ c¡3 d 3x¡

4

e 7z¡2 f 6c¡4 g y¡

12 h ¡2z¡

13

i (ab)¡3 j (a13 x¡2)¡3 k 2x¡1 3

p y2 l ( 1

16 a12b¡8)14

m 2y

3p

z2 n a¡1b¡1c¡1 o

3

x¡2 p

µ32a¡5

b¡10

¶15


a 25

1

2 b 144

1

2 c 8

1

3 d 27

1

3

e 3215 f 25¡

12 g 81¡

14 h (12 £ 3)

12

i 36¡

12 j 27¡

13 k 64¡

16 l 0

12


a 823 b 8

43 c 289¡

12 d 4

32

e 2723 f 729¡

23 g 32

35 h 25¡

32

i 256¡

38 j 81¡

14 k (12

£3)

12 l 625

34

6 Write using radicals.

a 823 b 4

32 c b

13 d 4b

13

e 6¡

12 f d¡

13 g 16

32 h a

23

i g¡

12 j (2a)

34 k a¡

23 l 5b¡

52

7 Write without a radical sign. Simplify if possible.

a p

a bp

a3 cp

a6 d 3p

t

e 3p h2 f 5p a5 g p 3d2 h p (3d)3

i 3p

8x4 j 1p

h k

33p

2y l

4p 4a4

8 Simplify, without using a calculator.

a 50 b d0 c (3c)0 d 1612

e 2713 f 32

15 g 2¡3 h 3¡2

i 5¡3 j 2 £ 3¡2 k 823 l 4

32

m 27

23 n

32

45 o

4

¡

12 p

9

¡

12



HISTORICAL NOTE +


q 16¡

54 r a

34 ¥ a

14 s 2x

54 ¥ x

14 t (a2)

12

u (a3)13 v a2 ¥ a

12 w

1

2¡2 x ( 1

3 )¡1

y 25¡

12 z 3

12 £ 3¡

12

9 Simplify, expressing the answer using radical.

a x12 £ x

13 b a

32 £ a

14 c b

23 £ b

12 d a

13 £ a

14

10 Simplify

a 3n £ 9n b 3n £ 92n c 43n £ 162n d 4n £ 8n

e 2n £ 4n £ 8n f 2n £ 4n+1

8n¡2 g 5¡n £ 252n¡2 h 83n £ 162n £ 32n

11 Use index laws to determine which of the following are equal.

Number Power Log 2 (Number)

2 21 1

4 22 2

8 23 3

16 24 4

32 25 5

Number Power Log 2 (Number)

64 26 6

128 27 7

256 28 8

512 29 9

1024 210 10

LOGARITHMS D

In the late th century, astronomers spent a large part of their working lives

doing the complex and tedious calculations of spherical trigonometry needed tounderstand the movement of celestial bodies.

A Scotsman, , discovered a method of simplifying these calculations using loga-rithms. So effective was Napier’s method that it was said that Napier’s discovery effectivelydoubled the life of an astronomer by reducing the time required to do these calculations.

16

John Napier

A Year 11 Mathematics B student said to his teacher, “I was browsing at a Lifeline book sale,

and found this tattered old book of tables called A Table of Four Figure Logarithms. What are

logarithms?”

The teacher replied, “Before calculators were invented, logarithms were used to perform complex

calculations. We will demonstrate the principle using the powers of two. This table relates powers

of two to its index. The log2 of a number is just the index when a number is written as a power

of 2.”

16¡

1

4 1

161

4(¡16)

¡

1

4 1

1644r 1

16

14p 16




Here are two examples of using powers to simplify calculations.

Calculation 1: Multiply 8 £ 32

Using the above table and index laws, we can calculate as follows:

8 £ 32

= 23

£25

= 23+5

= 28

= 256

Instead of multiplying 8 £ 32, we only had to add 3 and 5. Much easier!

Calculation 2: Calculate 512 ¥ 32

512 ¥ 32

= 29 ¥ 25

= 29¡5

= 24

= 16

Instead of dividing by 32, we had to subtract, i.e., 9 ¡ 5. Easy!

We can set these calculations out more efficiently as follows.

Number Log 2

8 332 5 +

256 8

Number Log 2

512 932 5 ¡

16 4

Logs can be used to simplify other calculations as well. Here is a table showing related calculations.

Calculation with numbers Calculation with logs

multiplication addition

division subtraction

raise to a power multiplication

find the nth root divide by n

These days we use a calculator to divide 512 by 32, so this particular use of logarithms is no longer

important. But as you will soon see, logarithms have other uses.

The tables that the student found at the Lifeline book sale were tables of common logarithms, or

base 10 logarithms.

Generations of school children, scientists and engineers used tables such as these for arithmetical

calculations, where the base was 10 rather than 2 as in the example above. These powers of 10were first calculated by Henry Brigg, a mathematician and astronomer, about four centuries ago.

You can check them on your calculator.

For example, 100:3010

= 1:9999 (there is a slight rounding error).

LOGARITHMS TO THE BASE 10




1 = 100 6 = 100:7782

2 = 100:3010 7 = 100:8451

3 = 100:4771 8 = 100:9031

4 = 100:6021 9 = 100:9532

5 = 100:6990 10 = 101

To calculate 2 £ 3, we proceed as in the base 2 example given previously.

2 £ 3 = 100:3010 £ 100:4771

= 100:3010+0:4771

= 100:7781

= 5:999 Notice the rounding error.

These indices are called logarithms. The setting out was as follows

2 £ 3 Number Logarithm

2 0:30103 0:4771 +

6 0:7781

8 ¥ 2 Number Logarithm

8 0:90312 0:3010 ¡4 0:6021

32 Number Logarithm

3 0:4771£2

9 0:9542

Any positive number can be expressed as a power of 10, and hence as a base 10 logarithm.

For example,

3 = 100:4771 we write log10 3 = 0:4771

30 = 3 £ 10

= 100:4771 £ 101

= 101:4771 we write log10 30 = 1:4771

300 = 3 £ 100

= 100:4771 £ 102

= 102:4771 we write log10 300 = 2:4771

Note in particular that

1 = 100 ) log10 1 = 0

10 = 101 ) log10 10 = 1

100 = 102 ) log10 10 = 2...

1 000 000 = 106 ) log10 1 000 000 = 6

Tables of logarithms, and the mechanical equivalent, the slide rule, were used for arithmetical

calculations until the 1970s. Up to that time, our buildings, bridges and aircraft were designed

almost entirely using tables of logarithms or slide rules.




THE THEORY OF LOGARITHMS

Although the above examples use base 2 and base 10, logarithms can be formed for any base greater

than zero.

For example,81 = 34 ) log3 81 = 4

25 = 52 ) log5 25 = 2

This leads us to our definition of a logarithm.

The logarithm of a number is the index to which the base is raised to equal the number.

In symbols, if N = ax, a > 0 then x = loga N

Note: The value of a logarithm may be positive or negative, but the base must be positive.

a Write in log form i 125 = 53 ii 1

9

= 3¡2

b Write in index form i log7 49 = 2 ii log101

10 = ¡1

a i log5 125 = 3 ii log319 = ¡2

b i 72 = 49 ii 10¡1 = 110

EXAMPLE 7.4

Find

a log4 16 b log16 4 c log41

16 d log1614

a log4 16 = x

) 16 = 4x

) 42 = 4x fWrite both sides in the same baseg) x = 2 fEquate indicesg

b log16 4 =

y

4 = 16y

) 161

2 = 16y fWrite both sides in the same baseg) y = 1

2 fEquate indicesg

c log41

16 = w

116 = 4w

) 4¡2 = 4w fWrite both sides in the same baseg) w = ¡2 fEquate indicesg

EXAMPLE 7.5




d log1614 = z

14 = 16z

) 4¡1 = (42)z fWe need to write both sides in the same baseg) 4¡1 = 42z

) ¡1 = 2z fEquate indicesg) z = ¡1

2

If a > 0, then an > 0 for all n. Since an = y can be expressed as loga y = n it follows

that we can only find the logarithm of positive numbers, if we are only interested in real number

solutions. (There is a solution involving complex numbers).

Confirm this with your calculator by keying in log(¡2). If you graph y = log x with your

graphics calculator, you will see that the log function only exists for positive values of x.

There are two important bases for logarithms. One is 10 and the other is e, which is an irrational

number approximately equal to 2.718281828459.

You will learn in Year 12 why we use this strange number as a base for logarithms.

Log10 is called the common log and we just write log (without writing the base) while loge is

called a natural log and is written as ln.

So, log10

10 = 1 and ln e = 1. Note that this terminology is not universal. In many university

mathematics textbooks, log2 means loge 2:

1 Write each of the following in the equivalent logarithmic form.

a 8 = 23 b 25 = 52 c 64 = 26 d 343 = 73

e 16 = 24 f 18 = 2¡ ¡3 g 1 = 60 h 1

12 = 144 1

2

i 12 = 8

¡

1

3 j 10 = 101

2 Write each of these in the equivalent logarithmic form.a 32 = 25 b 125 = 53 c 256 = 28 d 64 = 43

e 14 = 2¡2 f 1

32 = 2¡5 g 15 = 5¡1 h 9 = 91

i 12 = 2¡1 j

1

n = n¡1 k

1

n3 = n¡3 l

1

ab = a¡b

3 Write each of the following in the equivalent index form.

a log2 128 = 7 b log36 6 = 12 c log5 25 = 2 d log1000 10 = 1

3

e log10 1000 = 3 f log7 49 = 2 g log8 2 = 13 h log2

12 = ¡1

i log3

1

27 = ¡3 j log2 1 = 0

SOME IMPORTANT RESULTS

Since 1 = a0

loga 1 = 0

Since a = a1

loga a = 1

EXERCISE 7D




4 Write each of the following in the equivalent index form.

a log2 1024 = 10 b log3 81 = 4 c log4 16 = 2 d log9 729 = 3

e log21

32 = ¡5 f log51

125 = ¡3 g log41

64 = ¡3 h log616 = ¡1

i log16 2 = 14 j log64 2 = 1

6 k log27 3 = 13 l log16 8 = 3

4

5 Find the value of

a log2 64 b log3 81 c log5 1 d log3 9

e log21

1

2

2

f log4 8 g log10 10x h loga ax

i log 4

Simplify without using a calculator

a log6 4 + log6 9 b 2 log 2 + log 25 c log 72 + 2 log5 ¡ log 7

8

d 2log3 ¡ 1 e log8

log4

THE LAWS OF LOGARITHMS E

You saw earlier that:

² to multiply numbers, we can add their base 10 logs

² to divide numbers, we can subtract their base 10 logs

² to raise a number to a power, we can multiply two powers by the base 10 log of the number.

These can be generalized into the laws of logarithms.

Log Law 1 loga MN = loga M + loga N

Log Law 2 loga

M

N = loga M ¡ loga N

Log Law 3 loga M b = b loga M

EXAMPLE 7.6

We will prove the Log Law 1, and leave the rest as exercises.

Prove: loga MN = loga M + loga N

Proof: Let M = ax and N = ay and so loga M = x and loga N = y

Now MN = ax £ ay fsubstitutinggMN = ax+y fIndex Law 1g

) loga MN = x + y fusing definition of a logg) loga MN = loga M + loga N fsubstitutingg

Proofs of Log Law 2 and Log Law 3 are similar and may be found on the website.




a log6 4 + log6 9

= log6(4 £ 9)

= log6 36

= log6 62

= 2

b 2 log 2 + log 25

= log 22 + log 25

= log 4 + log 25

= log 100

= 2

c log 72 + 2 log5 ¡ log 7

8

= log 72 + log 52 ¡ log 7

8

= log 72 + log 25 ¡ log 7

8

= log( 72

£25

¥ 78 )

= log( 72 £ 25 £ 8

7 )

= log 100

= 2d 2log3 ¡ 1

= log 32 ¡ log10

= log 910

e log8

log4

= log23

log22

= 3 log 2

2log2

= 32

1 Evaluate the following.

a log2 12 ¡ log2 3 b log3 36 ¡ log3 4

c log3 54 ¡ log3 2 d log2 144 ¡ log2 36

e log3

p 18 + log3

p 24 ¡ log3 12 f log 500 + log 2

g h log6 144 ¡ log6 4

i log 35 + 2 log 5

2 ¡ log 536 ¡ log 27 j 2 log 3 + 1

2 log 49 + 2 log 53 ¡ log 7

4

2 Express as a single logarithm.

a log8 ¡ log2 b log 4 + log 5 c log40 ¡ log8

d 3 log2 + 2 log 3 e log 2 + log 3 + log 4 f 12 log 4 + log 3

g 3 ¡ log2 ¡ 2log5 h 1 + log 3 i 2log5 ¡ 3log2

j log 12 + log 4 ¡ log3

3 Write as a single logarithm.

a log 8 + log 2 b log 35 ¡ log7 c log 6 + log 4

d log12 ¡ log 12 e log 2 + log 6 ¡ log4 f log4 ¡ log16

g 3log2 ¡ log4

h 2 log6 + log 2

i 1 + log 2

j 3 ¡ log5

4 Simplify without a calculator

a log8

log2 b

log27

log9 c

log32

log16 d

log2

log16 e

log3

log9

f log5

log 125 g

log2

log 12

h log8

log 12

i log8

log 14

5 Simplify

a loga

a3 b logx

x¡3 c logb

b1

2 d loga

a¡2

EXERCISE 7E

log2 6 + log232 ¡ 2log2

34




Solve for x. a log(1 ¡ 2x) = ¡1 b log x + log(x ¡ 3) = log 18

a log(1 ¡ 2x) = ¡1

) 1 ¡ 2x = 10¡1

) 1 ¡ 2x = 110

) ¡2x = ¡ 910

) x = 920

b log x + log(x ¡ 3) = log 18

) log[x(x ¡ 3)] = log 18

) x(x ¡ 3) = 18

) x2 ¡ 3x ¡ 18 = 0

) (x ¡ 6)(x + 3) = 0

) x = 6 or x = ¡3

However, log x is defined only for x > 0.

) x = ¡3 is not a solution.

The solution is x = 6.

1 Solve for x.

a log2x = 3 b log(2x + 3) = 0

c log 1

x = ¡1 d log(3x ¡ 1) = 1

e log(x ¡ 1) = ¡ log2 f log(1 ¡ 2x) = ¡ log3

g log(x + 2)

¡log x = 1 h log2x

¡log(x

¡1) =

¡1

i log x + log(x + 3) = log 4 j log x + log(x ¡ 7) = log 8

k log(x + 1) + log(x ¡ 2) = 1 l log2x ¡ log(1 ¡ x) = ¡2

2 Solve for x.

a log4x = 2 b log(3x ¡ 2) = 0

c log

µ1

x

¶ = ¡2 d log(1 ¡ 2x) = 1

e log x + log(x + 1) = log 2 f log x + log(2x ¡ 3) = log2

For various reasons it is necessary to work with logarithms in other bases.

In general, let x = loga M . We can change the base to any other base, say b, as follows:

x = loga M

) M = ax frewrite as an index equationg) logb M = logb ax ftake the logb of both sidesg) logb M = x logb a fapply Index Law 3g

) x = logb M

logb a fsolve for x

g

SOLVING LOGARITHMIC EQUATIONS F

EXERCISE 7F

CHANGE OF BASE OF A LOGARITHMG

EXAMPLE 7.7




This derivation gives us the change of base rule:

For positive values of a, b and M loga M = logb M

logb a

1 Use a calculator to evaluate

a log7 9 b log2 7 c log100 50

d log2 5 e log2 10 f log5 2

g log3 7 h log2 6 i log8 5

2 Evaluate without a calculator

a log3 7 £ log7 3 b log3 8 £ log5 9 £ log2 5

c log6 4£

log4 36 d logb a

£loga b

a Evaluate log1:7 3.36 using first principles.

b Evaluate log1:7 3.36 using the change of base rule.

a Let x = log1:7 3:36

) 1:7x = 3:36

) log1:7x = log3:36

) x log1:7 = log3:36

) x = log3:36

log1:7) x = 2:2840

b log1:7 3:36

= log10 3:36

log10 1:7

= 2:2840

Find the value of log2 3 £ log3 4 £ log4 5 £ log5 6 £ log6 7 £ log7 8

log2 3 £ log3 4 £ log4 5 £ log5 6 £ log6 7 £ log7 8 fChange all bases to base 10g

= log3

log2 £ log4

log3 £ log 5

log4 £ log6

log5 £ log7

log6 £ log 8

log7

= log8

log2

= log23

log2

= 3 log 2

log2

= 3

EXAMPLE 7.8

EXAMPLE 7.9

EXERCISE 7G




3 a Simplify logb b2 £ logb2 b3 b Calculate y if log2 10 £ log10 2 = loga y

4 a If log2 3 = a and log2 7 = b, express the following in terms of a and b.

i log8 72 ii log21 28

b If log3 2 = p and log3 13 = q , express log78 52 in terms of p and q .

c Evaluate i logb

a3

£log

a

b3 ii loga

b2

£log

b

c2

£log

c

a2

Indicial equations are equations in which the index is the unknown.

Examples of indicial equations are 5x = 10 and (1:016)n = 4. Logarithms are used to solve

such equations.

Consider the very simple equation 4x = 16 which we know has the solution x = 2.

Confirm this solution by solving this equation using logarithms.

4x = 16

) log4x = log16 ftake the log of both sidesg) x log 4 = log 16 fapply Log Law 3g

) x = log16

log4 fsolve for xg

) x =

log42

log4

) x = 2 log 4

log4 fLog Law 3g

) x = 2 fcancellingg

Solve (1:7)x = 4:2

Using the same method as above, (1:7)x = 4:2) log(1:7)x = log4:2

) x log1:7 = log4:2

) x = log4:2

log1:7

) x = 0:6232

0:2304

) x = 2:70

Confirm this solution using your calculator.

EXAMPLE 7.10

EXAMPLE 7.11

APPLICATIONS OF INDICES AND LOGARITHMS H




1 Solve for x.

a 2x = 6 b 3x = 12 c 3x = 2 d 3x = 1

e 3x = 0:12345 f (0:5)x = 0:1 g ¼x = 1 h 2x = ¡1

2 Solve for x.

a 3x = 10 b 5x = 20 c 4x = 3 d 100x = 101

e 2x = 2:1 f 2x = 1:9 g (0:2)x = 0:1 h ex = ¼

3 Solve for x.

a 2x + 8 = 13 b 3x¡1 = 7 c (3:3)x¡1 = 14 d 2x+1 = 10

e (p

2)x = 10 f 1x = 2 g 3x+1 = 5x h 10

2x = 4x

EXERCISE 7H.1

COMPOUND INTEREST (AN APPLICATION OF INDICES AND

LOGARITHMS)Recall thecompound interest formula: A = P (1+ i)n where

P = the principalA = the amount to which the principal grows

i = the interest rate per compounding periodn = the number of compounding periods

For example, if $1000 is invested at an interest rate of 9% per annum compounded monthly for

five years, the final value of the investment can be found as follows:

P = 1000, I = 0:09 ¥ 12 = 0:0075 and n = 12 £ 5 = 60. Find A.A = P (1 + i)n

= 1000 £ (1 + 0:0075)60

= 1565:68

The investment grows to $1565.68 after five years.

I have $20000 to invest, and want it to grow to $30000 within four years.

What must be the annual interest rate for this to occur?

A = 30000, P = 20000 and n = 4. Find i.

A = P (1 + i)n

) 30 000 = 20 000(1 + i)4

) (1 + i)4 = 1:5

) 1 + i = 4p

1:5

) 1 + i = 1:1067

) i = 0:1067

My investment needs to earn 10.7% compounded annually.

EXAMPLE 7.12




I invest $8000 at an interest rate of 8% p.a., compounded quarterly.

How long will it take for my investment to grow to $12000?

Since n is an index, we will use logarithms to solve for n.

A = 12000, P = 8000, i = 0:08 ¥ 4 = 0:02 Find n.

A = P (1 + i)n

) 12 000 = 8000 £ (1:02)n

) (1:02)n = 1:5

) log(1:02)n = log1:5

) n log(1:02) = log 1:5

) n = log1:5

log(1:02)

) n = 20:5It will take 21 quarters, or 5 years.

Note that we use logarithms to solve equations when the unknown is an index.

1 Find the value of the unknown

A P i n

a $20 000 5% p.a. compounded annually 12 years

b $100 000 8:3% p.a. compounded daily 6 years

c $2000 9:2% p.a. compounded monthly 20 years

d $75 000 3:5% p.a. compounded annually 15 years

e $40000 $30000 ...... compounded annually 7 years

f $4000 $550 ...... compounded monthly 5 years

g $2500 $1000 6% p.a. compounded annually

h $1 million $300000 11:5% p.a. compounded monthly

2 It is anticipated that an investment will grow to $160000 after 5 years, with an annual interest

rate of 8% p.a. If interest is compounded annually, what initial amount must be invested?

3 Six thousand dollars deposited eight years ago into an account that pays interest annually has

grown to nine thousand dollars. Calculate the value of i, the annual interest rate.

4 Forty thousand dollars was deposited in an account that pays interest compounded monthly. It

has grown to $80000 in 6 years. Calculate the value of i, the annual interest rate.

5 How many years will it take for an investment of $30000 earning 11:5% p.a. compounded

annually to grow to $50000?

6 How many years will it take for an investment of $30000 earning 11:5% p.a. compounded

monthly to grow to $50000?

EXAMPLE 7.13

EXERCISE 7H.2

variable.



ACTIVITY

BENFORD’S LAW


7 The Rule of 72 is a quick way of estimating how many years it takes for an investment to

double, compounded annually.

The formula is

Number of years to double + 72

annual interest rate

For example, with an interest rate of 9% p.a. compounded annually, the investment will double

in 8 years.

a The Rule of 72 is only approximate. Calculate precisely how long it takes for an invest-

ment to double at 9% p.a. compounded annually. What is the percentage error in the

Rule of 72?

b Repeat the above for 4% p.a.. Is the percentage error greater or less, compared to that

for 9% p.a.?

c Repeat for 24% per annum interest. Does the Rule still hold for large rates of interest?

8 The formula for compound interest can be applied in other situations that exhibit a similar

pattern of growth (which is called exponential growth). For example, inflation exhibits ex-

ponential growth.The formula for growth due to inflation is F V = P V (1 + i)n where

F V = the future value (cost of a commodity in the future)

P V = present value (cost of a commodity now)

i = the annual percentage increase of the cost, as a decimal (for example, 4% = 0:04)

n = the number of years

Note that this is the compound interest formula, using different names.

a A house is worth $150000 today. If its cost increases by 4% per year, what will it cost

in ten years?

b The cost of a unit in Noosa rose from $300000 in 1994 to $1000000 in 2002. What was

the annual percentage increase in its cost?

c A sales representative claims that $100 000 invested today in developing a new type of

cow that can breathe underwater will grow to $600000 in just three years. What annual

percentage return is he claiming this investment will earn?

d A loaf of bread doubles in price over an eight year period. What was the annual percentage

increase in its cost?

e How many years will it take a valuable painting that is increasing in value by 25% per

year to grow in value from $200000 to $1000000?

In general, the “law” says that the probability of the first digit being a d is

P (d) = log(1 + 1

d)

Benford’ aw Simon Newcombs L (which was first stated by in states thatif you randomly select number from table of physical constants or statistical data,the probability that the first digit will be is about rather than as we mightexpect if all digits were equally likely

1881

301

)a a

a “ ” . , ..

1 0 0 1




1 Use this formula to find the probability of the first digit being a 1, 2, ..., 9.

2 Using a table of physical or statistical data that contains at least 100 data values (an

atlas would be useful here), determine the number of leading digits that are a 1, a 2, etc.

Compare your results with the results expected if Benford’s Law applies.

3 Visit the website to learn how Benford’s Law is used to detect corporate and scientific

fraud.

An important application of logarithms results from the Log Law 3. It is used by researchers when

two variables are related by a power function, which is a function of the form y = axk where

a and k are non-zero real numbers.

The underlying theory is simple. Consider a power function where a = 1:

y = xk

) log10 y = log10 xk ffind logs of both sidesg) log10 y = k log10 x .......... (1)

As x and y are both variables, log x and log y are also both variables, so equation (1) can be written

as p = kq where p = log10 y

and q = log10 x

This is of the form y = mx which is the equation of

a linear function that passes through the origin.

If two variables are related by a power function of the form y = xk, then

by taking logarithms of both sides, the function is transformed to a linear

function of the form p = kq .

Consider y = x2 for x > 0

Here is a table of values. Enter the numbers 1 to 10 into

List 1 of your calculator, then make List 2 = (List 1 2.)

x 1 2 3 4 5 6 7 8 9 10y 1 4 9 16 25 36 49 64 81 100

If you draw the scatterplot of this data, the points will lie

on the “right half” of a parabola.

What to do:

A FURTHER APPLICATION OF LOGARITHMS I

This implies that number in table of physical constants is more likely to begin withsmaller digit than larger digit. It was published by Newcomb in paper entitled “Note on theFrequency of Use of the Different Digits in Natural Numbers”, which appeared in The Ameri-can Journal of Mathematics It was re-discovered by Benford in and he pub-lished an article called “The Law of Anomalous Numbers” in Proc. Amer Phil. Soc.

a a aa a

( ). ,.

1881 1938




GRAPHING POWER FUNCTIONS

What to do:1 Repeat the above procedure on your graphics calculator for y = x3 0 < x 6 10:

a Use List1 for x. Then make

and

b Graph List 2 versus List 1

c Graph List 4 versus List 3

d Comment on your results.

2 Repeat this procedure with the function y = 5x2 0 < x 6 10:

In part 1 of the investigation, we looked at a function of the type y = x

k

.The logarithmic transformation mapped the power function onto a linear function whose

graph passes through the origin and has gradient k.

In part 2 of the investigation, we looked at a power function of the type y = axk (with a

and k constants). Notice the difference:

y = axk

) log y = log axk ftake logs of both sidesg) log y = log a + log xk fapply Log Law 1g) log y = log a + k log x ..... (2)

Now log y and log x are variables, since x and y are variables. But a is a constant, so log ais constanta as well. Therefore equation (2) may be written as

p = kq + c where p = log y

q = log x

c = log a

Hence, the transformed equation is a linear equation of the form y = mx + c.

If two variables are related by a power function of the form y = axk, then

by taking logarithms of both sides, the function is transformed into a linear

function with a gradient of k and a y-intercept of log a.


log x 0 0:3010 0:4771 0:6021 0:6990 0:7782 0:8451 0:9031 0:9542 1log y 0 0:6021 0:9542 1:2041 1:3979 1:5563 1:6902 1:8062 1:9085 2

Drawing the scatterplot of this data shows that the data lie

in a straight line. Either from the graph or from the table,you can confirm that the gradient of this line is k = 2.

List 2 = (List 1)3

List 3 = log(List 1)

List 4 = log(List 2)

Now make List log(List ) and List log(List ).3 1 4 2




Johann Kepler is a very famous man in the history of astronomy and mathematics. He used

data from observations of planetary orbits to show that these motions are not random, that

they obey certain mathematical laws, and that these laws can be written in algebraic form.

He took the Earth as his “base unit”, so the orbital periods are given as multiples of one

Earth year, and orbital radii as multiples of one Earth orbit. Some of his observational dataare given in this table.

Planet Mercury Venus Earth Mars Jupiter Saturn

Orbital period 0:241 0:615 1:000 1:881 11:862 29:457

Orbital radius 0:387 0:723 1:000 1:542 5:202 9:539

Kepler showed that this data could be summarised by the equation y = axk where a and k

are constants.

a Find values for a and k .

b Determine if this is an appropriate model for the data.

c Kepler stated the relationship as “The square of the planet’s year is proportional tothe cube of its distance from the sun.” Show that your solution is equivalent to this.

Put the Orbital Period data into List 1 and the Orbital

Radius data into List 2. Always look at the data first.

Draw a scatterplot of this data. It is not very helpful, as

the points for the first four planets are bunched near the

origin. However we can see that the data is not linear.

Kepler claimed the data follows a power function.

Make List 3 = log(List 1) and List 4 = log(List 2).

The lists are shown below.List 1 List 2 List 3 List 4

0:241 0:387 ¡0:618 ¡0:412

0:615 0:723 ¡0:211 ¡0:141

1 1 0 0

1:881 1:542 0:2744 0:188

11:862 5:202 1:074 0:716

29:457 9:539 1:469 0:980

Draw a scatterplot of List 4

versus List 3

. This is more

instructive. One effect of taking logs is to “unbunch” the

data. The points appear to lie in a straight line, which

implies that a power function is an appropriate model

for the data.

a The least squares regression line on the transformed linear data is

p = 0:6667q + 0:000793 with an r 2 value of 0.99998.

Hence the value of k is 0.6667, or 23 .

EXAMPLE 7.14




Now c = log a

so log a = 0:000793

) a = 100:000793

) a = 1:002

Allowing for rounding and measurement error,

the function y = x appears to fit the data

very well. The diagram alongside shows the

graph passing through the points on the scat-

terplot.

b The final step is to draw the residual plot of the

transformed data, to see if there are any strong

underlying patterns. The residual plot shows

that one data value appears to be too large, but

further investigation shows that the residual is

only 0.0043. If we had access to Kepler’s orig-inal data, we would certainly check it.

Given the large value of r 2 and no strong patterns in the residuals, we conclude

that y = x2

2

3

3

is a very good model for Kepler’s data.

c Our model is y = x2

3 , where x is the Orbital Period (year) and y is the Orbital

Radius (distance from the Sun). To write this without a fractional index (fractional

indices had not been discovered in Kepler’s day), cube both sides of this function:

y = x2

3

) y3 = (x2

3 )3

) y3 = x2

3£3

) y3 = x2

which in English is “The square of the planet’s year is proportional to the cube

of its distance from the sun.”

1 During wet weather, mushrooms grow very quickly. The diameter (in mm)

is measured every three hours.

Time (hours) 3 6 9 12 15 18 21

Diameter (mm) 2:99 6:87 11:20 15:78 20:66 25:67 30:89

Assume that the relationship between the diameter of the mushroom d and

the time t is in the form d =

atk. Find the value of a and k .

Note: It is hard to overstate the importance of the discovery of this simple pattern. Isaac Newtonasked the question, “What kind of force on planet would be needed to produce thismotion?”, and in answering it discovered the Law of Gravity

a.

EXERCISE 7I



MODELLING

V


2 When left in damp, salty conditions, motor vehicles rust. Once a

spot of rust develops, it will grow. Initially a spot of rust develops

and gradually grows into an area of rust that is roughly circular. The

diameter of the rust spot after t months is given in the table.

Time (months) 1 6 11 16 21 26 31

Diameter (mm) 14:8 21:2 23:9 25:8 27:2 28:4 29:4

Find a function of the form d = atk, where t is the time in months

and d is the diameter of the circular area.

1 Gathering data on certain characteristics of

wild animals can be difficult, if not dan-gerous. Take alligators, for example. The

length of an alligator can be estimated quite

accurately from aerial photographs or from

a boat. However, the alligator’s weight is

much more difficult to determine. The table

contains data on the length (in inches) and

weight (in pounds) of alligators captured in

central Florida.

a Use this data to develop a model of

the form y = axk from which the

weight of an alligator can be predictedfrom its length. Comment on the effec-

tiveness of the model.

Length Weight Length Weight

94 130 86 8374 51 88 70

147 640 72 6158 28 74 5486 80 61 4494 110 90 10663 33 89 8486 90 68 3969 36 76 4272 38 114 197

128 366 90 102

85 84 78 5782 80

b Three of the alligators are significantly larger than the others and may have undue

influence on our model for predicting weight from length. Remove these three sets

of paired data, and carry out the above analysis again. Does this model do a better

job of predicting the weights of the smaller alligators?

2

B = R

1 + 3:332log N

Chapter 3 contains a number of datasets. For those given below, comment on whether

Sturges’ Law gives a class interval which gives a good picture of the dataset.

a Example 3.7 (metric data set) on page 115

b Exercise 3G, question 2 (rainfall dataset) on page 116

c Exercise 3H, question 6 (Hitchcock dataset) on page 122

What to do:

As you learned in Chapter changing the class interval or bin width of histogram canmarkedly change the shape of the histogram. Much research has gone into finding an algo-rithm that selects an appropriate bin width for any set of data. An early attempt was made

by , who published his findings in . states that if dataset contains points that span range

3

L

, a

aa ,

Herbert Sturges Sturges’ aw1926N R then the bin width is given by:



PROBLEM SOLVING J


1 Simplify completely 4p

32a5

2 Without using a calculator, show that¡

12

¢ 1

2 =¡

14

¢ 1

4 :

3 Evaluate logc b2 £ logb c3

4 A steak left lying in the sun initially has

500000 bacteria on it. If the number of bacte-

ria doubles every 20 minutes, how long will

it take for the number of bacteria to reach

500000000?

5 In 1998, the population of Australia reached 19 million. The annual growth rate of the

population is 1.6%. In what year might you expect the population of Australia to reach 30million people?

6 Many graphics calculators have a built-in function called DeltaList for computing the difference

between consecutive numbers in a list.

For example,

if List 1 contains f1, 3, 8, 4, 1g then DeltaList(List 1) – > List 2 stores f2, 5, ¡4, ¡3g in List 2.

A teacher asked the following question in an email, “If I have a list entered on my [graphicscalculator], is there a way to divide consecutive values in the list in order to find the ratio

between values?”

Use your knowledge of logarithms and the DeltaList function to help this teacher. Write an

email to this teacher explaining how this might be done.

7 If log 10251024 = a and log 2 = b, prove log 4100 = a + 12b.

8 If x, y and z are positive numbers, prove that

1

logx(xyz)

+ 1

logy(xyz)

+ 1

logz(xyz)

= 1

9 Without using a calculator, evaluate 2log2

3 + 9log3

2.

EXERCISE 7J


base common log exponent index

logarithm natural log power reciprocal




It is known that good wine increases invalue with time. The value of bottle of fine wine over period of years is givenin the following table.

One method of modelling the relationship between the value of the wine and the time in

years is with function of the type: where and are constants.

Using the values given in the table, find and and hence find the value of the wine inyears as predicted by the model.

aa

a ( )V t a k

a k

= at k > 10

100


1 Express with a single index.

a (b4)3 b (b3)4 c (b1

4 )4 d (b1

3 )3

5

2 Express with positive indices.a a¡3 b b¡

1

2 c 2c¡

1

4 d 4d¡2e3

e 1

4e¡2 f

3g¡3h2

5 j2k¡4

3 Express with radical signs (and positive indices under the radicals).

a a3

4 b b¡

1

2 c 2c¡

1

2 d 2

d¡

3

4

4 Simplify

a µa

2

b3

¶ 1

3

£µ b

2

a3

¶ 1

2

b µ16x

2

y¡2

¶¡

1

4

c µ27x

3

8a¡3

¶¡

2

3

5 Simplify and express with positive indices.

a 5¡n £ 252n¡2

53n¡2 £ 10¡1 b

2n+4 ¡ 2 £ 2n

2n+2 £ 4

6 Express as a single logarithm.

a log 4 + log 5 ¡ log2 b 4log2 ¡ log8 c 2 ¡ 12 log4 ¡ log5

7 Prove that loga b = 1

logb a

:

8 Solve for x.

a 3x¡2 = 5 b 5x = 1000 c 55¡3x = 2x+2

d log(3x + 1) = 2 e log x + log(x + 1) = log 6 f log(2x ¡ 19) + log x = 1

9

10 (t) 10 20 30 40Value $V 340 660 900 1230

a

b

“The Australian” newspaper on January reported that loaf of bread incost cents penny in the currency of ). An equivalent loaf of bread costin Find the average annual rate of inflation for the period assumingthat the price change is all due to inflation.

is invested at p.a. It is compounded quarterly and amounts to Find

the number of years for which the money is invested.

26 2001 19011901 30

2001 1901 2001

1500 8% 2229

a( $

. - ,

$ $ .

2 1 2:

Time in years






1 Simplify

a 3a4 £ ¡2a6 b ¡2m4n2 £ ¡2mn8 c (¡3k)3 £ (¡k)

2 Evaluate, without using a calculator.

a 272

3 b 64¡

1

2 c 8¡

5

3

3 Express with positive indices.

a 5¡3 b (a2

3 c¡3)¡2 c

µ8a¡5

b¡6

¶ 1

3

4 a Define loga b. b Without using a calculator, find the value of each of the following:

i log8 64 ii log64 8 iii log81

64 iv log6418

5 a State the three log laws.

b Find, without a calculator

i log7 15 + log7 16 ¡ log7 240 ii logp 2 8 + logp 2 28 ¡ logp 2 240

iii log16 ¡ log 25 + log 4

log 8 + log( 15 )

iv log3 6 ¡ log3 2 + log3 7

log3 105 ¡ log3 5

6 If log y = 23 log x + 1

a express y in terms of x

b find y when x = 8, without using a calculator.

7 a State the rule for change of base of a logarithm.

b Use a calculator to determine which is greater: log13 17 or log17 13.

c Simplify without using a calculator: log81 64 £ log16 243 £ log32 27 £ log9 16

11

Heron’s formula is A =p

s(s ¡ a)(s ¡ b)(s ¡ c) where a, b, and c are the side lengths

and s is half of the perimeter.

Show, from the formula, that log A = 12 [log s + log(s ¡ a) + log(s ¡ b) + log(s ¡ c)].

12 Here is novel way to travel to the moon. Get big piece of paper Stand on it. Assumingthat the paper is mm thick, you are now mm closer to the moon! Now fold the pa-

per in half and stand on it. ou are now mm closer to the moon. Fold the paper againand you are mm closer again and you are mm closer and so on. How many timesdo you have to fold the paper in half before you reach the moon, assuming the moon isabout km away?

a a .

Y, ,

0 1 0 10 2

0 4 0 8

: ::

: :

400 000

A formula used to find the area of a triangle is called Heron’s formula.





V252 INDICES AND LOGARITHMS (Chapter 7)

The scale used in modern music is called an equal temperament scale. An octave, shown above

as part of a piano keyboard, contains twelve notes. The note A has a frequency of 440 hertz

(i.e., cycles per second).

A mathematical model of the frequencies of the notes of an octave divides the octave into twelveintervals, called semitones. The ratio of the frequency of one semitone to the frequency of the

next semitone is 1 : 12p

2, or in index notation, 1 : 2 1

12 .

1 Use this model to calculate the frequencies of the 12 notes of the octave.

2 What is the ratio of the frequency of upper C to that of middle C?

3 You will notice that the frequencies that you calculated in 1 are not whole numbers. Check

your calculated frequencies with those of the tuning forks in your school’s Music Depart-

ment or Science Department.

4 How accurate is this mathematical model of a musical scale?

C D E F G A B C

C# D# F# G# A#

PIANO KEYS



Polynomials

SUBJECT MATTER

relationship between the graph of and thegraphs of and for both positive and negative values of the constant

eneral shapes of graphs of polynomial functionsup to and including the fourth degree

composition of two functions

f xf x a f x a x f

a

( )( )+ ( + ) ( ) ( ), ,

g

af ax

CHAPTER 8



INVESTIGATION 1

MAXIMISING VOLUME

HISTORICAL NOTE +

A sheet of A4 paper has dimensions 29.7 cm £ 21cm. You have been given the task of

making an open rectangular box from this sheet of paper by cutting squares from each of the

corners, folding up the sides, and taping the sides together.

If you cut small squares from each corner, the box will have a large base, but be very shallow.

If you cut large squares, the box will be tall, but have a relatively small base.

Height Length Breadth Volume

1 27:7 19 526:3

1:5

2

etc.V = LBH

= 27:7 £ 19 £ 1

= 526.3 cm3

By making a table of values, find, as accurately as you can, the length of the side of the

square that will result in the box with maximum volume.

The method of to solve quadratic equations has been knownsince the Babylonian times. Despite much effort, almost no progress on solving equa-tions involving higher powers of was made until the early th century, almost yearslater! In , solved equations of the type , for example, .

In , , known as Tartaglia, solved equa-tions of the type .

A general method of solving all (of the form ) was found by

in . The general solution to fourth degree equa-tions (of the form ) wasfound by soon after.

One night in , a Frenchman, , scribbleda brief outline of a proof that no general solution for fifthdegree (or higher degree) equations exists. The next day, he

was engaged in a duel over a romantic involvement. Galoiswas killed in that duel. He was years old at the time.

completing the square

third degree equations

xx mx n x x

x mx n

ax bx cx d

ax bx cx dx

16 20001500 + = 5 = 7

1530+ =

+ + + = 01732

+ + + + 3 = 0

1732

21

dal Ferro

Niccolo Fontana

Leonhard

Euler

Lodovico Ferrari

Evariste Galois

3 3

3 2

3 2

4 3 2

¡

For example, if the squares have a side length of cm then the height of the box will be cm, the base

of the box will have length cmand a breadth of cm.

We can find the volume as follows:

1 129 7 (2 1) = 27 7

21 (2 1) = 19

: :¡ £

¡ £

254 POLYNOMIALS (Chapter 8)

Evariste Galois



Consider the following functions:

f (x) = 3 constant function

g(x) = 4x + 3 linear function

h(x) = x2 + x ¡ 2 quadratic function

We can extend this pattern by adding higher powers of x:

i(x) = x3 + 2x2 ¡ x ¡ 2 called a cubic function or a 3rd degree function

j(x) = x4 + 6x3 + 7x2 ¡ 6x ¡ 8 called a quartic function or a 4th degree function

k(x) = x5 + 3x4 ¡ 2x ¡ 5 usually we do not use special names for polynomials

of the 5th degree or higher

These are all examples of an important class of functions called polynomials.

A polynomial of degree n is a function of the form

f (x) = anxn + an¡1xn¡1 + an n¡2xn¡2+ :::::::: +a2x2 + a1x + a0, a 6= 0

where an, an¡1, ......, a0, are real numbers and the indices, n, n ¡ 1, .... , 1 are natural

numbers.

The symbols an, an¡1, an¡2, ...... , a2, a1 are called coefficients, while a0 is called the

constant term. This method of naming variables and constants is called subscript notation.

It is often used when the number of variables or constants is large, and they have a common pur-

pose, such as representing coefficients.

The first coefficient is an, which is pronounced “a sub n”, the second is an¡1, pronounced “a subn minus 1”. The constant term, a0, is pronounced “a sub zero”.

We would read the first term of the polynomial as “a sub n times x to the power of n” and the

second term as “a sub n minus 1 times x to the power of n minus 1”, and so on.

Following are the graphs of the above functions. We will be studying the graphs of polynomials in

detail later in this chapter.

f (x) = 3 g(x) = 4x + 3 h(x) = x2 + x ¡ 2

i(x) = x3 + 2x2

¡x

¡2 j(x) = x4 + 6x3 + 7x2

¡6x

¡8 k(x) = x5 + 3x4

¡2x

¡5

DEFINITION OF A POLYNOMIAL A

y

x

y

x

y

x

y

x

y

x

y

x

POLYNOMIALS (Chapter 8) 255



Based on the definition, which of the following functions are polynomials?

a f (x) = x3 + 4 b g(x) = 2x ¡ 4x¡3

c h(x) = 2x2 + x ¡ 1

x + 6 d i(x) = 3:6

¡0:5x4 + 3x

¡1:34x2

e j(x) = x2

3 ¡ ¼x +

p 5 f k(x) = x7

g l(x) = 0 h m(x) =p

x3

i n(x) = 2x3 +p

3x2 ¡ 3p

2x ¡ 12

a f (x) is a polynomial. Note that the coefficients on x and x2 are both zero, so these terms

are not written down.

b g(x) is not a polynomial, since the 2nd term contains a negative power of x.

c h(x) is not a polynomial, though it is a rational function in which both the numerator and denominator

d i(x) is a polynomial, though the terms are not written in the usual order. A polynomial

is usually written with the powers of x either consistently increasing or consistently

decreasing.

e j(x) is a polynomial. The coefficient of x2 is 13 , since

x2

3 can be rewritten as 1

3 x2, and

the other coefficients are both real numbers.

f k(x) is a polynomial, even though it consists of a single term. For this polynomial, all

of the coefficients of the other terms are 0.

g l(x) is a polynomial. Its degree is 0 and its graph is the x-axis.

h m(

x)

is not a polynomial. It can be re-written as m(

x) =

x3

2 , which has a fractional

index.

i n(x) is a polynomial. The coefficients and constant term can be any real number, including

roots.

You may wish to draw the graphs of these functions using your graphics calculator, just for interest.

We can use summation notation to represent the general polynomial as follows:

f (x) =nX

i=0

aixi

The first term is a0x

0

, then next term is a1x

1

, and so on. The last term is anx

n

. Note that in thisformat, the terms are added in reverse order to that usually given in the problem. It does not make

any difference, of course.

1 Which of the following are polynomials? For those that are not, explain why they are not

polynomials.

a f (x) = 4x ¡ 3 b f (x) = ¡3

c g(x) = 7¡

3x d h(x) = 1

x

EXAMPLE 8.1

EXERCISE 8A


are polynomials.



EXAMPLE 8.3



1 Evaluate each of the following polynomials for the given values of x.

a f (x) = 4x3 + 2x + 5 i f (¡3) ii f (0) iii f (5)

b g(x) = x4 ¡ 4x2 + 7 i g(¡1) ii g(2) iii g(¡3)

c h(x) = 5 ¡ 3x ¡ 5x2 + 2x3 i h(0) ii h(3) iii h(¡9)

2 Confirm that for the given values, each polynomial evaluates to be zero.

a f (x) = x2 + 5x + 6

i f (¡2) ii f (¡3)

b g(x) = x3 ¡ 4x2 + x + 6

i g(3) ii g(2) iii g(¡1)

c h(x) = x5 + 4x4 ¡ 7x3 ¡ 22x2 + 24x

i h(0) ii h(1) iii h(2) iv h(¡3) v h(¡4)

Polynomials can be added, subtracted and multiplied, and the answer is always a polynomial.

Polynomials can also be divided but the answer is often not a polynomial.

ADDING AND SUBTRACTING POLYNOMIALS

1 On your graphics calculator, graph the functions Y 1 = x2 ¡ 5x + 6 and Y 2 = x + 6.

2 Make Y 3 = Y 1 + Y 2, and graph Y 1, Y 2 and Y 3 on the same set of axes. Turn

on the [TRACE] function and toggle between Y 1, Y 2 and Y 3. What do you conclude?

3 Now make Y 3 = Y 1 ¡ Y 2, and graph Y 1, Y 2 and Y 3 on the same set of axes. Turn

on the [TRACE] function and toggle between Y 1, Y 2 and Y 3. What do you conclude?

EXERCISE 8B

OPERATIONS WITH POLYNOMIALS C

Given f (x) = x2 + 3x ¡ 2 and g(x) = 2x2 ¡ 4x ¡ 1, finda f (x) + g(x) b f (x) ¡ g(x)

You can only add or subtract like terms.

a f (x) + g(x)

= (x2 + 3x ¡ 2) + (2x2 ¡ 4x ¡ 1)

= 3x2 ¡ x ¡ 3

b f (x) ¡ g(x)

= (x2 + 3x ¡ 2) ¡ (2x2 ¡ 4x ¡ 1)

= x2 + 3x ¡ 2 ¡ 2x2 + 4x + 1

= ¡x2 + 7x ¡ 1

What to do:




1 i Add, and then ii subtract, each pair of polynomial expressions.

a x + 5 and 2x ¡ 6

b 2x + 2 and x2 + 3x ¡ 1

c 3x2 ¡ 4x + 5 and 5x2 ¡ 6x ¡ 1

d 3 ¡ 4x ¡ x2 and 4x3 ¡ 5x2 + 2x + 6

e ¡4x3 ¡ 7 and x3 + 7x2 ¡ 2x ¡ 5

f x4 ¡ 3x2 + 2 and x3 ¡ 3x

2 Expand each of these polynomial expressions.

a 4(x ¡ 3) b ¡2(5 ¡ 4x)

c (t + 1)(3t ¡ 5) d 2(x + 1)(x + 7)

e (2x + 3)(x + 1)(x ¡ 1) f x(x + 1)(x ¡ 2)(x + 3)

g t(t ¡ 1)2 h x(x ¡ 2)3

i (a + b)2 j (2t ¡ 1)(2t + 1)(3t ¡ 2)

k (a + b)3 l (a + b)4

3 For each of these polynomial expressions, without expanding write

i the degree of the polynomial ii the constant term, if any.

a (3 ¡ x)(5 ¡ x) b ¡(3t + 1)(4 ¡ t) c (t2 + 2t ¡ 4)(t + 1) d (2x2 + 3x ¡ 1)(2x2 ¡ 3x ¡ 4)

e (h4 + 2)(h3 ¡ 5h) f (2a3 ¡ 5a2)(4a2 + 2a4)

g x(x + 1)(x + 2)(x + 3) h ( 12 x2 + 3)( 1

4 x3 + 2x ¡ 13 )

4 Divide the following polynomials. State any restrictions on the domain.

a x2 + 5x + 6 by x + 2 b x2 ¡ 2x ¡ 15 by x + 3

c x2 ¡ x ¡ 12 by x ¡ 4 d x2 ¡ 11x + 24 by x ¡ 3

e 6x2 + 5x ¡ 6 by 2x + 3 f (x + 5)2(2x ¡ 1) by x + 5

g 6x2

¡3x

¡3 by 3x

¡3 h (x

¡3)2(x + 5)3 by (x

¡3)(x + 5)2

Divide x2 + 5x + 6 by x + 3:

(x2 + 5x + 6) ¥ (x + 3)

= x2 + 5x + 6x + 3

for x 6= ¡3

= (x + 3)(x + 2)

(x + 3)

= x + 2

Note that the domain of this function is x 6= ¡3, as we cannot divide by zero.

EXAMPLE 8.5

EXERCISE 8C.1




An important class of polynomials are those of the form f (x) = (x+a)n. This function is important

in the study of probability, and the study of calculus.

The expansion of such expressions is called the binomial expansion.

If you expand polynomials of the form f (x) = (x + a)n, for n = 0 to 5, you will get the following:

f (x) = (x + a)0 = 1 recall that b0 = 1, for any b

f (x) = (x + a)1 = x + a

f (x) = (x + a)2 = x2 + 2ax + a2

f (x) = (x + a)3 = x3 + 3x2a + 3xa2 + a3

f (x) = (x + a)4 = x4 + 4x3a + 6x2a2 + 4xa3 + a4

f (x) = (x + a)5 = x5 + 5x4a + 10x3a2 + 10x2a3 + 5xa4 + a5

The expansion of f (x) = (x+a)n is rich in patterns. Can we predict the expansion of (x +a)6?

Recalling the fact that a1 = a and a0 = 1, for all a, it is easy to see that for each successive

term the index of x reduces by 1 for each term, from n to 0, while the index of a increases by 1for each term, from 0 to n.

So, the expansion of (x + a)5 could have been written more fully as:

(x + a)5 = x5a0 + 5x4a1 + 10x3a2 + 10x2a3 + 5x1a4 + x0a5

The hardest pattern to find when expanding (x + a)6 is the pattern of the coefficients. Here are the

coefficients from the above expansions written in a triangular pattern.

This is called Pascal’s Triangle, after the famous mathematician of the 17th century, Blaise Pascal.

The last line of this triangle contains the coefficients of (x + a)5, in order.

It is possible to form each row from the one above it.

Look, for example, at the 2nd term of the 5th row, which is a 4. This value is found by adding the

numbers to the “northwest” and the “northeast” of it. Also, the first and last numbers in each row

are both 1.

The next row gives the coefficients for (x + a)6. These coefficients are derived from the last row

in the above table, as follows.

Hence, the expansion of (x + a)6 is x6 + 6x5a + 15x4a2 + 20x3a3 + 15x2a4 + 6xa5 + a6.

row 0 1row 1 1 1row 2 1 2 1row 3 1 3 3 1row 4 1 4 6 4 1row 5 1 5 10 10 5 1

POLYNOMIALS OF THE FORM f (x) = (x + a)n

1 5 10 10 5 11 6 15 20 15 6 1




1 Expand and simplify, using the binomial expansion.

a (x ¡ 1)3 b (x ¡ 4)3 c (x ¡ 2)5 d (2x + 1)5

e (3a ¡ 3)3 f (2c ¡ 5)3 g (2x ¡ 3)5 h (a + 1)7

i (5x ¡ 5)2 j (4 ¡ y)3 k (3 ¡ 2x)3 l (2 ¡ x)4

m (x ¡ 2

x2)3 n (a ¡ 1

a)4 o (x ¡ 1

2 )3 p ( 13 x2 ¡ 1

2 )4

2 Express in the standard form (x + a)n.

a x3 + 3x2 + 3x + 1 b x4 ¡ 4x3 + 6x2 ¡ 4x + 1

c d x3

+ 15x2

+ 75x

+ 125

Expand each of the following expressions.

a (x + 1)3 b (x ¡ 2)4 c (3x ¡ 4)3 d (2 + 1

x)4

a Substituting a = 1 into x3 + 3x2a + 3xa2 + a3 gives

(x + 1)3

= x3 + 3x2 + 3x + 1

b Substituting a = ¡2 into x4 + 4x3a + 6x2a2 + 4xa3 + a4 gives

(x ¡ 2)4

= x4 + 4x3(¡2) + 6x2(¡2)2 + 4x(¡2)3 + (¡2)4

= x4 ¡ 8x3 + 24x2 ¡ 32x + 16

c Substituting x = 3x and a = ¡4 into (x + a)3

= x3

+ 3x2

a + 3xa2

+ a3

gives(3x ¡ 4)3

= (3x)3 + 3(3x)2(¡4) + 3(3x)(¡4)2 + (¡4)3

= 27x3 ¡ 108x2 + 144x ¡ 64

d Note that this is not a polynomial, but the pattern still holds.

Substituting a = 2 and b = 1

x gives

(2 + 1

x)4

= 24 + 4 £ 23 £ (1

x) + 6 £ 22 £ (

1

x)2 + 4 £ 2 £ (

1

x)3 + (

1

x)4

= 16 + 32

x +

24

x2 +

8

x3 +

1

x4

= 16 + 32x¡1 + 24x¡2 + 8x¡3 + x¡4

EXAMPLE 8.6

EXERCISE 8C.2

8x3

+ 24x2

+ 24x

+ 8




The solutions (or roots) of the quadratic equation (x¡ 2)(2x + 1) = 0 are x = 2 and x = ¡12 .

These are called the zeros of the function y = (x ¡ 2)(2x + 1).

Graphically, they are the x-coordinates of the points where the graph of the function crosses the

x-axis.

This extends easily to polynomials of higher degree.

a Find the zeros of the function f (x) = (x + 1)(x ¡ 2)(x + 3).

b Draw the graph of the function using a graphics calculator and confirm that these

are the x-coordinates of the points where the graph crosses the x-axis.

c From your knowledge of f (x), what are the zeros of

g(x) = 2(x + 1)(x ¡ 2)(x + 3)? Without drawing it, what can you say about thegraph of g(x)?

a If f (x) = (x + 1)(x ¡ 2)(x + 3) then

x + 1 = 0 or x ¡ 2 = 0 or x + 3 = 0so x = ¡1, 2 and ¡3 are the zeros of this

function.

b The diagram on the right shows that the

graph crosses the x-axis at x = ¡1. The

other roots are found similarly.

c The zeros of g (x) are the same as those of f (x). The graph of g(x) crosses the x-

axis at the same point as f (x). From our knowledge of transformations, the graph

of g(x) is the graph of f (x) stretched vertically by a factor of 2.

1 Find the zeros of these polynomials.

a f (x) = x(2x + 1)(x ¡ 2) b f (x) = (x ¡ 2)(x + 1)(x + 5)(x + 9) c f (x) = x2 + 6x + 5 d f (x) = (3x ¡ 1)(x2 ¡ x + 20)

e g(x) = (x2 ¡ 2x ¡ 3)(x2 ¡ 6x + 8) f g(x) = x(x2 + 6x + 9)(x2 ¡ 13x + 22)

2 Find three polynomial functions whose graphs intersect the x-axis at x = ¡5, x = 0 and

x = 1.

3 Find the zeros of these polynomials.

a f (x) = (x ¡ 2)(x + 3) b f (x) = x(2x + 1)(3x ¡ 4)

c f (x) = (x ¡ 4)2 d f (x) = (x ¡ 2)2(x + 3)

e g(x) = x(2x + 1)(3x

¡4)3 f f (x) = (x + 3)5(x

¡4)2

ZEROS OF A POLYNOMIAL D

EXERCISE 8D

EXAMPLE 8.7




4 Given the graphs below, state the zeros of each function.

a b

y

x

y

x

GRAPHICS CALCULATOR EXERCISE

1 Using a graphics calculator, sketch the graphs of the following cubic functions.

Choose a window size that shows the critical points clearly.

a f (x) = x(x ¡ 2:25)(x ¡ 3:75) b g(x) = x(x ¡ 2:5)(x ¡ 3:5)

c h(x) = x(x ¡ 2:75)(x ¡ 3:25) d i(x) = x(x ¡ 2:95)(x ¡ 3:05)

2 Sketch by hand what you think the graph of i(x) = x(x ¡ 3)2 will look like.

Check your answer with a graphics calculator.

3 Sketch by hand the graphs of these functions. Check your answer using a graphics

calculator.

a f (x) = x(x + 1)2 b g(x) = x2(x

¡2)

c h(x) = (x ¡ 1)2(x + 3)2 d i(x) = x2(x ¡ 5)2

4 Explain in a few brief words what is meant by ‘repeated roots’.

REPEATED ROOTS

These screen shots from a graphics calculator show the characteristic shapes of polynomials of

degree 3 to degree 6.

If you wish to graph them yourself, the window is (¡4:7, 4, 7, 1; ¡3:1, 3:1, 1)

3rd degree equation 4th degree equation

y = 0:3(x + 2)(x

¡1)(x

¡3) y = 0:5x(x + 1)(x

¡2)(x

¡3)

GRAPHS OF POLYNOMIALS E



CLASS DISCUSSION


Note that for these functions, the number of zeros and the degree of the polynomial are equal. In

general, an nth degree polynomial can have up to n zeros, but it can have fewer than n zeros and

indeed no zeros at all.

Here are some examples:

² Is it possible to have a cubic function with no zeros?² Is it possible to have a quartic function with three zeros?

² Is it possible to have a function of degree five with no zeros?

POINTS OF INTEREST

When investigating the graph of a function, often a mathematician will start by finding the values

of the points that tell us something about the function, for example, the zeros of the function, its

y-intercept, and the local maxima and local minima.

5th degree equation 6th degree equation

y = 0:02x(x ¡ 2)(x ¡ 4)(x + 2)(x + 4) y = 0:008x(x + 4)(x + 3)(x ¡ 1)(x ¡ 3)(x ¡ 4)

3rd degree 4th degree

y = 0:1(x ¡ 2)(x + 3)2 y = 0:2x3 + 1

4th degree 6th degree

y = 0:3x(x2 + 1)(x ¡ 2)(x ¡ 3) y = 0:1x(x ¡ 2)2(x + 1)3 + 1

equation with zerostwo equation with zeroone

equation with zerostwo equation with zerosno




A local maximum is a point whose y-coordinate is larger than that of any nearby point.

A local minimum is a point whose y-coordinate is smaller than that of any nearby point.

The coordinates of such points can be found to reasonable accuracy using any graphics calculator,

either by zooming in on the point of interest or by selecting the option from a menu.

Use a graphics calculator to graph the following functions such that all points of interest are

visible in the viewing window. Find the values of these points to at least 3 decimal places.

a f (x) = x3 ¡ 3x2 ¡ 13x + 15 b g(x) = x3 + 4x2 ¡ 3x ¡ 4

a The viewing window (¡4, 6, 1; ¡30, 30, 10)

shows the necessary points. The zeros are at

(¡3, 0), (1, 0) and (5, 0). The local maxi-

mum has coordinates (¡1:309, 24:634) whilethe minimum is at (3:309, ¡24:634). As we

know that a polynomial of degree 3 can have

a maximum of 3 zeros only, it is not necessary

to expand the viewing window.

b The viewing window (¡5, 3, 1; ¡10, 20, 3)

shows the interesting points nicely. Using the

in-built programs, the zeros are at (¡4.471, 0),

(¡0:739, 0) and (1.210, 0). By substituting

x = 0, the y-intercept is at (0, ¡4). There

is a local maximum at (

¡3, 14) and a local

minimum at (0:333, ¡4:519).

1 Sketch the graphs of these polynomials using a graphing calculator. Adjust the window to

include all points of interest. State the coordinates of each of the zeros, the y-intercept, and

the local maximum and minimum values.

a y = x(x ¡ 4)(x + 1) b y = x3(x ¡ 1)2

c y = x3 + 3x2 ¡ 4x + 1 d y = x3 ¡ 4x2 + 9x ¡ 10

e y = x3

¡3x2 + 3x

¡1 f y = x4

¡3x2 + 1

EXAMPLE 8.8


ROOTS OF A SIXTH DEGREE POLYNOMIAL

a 5 zeros b 4 zeros c 3 zeros d 2 zeros e 1 zero?

On the previous page is sixth degree polynomial with six zeros, and another with no zeros.Can you find th degree polynomial that has

aa 6

EXERCISE 8E.1




2 Find the equation of a cubic function for which all of the interesting points lie in the window

(¡1, 1, 1; ¡1, 1, 1).

3 When graphing a polynomial with a graphics calculator, there is a useful theorem, proved by

Cauchy in 1829, that establishes upper and lower bounds on the roots. If you use these bounds

as your values for X min and X max on your graphics calculator, you will be sure to capture all

of the roots.

Let f (x) = xn + an¡1xn¡1 + an¡2xn¡2 + :::: + a1x + a0.

Note that the leading coefficient of f must be 1. If you have a polynomial whose leading

coefficient is not 1, divide your polynomial by its leading coefficient, and you will get a

function with leading coefficient 1 that has the same roots as your original polynomial.

Theorem: Let f be as indicated above. Then all the zeros of f lie in the interval:

jxj < 1 + maxfja0j, ja1j, ....., jan¡2j, jan¡1jg.

Use this theorem to find the upper and lower bounds of the roots for the following polynomials.

Use your graphics calculator to check that your window has captured all of the roots.

a f (x) = x2

+ x ¡ 2 b f (x) = x3

¡ 54 x

2

¡ 47

4 x + 3 c f (x) = 3x3 + 27

10 x2 ¡ 6310 x + 3

5

At a sufficiently large scale, the graphs of these functions are nearly identical. The reason is that

for large positive and large negative values of x, the largest power of x controls the basic shape

of the graph, as its value is so much greater than that of the other terms combined. For each of

these graphs, the largest power of x is x3. The other terms have their strongest effect on values of

x close to the origin.

THE OVERWHELMING INFLUENCE OF THE TERM WITH THEHIGHEST DEGREE

Below are the graphs of three cubic equations, drawn with a graphics calculator with the viewing

window (¡5, 5, 1; ¡15, 15, 3).

y = x3 y = (x ¡ 1)(x + 2)(x ¡ 4) y = (x ¡ 1)(x2 + x + 3)

Viewed ‘from a distance’, the same functions present us with a quite different picture. These are

graphs of the same equations, using the viewing window (¡60, 60, 10; ¡50000, 50000, 10000).

y = x3 y = (x ¡ 1)(x + 2)(x ¡ 4) y = (x ¡ 1)(x2 + x + 3)




Now consider the graphs of f (x) = x3, g(x) = 3x3 and h(x) = 12 x3, shown in the window

(¡40, 40, 10; ¡50000, 50000, 10000).

The functions g(x) and h(x) are vertical stretches of

f (x), by factors of 3 and 12 respectively. The effect of

these transformations will be visible at any scale factor.

In many real life situations one variable may depend upon a second variable which in turn depends

upon a third variable.

Consider a stone which is dropped onto the surface of

still water. Ripples are formed, which move out from

the point of impact in a circular form. The radius r

of the disturbance increases with time t, i.e., the

radius is a function of time. Consider a simple case

where r = 2t.

Now as r increases the area of the circle increases

as well, since area is a function of the radius, i.e.,

A = ¼r2:

So area is a function of the radius where the radius is a function of time. Mathematicians say that thearea is a function of a function of time. In particular, if A = ¼r2 then A = ¼(2t)2 = 4¼t2.

Or consider a hot air balloon. The velocity at which

it is rising, v, is a function of the temperature of

the air, T . The temperature of the air is a function

of the time, t, that the burner has been operating.

Hence, the velocity at which the balloon is rising

is a function of a function of the time the burner

has been operating. We can write v = f (T ) and

T = g(t) so v = f [g(t)].

We say “v = f of g of t”.

1 Sketch the graphs of the following simultaneously in the window (¡3, 3, 1, ¡20, 20, 4).

a y = x4 b y = x(x ¡ 3)(x +2)(x ¡ 1) c y = (x2 + x + 3)(x2 + x + 3)

Now change the window to (¡40, 40, 4; ¡50000, 50 000, 10000). Comment on what the two

different views of these functions show you.

2 Sketch the graphs of the following simultaneously in the window (

¡2, 2, 1,

¡1, 3, 1).

a y = x4 b y = 5x4 c y = 15 x4

Now change the window to (¡30, 30, 10; ¡10000, 50000, 5000). Comment on what these

two different views of these functions show you.

EXERCISE 8E.2

COMPOSITION OF FUNCTIONS F




Here is an algebraic example:

Let u = x2 ¡ 3x and y = u10. Now u is a function of x and y is a function of u, and

therefore y is a function of a function of x. In particular, y = (x2 ¡ 3x)10. Note that y is still a

function of x.

Solve the equation (x ¡ 3)2 ¡ 5(x ¡ 3) + 6 = 0

By substituting c for (x ¡ 3), we have the equation: c2 ¡ 5c + 6 = 0, which factorises as

(c ¡ 3)(c ¡ 2) = 0. Its solutions are c = 3 and c = 2.

By substituting x ¡ 3 back in place of c, we have: x ¡ 3 = 3 and x ¡ 3 = 2:

The solutions are x = 6 and x = 5. Check by substituting in the original equation.

While we often substitute a number for x, we can just as easily substitute an algebraic expression.

Consider f (x) = x2 + x ¡ 5.

Find an expression for a f (a) b f (3c) c f (x ¡ 2)

a f (a) = a2 + a ¡ 5 b f (3c)

= (3c)2 + (3c) ¡ 5

= 9c2 + 3c ¡ 5 c f (x ¡ 2)

= (x ¡ 2)2 + (x ¡ 2) ¡ 5

= (x2 ¡ 4x + 4) + (x ¡ 2) ¡ 5

= x2

¡3x

¡3

Form the composite function y = f (x) if

a y =p

u and u = 2x ¡ 5 b y = 2

u2 and u =

p x

a y =p

2x ¡ 5 b y = 2

(p

x)2 =

2

x

Now consider the following function: y = (x ¡ 3)2 ¡ 5(x ¡ 3) + 6.

If we substitute c for (x ¡ 3), the function can be written as y = c2 ¡ 5c + 6 which is obviouslya quadratic function.

We would say that “y is a quadratic function on x ¡ 3”.

This complex function can be thought of as the composition of two simpler functions, one being a

quadratic function and the other, a linear function.

EXAMPLE 8.9

EXAMPLE 8.10

EXAMPLE 8.11




b f (x) = x3 ¡ 4x + 1 and g(x) = 3 ¡ x. Find

i f ± g ii g ± f iii g ± g iv f ± f

6 Solve the following equations.

a (x + 3)2 ¡ 2(x + 3) ¡ 3 = 0 b (x ¡ 5)2 ¡ 16 = 0

c (x2

¡x + 1)2

¡(x2

¡x + 1)

¡2 = 0 d 3(4x2

¡9)

¡2 = 19

The knowledge you have gained in an earlier chapter about transformations of quadratic functions

applies equally well to transformations of polynomials (and to other functions you will study later

in this course).

TRANSFORMATIONS OF POLYNOMIALS G

EXAMPLE 8.13

y

x

y

x

y

x

The graph of y = x3 is shown.

Sketch by hand the graphs of

a y = x3 ¡ 4

b y = 2x3

c y = (x ¡ 4)3

d y = 2(x + 4)3

e y = ¡3(x ¡ 2)3 + 3

a The graph of y = x3 ¡ 4shifts the graph of y = x3

vertically 4 units down.

b The graph of y = 2x3

stretches the graph of y = x3

vertically by a factor of 2.




y

x

y

x

y

x

a g(x) = 2f (x)

b h(x) = f (x ¡ 2)

c i(x) = f (x) ¡ 3

The graph shown is the graph of particular function, Draw the graph of each of thesefunctions, along with the graph of

a .:

f xf x

( )( )

EXAMPLE 8.14

y

x

f(x)

c The graph of y = (x ¡ 4)3

shifts the graph of y = x3

horizontally 4 units to the right.

d The graph of y = 2(x + 4)3

shifts the graph of y = x3

horizontally 4 units to the left, and

stretches it vertically by a factor of 2.

e The graph of y = ¡3(x ¡ 2)3 + 3shifts the graph of y = x3 hori-

zontally 2 units to the right, flips it

about the x-axis, stretches it verti-

cally by a factor of 3, and then shiftsthe graph vertically upwards a dis-

tance of 3 units.




1 Describe in words how the graph of each of these functions is a transformation of the graph

of f (x).

a ¡3f (x) b f (x ¡ 3) c f (x) ¡ 1

d 2f (x + 3) e ¡f (x ¡ 1) + 3

2 If f (x) = x3, sketch by hand the graphs of each of the functions in question 1. Check your

answers with a graphics calculator.

3 If f (x) = x4, sketch by hand the graphs of each of the functions in question 1. Check your


y

x

g (!)

ƒ(!)

y

x

h(!)ƒ(!)

y

x

i(!)ƒ(!)

EXERCISE 8G

a The graph of the function g(x) = 2f (x) is

the graph of f (x) stretched vertically by

a factor of 2. Note that the zeros of f (x)and g(x) = 2f (x) are the same. Stretch-

ing the function does not change the zeros

of the function.

b The graph of the function

h(x) = f (x ¡ 2) is the graph

of f (x) shifted 2 units to the right.

c The graph of the function

i(x) = f (x) ¡ 3 is the graph

of f (x) shifted 3 units down.



Consider the earlier problem of maximising the volume of anopen box made from a sheet of A paper. We found an ap-

proximate solution by constructing a table of values. Now letus take an algebraic approach and see if we can do any better.

4


4 The diagram shows the graph of

h(x) = 15 (x ¡ 1)(x + 2)(x ¡ 3).

Sketch by hand the graphs of

a ¡2h(x) b h(x ¡ 1)

c h(x) ¡ 1 d 2 ¡ h(x)

5 Consider the quadratic function f (x) = (x + 1)(x ¡ 5). Without drawing the graphs,

a find the zeros of this function

b find the zeros of g(x) = 3(x + 1)(x ¡ 5)

c find the zeros of f (x ¡ 3)

6 Consider the graph of the polynomial f (x)alongside.

a What are the zeros of this function?

b What are the zeros of g(x) = ¡2f (x)?

c What are the zeros of h(x) = f (x + 1)?

d What is the y-intercept of g(x)?

7 a A function g(x) = f (x) + 3. What do you know about the function g(x)?

b A function h(x) = ¡4f (x). What do you know about the function h(x)?

c A function j(x) = f (x ¡ 2). What do you know about the function j(x)?

y

x

We can now substitute:

V = LBH

V = (29:7 ¡ 2x)(21 ¡ 2x)(x) or rearranging,

V = x(21¡

2x)(29:7¡

2x)

APPLICATIONS AND MODELLINGUSING POLYNOMIALS

H

EXAMPLE 8.15

Optimising volume

From the diagram it should be clear that if a square of side is cut from each corner of the paper, then the length of the base is given by and the breadth by .Also, the height of the box is .

xL : x B x

H x = 29 7 2 = 21 2

=¡ ¡




During periods of drought, it is occasionally sug-

gested in the press that Australia should consider

towing an iceberg up from Antarctica. Icebergs are

large frozen masses of fresh water, and could provide

a steady source of fresh water in a time of need. The

problem is that while an iceberg is being towed, it

is melting. Assuming that for the plan to be feasible

the iceberg needs to contain at least 1 billion litres

of water when it arrives, should such a scheme be

given serious consideration?

Deciding if such a scheme is feasible is a highly complex problem, and involves knowledge

of meteorology, oceanography, the capabilities of the towing ships, and much more. A very

simple mathematical model is probably a good way to start.

Assume the iceberg is a sphere of radius 100 m, and that the iceberg melts uniformly at a rateof 15 cm of depth per day. We will need to calculate the maximum number of days available

for the towing such that the iceberg is still of sufficient size when it arrives. The towing

experts can then decide if such a deadline could be met.

Recalling the metric conversion 1 m3 = 1000 L, the iceberg still has to have a volume of one

million m3 when it reaches Australia.

The volume of a sphere is given by the function V = 43 ¼r3.

The radius on day t, in metres, is given by the function r = 100 ¡ 0:15t.

Hence, V is a function of r, and r is a function of t, so V is a function of a function of t.

By substitution, the volume of the iceberg on day t, in

cubic metres, is given by V (t) = 43 ¼(100 ¡ 0:15t)3.

We can use a graphics calculator to determine the value

of t when V = 1000000 m3. From the screen shot

we can see that if we can tow the iceberg to Melbourne

within 250 days then we will meet the requirements.

Note that this answer may be wildly inaccurate! Icebergs are not spheres, and the actual rate

of melting would depend on many factors, such as the shape of the iceberg, the air and sea

temperatures and the speed at which the iceberg is moving. Large chunks of the iceberg may

break off during the tow.

The graph of as a function of is given alongside.We find, using a graphics calculator, that at the pointwhere the graph reaches a maximum value, the maxi-mum volume is . cm , and this occurs when thesquare has a side length of . cm.

V x

1128 54 04

3

y

x

Towing an iceberg

EXAMPLE 8.16




However, the mathematical model has given us a start, and some indication as to how we can

refine our solution to obtain a more accurate answer. Such calculations are fun to do, and can

give rise to some enjoyable discussions.

A few years ago your favourite aunty started depositing $1000 each year, on your birthday,

into an account that pays a fixed annual rate of interest, compounded annually. You are

allowed to withdraw the money on your 21st birthday, just after she has made her deposit on

that day.

a Let x = the fixed interest rate that the account is earning, expressed in decimal

format. Express as a polynomial the amount of money that is in the account three

years after the first deposit, assuming that the latest deposit (i.e., the fourth deposit)

has just been made.

b How much money is in the account just

after the 4th deposit is made, if the interest

rate is 9% p.a., compounded annually?

c If there is $4300 in the account just after the fourth deposit is made, what interest

rate is the account earning?

We need to use the compound interest formula A = P (1 + i)n.

When deposits are made annually and interest is compounded annually, then A is the amount

that a deposit of P dollars grows to in n years, at an annual interest rate of i, expressed as a

decimal. For example, for an annual interest rate of 6% p.a., i = 0:06.

a The total amount in the account is given by

T (x) =the amount the

first deposit

has grown to

+the amount the

second deposit

has grown to

+the amount the

third deposit

has grown to

+ $1000 just

deposited

Using the above formula,

T (x) = 1000(1 + x)3 + 1000(1 + x)2 + 1000(1 + x) + 1000

T (x) = 1000[(1 + x)3 + (1 + x)2 + (1 + x) + 1] fcommon factor gT (x) = 1000[(1 + 3x + 3x2 + x3) + (1 + 2x + x2) + (1 + x) + 1] fexpandinggT (x) = 1000[x3 + 4x2 + 6x + 4]

fcollecting like terms

g b Substitute 0:09 for x.

T (0:09) = 1000[(0:09)3 + 4(0:09)2 + 6(0:09) + 4] = 4573:13

There is $4573.13 in the account.

c We need to solve T (x) = 4300. We can solve this

equation to suitable accuracy by graphing the func-

tions T (x) and f (x) = 4300 on the same set of

axes and looking for the point of intersection be-

tween 0 and .09. The screen shot gives the answer

of x = 0:0484, or an annual interest rate of 4.84%.

Savings accounts

EXAMPLE 8.17

1 2 3 4

0 1 2 3

deposits

years




EXERCISE 8H

800600400200

120

100

80

6040

20

y

x

1 W. F. Weeks and W. J. Campbell in the Journal of Glaciology used a more complex cubic

model for towing a flat iceber g from Amery Ice Shelf in Antarctica to Australia. The equation

is V = 7:75 ¡ 0:35x + 0:0625x2 ¡ 0:0208x3, where V is the number of cubic kilometres

of ice remaining and x + 3 is the number of thousand kilometres travelled by the iceberg,

0 6 x 6 4. The volume of ice changed very little in the first 3000 km of the tow, as it wasstill in Antarctic conditions.

a What was the initial volume of ice?

b What was the volume of ice remaining after the iceberg had been towed 4000 km (that

is, when x = 1)?

c If the total length of the journey was 7000 km, what percentage of the original volume

of ice still remained?

d How many kilometres was the iceberg towed before it lost 10% of its volume?

2 In a coal mine, there are x men in each shift at the coalface, and the output of coal is

given by 1

60 x2

(36 ¡ x) tonnes of coal.a What is the maximum domain for this function?

b What is the optimum number of men per shift? Use a graphics calculator to assist.

3 A firm sells all units it produces at $4 per unit. The firm’s total cost, C , for producing x units

is given in dollars by C = 50 + 1:3x + 0:001x2

a Write an expression for the total profit as a function of x.

b How many items should be produced so that the profit is a maximum?

4 The owner of a small business estimates that the profit from producing x items is given by

the function P (x) = 0:003x3

¡1:5x2 + 200x

¡1000. This function is based on current

production levels, which cannot exceed 350 items due to limited space and resources. Howmany items should be produced to maximise the profit?

5

a The equation for this graph is of the form f (t) = kt(t ¡ a)2. From the graph, what is

the value of a? What does it represent?

b What is the domain?

c If the ideal crash barrier is depressed by 85 mm after 100 milliseconds, find the value of

k, and hence find the equation of the graph given.

d What is the maximum amount of depression, and when does it occur?

A T a.

scientist working for Crash est Barriers, Inc. is trying to design crash test barrier whoseideal characteristics are shown graphically below The independent variable is the time after impact, measured in milliseconds. The dependent variable is the distance that the barrier has

been depressed because of the impact, measured in millimetres.






4 A ladder of length 10 metres is leaning up against a wall

such that it is just touching a cube of edge length one

metre that is resting on the ground against the wall.

What height up the wall does the ladder reach?

5 Suppose that the annual interest rate at your bank is some fixed rate, i, (written as a

decimal). To calculate how much you have in your account after you put in $7000 for 1year compounded yearly, you use the formula A = 7000 £ (1 + i):

To simplify some more, let us call (1 + i) = x . So if you put in $7000 the first year,

your total at the end of the year is 7000x. Now, you probably want to keep on saving and

adding to this account. Suppose that at the beginning of the next year, you put in $3000.

Your total is now 3000 + 7000x:

a Assuming that the interest rate stays the same, show that at the end of year 2, the

amount has grown to 3000x + 7000x2 , by the miracle of compound interest. b

binomial expansion local maximum reciprocal function

composite function local minimum repeated rootscubic function Pascal’s Triangle roots of a function

degree of a polynomial polynomial subscript notation

expanded form quadratic function summation notation

factorial notation quartic function zeros of a function

factorised form rational function

1 Explain why each of the following is not a polynomial.

a f (x) = 1

x b g(x) = 3x3 ¡ 2

p x + 6 c h(x) =

x2 ¡ 1

x + 2

2 Rewrite in standard polynomial form: a

5Xi=1

1

i xi b

3Xi=1

i(x + 1)i

3 a Given T (x) = x4 + 3x3 + x2 ¡ 4x, U (x) = x + 5, find

i 2T (x) + 3U (x) ii T (x) ¡ 4U (x) iii T (x):U (x) iv T (x)

x

x 10 m

1 m

Now assume that the interest rate remains fixed, and you made the following deposits: At

the start of the third year you deposited At the start of the fourth year you depos-

ited At the start of the fifth year you deposited If you hope that this account

has at the end of the fifth year what is the fixed interest rate that you require?

, $ . ,

$ . , $ .

$ ,

45004000 900035000






b Express P (x) = (2x +1)(x¡ 2)(x + 2)¡ (x +4)(x ¡ 5) in polynomial form and hence

evaluate P (¡3).

4 If f (x) = x2 + 1 and g(x) = x3, find a f (g(x)) b g(f (x))

5 a Expand (2x ¡ 3)5. b Find the first three terms of the expansion of (3x + 1)4.

6 a Without using a graphics calculator, find the zeros of i y = x2 + x ¡ 6 ii y = (x + 1)(x ¡ 1)(x ¡ 2)2

b Find the points of interest of

i y = x3 ¡ 2x2 ¡ x + 2 ii y = x4 ¡ 4x + 3 iii y = x4 ¡ 2x3 + x2

7 A piece of wire 60 cm long is bent to form a rectangle. If one of its dimensions is x, express

the area as a function of x. Hence, find the maximum area of the rectangle.

8

Find the volume of the box as a function of x and hence find the maximum volume.

9 The diagram shows two ladders of length 20 feet and

30 feet (in “old” imperial units) placed between two buildings. The ladders meet 8 feet above the ground.

If x is the height of one building as shown, show that

x4 ¡ 16x3 + 500x2 ¡ 8000x + 32 000 = 0.

If D is the width of the street, find D.

1 Which of the following are polynomials?

a y = x2 + 2x¡1 b y = 5 c x = 2 d y = x3 ¡ 4x2 + 3x

x

2 Write in expanded form:

4Xi=1

ix2. Simplify if possible.

3 Expand using the binomial expansion: (2a + 3)4

4 For the function y = 5 ¡ 3x ¡ 6x2 ¡ x3, use a graphics calculator to find the

coordinates of the local maximum in the vicinity of the y-axis.

5 If f (x) = 3x2 and g(x) = x2 + 5, find f ± g .

6 Sketch by hand on the same set of axes: f (x) = x3 g(x) = ¡2(x ¡ 1)3

A a aclosed box (like pizza box) is to be formed from sheet of cardboard cm by cm by cutting equal squares, of side cm, from two corners of the short side, and two equalrectangles of width cm from the other two corners and folding along the dotted lines asshown in the diagram.

64 40x

x

40

64 x

x x

lid

D

x

30

208




Furtherfunctions andrelations

SUBJECT MATTER

distinction between functions and relations

,

distinction between continuous and discontinu-ous functions

practical applications of the reciprocal functionand inverse variation

relationship between the graph of and thegraphs of and for both positive and negative values of the constant

f xf x a f x a af

a

( )( )+ ( + ) ( )x

CHAPTER 9




RECIPROCAL FUNCTIONS

A rich old man left a fortune of $12 million dollars to be divided equally amongst all of his

grandchildren.

Let x = the number of grandchildren, and

y = the amount that each grandchild receives.

We can start by making a table:

x (number of grandchildren) 1 2 6 12 24y ($millions) 12 6 2 1 0:5

From the table we can draw a graph. Following the con-

vention, we will join the points with a smooth curve, even

though we cannot have fractions of a grandchild.

To find an algebraic equation, we note that the product of

the x and y coordinates is always 12, so we have

xy = 12 or, solving for y ,

y = 12

x which is our required function.

This is an example of a class of functions called reciprocal functions. The name of this graph is

a rectangular hyperbola.

The simplest reciprocal function is y = 1

x, whose

graph is given alongside.

This graph has two branches. The graph approaches butdoes not intersect the two axes. Such lines are called

asymptotes, which in Greek means “never touching”.

Geometrically it is as though the curve has tangents whose

point of contact has gone to infinity. The function is not

continuous at x = 0, i.e., you cannot draw its graph

without taking your pencil from the paper at x = 0.

The basic form of a reciprocal function is y = a

x where a is a non-zero real number.

On your graphics calculator, draw the reciprocal function for values of a from ¡6 to 6.

What effect does a have on the graph of the function?

RECIPROCAL FUNCTIONS A

number of grandchildren

$millions

y

x

The reciprocal function is an example of a discontinuous function. A discontinuous function may

be thought of as a function that cannot be drawn without lifting the pen from the graph paper.

Section B contains further examples of these functions.

282 FURTHER FUNCTIONS AND RELATIONS (Chapter 9)



TRANSFORMATIONS OF y = 1

x

Use a graphics calculator to sketch the following graphs. Choose a viewing window that

illustrates the features of each graph.

a f (x) = 1

x b g(x) =

2

x c h(x) =

1

2x

The function g(x) = 2

x can be rewritten as g(x) = 2(

1

x), or g(x) = 2f (x).

The function h(x) = 1

2x can be rewritten as h(x) = 1

2

µ1

x

¶, or h(x) = 1

2 f (x).

Its graph is the graph of f (x) “stretched” vertically by a factor of 12 .

a b c

1 If f (x) = 1

x, sketch by hand the graphs of each of these functions.

You may check your answers with a graphics calculator.

a ¡3f (x) b f (x ¡ 3) c f (x) ¡ 4 d ¡f (x ¡ 1) + 3

2 Use your knowledge of transformation of functions to sketch the following sets of graphs byhand. Sketch each set of functions on the same set of axes.

Use a graphics calculator to check your solution.

a y = 1

x y =

3

x y =

¡1

x

b y = 1

x y =

1

3x y = ¡ 1

3x

c y = 1

x

y = 1

x ¡ 1

y = 1

x + 3

EXAMPLE 9.1

EXERCISE 9A

Its graph is just the graph of stretched vertically by factor of urning on the racefeature of graphics calculator and toggling back and forth between the graphs of and

illustrates that the coordinates of are twice those of

f xf x

g x y g x f x

( ) 2( )

( ) ( ) ( )

a . T Ta

.

FURTHER FUNCTIONS AND RELATIONS (Chapter 9) 283

What you have learned about transformations of functions in applies to reciprocalfunctions.

Chapter 2



d y = 1

x y =

1

x + 1 y =

1

x ¡ 2

e y = 1

x y =

1

x ¡ 1 + 2 y =

1

x + 3 ¡ 1

f y = 1

x y =

¡2

x ¡1

g y = 1

x y =

1

2x + 3

3 A large plot of land has an area of 1200 hectares. The owner is hoping to subdivide the

land into equal-sized smaller plots. The independent variable is the number of plots, and the

dependent variable is the size of each plot.

The graph shows the values of the interest rates set by the Reserve Bank of Australia from January

1998 to December 2000. The values jump on certain dates. If you were paying off a loan, the

amount you repay would move up and down at the time when your lending institution changed its

rates in response.

CONTINUOUS AND DISCONTINUOUS FUNCTIONS B

0 10 20 30 40 50 600

1

2

3

4

5

6

7

8

numberof months after January 1996

p e r c e n

t a g e

i n t

e r e s

t r a

t e

This function has a number of jump discontinuities, where the value of the function changes

abruptly. There are many examples of discontinuities in the natural world.

² The wake of a ship has a boundary which separates smooth water from the turbulent

water within the wake.

² A sonic boom is heard when an aircraft exceeds the speed of sound. The boom is due

to a discontinuity in air pressure across a shock wave caused by the flight of the aircraft.

² Weather maps show fronts; these are almost always due to a discontinuity of some

meteorological phenomenon, rain for example.


a Find a function that gives the plot size as a function of the number of plots, where area

is measured in m2.

b Sketch the graph of the function.

c If each plot is to be as close to 700 m2 as possible, how many plots will there be?



First half hour $3:30

Second half hour $2:20 additional

Next hour $2:20 additional

Each succeeding $1:10 additionalhour up to 24 hours

a Plot the graph of the cost.

b What is the domain and range of the function?

a

b The range is the four different rates, from $1:10 to $6:60. The domain is 0 to 24The graph plot is truncated at 6 hours.hours.

1

100 units (kwh) 17:116

next 300 (kwh) 11:627

remaining units 10:373

2 For long term parking at an airport the following

charges apply. Determine the rate for each day and

plot the rates versus the number of days parked.

EXAMPLE 9.2The cost of parking in the short term car park atBrisbane airport is shown in the table. Accordingto the table, the cost is charged for the full periodin which the length of the stay falls. For example,

minutes counts as hour stay hours

minutes counts as hours and so on.

40 10a ,1 3

4

0 1 2 3 4 5 60

2

4

6

8

10

12

number of hours in car park

c o s

t ,

i n

d o

l l a r s

EXERCISE 9B


Domestic electric power rates are charged according

to the table. Plot a graph of the rates versus the

power consumed.

1 day 2 days 3 days 4 days 5 days

$19:70 $34:70 $49:70 $61:70 $73:70

6 days 7 days 8 days 9 days 10 days

$85:70 $97:70 $$ 115106 :: 7070 $124:70



3 The investment rates of banks varies so that higher sums attract higher interest rates. A well

known bank offers the interest rates shown.

a Plot the interest rate against the sum invested.

b Plot the amount of interest gained over a period

of 12 months against the sum invested.

In Chapter 1 you studied direct variation, in which one variable is proportional to another.

If y varies directly as x, then y = kx where k is the constant of proportionality.

The graph of this function is a straight line passing through

the origin.

If y varies inversely as x, then y = c

x.

The graph of inverse variation is a rectangular hyperbola

centered at the origin.

We can also write this as xy = c which states that the

product of the two variables equals the constant, c.If two variables vary directly, doubling the value of the

independent variable doubles the value of the dependent

variable. With inverse variation, if the value of the in-

dependent variable doubles, the value of the dependent

variable halves.

Two men can paint a house in 15 days. How long does it take three men to paint the house,

if they work at the same rate?

If the number of painters is doubled, the time to complete the job is halved, so this is an

example of inverse variation.

Let x = number of men

y = days each man works

Since xy = c

we have 2 £ 15 = 30

so c = 30

It takes 30 man days to paint the house.

INVERSE VARIATIONC

EXAMPLE 9.3

x

y

xy chalf

x

y

y kx


$5000 ¡ $19 999 1:85% p.a.

$20000 ¡ $49 999 3:85% p.a.

$50000 ¡ $69 999 4:15% p.a.

$70000 ¡ $100 000 5:2% p.a.

4 Give three other real world examples of functions that contain jump discontinuities.



If there are 3 men, we have 3y = 30

so y = 10

It will take 10 days for 3 men to complete the task.

1 A roller with a diameter of 20 cm is running at 1520revolutions per minute (rpm), and drives a conveyor

belt. There is another roller, with a different diameter,

at the other end of the conveyor belt.

a The speed of the second roller (measured in rpm)

is inversely proportional to its diameter. Explain

why this must be true.

b Find the speed, in rpm, of the second roller if its

diameter is 30

cm.

2 The frequency of a radio wave is inversely proportional to the wave length. If a wave of

300 metres has a frequency of 1000 kilocycles per second, find the length of a wave with a

frequency of 750 kilocycles per second.

3 Two meshed gears have 36 and 56 teeth respectively. If

the speeds are inversely proportional to the number of

teeth, at what speed should the second gear be driven

in order to make the first gear run at 1296 rpm?

4 According to the universal gas law, known as Boyle’s

Law, the pressure of a gas, under constant conditions,

varies inversely with its volume. Suppose that a gas

with a volume of 400 cm3 results in a pressure of

1:2 kg/cm2. If the same gas is put into a container

with a volume of 250 cm3, under the same constant

conditions, what will be the pressure?

5 According to Newton’s Second Law of Motion, given a constant force, the acceleration of an

object varies inversely with its mass. If such a force causes a 20 kg object to accelerate at 4m/s2, what would be the acceleration of a 15 kg mass?

6 For a rectangle with a given area, the length of the rectangle varies inversely with the width.

a Explain why this is true. b A rectangle with a length of 40 cm has a width of 25 cm. If the area is to remain constant,

what is the length if the width is 4 cm?

7 Farmer Brown has 30 horses, and enough hay to

feed them for 60 days. But on the 20th day, he

buys another 20 horses. Assuming that each horse

eats the same amount of hay, how many more days

will the hay last?

EXERCISE 9C




A RECIPROCAL MODEL – THE UNIVERSAL GAS LAW

T

48"

24"Initiallevel of mercury

Shorter legwith scale

29\\Qq _Qy _ "

Mercury columnincreased by pouring mercuryin at T

48

46

44

42

40

38

36

3432

30

28

26

24

29 216

30 916

31 816

33 516

35

37 516

39 1016

41 3

16

44 116

47 116

50 516

54 516

58 1316

29 216

30 616

31 1216

33 17

35

36 1516

38 78

41 2

17

43 1116

46 35

50

53 1013

58 28

23

22

21

20

19

18

17

1615

14

13

12

61 516

64 116

67 116

70 1116

74 216

77 1416

82 1216

8714

16

93 116

100 716

1071316

117 916

60 1823

63 611

66 47

70

73 1119

77 23

82 417

873

8

93 15

99 67

107 713

116 48

A A B (in.) B (in.)C (in.) C (in.)


a Choose three or four rows of this table, at random, and multiply the value in column A

by the value in column B. What did you expect to find? What did you find?

b calculator. Plot List 2 (y) vs List 1 (x).

function?

c We want to determine the least squares regression line for this data. Most graphics

calculators do not have reciprocal regression as an in-built menu item, so we must do the

calculations ourselves.

First we have to “straighten” the data, by plotting Height vs 1

Pressure.

i Make List 3 = 1 ¥ List 2:

ii Draw a scatterplot of List 3 (y) vs List 1 (x).

iii If Boyle’s Law is correct, the data should be linear. Does the data appear to be

linear?

Enter the data from column A into List 1 and column B into List 2 of your graphicsDoes the data have the shape of recipriocala

R

s L

obert Boyle

Boyle’ aw

announced his discovery of the uni-versal gas law now known as (see

question above) in an article entitled “A Defence of the Doctrine ouching the Spring And eight of theAir..”, published in

His original data is given in the table. Column

gives the height of cylinder of air This is equiva-lent to the volume, as the height of cylinder is di-rectly proportional to its volume. The values in this

column are the number of equally spaced tick marksmeasured from the top of the cylinder Columngives the pressure, measured in inches of mercuryColumn gives the pressure predicted by this theory

our task is to confirm Boyle’ Law using his origi-nal data.

, ,

T W.

A

a .a

. B.

C .

Y s ,

1662

4



d As we want the graph of the linear regression equation to pass through the origin, the

equation has the form y = ax, where the data for y is in List 3, the data for x is in

List 1, and

a =

Pxy

Px2

(this formula is discussed in Section 5.6)

i Find the value of a. Carry out the usual analysis, i.e., find the value of r2,

calculate the residuals, and do a residual plot.

ii Use this analysis to decide if the transformed data can be modelled by a linear

function.

e ‘Back transform’ the linear regression equation to find the corresponding reciprocal func-

tion.

f Draw the reciprocal function on a scatterplot containing the original data. It should fit

beautifully!

g Column C of the data contains the predicted values according to Boyle’s hypothesis.

Compare these to the values predicted by your least squares regression line.

NON-LINEAR MODELS

The above example illustrates a sound approach to fitting a non-linear function to a set of paired

data.

² Enter the data into lists on a graphics calculator (or a spreadsheet or a statistics program).

² Look at the data, by drawing the scatterplot. What did you expect to find? Are there any

unexpected patterns, or anomalies? Does the data appear to have errors?

² Based on previous knowledge, choose a function that you believe to be an appropriate

model for this data.

² Apply the algebraic transformation that “straightens” the data.

² Draw a scatterplot of the transformed data. Does it look linear? Are there any anomalies?

² If the data does not look linear, you might try another function. If it does, find its least

squares regression equation.

² Draw a residual plot of the linearised data. Are there any strong underlying patterns? If

so, can they be explained?

² Note the value of r 2. This gives you an indication of how useful the model will be for

prediction.

² Back transform the linear model to obtain a non-linear model of the original data.

² Plot the non-linear model on the scatterplot of the original data. How well does it fit?


If you are not sure which type of function to choose, you might fit a few different functions, and

choose one based firstly on the residual plot and secondly on the value of r2.

The web page contains some of RobertBoyle’ original notes about how he conducted the experiment. It is worth visit!s a

http://dbhs.wvusd.k12.ca.us/GasLaw/Gas-Boyle-Data.html





² Pushing a regression button on a calculator does little to assist in understanding the under-

lying mathematics.

² A student can learn much about the appropriateness of the chosen function by viewing the

scatterplot of the straightened data.

² A function that is a possible model for the paired data may not be available on the graph-

ics calculator. Presenting a subset of the possible functions implies that these are the only

functions that one should try.

The Pythagoreans of ancient Greece believed that all

numbers could be expressed as the ratio of two integers,

i.e. that all numbers were rational numbers. They were

inconsolable when it was proved that the diagonal of a unit

square (a square with a side length of one unit), which

we now symbolise asp

2, was not rational.

Let us consider a simple case, a circle of radius r units, centred at the origin.

From the diagram, and using the fact that for any right-angled triangle, a2 + b2 = c2, it is

obvious that:

The equation of a circle of radius r, centred at the origin, is x2 + y2 = r2:

To find the equation of a circle with radius r at centre (h, k), we translate the circle h units

horizontally and k units vertically. In the equation, we replace x with (x ¡ h) and y with (y ¡ k).Hence:

The equation of a circle with centre (h, k)

and radius r is (x ¡ h)2 + (y ¡ k)2 = r2.

CIRCLES AS RELATIONS D

x

yr

x

y

( ) x, y

r

h k ,

x

y

Partly because the Greeks lacked algebraic notation, they

used plane geometry (Euclidean geometry) to advance

their understanding of mathematics.

Rene Descartes was able to marry algebra and geometry with his discovery of coordinate geometry,

and many of the geometric figures such as the line, parabola and hyperbola were able to be studied

from an algebraic viewpoint.You have already learned about the equations of lines and parabolas. Is there an equation for a

circle?


Most graphics calculators have a variety of non-linear regression models built into their software.

A student might ask, “Why not just use those? It is much easier.” Three arguments against this

are:



a centre (0, 0), radius 3 b centre (¡2, 0), radius 1

c centre (3,

¡2), radius 2 d 3x2 + 3y2 = 48

) x2 + y2 = 16So, centre (0, 0), radius 4

State the equation of each graph:

a b c

a centre (0, 0), radius 4

) x2 + y2 = 42

i.e., x2 + y2 = 16

b centre (0, ¡1), radius 4

) x2 + (y + 1)2 = 42

i.e., x2 + (y + 1)2 = 16

c centre (¡2, ¡2), radius 3

) (x + 2)2 + (y + 2)2 = 32

i.e., (x + 2)2 + (y + 2)2 = 9

y

x

y

x

y

x

y

x

y

x

y

x

y

x

EXAMPLE 9.5

Sketch the graph of each equation:

a x2 + y2 = 9 b (x + 2)2 + y2 = 1

c (x ¡ 3)2 + (y + 2)2 = 4 d 3x2 + 3y2 = 48

EXAMPLE 9.4





c What other function must you draw, to complete the other half of the circle?

d Draw the second function on the same screen.

DRAWING A CIRCLE WITH A GRAPHICS CALCULATOR

The circle is a relation that is not a function.

For the equation x2 + y2 = 25, verify by substitution that if x = 4, then y = 3 or y = ¡3.

A graphics calculator only graphs functions, so how can we sketch the graph of a circle using a

graphics calculator?

Note that the top half of a circle is a function, as is the bottom half. To sketch a circle we need to

find the equations for each of these semicircles.

Sketch the graph of x2 + y2 = 2 on a graphics calculator.

x2 + y2 = 25

) y2 = 25 ¡ x2

) y = §p 25 ¡ x2

The top half of the circle is the function y =p

25 ¡ x2

while the bottom half of the circle is the function

y = ¡p 25 ¡ x2.

To sketch these two functions on your graphics calculator, the vertical and horizontal scales

must be the same for the graph to look circular.

Because of the algorithm used by graphics calculators to draw graphs, the parts of the circle

that have very steep gradient (the left and right ‘edges’) may not be drawn.a

If the entire circle is not drawn, sometimes changing the window slightly will improve the

graph.

EXAMPLE 9.6

THE GENERAL EQUATION OF A CIRCLE

Recall that there are two common forms for the equation of a straight line:

y = mx + c the gradient-intercept form of a straight line

ax + by + c = 0 the general equation of a straight line

Similarly there are two forms for the equation of a circle:

(x ¡ h)2 + (y ¡ k)2 = r2 the centre-radius form of a circle

x2 + y2 + 2gx + 2f y + c = 0 the general equation of a circle

We need to be able to convert from one form to another.




1 Write each of these equations of a circle in the general form.

a (x + 3)2 + y2 = 16 b (x ¡ 1)2 + (y ¡ 5)2 = 9

c x2 + (y + 3)2 = 3 d 2(x ¡ 3)2 + 2y2 = 50

e (x + 1)2 + (y ¡ 1)2 = 10 f 3x2 + 3(y + 3)2 = 12

2 Write each of these equations in the general form.

a (x ¡ 2)2 + y2 = 4 b x2 + (y ¡ 5)2 = 9 c (x + 2)2 + (y + 3)2 = 1

d (x + 3)2 + (y + 3)2 = 6 e x2 + (y

¡5)2 = 1 f y2 = 12

¡(x

¡2)2

3 Write each of these equations of a circle in the centre-radius form.

a x2 + y2 ¡ 4x + 2y ¡ 11 = 0 b x2 + y2 + 6x ¡ 16y ¡ 27 = 0

c x2 + y2 ¡ 4y + 3 = 0 d x2 + y2 + 8x ¡ 2 = 0

e x2 + y2 ¡ 3x + 2y = 0 f x2 + y2 ¡ x ¡ y ¡ 1 = 0

4 Write each of these equations of a circle in the centre-radius form.

a x2 + y2 ¡ 6x + 10y + 33 = 0 b x2 + y2 ¡ 4y ¡ 5 = 0

c x2 + y2 ¡ 12x + 29 = 0 d x2 + y2 ¡ x + 23 y ¡ 23

36 = 0

AN AMAZING EQUATION

The ancient Greeks spent a lot of time studying a collection of

geometric objects called conic sections - the parabola, the circle,

the ellipse and the hyperbola. You have met all of these except

the ellipse. An ellipse (shown alongside) can be thought of as a

“stretched circle”.

After Rene Descartes invented analytical geometry mathematicians were able to give a single

equation to give rise to all of these graphs:

ax2 + by2 + 2hxy + 2gx + 2f y + c = 0, a > 0

EXERCISE 9D.2

a Convert (x + 3)2 + (y ¡ 2)2 = 16 to the general equation of a circle.

b Convert x2 + y2 ¡ 2x + 6y + 6 = 0 to the centre-radius form of a circle.

a (x + 3)2 + (y¡

2)2 = 16

) x2 + 6x + 9 + y2 ¡ 4y + 4 ¡ 16 = 0 fexpandg) x2 + y2 + 6x ¡ 4y ¡ 3 = 0 fcollect like termsg

b Re-arrange and complete the square.

x2 + y2 ¡ 2x + 6y + 6 = 0

) (x2 ¡ 2x + ) + (y2 + 6y + ) + 6 = 0 fre-arrange the termsg) (x2 ¡ 2x + 1) + (y2 + 6y + 9) + 6 ¡ 1 ¡ 9 = 0 fcomplete the square, twiceg

) (x ¡ 1)2 + (y + 3)2 = 4 fre-write in factorised formg

EXAMPLE 9.7




2x2 + 6x + 3y ¡ 2 = 0 is a parabola

2x2 + 2y2 ¡ 3x + 4y ¡ 5 = 0 is a circle

2x2 + 5y2 ¡ 3x + 4y ¡ 5 = 0 is an ellipse

2x2 ¡ 3y2 ¡ 3x + 4y ¡ 5 = 0 is an hyperbola

7x + 2y ¡ 4 = 0 is a straight line

Examples of two special cases are2x2 ¡ 3y2 = 0 is a pair of intersecting lines through the origin

x2 + y2 + 4 = 0 has no real solution and hence has no graph.

On a graphics calculator, sketch the graph of ax2 + by2 + gx + f y + c = 0, where

a = 4, b = 9, g = ¡8, f = 0, c = ¡32:

The equation is 4x2 + 9y2 ¡ 8x ¡ 32 = 0.

In order to graph it, we need to solve this equation for y.

4x2 + 9y2 ¡ 8x ¡ 32 = 0

) 9y2 = 32 + 8x ¡ 4x2

) y2 = 32 + 8x ¡ 4x2

9

) y = §r

32 + 8x ¡ 4x2

9

Graphing these two equations in the window (¡3, 5, 1; ¡3, 3, 1) shows that the relationis an ellipse centred at ( , ).0 1

EXAMPLE 9.8

Sketch the graph of x2 + y2 ¡ 2x + 4y ¡ 4 = 0 on a graphics calculator.

Solve for y: x2 + y2 ¡ 2x + 4y ¡ 4 = 0

) (x2 ¡ 2x + 1) + (y2 + 4y + 4) = 4 + 1 + 4

EXAMPLE 9.9

) (x ¡ 1)2 + (y + 2)2 = 9

) (y + 2)2 = 9 ¡ (x ¡ 1)2

)p

(y + 2)2 = §p 9 ¡ (x ¡ 1)2

) (y + 2) = §p

9 ¡ (x ¡ 1)2

) y =p

9 ¡ (x ¡ 1)2 ¡ 2

Sketch these two equations in the window (¡4, 5, 1; ¡5, 1, 1).




1 Sketch the graph of these circles on a graphics calculator.

a x2 + y2 ¡ 2x ¡ 15 = 0 b x2 + y2 ¡ 4x + 2y ¡ 11 = 0

c 2x2 + 2y2 ¡ 12x + 8y + 24 = 0 d x2 + y2 ¡ 16x ¡ 6y + 48 = 0

e x2

+ y2

¡ 6x ¡ 5y + 1114 = 0 f x

2

+ y2

+ 32 x + y ¡

316 = 0

EXERCISE 9D.3

Concurrent lines are three or more lines that intersect in a single point.

To show algebraically that three lines are concurrent, we find the point of intersection of any two

of them, and then show that the coordinates of the point of intersection satisfy the third line.

Prove algebraically that the lines with equations y = 2x + 1, y = 4 ¡ x and

3x ¡ 2y + 3 = 0 are concurrent.

We will use the substitution method.

2x + 1 = 4 ¡ x fsetting the first two equations equal to each other g) 3x = 3

) x = 1 fsolving for xg) y = 3 fsubstitute to solve for yg

The coordinates of the point of intersection are (1, 3). Substitute these values into the third

equation.LHS = 3 £ 1 ¡ 2 £ 3 + 3

= 0

= RHS

The coordinates of the point satisfy the equation, therefore the lines are concurrent.

FINDING POINTS OF INTERSECTION OF LINES, PARABOLAS AND CIRCLES

E

EXAMPLE 9.10

a Find the coordinates of the points of intersection of the line x + y = 4 and the

circle x2 + y2 = 8.

b Find the coordinates of the points of intersection of the line y = 3x and the

parabola y = 4 ¡ x2.

c Find the coordinates of the points of intersection of the circle with equation

(x ¡ 1)2 + (y ¡ 2)2 = 9 and the parabola with equation y = 8 + 2x ¡ x2.

EXAMPLE 9.11

A aa a a

line and circle intersect in at most two points. When we solve for the points of intersection, wewould expect to get quadratic equation. Similarly line and parabola intersect in at most two

points.




a y = 4 ¡ x fre-write, making y the subjectg) x2 + (4 ¡ x)2 = 8 fsubstitute into the equation of the circleg) x2 + 16 ¡ 8x + x2 = 8 fexpandg) 2x2 ¡ 8x + 8 = 0 fsimplifyg) x2

¡4x + 4 = 0

fdivide both sides by 2

g) (x ¡ 2)(x ¡ 2) = 0 ffactoriseg) x = 2 fsolveg) y = 2 fsubstitute and solvegAs there is a single solution, the line must be a

tangent to the circle and touch it at the point (2, 2).

Graphing the line and circle on the same set of

axes provides a worthwhile check.

b 3x = 4 ¡ x2

) x2 + 3x ¡ 4 = 0 fgather terms on one side of the equationg) (x + 4)(x ¡ 1) = 0 ffactoriseg) x = ¡4 or 1 fsolvegwhen x = ¡4, y = ¡12 fsubstitute and solvegand when x = 1, y = 3

The coordinates are (¡4, ¡12) and (1, 3).

The graphs verify our result.

c A quick sketch of a parabola and a circle shows

that it is possible to have up to four points of

intersection.

Substituting for y in the equation

(x ¡ 1)2 + (y ¡ 2)2 = 9

gives (x ¡ 1)2 + [(8 + 2x ¡ x2) ¡ 2]2 = 9.

y

x

y

x

x

y

2

2

4

4

6

8

-2

-2-4

fequate the expressions for yg




4 Use a graphics calculator or graphing software to find the points of intersection of the following

functions:

a the circle (x ¡ 1)2 + (y ¡ 4)2 = 12 and the parabola y = 2x2 + x ¡ 4

b the circle (x + 2)2 + (y ¡ 3)2 = 16 and the parabola y = x2 + 3x ¡ 5

Recall from Chapter 1 that a relation is a set of ordered pairs. You might be surprised to know

that relations are fundamental in the study and design of computerised databases. In fact the name

given to a database with multiple inter-connected tables is a relational database.

Prior to studying the equations of circles, all relations we studied were also functions. Each value

of x in the domain is paired with a single value of y.

In this chapter we studied a relation that is not a function, i.e., the circle.

Now we will study relations in general. We will consider four cases. In each case we define the

relation, explain how each type of relation can be determined from its graph, and give an example.

a one-to-one mapping

Each value in the domain maps to a single

value in the range, and each value in the range

is mapped from a single value in the domain.

Line test: Any horizontal or vertical line will

intersect the graph in at most one point.

Example: A linear function.

a many-to-one mapping

Each value in the domain maps to a single

value in the range At least one value in the

range is mapped from two or more values in

the domain.

Line test: Any vertical line will intersect the

graph at at most one point. At least one hor-

izontal line will intersect the graph in two or

more points.

Example: A cubic equation with three roots.

Both of these relations are also functions, since they pass the vertical line test.

y

x

y

x

MORE ON RELATIONS AND MAPPING F



a many-to-many mapping

At least one value in the domain maps to two

or more values in the range, and at least one

value in the range is mapped from two or

more values in the domain.

Line test: At least one horizontal line will

intersect the graph in two or more points. At

least one vertical line will intersect the graph

in two or more points.

Example: A circle.

a one-to-many mapping

At least one value in the domain maps to two

or more values in the range, and each value

in the range is mapped from a single value in

the domain.

Example: The graph of x = y2.

The mathematics of relations extends to situations that are not normally thought of as mathematics.

a the relation x “is the father of” y b the relation x “is the brother of” y

c the relation x “is married to” y d the relation x “has the sister” y

We can illustrate each of these relations with a mapping.

a b

c d

a Ed is the father of three children, so the relation is one-to-many.

b Both Rocket and Charles are the brother of two others, so the relation is many-to-many.

c

d Rocket and Charles each have a single sister, so the relation is many-to-one.

y

x

EdRocketCharlesMoonbeam

MoonbeamRocketRocketCharlesCharles

RocketMoonbeam

CharlesEd Martha

y

x

EXAMPLE 9.12

Consider family consisting of a married couple, Ed and Martha, with two boys, Rocket andCharles, and one girl, Moonbeam. Classify each of these relations as one-to-one, one-to-manymany-to-one, many-to-many Justify your answer in each case.

a ,.

Ed is the only element in the domain, and only married to Martha, so the relation isone-to-one.


Line test: At least one horizontal line will

intersect the graph in at most one point. At

least one vertical line will intersect the graph

in two or more points.

Note: Neither of these are functions, as they fail the vertical line test.



1 Classify each of these relations as one-to-one, many-to-one, one-to-many or many-to-many.

a b c

d e f

2 Consider a family consisting of Nanna, Grandpa, their daughter Claire, who is a single mum,

and her two daughters Alicia and Portia. Classify each of these relations as one-to-one, many-to-one, one-to-many or many-to-many. Justify your answers.

a x “is the daughter of” y b x “is married to” y

c x “has the child” y d x “has the grandfather” y

e x “has the granddaughter” y f x “has the mother” y

EXERCISE 9F

y

x

P (3, )t


1 The cost C , measured in thousands of dollars, of removing a given percentage P , of

pollutants from an abandoned tailings pond, is given by C = 10000P

100 ¡ P .

a Sketch this function using a graphics calculator, for 0 6 P < 100.

b What is the cost of removing

i 10% ii 50% iii 90% iv 99% v 99:5%

of the pollutants?

c What percentage of the pollutants can be removed if the budget is $2000000?

2 Point P lies on the circle as shown. Find t.

3 Consider the class of functions y = a

x, a > 0 and the function y = x.

Show that for all a, the graphs of these functions meet at (p a, p a).

PROBLEM SOLVING AND MODELLINGG

EXERCISE 9G





4 Find the equation of the circle that passes through the origin and the points (0, 4) and (4, 0).

1 Use your knowledge of transformations to sketch the functions y = 1x

and

y = 2

x + 1 Use a graphics calculator to check your solution.

2 The cost of Overnight Express bags from Australia

Post is based on the weight of the contents as follows:

0 kg to 1 kg $4:50

over 1 kg to 3 kg $9:00

over 3 kg to 6 kg $12:00

Draw a graph showing the cost of posting articles

weighing between 0 kilograms and 6 kilograms.

3 Two meshed gears, A and B, have 32 and 20 teeth respectively. B is also meshed with C,

which has 16 teeth. At what speed should A be driven in order to make C run at 6000 rpm?

4 Given that x2 +y2 ¡6x+ 8y +24 = 0 is the equation of a circle, find its centre and radius.

5 Find the equations of the circles whose graphs are given below.

a b


asymptotes circle ellipse

inverse variation many-to-many mapping many-to-one mapping

one-to-many mapping one-to-one mapping reciprocal function

rectangular hyperbola

on the same set of axes.

y

x

y

x






6 Find the points at which the line with equation y = 6 ¡ 8x intersects the parabola with

equation y = 2x2 + x ¡ 5.

7 Classify each of these relations as one-to-one, many-to-one, one-to-many, or many-to-many.

a b c

8 Use a graphics calculator to find the points of intersection of the two circles in question 5 .

9 Prove that the straight lines 3x ¡ y = 9, x ¡ 2y = 8 and 2x ¡ 3y = 13 are concurrent.

10 Find the distance between the centres of the circles

x2 + y2 ¡ 2x + 4y = 0 and 2x2 + 2y2 + 8x ¡ 4y ¡ 1 = 0

11 Find the value or values of k for which the three lines x ¡ y ¡ 1 = 0, 2x ¡ y + k = 0 and

kx ¡ 2y + 2 = 0 are concurrent.

12 Find the coordinates of the points where the circle x2 + y2 ¡ 4x ¡ 2y ¡ 5 = 0 cuts the

x-axis.

13 Find the points of intersection of the two circles x2+y2 = 1 and x2+y2¡4x¡2y+1 = 0:

1 If f (x) = 1


You may check your answers with a graphics calculator.

a 2f (x) b f (x + 1) ¡ 2

2 Two meshed gears have 24 and 40 teeth respectively. If the speeds (measured in

revolutions per minute) are inversely proportional to the number of teeth, at what

speed should the first gear be driven in order to make the second gear run at 360 rpm?

3 Write this equation of a circle in general form: (x + 4)2 + (y ¡ 2)2 = 16.

4 Write in centre-radius form x2 + y2 ¡ 6x ¡ 4y ¡ 24 = 0:

5 Sketch the graph of x2 + y2 ¡ 10x = 0 on a graphics calculator.

6 Find the points of intersection of the line x + y = 2 and the curve

x2 + xy ¡ y2 = ¡31.

7 Find the equation to a circle concentric with x2 + y2 + 8x ¡ 2y ¡ 15 = 0 and with radius

3 units.

y

x

y

x





V304 FURTHER FUNCTIONS AND RELATIONS (Chapter 9)

THE COST OF RUNNING A CAR

The running cost of a car depends upon many factors such as the distance driven and whether

the car is driven in town or on the highway. In this modelling activity, we will confine our investigation to fuel efficiency, which will be measured in kilometres per litre.

Assume a car is driven 15 000 kilometres per year and that petrol costs $1:00 per litre.

1 a What is the annual fuel cost if your fuel consumption is 10 km per litre?

b What is the annual fuel cost if your fuel consumption is 12 km per litre?

c What is your annual savings with this improved fuel consumption?

2 What is your annual savings if you improve your fuel consumption from 12 km per litre

to 14 km per litre?

3 Develop a general algebraic model for a person who travels 15000 km per year, and

improves the fuel consumption by 2 kilometres per litre. Let x represent the initial fuel

consumption, and y represent the annual savings on fuel costs, in thousands of dollars.

Choose a sensible domain for x.

4 Sketch the graph of your model on a graphics calculator. Comment on the shape of the

graph.

5 What is the predicted savings if you improve your fuel consumption from 6 km per litre

to 8 km per litre?

6 This model gives rise to a rational function. The graphs of rational functions are often very

interesting. For the above function, extend the view window of your calculator to (¡5, 5,

1; ¡60, 40, 10). Discuss what you see.

7 In this modelling activity, we investigated the savings if you were able to improve your fuel

consumption by 2 kilometres per litre. Extend your model to the general case, in which

you are able to improve your fuel consumption by n kilometres per litre.



Rate

SUBJECT MATTER

concept of rates of change

calculation of average rates of change in both practical and purely mathematical situations

interpretation of average rates of change as thegradient of the secant

intuitive understanding of a limit

CHAPTER 10



HISTORICAL NOTE +

A typist is often judged on both typing speed, measured in words per minute, and accuracy, measured

in errors per document. Both batsmen and bowlers in cricket have averages measured in runs per

wicket. Many people measure their pay in dollars per hour. All of these are examples of rates.

A rate is simply a comparison of two quantities of different kinds.

The concept of a rate is fundamental to one of the greatest mathematical discoveries in history,

known as the differential calculus.

Over the next two chapters you will learn why rates are important to mathematics.

RATES A

EXAMPLE 10.1

Zeno was a philosopher who was born about 490 BC. He is famous for four para-doxes, in which he argued that motion is not possible. Here is one of them.

Assume that Archilles and the Tortoise are racing over 100 metres, and the Tortoise is given a 50

metre start. Zeno argued that no matter how fast Archilles runs, he cannever overtake the Tortoise.

He argued as follows: call the point where the Tortoise starts .After a period of time, Archilles reaches the point . But by

this time the Tortoise has already

p p

0

0

moved forward, say to point . By the time Archilles reaches , the Tortoise has moved a bitfurther, to point . By the time Archilles reaches , the Tortoise is at . And so on, and so on.

Each time Archilles reaches a point previously occupied by the Tortoise, the Tortoise has al-ready moved ahead. Hence, Archilles can never catch the Tortoise.

Now we all know that Archilles can catch the Tortoise, so there must be a flaw in Zeno’s rea-soning. Can you figure out what it is?

p p p p p

1 1

2 2 3

306 RATE (Chapter 10)

Stephen typed 240 words in 3 minutes, with 2 errors. Samantha typed 420 words in6 minutes, with 3 errors.

a Who is the faster typist? b Who is more accurate?

a Stephen’s typing speed

= 240 words

3 minutes

= 80 words/min

Samantha’s typing speed

= 420 words

6 minutes

= 70 words/min

) Stephen is the faster typist.



Stephen’ b s error rate was

= 2 errors

240 words

= 1 error per 120 words

while Samantha’s error rate was

= 3 errors

420 words

= 1 error every 140 words

) Samantha is the more accurate typist.

a Using the information from Example 10.1 convert Stephen’s and Samantha’s error

rates to errors/100 words.

b For large areas, the coverage rate of a weed killer is given in kg/ha, while for

suburban blocks it is given in g/m2. Convert 120 kg/ha to g/m2.

c The common speed limit on the highway is 100 km/h. Convert this speed to m/s.

a Stephen’s error rate

= 1 error

120 words

= 0:0083 errors/word

= 0:83 errors/100 words

Similarly Samantha’s error rate

= 1 error

140 words

= 0:0071 errors/word

= 0:71 errors/100 words

b We convert kilograms to grams by multiplying by 1000, and hectares to square

metres by multiplying by 10000.

120 kg/ha

= 120 kg

1 ha

= 120 £ 1000 g

10000 m2

= 12 g/m2

c An alternative method of converting rates is to multiply by factors that equal 1.

Note that the units appear to ‘cancel’ leaving metres over seconds.

100 km

1 h £ 1 h

3600 s £ 1000 m

1 km

= 100 £ 1000

3600 m/s

= 27:8 m/s

or 100 km/h

= 100 km

1 h

= 100 £ 1000 m

3600 sec

= 27:8 m/sec

So, 100 km/h = 27:8 m/sec

EXAMPLE 10.2

RATE (Chapter 10) 307



1 a James worked 18 hours for $189, Jeda worked 22 hours for $247:50, while Trixie earned

$75:12 for 6 hours work. Who had the best hourly wage?

b I drove from Rockhampton to Mackay, a distance of 354 km, in 3 hours and 45 minutes.

What was my average speed?

c The new tyres on my car had a tread depth of 12 mm.

After I drove for 60000 km, the tread had reduced

to 2:5 mm, so I replaced them. What is the rate at

which the tread depth is decreasing? Choose units

for this rate that are easy to understand.

d The distance from the Earth to the Sun is about 150 million kilometres. Assuming that

the path of the earth is a circle, what is the speed of the earth around the sun?

e How earth travel in 10 years?

f Joshua sold 120 punnets of strawberries at $1:50 per

punnet, and the remaining 70 punnets at $1:00 per punnet. What was his average cost per punnet?

g Samantha made 14 typing errors in a 12 page document. If a typical page contains 450words, express Samantha’s error rate in errors/100 words.

h A typical rate for a heartbeat is 60 beats/min. How many times will the heart beat in a

lifetime?

2

a Explain how Claire calculated the overall percentage.

b Explain why her method is not correct.

c Calculate the correct overall percentage.

3

4

5

6 Convert these rates.

a 60 km/hr to m/s b 100 m/s to km/h

c 12 g/m2 to kg/ha d 80 c/kg to dollars/tonne

EXERCISE 10A


Claire received the following marks on her last two assessment items: 30 out of 50 marks,

and 8 out of 10 marks. She converted these to percentages, then averaged the percentages,

and found that she had an overall percentage of 70%:

List 5 rates, other than speed, that you are familiar with from everyday life.

depth of tread

tyre cross-section

A car travelled from Nambour to Gympie at an average speed of km/h. What speed must itaverage on the return journey so the overall average speed is km/h?

4080

The SP factor for sunscreen lotion is based on

the sunscreen being applied mm thick.

What is the coverage rate, measured in

/cm for sunscreen?What surface area could expect tocover with my mL bottle?

q _Qp _p _

mI

2 3

500

a

b

1100

far does the



7 The old imperial method of measuring petrol consumption for cars was miles per gallon (mpg).

The metric measure is litres per 100 kilometres.

Knowing that 1 mile = 1:61 kilometres, and 1 gallon = 4:54 litres, convert these rates

to litres per 100 km: a 30 mpg b 12:7 mpg

to mpg: c 6:5 litres per 100 km d 12:1 litres/100 km

8 a

b

c A geo-stationary satellite circles the earth at exactly the same speed as the earth rotates,

so it remains above the same spot on the earth. All geo-stationary satellites are 24600km above the earth, and the radius of the earth is 6400 km. Express the speed of a

geo-stationary satellite in m/s.

d Typically hair grows about 15 cm per year. Express this in km/h.

9 Research the facts to complete the sentences and then find the answer.

The fastest Melbourne Cup in history was run by ...................... in a time of ................. .

If the length of the race was ................., what was the average speed, in km per hour?

10 The world records in some track events, as of September 1998, are given below. Convert each

record into an average speed, measured in km/h.

Men’s Records

Race Time Runner Country Date City

100 metres 9:84 Donovan Bailey Canada 27 Jul 96 Atlanta

400 metres 43:29 Butch Reynolds United States 17 Aug 88 Zurich800 metres 1:41:11 Wilson Kipketer Denmark 24 Aug 97 Cologne

1500 metres 3:26:00 H. el Guerrouj Morocco 14 Jul 98 Rome

Women’s Records

Race Time Runner Country Date City

100 metres 10:49 F. G. Joyner United States 16 Jul 88 Indianapolis

400 metres 47:60 Marita Koch East Germany 06 Oct 85 Canberra

800 metres 1:53:28 J. Kratochvilova Czechoslavakia 26 Jul 83 Munich

1500 metres 3:50:46 Qu Yunxia China 11 Sep 93 Beijing

11 a The Boston Marathon is thought by many to

be the premiere marathon in the world. The

world record holders at the time of writing are

Cosmas Ndeti from Kenya who ran the race

in 2:07:15 in 1994. The women’s record is

held by Uta Pippig from Germany who had a

time of 2:21:45 in 1994. Calculate the aver-

age speed of each of these runners, given that

the official length of a marathon is 26 miles,

385 yards. (Note: 1 yard = 3 feet. See ques-

tion 4 for other useful conversion factors.)


A typical highway speed in the US is 65 miles/h. Convert this to km/h, given that 1 inchequals 2:54 cm, there are 12 inches in a foot, and 5280 feet in a mile.

The QEII is a large trans-oceanic passenger liner. Its fuel consumption is 42 feet per

gallon of diesel. Convert this into equivalent metric units. One gallon equals 4:54 litres.



AVERAGE RATES OF CHANGE B

b The world record for any marathon was set on September 20 1998 in Berlin by Brazilian

Ronaldo da Costa, running only his second marathon, in a time of 2:06:05. Convert this

speed to kilometres per hour.

c In what time would a marathon have to be run for the average speed to be 20 km/h?

d What would be the average speed required for a marathon runner to break the 2 hour

barrier?

Distances and times for the journey of a

truck from Brisbane to Rockhampton

Town or suburb Distance travelled AEST

Brisbane, Windsor 0 8:00 a.m.

Brisbane, Chermside 6 8:16 a.m.

Brisbane, Gateway Arterial 16 8:36 a.m.

Caboolture 45 8:58 a.m.

Nambour 107 9:56 a.m.

Gympie 176 10:38 a.m.

Childers 326 12:13 a.m.

Gin gin 382 1:53 p.m.

Miriam Vale 478 3:01 p.m.

Gladstone 542 3:40 p.m.

Rockhampton 649 5:05 p.m.

The times and distances for the journey of a truck from Brisbane to Rockhampton is plotted on the

graph above. From the table we can work out the average speed between any two places on the

journey.

time (min)

distance (km)

0

100

200

300

400

500

600

1000 200 300 400 500

In France, long distance trucks are fitted with a tachograph which keeps a record of the distance

travelled at all times on a journey. From this record it may be established when the truck was

stationary, the average speed for different parts of the trip and whether the driver stayed within the

legal speed limits on the trip. This information can be deduced for any journey from the graph of

distance versus time. We shall see later how to deduce this information with the application of the

methods of calculus. Consider a trip from Brisbane to Rockhampton.




a Find the average speed from Brisbane to Caboolture, in

i km per hour ii m/s

b Find the average speed from Miriam Vale to Gladstone in km/h.

c How could you have told from the graph that the average speed from Miriam Vale

to Gladstone was greater than the average speed from Brisbane to Caboolture?

a Speed = Distance

Time (algebraically S =

D

T ),

i Average speed

= 45 km

58 min

= 45 km

5860 hr

= 45 £ 6058

kmh

= 46:6 km/h

ii Average speed

= 45 km

58 min

= 45 £ 1000 m

58 £ 60 s

= 12:9 m/s

b The distance travelled is D = 542 ¡ 478 = 64 km, while the time is 39 min.

Average speed, S = 64 km

39 min

= 64 km

3960 h

= 64 £ 6039

kmh

= 98:5 km/h

c The units on the gradient of a line segment in the graph are km/min, (which could be

changed to km/h). Hence, the gradient of a line segment represents the average speed

during that part of the journey. The greater the gradient (i.e., the steeper the line) the

greater the speed. The graph is steeper between Miriam Vale and Gladstone, so the

average speed is greater.

In everyday usage, speed and velocity have the same meaning, namely the rate at which distance is

changing over time. However, often it is important to know not just how fast an object is travelling,

but in what direction.

A cannonball coming towards you at 50 m/s is a different proposition to one going away from you

at the same speed. Similarly it can be important to know not only how far an object is, but in what

direction. A competitor in a race who is 2 metres in front is not the same as a competitor who is

2 metres behind.

EXAMPLE 10.3

DISTANCE, DISPLACEMENT, SPEED AND VELOCITY

C




a How far is the corner store from my house?

b What is the gradient of the line segment for the first 4 minutes of the walk?

c What was my average walking speed during the first 4 minutes, in m/min?

d What is the physical representation of the gradient in this problem?

e What was my average walking speed during the first 4 minutes, in km/h?

f Make a guess as to what I was doing 6 minutes after I left home.

g How many minutes did I spend in the corner store?

h What was my average speed on the return journey, in km/h? i What was my average velocity on the return journey, in km/h?

j What was the total distance that I travelled?

k

a 800 metres b gradient

= 500 m

4 min

= 125 m/min

c 125 m/min

d The gradient represents the speed of the walker.

e average speed

= 500 m

4 min

= 500 ¥ 1000 km

4 ¥ 60 h

= 7:5 km/h

f Stopped, and chatted to a neighbour.

g 8 min

EXAMPLE 10.4

4

1

5

7

2

6

8

3

0164 208 2412 28 32

1 0 0

s o

f m

e t r e s

minutes

I walk to the corner store to get the pa-

per each morning. The graph alongside,

called a travel graph, shows my distance

from home during my walk. Use the travel

graph to answer the questions.

What was my displacement?


If direction is not important then we use the terms distance and speed.

If direction is important we use the terms displacement and velocity. Displacement and velocity

can be both positive and negative.

Travelling away from home, both displacement and velocity are usually considered to be positive.

Travelling towards home, displacement is still positive, but decreasing, while velocity is negative.




h average speed

= 800 m

12 min

= 800 ¥ 1000 km

12 ¥ 60 h

= 8001000

£ 6012

kmh

= 4 km/h

i ¡4 km/h (towards home, so the direction is negative)

j 1600 metres (800 m to the store, then 800 m home)

k 800 m + ¡800 m =

The CBR and Data Logger are two devices that attach to a graphics calculator,

The data are stored in lists, and displayed graphically as a travel graph.

Using one of these peripherals, try to create each of the travel graphs given below.

1 From the table on page 310, determine the average speed for the section from:

a Nambour to Gympie b Gladstone to Rockhampton

2 The table shows the population of Queensland

and NSW taken from various census figures

over the last 17 years. The data is from the

Queensland Year Book and Census Charac-

teristics of Australia.

Calculate the annual rate of growth of both

Queensland’s population and NSW’s popula-

tion, over the periods:

Queensland New South Wales

Year Population Year Population

1982 2 425 000 1981 5 125 0001986 2 624 000 1986 5 402 0001991 2 830 000 1991 5 732 0001996 3 369 000 1996 6 039 000

a 1991-1996

b 1982-1996 (for NSW the first time period will be from 1981-1986)

EXERCISE 10C

and measure distance and time.

a displacement of 0 metres




Year Population Year Population

1890 2447957 1962 2 8111711910 2 888110 1968 27900011926 2 8 71 429 1975 2 2 96 9451936 2 829746 1982 2 1889181946 2 725374 1990 21524231954 2850189

4 Since the beginning of March, a local reservoir has been losing water at a constant rate. The

Water Resource Department estimates that on 12th March, the reservoir held 200 million kL

of water, and on the 21st March, it held 164 million kL.

a What is the average rate of water loss, measured in million kL per day?

b Hence determine a linear equation relating Q, the volume of water remaining (in million

kL), and n, the number of days since the beginning of March (when the reservoir was

full).

c What is the maximum capacity of the reservoir?

5 A tanker hit a reef in the Barrier Reef, and an

oil slick appeared. Some days after the accident,

it was reported that the slick was roughly rectan-

gular in shape and covered an area 20 km long

and 11 km wide. Four days later it was estimated

that the slick had increased to 50 km long by 15km wide.

What was the average rate of change of the area

of the slick with respect to time over this period?

6 In a dry climate, the rate of evaporation is a veryimportant statistic. The graph shows the depth

of water in a shallow dish during one hot day in

Mackay, over a 24 hour period.

Use the graph to answer the following questions.

Show how you arrived at your answer.

a What was the average rate of evaporation

between 9 a.m. and 3 p.m.?

b Use the graph to estimate the rate of evapo-

ration at 2 p.m.

7

Write a story for this travel graph. Include nu-

merical details such as the speed for various

stages.

8

2

10

14

4

12

6

03 a.m. 6 a.m. 9 a.m. noon 3 p.m. 6 p.m. 9 p.m. midmid.

mm

20

5

25

35

10

30

40

15

0

41 52 63 7 8

k m

h


3 This table gives the City of Paris population

from 1890 to 1990.

a Find the first period in which Paris had

a negative growth in population.

b Find the population growth for the period

1890 - 1990.



12 The graph alongside shows the velocity of a car

over an 8 minute period. Note that a change in

velocity over time is called acceleration.

a Describe what the car was doing in the first

minute.

b What was the average velocity during this

time? c Describe the motion of the car between the

1st minute and the 5th minute.

d Describe what may have happened between

the 5th minute and the 8th minute.

e What was the average acceleration between the 1st and 5th minute?

f What was the total distance travelled between the 1st and 5th minute?

g What was the total distance travelled between the 7th and 8th minute?

13 Draw a velocity-time graph for each of these stories.

a John was walking to the bus stop when he saw the bus come around the corner. He ran tocatch the bus, but he was not even close. As the bus pulled away, he stopped and caught

his breath. He then continued walking to the bus stop to wait for the next bus.

b A car is waiting at a red light. The light turns green, and the car accelerates smoothly

until it reaches the speed limit of 60 km/h.

c A boat is at a mooring, moving up and down as the tide first rises, and then falls. For this

exercise, we are interested in the distance the boat moves vertically, and not horizontally.

14

15 For each travel graph given below, draw the corresponding speed vs time graph.

a b

20

5

25

35

10

30

40

15

041 52 63 7 8

k m

/ h

min

4

1

5

7

2

6

8

3

0164 208 2412 28 32

1 0 0

s o

f m

e t r e s

minutes

I

.

walk to the corner store to get the paper eachmorning. This travel graph shows my distancefrom home during my walk. Draw the speed-time

graph for this journey

4

1

5

7

2

6

8

3

041 52 63 7 8

m

sec

4

1

5

7

2

6

8

3

041 52 63 7 8

m

sec




VARIABLE RATES OF CHANGE D

80

20

100

40

60

041 52 63 7 8

m

sec

80

20

100

40

60

041 52 63 7 8

m

sec

EXAMPLE 10.5


The previous examples either all exhibited a

steady rate of change over a given length of

time, or we studied the average rate of change

over a particular time period. More interestingare situations where the rate of change varies

continuously over time. Consider the graph

shown of a car accelerating away from an in-

tersection.

Note that the average speed of the car over the first 8 seconds is 12:5 m/s, since the car has gone

100 metres in 8 seconds. At first the car is travelling much slower than that, and after 8 seconds it

is going considerably faster. How can we determine the speed of the car, say after 4 seconds? The problem is that the graph is curved. How do we find the gradient of a curved line?

By convention, mathematicians draw the tan-

gent line to the curve at x = 4, and say that

the gradient of the curve at that point is equal

to the gradient of the tangent line. The tangent

line passes through the points with coordinates

(7, 40) and (2, 0), so the gradient of the

tangent line is 40 ¡ 0

7 ¡ 2 = 8 m/s at time t = 4 s.

The graph given shows the height of a

hot air balloon (in metres) for the first

50 minutes of its flight.

a At what speed was the bal-

loon rising initially?

b When was the balloon neither rising nor falling?

c What was the greatest rate of

descent, and when did this

occur?

d How fast was the balloon ris-

ing in the 50th minute?

e What was the average rate at which the balloon rose during the first minutes?15

0

200

400

600

800

1000

1200

1400

1600

0 10 20 30 40 50

height (m)

time (min)




a

b

c

m = (730 ¡ 1400)(35 ¡ 15)

= ¡33.5 m/min

d At t = 50, the rate of rise is given by

the slope of the tangent at (50, 1500).

The tangent passes through (40, 0) as

well.

) rate = 1500 ¡ 0

50 ¡ 40 = 150 m/min

e At t = 15, height = 1240 m

) average rate of rise

= change in position

change in time

= 1240 ¡ 0

15 ¡ 0

+ 83 m/min

Consider the graph of the function y = x2

¡4x + 7, together with the tangent line at (3, 4)

shown in Figure 1.

In Figure 2, we have zoomed in on the graph near the point (3, 4). The curved graph appears to

be fairly straight in the neighbourhood of this point.

0

200

400

600

800

1000

1200

1400

1600

0 10 20 30 40 50

height (m)

time (min)16 37

The tangent line passes throughand ), so the gradient

m/min.

( , )( ,

0 05

=900

180m

The greatest rate of descent occurredaround the th minute, as that iswhen the graph has the greatest nega-tive gradient. The tangent passesthrough the points and

so the gradient

26

15 140035 730

( , )( , )

A

,,

,

graph is neither rising nor fallingwhen the gradient of the tangent lineis i.e., the tangent line is horizontal.This occurs at time and also attime i.e., the balloon was nei-ther rising nor falling at the th and

th minutes.

0=

=t

t16

3716

37

0

200

400

600

8001000

1200

1400

1600

0 10 20 30 40 50

height (m)

time (min)

(15' 1400)

(35' 730)

0

200

400

600

800

10001200

1400

1600

0 10 20 30 40 50

height (m)

time (min)

(15' 1240) (50' 1500)

(40' 0)

Figure 1 Figure 3 Figure 3




In Figure 3 we have zoomed in even further, and the graph appears to be perfectly straight. The

more we zoom in, the straighter the line appears. Very close to the point, the curved graph and

the tangent line appear to be identical. Using the gradient of the tangent line at a point to find the

gradient of the function at that point is reasonable.

Drawing an accurate tangent line can be difficult. If five students were asked to draw such a line,

not every line would be the same. We need an analytical method of finding the gradient of a

function at a given point.

Consider again the function

y = x2 ¡ 4x + 7. We will

draw a secant at this point,

from (3, 4) to (4, 7). A secant

is a line that joins two points

on the graph of a function. The

gradient of this secant can be

calculated to be m = 3. The

secant does not have the same

gradient as the tangent line, but

it is reasonably close.

We can do even better by choosing a point that lies on the graph that is closer to (3, 4), say

(3:5, 5:25). The gradient of this secant is m = 1:5. The table below shows the gradient of the

secant as the second point gets closer and closer to the point (3, 4).

x-coord y-coord gradient of secant

4 7 3

3:5 5:25 2:5

3:2 4:44 2:2

3:1 4:21 2:1

3:01 4:0201 2:01

3:001 4:002 001 2:001

3:0001 4:0002 2:0001

It appears that as the secant line gets shorter, the

gradient of the secant line approaches the value

m = 2

. We might guess that the gradient of the

tangent is exactly m = 2. This turns out to be

correct. You will learn how to calculate this in the

next chapter.

Since the secant gets closer to the tangent as points

that are closer and closer together are chosen,

mathematicians say, “The tangent is the limit of

the secant”.

Use the above method to determine the gradient of y =

x2

¡ 4x

+ 7 at the point (

0, 7

).

We make a table showing the gradient of

the secant for points closer and closer to

the point (0, 7). Note that a spreadsheet

or lists on a graphics calculator are par-

ticularly useful here. From the table, the

gradient of the tangent when x = 0 ap-

pears to be m = ¡4.

x-coord y-coord gradient of secant

1 4 ¡3

0:5 5:25 ¡3:5

0:2 6:24 ¡3:8

0:1 6:61 ¡3:9

0:01 6:9601 ¡3:99

0:001 6:996001 ¡3:999

0:0001 6:9996 ¡3:9999

function

secant

tangent

EXAMPLE 10.6



2

a

b

c

d

e

3

a

b

c

d

e

4

A safe is dropped from the top floor

of the MLC building, with the in-

tent of cracking it open. The travelgraph for the safe is given below,

with the height measured in metres

and the time in seconds. Some use-

ful tangent lines have been added.

What is the speed of the safewhen it is initially dropped?

How fast is the safe travellingafter seconds?

When does the safe hit theground?

Where does the safe hit theground?

How fast is the safe moving when it hits the ground?

There is water in a water tank, but

unfortunately the tank is leaking

quite badly. The amount of water

left in the tank (measured in thou-

sands of litres) after hours is

given in the graph alongside.

How much water was in thetank originally?

How much water was in thetank after hour?

How quickly was the tank los-ing water initially?

How quickly was the tank los-ing water after hour?

How quickly was the tank losing water after hours?

Explain in your own words the meaning of the phrase, “The tangent is the limit of the secant”.

2

1

1

3

x


1 On graph paper, accurately sketch each of the functions given below.

Draw a tangent line to the function at the given point.

Find the gradient of the tangent line, and hence the gradient of the function at that point.

Compare your answer to that of your classmates.

Did everyone get the same answer?

Function Coordinates

a f (x) = x2 ¡ 4 (¡1, ¡3)

b g(x) = x2 + x + 1 (2, 7)

c h(x) = x3 (1, 1)

d i(x) = x3 ¡ 2x2 + x ¡ 1 (2, 1)

EXERCISE 10D

y

x

h e

i g h t ( m e

t r e s )

time (sec.)




5 For each function in question 1, make a table of secants, and hence determine the gradient of

the function at that point. Compare these answers to those you found by drawing the tangent

line.

6 Consider the function y = x2. Find the gradient of the function at each of these points:

a (0, 0) b (1, 1) c (2, 4) d (3, 9) e (4, 16)

Can you guess what the rule for finding the gradient at any point might be?

7 Consider the function y = x3. Find the gradient of the function at each of these points:

a (¡2, ¡8) b (¡1, ¡1) c (0, 0) d (1, 1) e (2, 8)


In this last section, we developed a method of finding the gradient of a function at a given point,

or, more accurately, a method that allows us to make a good guess about the value of the gradientat the point. We have not yet shown that these guesses are correct.

In question 6 in the above exercise set, you were asked to use these guesses to surmise what the

gradient function of y = x2 might be. A good guess is m = 2x, since the gradient appears

to be twice the x-coordinate, at every point.

From question 7 , you may have guessed that the gradient function of y = x3 is m = 3x2.

But again, none of this has been proven.

For these simple functions, it is possible to guess the corresponding gradient function. But guessing

the gradient function for y =p

2x + 1 is almost impossible. A more powerful tool is needed to

solve such problems. That tool is differential calculus, which is introduced in the next chapter.

1 Water is flowing into each of the three containers below at a constant rate.

For each container, sketch the graph of the height of the water with respect to time.

a b c

2 The following are four travel graphs of a car. For each graph, describe the motion of the car.

a b c d

FINDING THE GRADIENT FUNCTION E

PROBLEM SOLVING F

y

x

d i

s t

a n c e

time

y

x

d i

s t

a n c e

time

y

x

d i

s t

a n c e

time

y

x

d i

s t

a n c e

time

EXERCISE 10F





3 The graph alongside is is a velocity vs

time graph. Draw the related travel graph,

and hence determine the total distance the

object travelled.

4 Find the gradient function for the function f (x) = 1

x3. Justify your decision.

80

20

100

40

60

041 52 63 7 8

m/sec

sec


acceleration rate tangent displacement secant

velocity gradient

1

2

3

a

b

c

d

Water is flowing into a circular swimming pool at the rate of litres/minute. The pool has a diameter of metres. What is the rate

at which the height of the water in the pool is increasing?The cubit was a measure of length used in ancient Egypt. The Royal Cubit, a piece of black marble about cm long using today’s measurement, was the standard measure. If amessenger can run one kilometre in five minutes, convert that to cubits per day.

The graph alongside refers to awater tank in the shape of acuboid with a base of area

cm which is filled bytwo taps which can be turned onor off independently. Tap A runs

by itself for a short time and

then tap B is turned on also and both taps fill the tank until theyare both turned off simultane-ously. A short time later a con-crete block falls into the tank and subsequently the water isdrained out to recover the block.

At what rate in litres per minute is the tank initially filled by tap A?

At what rate in litres per minute is the tank filled by tap B?

What is the volume of the concrete block?

What is the rate, in litres per minute, at which water drains out of the tank?

15 4

52

10 000 2

height of water (cm)

time (min)

2 4 6 8 10 12 14 160

4

8

12

16

20

24

28





4 This is a distance vs time graph for a Year 8 girl

running the 100 metre sprint.

a What was her time for the race?

b How many seconds into the race did she

reach her greatest speed?

c From the graph, estimate her greatest speed.d Estimate her speed after 4 seconds.

e During which parts of the race

i was she accelerating

ii was her speed almost constant

iii was she decelerating?

5

6 In Western Queensland the highway

follows the railway. At 12 noon, the

Express train (The Midlander) passes

through a town at 90 km/hr without

stopping.

At the same time a car, which had

stopped to fill up with petrol, leaves

the town.

At what time will the car overtake the

train?

7 Make a table of secants to determine a possible value for the gradient of

y = 3x2 ¡ x + 1 at the point (¡2, 15).

00

20

40

60

80

100

120

2 4 6 8 10 12 14 16

distance (m)

time (s)


1 A jet plane can fly at a speed of 950 km/h. Convert this to metres per second.

2 I drove to my Auntie’s place last Sunday. Here is a travel graph of my journey to her house.

a What was my average velocity for the

first half hour?

b What was my

i speed

ii velocity in the second half hour?

c What was the

i distance travelled

ii displacement in the first hour?

80

20

100

140

40

120

160

60

021 3 4

km

h

Sketch velocity time graph for this story The metre sprinter came out of the blocksfast, and maintained that pace for metres. She then reduced her speed by ten percent,and maintained that constant pace until the metre mark. She then sprinted home in thelast metres.

a . 800200

700100

12:00 12:10 12:20 12:30 12:40 12:50 1:00 pm

25

50

75

100

time

velocity (km/h)car (100 km/h)

train (90 km/h)





EXTENDED INVESTIGATION

LOCAL LINEARITY


d What was my average speed for

e What was my fastest average speed for any section of the journey?

f Draw a velocity time graph for this trip.-

3 Consider the function y = 1

x

.

the entire trip?

By considering the gradient of the secant as the secant approaches the tangent, find a possible value for the gradient of the function at the point .(2, 1

2 )

Earlier in this chapter, we zoomed in on the graph of y = x2

¡4x + 7 near the point (3, 4),

and saw that the graph looked more and more like a straight line as we zoomed in. This feature

of a graph is called local linearity, which means that near any point on this graph, if we zoom

in sufficiently close, the graph of the function will appear to be linear.

1 On your graphics calculator, graph Y 1 = 3x2.

2 Choose any point on the curve, and zoom in on that point until the graph looks linear.

3 Turn on TRACE to find the coordinates of a point near the point you have chosen.

4 Use m = y2 ¡ y1

x2 ¡ x1to find the gradient of a secant near your chosen point.

What do you think the gradient of the tangent at your chosen point might be?

5 Repeat for at least four other points on the curve.

6 Make a conjecture about the function that gives the gradient of Y 1 = 3x2 at any point

on the curve.

7 Test your conjecture on at least one other point on the curve.

Repeat this process for at least two other functions of your choosing.

Can you find a function that does not exhibit local linearity at all points on its graph?

In the next chapter you will learn how to justify the conjectures you have made in thisinvestigation.

,



Instantaneousrates of change– differential

calculus

SUBJECT MATTER

interpretation of the average rate of change as thegradient of the secant

intuitive understanding of a limit

definition of the derivative of a function at a point

derivative of simple algebraic functions from first principles

rules for differentiation

evaluation of the derivative of a function at a point

interpretation of the derivative as an instantaneousrate of change

interpretation of the derivative as the gradientfunction

CHAPTER 11



HISTORICAL NOTE +

In this period scientists were interested in four

major problems. The first was the calculation

of the orbits of celestial bodies, and in particular

finding the velocity and acceleration knowing the

displacement. Also during this period the science

of optics was being developed, largely for the

construction of telescopes. Students of physics

knew that materials have a refractive index. Tocalculate the refractive index scientists needed to

calculate the angle between the ray of light and

the normal. The normal is a line perpendicular to

the tangent; hence there was interest in tangents

and normals.

THE IMPORTANCE OF CALCULUS TODAY

There is a set of rules routinely used by many engineers and scientists in their daily work. They un-

derpin the design and manufacture of motor vehicles, aircraft, the building of bridges and skyscrapers

and many other products we take for granted in everyday life. They were developed by Sir Isaac

Newton in the seventeenth century, and are called Newton’s Laws of Motion. Newton developed

them to understand Kepler’s laws of planetary motion. His genius was to realise that these laws

governed not only the motion (i.e., displacement, velocity and acceleration) of the solar system, but

that they also described the motion we observe in our daily lives. This is why they are regarded as

fundamental in applied mathematics.

Differential and Integral Calculus are such powerful mathematical tools that it isimpossible to imagine a modern world in which calculus does not exist. They haveapplications in science, engineering, economics and finance, to name some areas.

was developed inthe period to as ascientific tool, primarily by

and.

Calculus

Isaac Newton Gottfried

Leibnitz

1650 1760

Isaac Newton Gottfried Leibnitz.

This period also saw a rapid development in the instruments of war. Calculations of maximumrange of a cannon or the maximum height of a cannonball were problems to which calculuscould be applied. The other major application of calculus in this era was the calculation of arclengths, volumes and surface area, all of which yield to the power of . Youwill study this topic in Year .

Integral Calculus

12

tangent

normal

ray

326 INSTANTANEOUS RATES OF CHANGE – DIFFERENTIAL CALCULUS (Chapter 11)



CLASS DISCUSSION

INSTANTANEOUS VELOCITY A

0

A

1

B

2

C

3

D

4

E

5

F

6

G

7

H

8

I

9

J

10

K

INSTANTANEOUS RATES OF CHANGE – DIFFERENTIAL CALCULUS (Chapter 11) 327

Newton’s second law relates the rate of change of momentum and the forces present. The momen-

tum of a moving object is obtained by multiplying mass and velocity; thus heavy objects can have

a lot of momentum at slow velocities whereas a light object has to move much faster to have the

same amount of momentum.

The key notion of this chapter is that of “instantaneous rate of change”. If Newton’s second law is

to be made precise we have to define what we mean in mathematical terms.

Consider the following situation:

A grooved piece of timber is 10 metres long. It has very fine lines drawn across at a spacing of

exactly 1 metre. A marble is released at A.

The timber is inclined and made of a material such that the distance of the marble from A at any

time is equal to the square of the time for which it has been rolling.

If s equals the displacement of the marble, measured from A in metres, and t equals the time the

marble has been rolling, measured in seconds, then s = t2.

We can show its motion in the following table: t 0 1 2 3 .......

s 0 1 4 9 .......

We can calculate the average velocity from A to B as follows:

average velocity = distance travelled

time taken

= 1 metre

1 second

= 1 m/s

What would the speedometer read at the instant that it passes over one of the lines, say the line at

E ? If we assume that the width of the line is zero and hence the time taken to cross the line is 0,

then when we put these numbers into our above formula we get average velocity = 0

0 .

Now assume this marble had speedometer attached which we could read.What would be the speed at the point halfway between and B? What would bethe speed just to the left of this midpoint? Just to the right? What is the speed at B?

aA




The expression 00 has no meaning in the real number system, so we need to try another approach.

One way to get some idea of the instantaneous velocity at E is by looking at what happens just

after the marble crosses E .

Consider where the marble is after 2 seconds, and after 2.1 seconds.

After 2 seconds the marble is at point E, with displacement 4 metres. After 2.1 seconds the marble

is at P, with displacement of 4:41 metres, found by squaring 2:1. Therefore,

average velocity = (4:41 ¡ 4) m

(2:1 ¡ 2) s

= 0:41 m0:1 s

= 4:1 m/s

Note that this is the average velocity near E , and not necessarily the speed the marble is travelling

when it crosses E. We can zoom in even closer to E . Consider where the marble is at 2 seconds

and at 2.01 seconds.

Calculating the average velocity:

average velocity = (4:0401 ¡ 4) m

(2:01 ¡ 2) s

= 0:0401 m

0:01 s

= 4:01 m/s

It appears that as the time interval shrinks to 0, the average velocity approaches 4 m/s. This still is

not proven. It is only a guess at this stage.

time

displacement

0

0

1

1

2

2

3

3

4

4

5

5 metres

seconds2.1

4.41

E P

time

displacement

0

0

1

1

2

2

3

3

4

4

5

5 metres

seconds2.01

4.0401

E Q

INVESTIGATION 1

Zoom in closer and closer to the point C, using times of sec, and so on.What can you conclude? Are you absolutely sure?

2.001

CLOSER APPROXIMATIONS




1 Consider the velocity of the marble rolling down the incline when it is at J, i.e., after 3 seconds.

By considering the average velocity for small time intervals just after 3 seconds, determine a

likely value of the instantaneous velocity of the marble at J.

2 Assume that the ramp is made steeper, so the displacement s at time t is given by s = 2t2

.a Where is the marble after 2 seconds?

b What is a likely value of the instantaneous velocity at this time?

3 If the displacement s at time t was given by s = 3t2, what is a likely value of its instanta-

neous velocity after 1 second?

4 If the displacement of the marble s at a time t was given by s = 12 t2,

a where would it be after 4 seconds?

b Determine a likely value for the instantaneous velocity at this time.

5 Assume that an additional force was acting on the marble, so its displacement was given bys = t3. Determine a likely value for the instantaneous velocity at these times:

a 1 sec b 1 12 sec c 2 sec

6 Assume that the force on the marble was such that the displacement was given by s =p

t.

a Determine a likely value for the instantaneous velocity after 2 seconds.

b How accurate might this value be?

SMALL CHANGES

What we are doing is considering small changes in time and small changes in displacement.

Note that ¢t does not mean ¢ £ t.

¢t is a single variable indicating a small change in the variable t.

What we have been calculating above is the average speed, which is equal to ¢s

¢t.

Then we considered the process of letting both ¢t and ¢s get smaller and smaller.

We found that as ¢t approaches 0, ¢s

¢t approaches the instantaneous velocity v .

Mathematicians say that “v equals the limit as delta t approaches zero of delta s on delta t”,

and write v = lim¢t!0

¢s

¢t.

Before we can proceed with the formal study of calculus we need to understand what is meant by

a limit.

EXERCISE 11A

Mathematicians have neat way of indicating small changes in variable. They write

(pronounced “delta ”) indicating small change in time and indicating small change in

displacement.

a a

a a

¢¢

t

t s




Consider the function f (x) = 3x + 1. What is limx!2

3x + 1?

In other words, what value does 3x + 1 approach,

as x approaches 2?

We can make a table, and see what happens.

x 1:9 1:99 1:999 2 2:001 2:1 2:01

f (x) 6:7 6:97 6:997 ::: 7:003 7:3 7:03

From both below and above, f (x) is approaching 7.

This is hardly surprising, since f (2) = 3 £ 2 + 1 = 7.

Many functions in this course have a graph which is a continuous curve, and for such functions:

If f (a) exists, then limx!a

f (x) = f (a):

In words, for functions for which f (a) exists, the limit of the function as x approaches a can be

found by substituting a for x, and then evaluating.

Now consider the function f (x) = x2 ¡ 4

x ¡ 2 at x = 2.

Substituting gives us f (2) = 00 :

There is no real number which equals 00 . Mathematicians

say that the function is not defined at x = 2, so we cannotfind the limit by substituting and evaluating.

Whilst we cannot evaluate this function at this point, we

can still find the limit of f (x) at x = 2 if we let x approach

2 from both above and below:

x 1 1:5 1:9 1:99 2 2:01 2:1 2:5

x2 ¡ 4

x ¡ 2 3 3:5 3:9 3:99 ? 4:01 4:1 4:5

Except for x = 2, this looks remarkably like the function f (x) = x + 2. In fact, if we factorise thenumerator and simplify, we have the following:

f (x) = x2 ¡ 4

x ¡ 2

= (x + 2)(x ¡ 2)

x ¡ 2

= x + 2 with the domain x 6= 2.

So, the function f (x) = x2 ¡ 4

x

¡2

is identical to the function f (x) = x + 2 except at the point

where x = 2.

THE CONCEPT OF A LIMIT B

2

4

6

8 y

x

2 4 6

y x 3 1

2

2

4

4

2

2

4

4 x

y

2

42

x

x y




We can substitute 2 for x into x + 2, and hence find that the value of f (x) approaches 4 as x

approaches 2.

We can write limx!2

x2 ¡ 4

x ¡ 2 = 4.

RULES FOR MANIPULATING LIMITS

Consider limx!3

x2 + 3x. We can find the limit by substituting 3 for x:

limx!3

x2 + 3x

= 32 + 3

£3

= 18:

Consider limx!3

x2 + limx!3

3x: We can find the value of this expression by substituting:

limx!3

x2 + limx!3

3x

= 32 + 3 £ 3

= 18:

What we have just shown is a special case of:

limx!

a[f (x) + g(x)] = lim

x!

af (x) + lim

x!

ag(x)

When we have a complicated function and we want to investigate its limit it is a good idea to break

it down into a number of simpler functions and find their limits separately. Then we combine the

separate answers. The rules for working with limits are intuitively obvious. Their proofs can be

found on the accompanying website.

Here are the Limit Laws that we will use in this course:

² limx!a

[f (x) + g(x)] = limx!a

f (x) + limx!a

g(x)

² limx

!a

[f (x) ¡ g(x)] = limx

!a

f (x) ¡ limx

!a

g(x)


1 On your graphics calculator, draw the graph of y = x2 ¡ 4

x ¡ 2 :

Zoom in on the point where x = 2. What do you expect to find? What do you find?

2 You can set a “friendly” window on any graphics calculator by carefully choosing your window

size. In a friendly window, when the trace is turned on, the x-coordinates have decimal values.

Some calculators have the friendly window as a menu choice, such as (Zoom, Decimal) or

(Zoom, Integer). Does the graph change when you set a friendly window?

3 Use the table command on your graphics calculator to compare the functions

y = x2 ¡ 4

x ¡ 2 and y = x + 2 for x starting at 0 with an increment of 1. Interpret the result.

What to do:




² limx!a

[f (x) £ g(x)] = limx!a

f (x) £ limx!a

g(x)

² limx!a

·f (x)

g(x)

¸ =

limx!a

f (x)

limx!a

g(x), if g(a) 6= 0

These rules say that if you add two functions you can add their limits. They also say that you cansubtract and multiply them and you can divide them provided the function in the denominator is

not zero at x = a.

We will verify the first Limit Law. Let f (x) = x2, g(x) = 3x and a = 2.

limx!a

[f (x) + g(x)]

= limx!2

(x2 + 3x)

= 22 + 3 £ 2

= 10:

limx!a

f (x) + limx!a

g(x)

= limx!2

x2 + limx!2

3x

= 22 + 3 £ 2

= 10:

Verifying the other Limit Laws will be left as an exercise.

Find a limx!1

3x ¡ 8

x + 5 b lim

x!13x2 ¡ 8

x + 5

As infinity is not a number, we cannot substitute 1 for x. Instead we will determine the value

this expression approaches as x becomes very large.

a Let Y 1 = 3x ¡ 8

x + 5 , and make a table in your graphics calculator that starts atx = 0, with

an increment of x = 1000. The value of Y 1 approaches 3, and hence limx!1

3x ¡ 8

x + 5 = 3:

This could have been determined easily without using a table. As x becomes very large,

the terms ¡8 in the numerator and +5 in the denominator become insignificant compared

to the terms 3x and x.

Hence the expression approaches the expression 3x

x , which simplifies to 3.


A VERY IMPORTANT LIMIT Set your calculator to radians. Consider the function f (x) =

sin x

x .

1 Evaluate this function at x = 0. Explain the result.

2 Graph the function. What happens near x = 0?

3 Make a table of values, as x approaches 0 from both the positive and negative sides.

4 What is the value of limx!0

sin x

x ?

What to do:

EXAMPLE 11.1




b Here, as x becomes very large, the expression approaches 3x2

x , which simplifies to 3x.

And as x ! 1, 3x ! 1, so limx!1

3x2 ¡ 8

x + 5 = 1

1 Evaluate

a limx!3

x2 + 2x + 5 b limx!1

x + 4

x + 1 c lim

x!¡25

d limx!2

x ¡ 2

x + 2 e lim

x!¡22x3 ¡ 4x2 + x ¡ 3 f lim

x!1

2x + 1

x3 ¡ 3x + 1

2 On page 332, the identity limx!a

[f (x) + g(x)] = limx!a

f (x) + limx!a

g(x)] was verified using

f (x) = x2, g(x) = 3x and a = 2.

Using these functions, verify the remaining Limit Laws.

3 By first simplifying each of the rational expressions, evaluate each of the following limits.

Use a graphics calculator to check your answers.

a limx!1

x ¡ 1

x2 ¡ 1 b lim

x!2

x2 + 3x ¡ 10

x + 5 c lim

x!3

x2 ¡ 6x + 9

x ¡ 3

d limx!¡2

x4 ¡ 16

x2 ¡ 4 e lim

x!3

x2 ¡ 2x ¡ 3

x ¡ 3 f lim

x!2

6x2 + x ¡ 12

3x ¡ 4

4 Evaluate

a limx!1

5x + 1x ¡ 2

b limx!1

6 ¡ 2x

x ¡ 2 c lim

x!1x + 19 ¡ x

d limx!1

7x + 34x + 1

e limx!1

x2 + 1

x ¡ 6 f lim

x!1x3 + 1

x2 ¡ 2 g lim

x!12x ¡ 3

4x2 ¡ 4 h lim

x!15x2 + 1

x5 ¡ 2

5 By expanding and then simplifying, find lima!0

(x + a)3 ¡ x3

a .

6 Suppose I have a ruler of length 1 metre. I cut it in half and put one half aside. I take the

other piece and cut it in half again. I put one of these two pieces with the piece already aside.

a What is the total length of the two pieces I have put aside?

b I continue, cutting the piece I have left into two more equal length pieces and adding one

of them to the pieces I have set aside. Write down the total length of pieces set aside as

a sum of the individual lengths, putting the lengths in a decreasing order of size, after 5cuts have been made.

c Now consider this process repeated many times, a number n times say. Write down the

total length of n pieces, using the case of n = 5 to see the pattern.

d Finally if this process is continued indefinitely, what is the value that the sum of the

lengths of the pieces approaches as we continue?

e Is it possible to add an infinite number of terms, and get an answer that is a real number?

EXERCISE 11B




7 The value of ¼ can be calculated by ¼

4 =

µ1 ¡ 1

3 + 15 ¡ 1

7 + :::: + (¡1)n

2n + 1 + ::::

¶If you approximate the value of ¼ by calculating the sum of the first n terms in the series

above, the error will not be greater than the (n + 1)th term, i.e., the term following the last

one used in the sum.

For example, if you sum the first 3 terms then the error in using the sum of these first three

terms will be less than the next term, 17 .

Calculate this sum for n = 10 and n = 20. Verify that the approximation to the value of ¼

obtained has an error consistent with this result.

8 It can be shown that ¼4 = 4f ( 1

5 ) ¡ f ( 1239 ) where f (x) = x ¡ x3

3 +

x5

5 ¡ x7

7 + ::::::

Calculate ¼ from the formula above using f (x) to the term in x7.

Compare your value of ¼ with that found in question 7 .

You have seen in the previous chapter that the gradient of a curve at a given point is defined to be

the gradient of the tangent at the point of contact. The gradient of a straight line is constant as we

move along the line. However, the gradient of a curve changes from point to point along the curve.

Consider the graph of y = x2.

To the left of the origin the graph of the function

y = x2 has a negative gradient.

To the right of the origin the gradient of the curveis positive.

At the origin the curve “sits on its tangent” so

the gradient at the origin is 0.

The problem of calculating the gradient of a curve is related to the problem of calculating velocities.

Consider the curve y = x2. We wish to find

the gradient of the curve at the point (2, 4).

HOW TO CALCULATE THE GRADIENT OF A CURVE

THE GRADIENT OF A CURVE C

1 2

x

y

2

1

3

4

12

P (2, 4)

x

y

4

2




In the last chapter we looked at what

happens as the secant approaches

the tangent, and we found what we

thought to be the correct answer. We

can now use limits to prove that our

answer is indeed correct.

Consider a secant joining P and Q.Let Q be a point very close to P -

the technical term is a neighbouring

point.

Let MN = ¢x, which represents a small change in the horizontal direction, from P to Q.

The x-coordinate of P is 2, so the x-coordinate of Q is 2 + ¢x.To find the y-coordinate of Q we substitute 2 + ¢x in place of x in the equation y = x2.

The y-coordinate of Q is (2 + ¢x)2.

Therefore, the gradient of the secant is

m = ¢y

¢x

= y2 ¡ y1

x2 ¡ x1

= (2 + ¢x)2 ¡ 4

2 + ¢x ¡ 2

= 4 + 4¢x + (¢x)2 ¡ 4

¢x

= ¢x(4 + ¢x)

¢x

= 4 + ¢x

= 4 + ‘a very little bit’.

f(x1, y1) are the coordinates of P and (x2, y2)

are the coordinates of Qg

fsubstitute

gfexpandg

fthe 4s subtract out, and then factoriseg

fwe can divide out ¢x, as ¢x 6= 0g

Now we let the secant approach the tangent by allowing Q to approach P along the curve.

As this occurs, ¢x approaches 0 (we write “¢x ! 0”) and ¢y¢x

! 4:

We write: lim¢x!0

¢y

¢x = 4:

The gradient of the function y = x2 at x = 2 is 4.

Note that the limit does not approach 4, the limit equals 4.

While the process of ¢x approaching 0 makes us think of calculating a sequence of values, it is

important to realise that the limit is a real number.

y

x x

M N

P (2, 4)

*)2(,2&Q 2 x x

ƒ(!)=!X



1

a y = 2x2 at the point (2, 8) b y = x3 at the point (2, 8)

c y = x2 + 3x at the point (1, 4) d y = ¡2x2 at the point (¡1, ¡2)

e y = 3x ¡ x2 at the point (¡2, ¡10) f y = x2 + 1 where x = 2

g y = 3x2 ¡ 5x + 3 where x = 1 h y = 4 ¡ 3x2 where x = 0

In the previous section, we found the gradient of the function y = x2 at a specific point.

Let us now generalise this to find the gradient of the function at any point on the curve y = x2:

Find the rule for the gradient at any

point on the graph of y = x2.

P is the point with coordinates (x, x2) and

Q is the point with coordinates .

The gradient of the secant is ¢y¢x

= y2 ¡ y1

x2 ¡ x1

= (x + ¢x)2 ¡ x2

x + ¢x ¡ x

= x2 + 2x¢x + (¢x)2 ¡ x2

¢x

= ¢x(2x + ¢x)

¢x= 2x + ¢x

fsubstituteg

fexpandg

fsubtract x2, and factorise

gfwe can divide out ¢ ¢x, as x 6= 0g

As Q ! P, i.e., as the secant approaches the tangent, ¢x ! 0

¢y

¢x ! 2x

and .lim¢x!0

¢y

¢x = 2x

Therefore, the gradient of the function at any point on the curve is given by the expression 2x. At

every point along the graph of x2 the gradient is twice the x-coordinate.

EXERCISE 11C.1

EXAMPLE 11.2 y

x

*) x(, x&Q 2 x x

x

M N

ƒ(!)=!X

),(P 2 x x

³x + ¢x; (x + ¢x)

2

´

Find the gradient of the function using the method explained on the previous page.




We use dy

dx as a shorthand notation for lim

¢x!0

¢y

¢x:

That is, dy

dx = lim

¢x!0

¢y

¢x

1

a y = 2x2 at the point (x, 2x2)

b y = x3 at the point (x, x3)

c y = x2 + 3x at the point (x, x2 + 3x)

d y = ¡2x2 at the point (x, ¡2x2)

e y = 3x ¡ x2 at the point (x, 3x ¡ x2)

f y = 6x2 ¡ 1 at the point (x, 6x2 ¡ 1)

g y = x2 + 2x ¡ 7 at the point (x, x2 + 2x ¡ 7)

h y = 9 ¡ 3x2 at the point (x, 9 ¡ 3x2)

FINDING THE GRADIENT WITHOUT THE “PICTURE”

In the last section you saw how to find the gradient of a curve, using a graph to show what was

happening geometrically. In many instances it is not easy to sketch the curve. Let us develop the

algebraic process of finding the gradient function, without a diagram to assist.

What is the gradient of the function y = x2 + 5x + 6 at any point on the curve?

Now the coordinates of point P would be (x, x2 + 5x + 6) and of

Q would be (x + ¢x, (x + ¢x)2 + 5(x + ¢x) + 6)

Therefore the gradient of the secant would be

¢y

¢x

= y2 ¡ y1

x2 ¡ x1

= (x + ¢x)2 + 5(x + ¢x) + 6 ¡ (x2 + 5x + 6)

x + ¢x ¡ x fsubstituteg

= x2 + 2x¢x + (¢x)2 + 5x + 5¢x + 6 ¡ x2 ¡ 5x ¡ 6

¢x fexpandg

= 2x¢x + (¢x)2 + 5¢x

¢x fsubtract termsg

EXERCISE 11C.2

EXAMPLE 11.3


Using the method of Example 11.1, find the gradient of the function



= ¢x(2x + ¢x + 5)

¢x ffactoriseg

= 2x + 5 + ¢x fwe can divide out ¢ ¢x, as x 6= 0g

As ¢x ! 0 (and as Q ! P), ¢y

¢x ! 2x + 5 and lim

¢x!

0

¢y

¢x = 2x + 5:

Therefore the gradient of the function f (x) = x2 + 5x + 6 is given by 2x + 5.

We can write .dy

dx = 2x + 5

Note: We started with the function y = x2 +5x +6. From it we derived the gradient function

2x + 5.

The process of finding the gradient function is called differentiation and 2x + 5 is called

the derived function or the derivative.

When we find the derivative from “scratch” so to speak, the process is called differenti-

ation from first principles.

Find the gradient function of y = x3 + 6x.

This solution has slightly different setting out.

y = x3 + 6x fthe y-coordinate of the point Pg) y + ¢y = (x + ¢x)3 + 6(x + ¢x) fthe y-coordinate of the point Qg

Now ¢y = (y + ¢y)

¡y

) ¢y = (x + ¢x)3 + 6(x + ¢x) ¡ (x3 + 6x) fsubstituteg) ¢y = x3 + 3x2¢x + 3x(¢x)2 + (¢x)3 + 6x + 6¢x ¡ x3 ¡ 6x fexpandg) ¢y = 3x2¢x + 3x(¢x)2 + (¢x)3 + 6¢x

) ¢y = ¢x(3x2 + 3x¢x + (¢x)2 + 6) ffactoriseg

) ¢y

¢x =

¢x(3x2 + 6 + 3x¢x + (¢x)2)

¢x

) ¢y

¢x = 3x2 + 6 + 3x¢x + (¢x)2 fwhen ¢x 6= 0g

As ¢x ! 0, ¢y

¢x ! 3x2 + 6 and lim¢x!0 ¢y

¢x = 3x2 + 6:

) the gradient function is 3x2 + 6, or dy

dx = 3x2 + 6:

EXAMPLE 11.4

EXERCISE 11C.3


1 Find the gradient function of each of these functions.

a y = 6x b y = 9x2 c y = ¡2x3

d y = 4x2 ¡ 2x + 1 e y = ¡7 f y = (2x + 1)(x ¡ 3)

g y = 2x3

¡ 4 h y = 6 ¡ 2x2

i y = x(x + 1)(x ¡ 2)



Now we will generalise this method even

further, and find the rule for the gradient

of any function. Consider the graph of the

function y = f (x).

P has coordinates (x, f (x))

Q has coordinates (x + ¢x, f (x + ¢x))

The gradient of the secant is

Now ¢y

¢x =

f (x + ¢x) ¡ f (x)

x + ¢x ¡ x

) ¢y

¢x =

f (x + ¢x) ¡ f (x)

¢x

) gradient = lim¢x!0

f (x + ¢x) ¡ f (x)

x + ¢x ¡ x

So, dy

dx = lim

¢x!0

f (x + ¢x) ¡ f (x)

¢x

For the function y = x4, find the derived function, i.e., the gradient of the curve from the definition.

dy

dx

= lim¢x!0

f (x + ¢x) ¡ f (x)

¢x fderived function definitiong

= lim¢x!0

(x + ¢x)4 ¡ x4

¢x fsubstituteg

= lim¢x!0

x4 + 4x3¢x + 6x2(¢x)2 + 4x(¢x)3 + (¢x)4 ¡ x4

¢x fuse the binomial expansiong

= lim¢x!0

¢x(4x3 + 6x2¢x + 4x(¢x)2 + (¢x)3)

¢x fsimplify then factoriseg

= lim¢x!0

4x3 + 6x2¢x + 4x(¢x)2 + (¢x)3 fdivide out ¢x, as ¢x 6= 0g

= 4x3

THE GRADIENT OF ANY FUNCTION AT ANY POINT D

EXAMPLE 11.5

y

x x

M N

P &!' ƒ(!)*

y ƒ(!)

( )& * x x f x x ,Q


fall terms containing a ¢x

go to zerog




For the function y = 1

x find the derived function, i.e., the gradient of the curve from

the definition.

dy

dx

= lim¢x!0

f (x + ¢x) ¡ f (x)

¢x fdefinition of the derivativeg

= lim¢x!0

1

x + ¢x ¡ 1

x¢x

fsubstitute 1

x for f (x), and

1

x + ¢x for f (x + ¢x)g

= lim¢x!0

x ¡ (x + ¢x)

x(x + ¢x)¢x ffind a common denominator g

= lim¢x!0

x ¡ (x + ¢x)

(¢x) x (x + ¢x) fsimplify using

a

bc

= a

bcg

= lim¢x!0

x ¡ x ¡ ¢x

(¢x) x (x + ¢x) fexpand the numerator g

= lim¢x!0

¡¢x

¢x (x2 + x ¢x) fsimplify the numerator g

= lim¢x!0

¡1x2 + x ¢x

fdivide by ¢x, as ¢x 6= 0g

= ¡1

x2 ftake the limit as ¢x ! 0g

Written as a rational expression: if y = 1

x then

dy

dx = ¡ 1

x2

Written in index notation: if y = x¡1 then dy

dx

=

¡x¡2

Note: In a similar way (and with a little bit of algebra) we can show that if

y =p

x = x12

dy

dx = 1

2 x¡

12

) dy

dx =

1

2p

x

EXAMPLE 11.6




Exercise 11D

1 For each of the following functions, find dy

dx from the definition of a derivative.

a y = x3 b y = 4x2 c y = x2 ¡ 4

d y = x2

¡6x + 5 e y = x3 + x + 1 f y = x5

g y = 2x3 h y = x6 i y = 6 ¡ x3

2 Challenge: Differentiate y = (x ¡ 3)2 from the definition.

3 Challenge: Differentiate y = x8.

Here are some results from the previous exercise.

Example Function Derived function

a y = x3 dy

dx = 3x2

b y = 4x2 dy

dx = 8x

c y = x2 ¡ 4 dy

dx = 2x

d y = x2 ¡ 6x + 5 dy

dx = 2x ¡ 6

e y = x3 + x + 1 dy

dx = 3x2 + 1

f y = x5 dy

dx = 5x4

2 g y = 2x3 dy

dx = 6x

h y = x6 dy

dx = 6x5

From some of the above, the following seems evident:

If y = xn then dy

dx = nxn¡1, where the index n is an integer.

From Example 11.6 and the note on page 340, the implication is that the rule can be extended to

any real values of n (although a general proof of this is outside the scope of this course).

That is,

If y = axn then dy

dx = naxn¡1, for all real values of n.

EXERCISE 11D

DIFFERENTIATION BY RULE E

Note: Some of the functions in this chapter have the domain of the function and/or

its derivative restricted.




Examples c , d and e lead us to conclude:

The derivative of a sum or difference equals the sum or difference of the derivatives.

In other words, if the function f (x) consists of the sum or difference of a number of terms, its

derivative can be found by simply finding the derivative of each term separately and adding (or

subtracting) them.

Finally two special cases should be noted.

Case 1:

If y = ax

then dy

dx = 1ax1¡1

) dy

dx = ax0

) dydx

= a

The derivative of a linear function is a constant,

and equal to the gradient of the function.

In symbols, if y = ax, then dy

dx = a.

This makes sense as the gradient of the straight line with

equation y = ax has a constant gradient of a.

Case 2:

The constant function, for example if y = 6, can be

thought of in two ways:

a If y = 6£

x0

then dy

dx = 6 £ 0 £ x¡1 = 0

b Geometrically, the equation y = 6 graphs as a horizontal line, which obviously has a

gradient of 0.

The derivative of any constant function is 0.

In symbols, if y = c, dy

dx = 0:

Find the derivative of each of these functions.

a y = x¡3 b s = 4t4 + 3t3 ¡ 5t2 ¡ 4t¡2 ¡ 8

c y = 2

3x2 d y =

1p x

a Applying our rule for differentiation, dy

dx = ¡3x¡4

EXAMPLE 11.7




b We need to find the derivative of each term, so ds

dt = 16t3 + 9t2 ¡ 10t + 8t¡3.

Using the variables s and t instead of x and y have no effect on the method used.

c Rewrite the function as y = 23 x¡2. Then

dy

dx = ¡4

3 x¡3

= ¡ 4

3x3

d Rewrite in index notation as y = x¡

1

2 . Then dy

dx = ¡1

2 x¡

3

2

= ¡ 1

2x3

2

= ¡ 1

2p

x3

NOTATION

Mathematicians who developed calculus used a variety of notations for the derived function, many

of which have survived to this day. You will need to be familiar with the following:

If y = xn

From Leibnitz dy

dx = nxn¡1 ........ (1)

Substituting xn for y into (1) gives us d

dxxn = nxn¡1 ........ (2)

From Bernoulli Dxy = nxn¡1 ........ (3)

Lagrange used function notation, if f (x) = xn f 0(x) = nxn¡

1 ........ (4)

In other words, dy

dx, Dxy and f 0(x) all have exactly the same meaning.

d

dx is a mathematical operator. Think of it as an instruction to find the derived function.

1 Differentiate by rule:

a y = x4 b y = x7

c y =¡

x5 d y = 3x2

Find d

dx(3x2 ¡ 6x + 2)

d

dx(3x2 ¡ 6x + 2) = 6x ¡ 6

EXAMPLE 11.8

EXERCISE 11E




e y = 5x f y = ¡12 x

g y = 5 h y = 0

i y = ¡x j y = 13 x3

k y = 2x2 ¡ 5 l y = 4x3 ¡ 4x2 + 7x + 6

m y = 12 x3 + 1

3 x2

¡ 14 x + 1

5 n y = ax3 + bx2 + cx + d

o s = t¡2

2 Differentate by rule:

a s(t) = 3t¡3 b s(x) = ¡15x¡2 c y(x) = 6x¡

1

2

d f (x) = x¡3 ¡ x¡4 e f (x) = x¡1 + x¡2 + x¡3 f f (x) = 5

x

g f (x) = 1

x2 h f (x) =

3

x4 i f (x) =

¡2

x3

j y = x +

1

x + 5 k y = x

1

3

l y = p x m y = ¡2

p x n y = 5 3

p x o y = 2x ¡ 4

p x

3 Find

a d

dx(

1p x

) b d

dx(

5

2p

x) c

d

dx(x +

1

x +

2

x2)

d d

dx(

3

x ¡ p

x) e d

dp( p2 +

1

p2) f

d

dx(

3x3 ¡ 5x2

x )

g ddx

( x3

¡ 3x

) h ddx

( 3p x) i ddx

( 42p

x3)

j d

dt(ut + 1

2 gt2) k d

dt

µ1

t ¡ 1

t2

¶ l

d

dx(4 ¡ x ¡ 3

p x)

m d

dt

µ 1p

t¡ 2

t2

¶ n

d

dt

µ 33p

t

¶ o

d

dt

µa

2t ¡ b

3t2

¶

4 If s = 10t ¡ 5t2,

a find dsdt : b What is the value of dsdt when t = 2?

c If v = 4 + 5t, find dv

dt.

5 If A = ¼r2,

a find dA

dr . Do you find anything interesting about this result?

b Show that dA

dr ¼62.8 when r = 10.




c The formula for the volume of a sphere is V = 43 ¼r3.

Find dV

dr . Do you find anything interesting about this?

Consider the function y = 2x3 ¡ 3x2 + 2x.

If we differentiate this function we get dy

dx = 6x2 ¡ 6x + 2:

Because dy

dx is a function of x, we may differentiate the function again and we get

the derivative of dy

dx = 12x ¡ 6:

For the derivative of

dy

dx we write

d

dx

dy

dx = d2y

dx2 using a pseudo algebra,

i.e., if y = 2x3 ¡ 3x2 + 2x the first derivative dy

dx = 6x2 ¡ 6x + 2

the second derivative d2y

dx2 = 12x ¡ 6

and we may continue d3y

dx3 = 12

d4y

dx4 = 0:

Alternatively we write f 0(x), f 00(x), f 000(x) etc. or Dxy D2xy D3

xy:

The second derivative has graphical and physical significance which you will learn about next year.

Find d2y

dx2 for the function a y = ax3 + bx2 + cx + d b y =

1

2p

x

a y = ax3 + bx2 + cx + d

dy

dx = 3ax2 + 2bx + c

) d2y

dx2 = 6ax + 2b

b y = 1

2p x = 1

2 x¡

12

dy

dx = ¡1

4 x¡

32

) d2y

dx2 = 3

8 x¡

52

) d2y

dx2 =

3

8x52

) d2y

dx2 =

3

8p

x5

THE SECOND (AND HIGHER) DERIVATIVES F

EXAMPLE 11.9




1 Find dy

dx and

d2y

dx2 for each of the following:

a y = x5

¡2x b y = x

p x

c y = 1

5x2 d y =

x

4 ¡ 1

2x

e y = 1

5p

x f y = (x2 ¡ 1)(x4 ¡ x + 3)

g y = x2 + 2p

x h y = (x2 ¡ 1)7

2 If y = x2 ¡ 2x prove that d2y

dx2 +

dy

dx + y = x2.

Consider a particle moving

After t seconds it is at A and its displacement measured from O is s.

After t + ¢t seconds the particle is at B, where OB is s + ¢s:

Therefore, ² average velocity over AB is ¢s

¢t and the

² instantaneous velocity at A is lim¢t!0

¢s

¢t

and so, v = ds

dt:

Likewise, if the velocity changes from v to v + ¢v in time ¢t, then

² the average acceleration over AB is ¢v

¢t and the

² instantaneous acceleration at A is lim¢t!0

¢v

¢t

and so, a = dv

dt.

Also, instantaneous acceleration may be written as a = d2s

dt2:

EXERCISE 11F

VELOCITY AND ACCELERATION AS DERIVATIVES G

O A B

s

s s

s

in straight line.a




1 The motion of a particle is given by s = t3 ¡ 6t2 + 9t where t is the time in seconds and s is

the displacement in metres. Find

a expressions for its velocity and acceleration at any time t

b when the particle is at rest, i.e., when v = 0

c when it is travelling with a constant velocity, i.e., when a = 0:

2 The motion of a particle is given by s = t2 ¡ 3t, where t is time in seconds and s is

displacement in metres.

a Where is the particle at time t = 0?

b Find expressions for the velocity and acceleration at any time t.

c Interpret the expression for acceleration.

d What is the initial velocity, i.e., at time t = 0?

e Does velocity ever equal 0? If so, when? Where is the particle at this time?

f Does the particle ever return to its starting position? If so, when?

g Sketch on the same set of axes:

i the displacement at time t, s(t)

ii the velocity at time t, v(t)

iii the acceleration at time t, a(t).

3 A particle is moving so that its displacement s metres from the origin after t seconds is given

by s = t2 ¡ 6t + 5.

Prove that it moves with a constant acceleration.

An electron is moving in a magnetic field such that its displacement from its source is given

by s = t3 ¡ 12t2 + 45t: Find

a its velocity at any time t b its acceleration at any time t

c when it is at rest d its acceleration when it is at rest

a v = ds

dt = 3t2 ¡ 24t + 45 b a =

dv

dt = 6t ¡ 24

c At rest v = 0. Hence, 3t2 ¡ 24t + 45 = 0

) t2 ¡ 8t + 15 = 0 fdivide both sides by 3g) (t ¡ 3)(t ¡ 5) = 0 ffactoriseg

) t = 3 s or t = 5 s fsolvegTherefore, it is at rest after 3 seconds and after 5 seconds.

d a = 6t ¡ 24when t = 3, a = ¡6 and when t = 5, a = 6 fsubstituting and simplifyingg

Note: If time is in seconds and displacement is in metres then velocity is measured in metres per

second (m/s) and acceleration is measured in metres per second per second or m/s2.

EXAMPLE 11.10

EXERCISE 11G




4 A test rocket’s height in metres above its launching point is

given by h = 45t2 ¡ t3 where t is the time in seconds

after launch.

a Find the velocity dh

dt of the rocket.

b At the top of its flight, the velocity of the rocket will

be zero. Find this maximum height.

5 A stone is thrown upwards such that its height above ground level at time t is given by

h = 2 + 6t ¡ 5t2 where t is given in seconds and h is given in metres.

a Was the stone thrown from ground level? Explain.

b When does the stone strike the ground? Answer to the nearest 0:1 sec.

c What was the initial velocity of the stone?

d When does the stone reach a velocity of 0 m/s?

e How high is the stone when it reaches this velocity?

f What is the velocity of the stone when it strikes the ground?

6 An astronaut on the moon drops a rock from the top of a crater onto the crater floor. The

height of the rock above the crater floor is given by h = 200 ¡ 0:8t2 where h is the height

in metres and t is the time in seconds.

a How high is the top of the crater above the crater floor?

b How can you tell from the velocity function that the astronaut did not throw the rock

downwards?

c Why is the velocity negative?

d How long did it take for the rock to reach the crater floor?

e What was its velocity at this time?

f The only force acting on the rock is the acceleration due to gravity. What is the accelerationdue to gravity on the moon?

g How fast is the rock travelling after 10 seconds? Where is it at this time?

h When has the rock reached a velocity of ¡19:2 m/s? Where is it at this time?

The slope of the tangent to y = f (x) at the point where x = a is the value of dy

dx evaluated

at x = a:

For the curve y = x2 ¡ 5x + 6

a find the gradient at the point P(1, 2)

b the equation of the tangent at P(1, 2)

TANGENTS

EQUATIONS OF TANGENTS AND NORMALS H

EXAMPLE 11.11




a dy

dx = 2x ¡ 5 which is the gradient at any point on the curve.

In particular at P(1, 2) dy

dx = 2 £ 1 ¡ 5

= ¡3:

Hence, the gradient of the tangent at P is ¡3.

b For the equation to the tangent at (1, 2)

y ¡ y1 = m(x ¡ x1)

) y ¡ 2 = ¡3(x ¡ 1)

) y ¡ 2 = ¡3x + 3

) y = ¡3x + 5

NORMALS

A line which is perpendicular to the tangent to a curve at the point of contact is called anormal to the curve.

Recall that:

Given that two lines have slopes m1 and m2, if they are perpendicular then m2 = ¡ 1

m1.

1 Find the equation of the tangent and the normal at the points where x has the given value.

a y = x2 x = 2 b y = x(6 ¡ x) x = 3

c y = x3 x = 12 d y = 3x2

¡8x + 5 x = 3

Find the equation of the normal to the curve y = x2 ¡ 5x + 6 at the point P(1, 2).

From the previous example, the tangent has slope ¡3:

Therefore, the gradient of the normal ¡ 1

¡3

= 13

The equation of the normal is y ¡ y1 = m(x ¡ x1)

) y ¡ 2 = 13 (x ¡ 1)

) y ¡ 2 = 13 x ¡ 1

3

) y = 13 x + 1 2

3

EXAMPLE 11.12

EXERCISE 11H

y x x 2 5 6

2 2

2

4 6

4

6

y

x

P

y x x 2 5 6

2 2

2

4 6

4

6

y

x

normal

tangent

P




e y = ¡x4 x = ¡1 f y = 6

x x = 3

g y =p

x x = 4 h y = 1p

x x = 4

2 Find the point(s) on the curve y = x3 ¡ 6x2 + 9x + 4 where the tangent is parallel to the

x-axis.

3 Find the point(s) on the curve y = x2 ¡ 3x + 1 where the tangent is parallel to the line

x + y ¡ 3 = 0:

4 Find the points on the curve y =p

x3 where the tangent is perpendicular to the line

y = 2 ¡ 13 x:

5 Find the coordinates of the point at which f (x) = 2x + 5 will touch

g(x) = ¡x2 + 10x ¡ 11. Show that f (x) is the tangent to g(x) at that point.

Recall from Chapter 8 that a composite function is a function of a function.

For example, if y = u3 and u = x ¡ 2, then we can form the composite function y = (x ¡ 2)3.

Consider the problem of finding the derivative of y = (x ¡ 2)3.

Recall that if f (x) = x3 we know f 0(x) = 3x2.

Now the graph of y = (x ¡ 2)3 is just the graph of y = x3 displaced 2 units to the right.

Consider their graphs as shown.

We can see that the gradient of the tangent

of y = (x ¡ 2)3 is the same as the gradient

of the tangent of y = x3, but at a point two

units to the right.

In other words,

if the gradient of y = x3 is 3x2, then the

gradient of y = (x ¡ 2)3 is 3(x ¡ 2)2.

This argument could be applied to any func-

tion of the form y = f (x + a), which gives

the result:

If y = (x ¡ a)n, then dy

dx = n(x ¡ a)n¡1

Soon we will devise a more general result which includes this one as a special case.

Find the derivative of y = 1

x + 4.

DIFFERENTIATION OF COMPOSITE FUNCTIONS I

EXAMPLE 11.13

y

x

y x 2 3

y x 3




y = 1

x + 4

) y = (x + 4)¡1 fwrite in index notationg

So,

dy

dx = ¡1(x + 4)¡2

fusing

dy

dx = n(x ¡ a)n¡1

g

) dy

dx =

¡1

(x + 4)2 fwrite as a rational functiong

THE COMPOSITE FUNCTION RULE OF DIFFERENTIATION

There is a simple rule for differentiating any composite function.

This rule is

If y = f (x) where u = g(x) then dy

dx =

dy

du

du

dx.

This is commonly known as the function of a function rule or the chain rule.

This is a fairly obvious rule when it is written in Leibnitz notation, and it follows from the fact that¢y

¢x =

¢y

¢u

¢u

¢x since the factor ¢u can be divided out of the right hand side.

Note that it is not so obvious in the Bernoulli notation: Dxy = Duy Dxu

This partly explains why a variety of notations are still in existence. Each notation is the best for

some purposes.

A proof of the composite function rule or function of a function rule or chain rule is given as

a formality. The proof uses the limit laws.

If a change of ¢x in x causes a change of ¢u in u which causes a change ¢y in y, then

¢y

¢x =

¢y

¢u

¢u

¢x

lim¢x!0

¢y

¢x = lim¢x!0

¢y

¢u

¢u

¢x ftake the limit of both sides

g) lim

¢x!0

¢y

¢x = lim

¢u!0

¢y

¢u lim¢x!0

¢u

¢x frule for the product of limitsg

) lim¢x!0

¢y

¢x = lim

¢u!0

¢y

¢u lim¢x!0

¢u

¢x fas ¢x ! 0, ¢u ! 0 alsog

) dy

dx =

dy

du

du

dx fdefinition of the derivativeg

Note that this proof is valid only if ¢u never equals 0. A more sophisticated proof is needed if

¢u

= 0 in any small interval around x, as

¢x goes to

0.




Differentiate

a y = (x2 ¡ 3x)10 b y =p

x2 ¡ 3x c y = 1

x2 ¡ 3x

a Let y = u10 where u = x2 ¡ 3x,

then dy

du = 10u9 and

du

dx = 2x ¡ 3:

Now dy

dx =

dy

du

du

dx ffunction of a function ruleg

= 10u9(2x ¡ 3) f= 10(2x ¡ 3)(x2 ¡ 3x)9 freplace ug

b Rewrite in index notation y = (x2 ¡ 3x)1

2 = u1

2 where u = x2 ¡ 3x

dy

du = 1

2 u¡

12

du

dx = 2x ¡ 3

Now dy

dx =

dy

du

du


= 12 u

¡

1

2 (2x ¡ 3) fsubstituteg

= (2x

¡3)

2u1

2 fas u

¡

1

2 = 1

u1

2 g

= 2x ¡ 3

2p

x2 ¡ 3xfreplace ug

c y = 1

x2 ¡ 3x = (x2 ¡ 3x)¡1

Let y = u¡1 where u = x2 ¡ 3x

dy

du

=

¡u

du

dx

= 2x

¡3

Now dy

dx =

dy

du

du


= ¡u¡

¡

2

2

(2x ¡ 3) fsubstituteg

= ¡2x ¡ 3

u2 fas u¡2 =

1

u2g

= ¡ 2x ¡ 3

(x2 ¡ 3x)2 freplace ug

EXAMPLE 11.14

substituteg




1 Differentiate each of the following functions.

a y = (x + 7)4 b y = 2(x ¡ 6)3 c y = (5x ¡ 2)3

d y = (3 ¡ x)6 e y = (x3 ¡ 3x2 + 1)9 f y = (ax2 + bx + c)7

g y = (3 ¡ 5x ¡ 4x2

)3

h y = (x ¡ 5)¡3

i y = (6 ¡ x)¡2

j y = 3

x ¡ 7 k y =

2

(x ¡ 5)4 l y =

p 5x ¡ 1

m y = 3p

2x + 4 n y =p

(x + 3)3 o y = 33p

3x + 1

p y = 1p

x + 3 q y =

1

x2 + 3x + 1 r y =

1p 3x2 ¡ 2x + 1

2 Differentiate each of the following functions using the chain rule.

a f (x) = (15x ¡ 23)

1

3 b f (x) = p 2 ¡ 3x ¡ 5

x c f (x) = 3

p 3x ¡ 6

d f (x) = 1p 3 ¡ 2x2

e f (x) = ¡ 23p

1 ¡ x2 f f (x) =

¡3p (x + 3)(x ¡ 3)

3 a Differentiate y = (2x ¡ 1)3 by using the function of a function rule.

b Expand (2x ¡ 1)3 by the binomial theorem, then differentiate the resulting expression.

c Show that these two results are the same.

4 The velocity of a raindrop which falls from a cloud at time t = 0 is given by

v = 2¡

1

(t + 1)2 metres per second.

Find the instantaneous acceleration at t = 2 seconds.

5 The concentration of a pesticide d days after application is given by C = 1500

d + 30 mg per m2.

Find the instantaneous rate of reduction on day 20.

6 If a tank holds 10000 litres of water which takes 50 minutes to

drain from the tank, then the volume of water remaining in the

tank after t minutes is given by

V = 10000(1

¡

t

50

)2; 0

·t

·50:

Find the rate at which the water is flowing out of the tank after

a 5 minutes b 10 minutes c 20 minutes.

7 The theory of relativity predicts that the mass of an object that is moving at a velocity close

to the velocity of light will increase, with the mass given by m = m0r

1 ¡ v 2

c2

where m0

is the mass of the object at rest, v is the velocity of the object and c is the velocity of light.

a Find dm

dv

. b What is the physical interpretation of dm

dv

?

EXERCISE 11I.1




8 If we have a function such that y is a function of u, u is a function of v, and v is a function

of x, then we can extend the chain rule as follows:

dy

dx =

dy

du

du

dv

dv

dx

Use this function of a function of a function rule to find the derivative of y = (3p

x ¡ 1)5.

CHAIN RULE “BY INSPECTION”

With practice, it is possible to find the derivative of many composite functions with only a few

lines of working.

If you can learn to do so, you will save time and minimise the number of errors.

Find the derivative of y = (4x + 1)5 by inspection.

We think: For y = (4x + 1)5 we let u = 4x + 1 so that y = u5.

The derivative of u5 is 5u4, or, in terms of x, 5(4x + 1)4.

The derivative of u = 4x + 1 is 4.

We write dy

dx =

dy

du

du


= 5(4x + 1)4 (4) fsubstitutingg= 20(4x + 1)4 fsimplifyingg

Find the derivative of y = 3p 4x ¡ 3

by inspection.

y = 3p 4x ¡ 3

= 3(4x ¡ 3)¡

1

2 fwrite in index notatong

) dy

dx

= 3(¡12 )(4x ¡ 3)¡

3

2 (4) fderivative of 3u¡

1

2 times derivative of (4x ¡ 3)g= ¡6(4x ¡ 3)¡

3

2

= ¡6p

(4x ¡ 3)3

1 Differentiate these composite functions, by inspection.

a y = (3x + 4)

2 b y = (2x + 1)

7 c y = (3 ¡ 4x)

2

EXAMPLE 11.15

EXAMPLE 11.16

EXERCISE 11I.2




d y = (2 + x2)3 e y = (x2 ¡ 2x + 1)6 f y = 1

3x ¡ 4

g y = ¡2

5 ¡ 3x h y =

3

x2 + 3 i y = ¡ 3

5 ¡ x ¡ x3

2 Differentiate these composite functions, by inspection.

a y = p 3x b y = p 5x + 7 c y = p 4 ¡ 5x3

d y = 3p

9xe y =

¡2p x2 ¡ 4x + 1

f y = 2

3p

3 ¡ 2x2

Differentiate y = (2x + 1)(5x ¡ 2) by

a using the product rule

b expanding and simplifying, and then finding the derivative of the resulting function.

a Here u = 2x + 1, so

du

dx = 2, and v = 5x ¡ 2, so

dv

dx = 5.

Because y is the product of u and v , dy

dx = u

dv

dx + v

du

dx

= (2x + 1)(5) + (5x ¡ 2)(2)

= 10x + 5 + 10x ¡ 4

= 20x + 1

b y = (2x + 1)(5x ¡ 2) = 10x2 + x ¡ 2

so, dy

dx = 20x + 1

THE PRODUCT RULE

An important class of functions are those that are the product of simpler functions.

For example, y = (2x + 1)(5x ¡ 2) is the product of u = 2x + 1 and v = 5x ¡ 2

y = (3x ¡ 4)2(2x + 7)3 is the product of u = (3x ¡ 4)2 and v = (2x + 7)3

y = xp

5 ¡ 3x is the product of .u = x and v =p

5 ¡ 3x

In Mathematics B we usually restrict the function to two factors although the theory can be expanded

to functions that involve three or more factors.

Here is the rule for differentiating products. It will be proved later in the chapter.

Product Rule: If y = u £ v where u = f (x) and v = g(x) then

dy

dx = u

dv

dx + v

du

dx

DIFFERENTIATION OF PRODUCTS J

EXAMPLE 11.17




This example verifies the product rule. Why does it not prove the product rule?

Differentiate y = (5x ¡ 2)3(3 ¡ x)6 using the product rule.

This is more difficult, as the factors of the product are composite functions.

We must apply the chain rule first to find the derivative of each factor.

Here u = (5x ¡ 2)3 and du

dx = 15(5x ¡ 2)2

v = (3 ¡ x)6 and dv

dx = ¡6(3 ¡ x)5

and dy

dx

= u dvdx

+ v dudx

= (5x ¡ 2)3[¡6(3 ¡ x)5] + ( 3 ¡ x)6[15(5x ¡ 2)2] fsubstituteg= (5x ¡ 2)2(3 ¡ x)5(¡30x + 12 + 45 ¡ 15x) fcommon factor of (5x ¡ 2)2(3 ¡ x)5g= (5x ¡ 2)2(3 ¡ x)5(¡45x + 57) fcollect like termsg= ¡3(5x ¡ 2)2(x ¡ 3)5(15x ¡ 19) fcommon factor of ¡3g

1 Differentiate using the product rule.

a y = (4x + 1)(3x¡ 2) b y = (2x + 5)(3¡ 2x)

c y = 2(x + 1)(3x¡ 7) d y = (4x2 ¡ 2x + 1)(x + 3)

e y = (x2 ¡ 3)(2x3 + 5x ¡ 1) f y = (x3 ¡ 3x2)(2x5 + 5)

g y = (3x + 1)4(2x¡ 3)3 h y = (x + 3)10(x¡ 4)8

j s = (3t¡ 4)2(5¡ 6t)3 i y = (5x + 3)4(3x¡ 2)3

2 Find dy

dx if

a y = xp x¡ 5 b y = xp 3x¡ 1

c y = ¡2x 3p

3x + 1 d y = (x + 2)3p

x

e y = (2¡ 3x)p

x + 1 f y =p

xp

4¡ x

g y = x(3x + 1)¡2 h y = (x + 3)2(2x¡ 5)¡3

i y = (x + 1)¡2(3x ¡ 3)¡3 j y = (3¡ 4x2)¡3(5x¡ 7)2

3 Students often ask why the product rule is not applied to functions such as y = 2x. It can

be! Show this, by differentiating using the product rule:

a y = 2x b y = x2 (think of this as the product x:x)

EXAMPLE 11.18

EXERCISE 11J

(see Exercise 11I.1, 1c)

(see Exercise 11I.1, 1d )




4 Find the derivative of

a f (x) =p

x + 1p

x + 2 b f (x) = (2x2 + 1)p

x

c f (x) = (3x ¡ 2)x¡1 d f (x) = (3x2 ¡ 2)¡3(2x ¡ 3)¡2

e f (x) = 1

x + 1

1

x + 2 f f (x) =

2p 1 ¡ x

¡2p x + 1

5 What do you think the product rule for the product of three terms might be? Test your

hypothesis by

a differentiating y = x2(2x + 1)(2x ¡ 1) using your product rule

b expanding y = x2(2x + 1)(2x ¡ 1) and differentiating the resulting expression.

PROOF OF THE PRODUCT RULE

For completeness, the proof of the product rule is given here.

for limits given on page 331.

y = uv where u = f (x) and v = g(x)

Let a change ¢x in x produce a change ¢u in u and ¢v in v and hence a change ¢y in y.

) y + ¢y = (u + ¢u)(v + ¢v)

) ¢y = (u + ¢u)(v + ¢v) ¡ uv fsince y = uvg) ¢y = uv + u¢v + v¢u + ¢u¢v ¡ uv

) ¢y = u¢v + v¢u + ¢u¢v

and ¢y

¢x = u

¢v

¢x + v

¢u

¢x +

¢u¢v

¢x fdividing by ¢xg

) dydx

= lim¢x!0

¢y¢x

fdefinition of derivativeg

) dy

dx = lim

¢x!0

µu

¢v

¢x + v

¢u

¢x +

¢u¢v

¢x

¶ fusing Limit Laws 1 and 2g

) dy

dx = u lim

¢x!0

¢v

¢x + v lim

¢x!0

¢u

¢x + lim

¢x!0

¢u

¢x

¢v

¢x¢x fmultiply

¢u¢v

¢x by

¢x

¢xg

) dy

dx = u

dv

dx + v

du

dx +

du

dx

dv

dx (0)

)

dy

dx = u

dv

dx + v

du

dx

This proof uses some of the rules

Quotients may be differentiated using the product rule.

Differentiate y = 2x + 1

x

¡7

:

DIFFERENTIATION OF QUOTIENTS K

EXAMPLE 11.19



y = 2x + 1

x ¡ 7 (2x + 1)(x ¡ 7)¡1

If u = (2x + 1) then du

dx = 2

and if v = (x ¡ 7)¡1 then dv

dx = ¡1(x ¡ 7)¡

2

so dy

dx = (2x + 1)[¡1(x ¡ 7)¡2] + (x ¡ 7)¡1(2) fsubstituteg

) dy

dx =

¡(2x + 1)

(x ¡ 7)2 +

2

x ¡ 7

) dy

dx =

¡(2x + 1) + 2(x ¡ 7)

(x ¡ 7)2 f g

)

dy

dx = ¡15

(x ¡ 7)2 f g

Consider y = u

v where u = f (x) and v = g(x)

) y = uv¡1

)

dy

dx = uµ¡1v¡2 dv

dx¶

+ v¡1 du

dx

) dy

dx =

¡u

v2

dv

dx +

1

v

du

dx

) dy

dx =

1

v

du

dx ¡ u

v2

dv

dx

) dy

dx =

v

v2

du

dx ¡ u

v2

dv

dx

)

dy

dx =

vdu

dx ¡u

dv

dxv2

If y = u

v then

dy

dx =

vdu

dx ¡ u

dv

dxv2

:

There is a more appropriate rule for differentiating quotients that applies, called the quotient rule.

THE QUOTIENT RULE

The quotient rule:

f gwrite as a sum of algebraic fractions

write as a single fraction

simplify the numerator

which can be written as


Differentiate y = 2x + 1

x

¡7

using the quotient rule.

EXAMPLE 11.20




Let u = 2x + 1 so du

dx = 2 and let v = x ¡ 7 so

dv

dx = 1.

dy

dx =

vdu

dx ¡ u

dv

dxv2

= (x ¡ 7)(2) ¡ (2x + 1)(1)

(x ¡ 7)2 fsubstituteg

= 2x ¡ 14 ¡ 2x ¡ 1

(x ¡ 7)2 fexpandg

= ¡15

(x ¡ 7)2 fcollect like termsg

1 Differentiate using the product rule.

a y = 5x ¡ 3

2x + 1 b y =

3x

2x2 + 1 c y =

x

x3 ¡ 3 d y =

(4x + 2)3

(3x ¡ 1)5

2 Differentiate using the quotient rule.

a y = 8x + 5

3x ¡ 7 b y =

(x ¡ 3)3

(2x + 1)2 c y =

x3 ¡ 2x

x2 + 1 d y =

(x ¡ 4)3

(5 ¡ x2)2

3 Differentiate, using whichever method you prefer.

a y = 2x + 4

3x

¡6

b y = (2x + 1)3

(x

¡4)2

c y = (2x + 1)3

x

¡4

d y = 3x + 1p

x

¡3

e y =p x + 1p

x + 2

4 Differentiate y = 3

(2x ¡ 1)2

a by rewriting as a product, and then using the chain rule.

b by rewriting as a product, and then using the product rule.

c using the quotient rule.

d Which method of differentiating this function do you prefer? Why?

5 Use your knowledge of algebraic fractions and index laws to first simplify these functions, and

then differentiate.

a y = 4x ¡ 3

x b y =

x2 + 5x + 6

x + 3 c y =

3x ¡ 7p

xp x

d y = 5x + 4

(5x + 4)3

6 Differentiate these functions, using any appropriate method.

a y = 3(2x ¡ 1)2 b y = (4x + 1)(3x ¡ 2)2 c y = ¡2

x ¡ 5

d y = 3xp

x ¡ 3 e y = (4x + 1)2

2x3 f y =

3x ¡ 5p x

g y = x(2x ¡ 2)(2x + 2) h y =

p 3 ¡ 2xp

4x2

+ 1

EXERCISE 11K




7 Differentiate these functions, using any appropriate method.

a y = 6x

3x2 b y =

p 1 ¡ x ¡ x2 c y =

p 2x + 1

p 2x ¡ 1

d y = x ¡ 1

1 ¡ x e y =

(3x ¡ 1)2

3x ¡ 1 f y =

(3x ¡ 1)2

3x

g y = x(x ¡ 2)(x + 3) h y =

2x

¡1

p x i y =

1

(3x + 4)2

j y =

r 2x + 1

x ¡ 3 k y =

2x3p

x + 1 l y =

p 5 ¡ x

3p

5 ¡ x

When using the chain rule, product rule and quotient rule for differentiation, students often forget

what a derivative is, and start to view differential calculus as just a collection of rules. It is easy

to forget why the rules are used. Recall that a derivative is the gradient function; it is the rule that

defines the gradient for every value of x in the domain.

In this last section, we will take a new look at the meaning of a derivative, by studying the graphs

of functions and their derivatives.

FUNCTIONS, DERIVATIVES AND GRAPHS L

FEATURES OF GRAPHS y

x

a zero

turning point(local minimum)

turning point(local maximum)

point of inflection

(the graph changesfrom concave up

to concave down)increase

in y decrease

in y

increasein x

increasein x

function increasing function decreasing

A graph of a function, f (x), is given.

y

x

(-5' 0)

(-3' 2)

(-\Qw _' -2)

(-1' 0)

(2' -4)

(6' 0)

EXAMPLE 11.21



EXAMPLE 11.22

y

x

m

m m

m

m

m

m

m


a What are the zeros of the function?

b What are the coordinates of the function’s turning points?

c For what values of x is the function

i positive ii negative iii zero?

d For what values of x is the function

i increasing ii decreasing iii stationary?e For what values of x is the derivative of the function

i positive ii negative iii zero?

f For what values of x is the derivative of the function

i increasing ii decreasing iii stationary?

a These are the values of x for which the function is zero. These are ¡5, ¡1 and 6.

b (¡3, 2) and (2, 4).

c The function is positive above the x-axis, negative below the x-axis and zero whereit crosses the x-axis.

i ¡5 < x < ¡1 and x > 6 ii x < ¡5 and ¡1 < x < 6

iii x = ¡5, x = ¡1 and x = 6

d If you think of someone walking along the function, from left to right, they are walk-

ing uphill to (¡3, 2), then downhill to (2, ¡4) and then uphill again.

i x < ¡3 and x > 2 ii ¡3 < x < 2 iii x = ¡3 and x = 2

e The derivative at a point is the gradient of the function at that point. The gradient

is positive when the function is increasing, negative when the function is decreasing,

and zero at the stationary points. Questions d and e are two ways of asking the same

question, so the answers are identical.

i x < ¡3 and x > 2 ii ¡3 < x < 2 iii x = ¡3 and x = ¡2

f This question requires careful thinking.

Consider how the gradient is changing, as

a point moves along the curve from left

to right. The diagram gives the gradient

at various locations along the curve. As

we move from left to right, we have these

values for m: 3, 1, 0, ¡1, 0, 1, 3

The gradient decreases from 3 to ¡

1, and then increases from¡

1 to 3.

At the point of inflection, with coordinates (¡12 , ¡2), the gradient reaches its mini-

mum value of ¡1.

i x > ¡12 ii x < ¡1

2 iii x = ¡ 12

A function has a gradient of zero at (¡1, ¡2) and (3, 2). The gradient is negative for

values of x < ¡1 and x > 3, and positive for ¡1 < x < 3. Draw a sketch of the function.




The graph has a minimum at (¡1, ¡2) and

a maximum at (3, 2). Here is a sketch that

fits the criteria.

y

x

(-1' -2)

(3' 2)

y

x

A a ( , a( , ) a

( , .a

function has gradient of at ),gradient of at and gradient of at ). Its zeros are at andSketch possible graph of this function.

2 1 30 2 6 1

5 3 = 3 = 7

¡¡

¡x x

EXAMPLE 11.23

CLASS DISCUSSION

What special point on the function corresponds to the minimum

the derivative?

value of


Almost all graphics calculators have the facility of graphing a function and its derivative on

the same set of axes. This is a powerful tool in the study of the graphs of derivative functions.

1 Let f (x) = 0:1(x + 4)(x ¡ 1)(x ¡ 3).

a On the same set of axes, sketch f (x) and f 0(x). Use a window with “nice” x-

coordinates, and with the same scale horizontally and vertically (for example, so acircle looks like a circle).

b Use Trace to explore the relationship between

i points on the function, and corresponding points on its derivative

ii the zeros of the derivative and the turning points of the function.

c Find the coordinates of the point of inflection.

2 Repeat for these functions (or ones of your own choosing).

a f (x) = 1

x b f (x) =

p 9 ¡ x2 c f (x) = jx ¡ 1j

What to do:

GRAPHS OF DERIVATIVES USING A GRAPHICS CALCULATOR




1 For each function below, for what values of x is the

i function zero ii function positive iii function negative

iv gradient zero v gradient positive vi gradient negative

a b

c d

2 For each function in question 1, for what values of x is the gradient of the function increasing?

3 Draw neat sketches of functions for which

a the gradient is positive and the gradient is increasing

b the gradient is positive and the gradient is decreasing

c the gradient is negative and the gradient is increasingd the gradient is negative and the gradient is decreasing.

4 For each of these functions, draw a neat sketch of the gradient function.

a b

c d

y

x

(-4' 0) (3' 0)

y

x

y

x

-1\Qw _ (-2' 0)

(-1' 0)

(2' 0)

(-\Qw _' -1)

x

y

x

EXERCISE 11L

y

x

y

x

(-3' 5)

(2' -3)

y

x

(0' 5) y

x

x

(0' 1)




5 a A function has a gradient of zero at (¡1, 4) and (0, 1). The gradient is positive for values

of x < ¡1 and x > 0, and negative for ¡1 < x < 0. Draw a sketch of the function.

b A function has a gradient of zero at (0, 0) and (2, 16). The gradient is negative for values

of x < 0, and positive for 0 < x < 2 and x > 2. Draw a sketch of the function.

c A function has a gradient of zero at (¡2, ¡16) and (0, 0). The gradient is positive for all

other values of x. Draw a sketch of the function.

d Draw a sketch of a function whose gradient is negative for all values of x.

e A function has a gradient that is constant and equals 3. The function passes through

(0, ¡2). Sketch the function.

6 a A function has a gradient of ¡3 at (¡2, 3), a gradient of 3 at (2, 3) and its only zero is

at the origin. Sketch a possible graph of this function.

b A function has a gradient of ¡3 at (¡2, 3), a gradient of ¡3 at (2, ¡3) and its only zero

is at the origin. Sketch a possible graph of this function.

c A function has a gradient that equals ¡1 for all values of x. Sketch a possible graph of

this function.

Here are some challenging questions of this type.

d A function has a gradient that is always positive. Its domain is all real numbers except

x = 0. Its graph has no x-intercept and no y-intercept. Sketch a possible graph of this

function.

e A function has the property that the gradient at any point on its graph equals the x-

coordinate of that point. Sketch a possible graph of this function.

f A function has the property that the gradient at any point on its graph equals the y-

coordinate of that point. Sketch a possible graph of this function.

7 a A function is positive for all values of x. What does this tell you about the gradient of

the function?

b The gradient of a function is positive for all values of x. What does this tell you about

the function?

8 Each of these news stories is about a rate, and how it changes. Sketch a graph to illustrate

each. Include a title, and label your axes.

a Our country is in danger of becoming overpopulated. There is some good news, though.

Although the population is still growing too fast, the growth rate has decreased signifi-

cantly each year for the last decade.

b While the number of incidents of SIDS is still on a downward trend, the rate at which it

is decreasing is slowing.

c There is good news in the fight against drink driving. The decline in the number of drink

drivers has been increasing each year.

9 For each of the graphs below, make up a news story to match.

a b c d population deaths frequency savings

year year month cost

x x x x

y y y y




1 P(1, 3) is a point on the curve with equation y = x(4¡

x). If the normal to the curve

at P cuts the curve again at point Q, find the equation to the tangent at Q.

2 f 0(x) is one symbol used for the derivative of a function.

f 00(x) is a symbol used when we differentiate a function twice.

We call f 00(x) the second derivative.

For example, if f (x) = x2 ¡ 5x + 6

then f 0(x) = 2x ¡ 5

and f 00(x) = 2:

If f (x) = ax + bx

, a, b 2 R, prove that f (x) = xf 0(x) + x2f 00(x):

3 The angle of intersection of two curves can be

defined as the acute angle between the tangents

to the curves at the intersection points. Find the

angle of intersection of the two curves

y = ¡x2 + 3

y = x2 ¡ 5:

MODELLING

V1 Alongside are two graphs showing the amount of

aspirin (measured in mg per litre) in the blood-

stream at a time t hours after the aspirin was

taken. One graph is of a slow-release tablet,while the other is a fast-acting aspirin.

a Which is the graph of the fast-acting tablet?

Explain how you arrived at your decision.

b What is the maximum concentration of the

i fast-acting tablet

ii slow-release tablet?

c For each tablet, estimate the rate at which the drug is entering the bloodstream after

one hour.

d For each tablet, estimate the rate at which the drug is entering the bloodstream after

two hours.e Which tablet will have the most effect two hours after the drug was taken? Explain.

f Which tablet will have the most effect four hours after the drug was taken? Explain.

t

y

amount

time (hours)

EXERCISE 11M

PROBLEM SOLVING M

x

y




4 A space vehicle is moving on a path which is

described by the equation y = d2

3 +2d

3 where

y is the vertical distance from the launch point

and d is the horizontal distance from launch point

in kilometres.

When the horizontal distance is 27000

km, the

space vehicle is to fire an exploratory rocket into

space tangentially on a straight line course for

Bega Nova, an abandoned space station which is

located at a horizontal distance of 93300 km and

a vertical distance of 1.3 £ 105 km from earth.

Will the rocket lay a course which will dock with

Bega Nova?

5 If pv1:5 = k (a constant), prove that dP

dv =

¡1:5P

v .

6 Newton’s Law of Gravitation states thatthe gravitational force between two parti-

cles of mass m1 and m2 at a distance r is

given by F = Gm1m2

r2 . Find the rate

of change of force between the particles

with respect to r (i.e., assume m1 and m2

are constant).

7 The time it takes a pendulum to complete one

“swing” is called its period. The period, T ,

is a function of its length, l, and is given by

T = 2¼

r l

g where g is the acceleration due to

gravity (which is constant). Find dT

dl .

8 Consider the following table of values: x 1 2 3 4 5y 3 8:485 15:59 24 33:54

If these values are from a function of the form y = axn, find the gradient of this

function when x = 2:5.

9 The frequency of vibrations of a vibrating violin string is given by f = 1

2L

r T

½ where

L is the length of the string, T is the tension in the string, and ½ is its linear density.

Find the rate of change of the frequency with respect to

a L, if T and ½ are constants

b T , if L and ½ are constants

c ½, if L and T are constants.

10 Prove that the derivative of an even function is odd, and that the derivative of an odd function

is even.

d

y

m1

m2

r

F

F






chain rule gradient normal

composite functions indeterminate product rule

derivative instantaneous velocity quotient rule

derived function limit secant

differentiation neighbouring point tangent

1 Differentiate using the definition: y = x3 ¡ 2x


a y = x +

1

x +

2

x2 b y = (3x

4

¡ 5)5

c y = 3x ¡ 4

2x + 5 d y = (x3 ¡ 2)2(x2 ¡ 2)

3 A ball is thrown vertically upwards and its height after t seconds is s metres where

s = 30t ¡ 5t2. Find

a its velocity at 2 seconds

b its acceleration

c when it is momentarily at rest

d the greatest height reached

e the time taken to reach the ground again.

4 a Find the equation of the i tangent ii normal to the parabola y = x2

4 at the

point (6, 9).

b Find the distance between the points where the tangent and normal meet the y-axis.

5 A company’s annual profit P is related to time t, by the equation P = 0:03t2 + 5, t 6 3where P is the profit in millions of dollars for the tth year the company has been operating.

a Find the rate of change of profit in the second year. Is the profit increasing or decreasing

at this point?

b If the rate of change of profit remains constant at and beyond the third year, find the

equation relating P and t for t > 3. Calculate the profit for the 6th year.

6 If y = x2 ¡ x prove that d2y

dx2 ¡

µdy

dx

¶2

+ 2dy

dx = 4(x ¡ y) ¡ 1.

7 If y = xm(xn + 1) prove that dy

dx = (m + n)xm+n¡1 + mxm¡1.

8 If h x( ) = a

x2 and h0(1) = 4 find the value of a.






1 a Differentiate from the definition of the derivative: y = 2x2 ¡ x + 3

b Draw a diagram to show why dy

dx ¼ ¢y

¢x

when ¢x is chosen close enough to 0.


a y = 3x2 + 2x ¡ 4

x2 b y = (3x2 ¡ 2)(x + 1) c y =

p 4x2 + 9

d y = (3x + 1)3(2x ¡ 4) e y = (1 ¡ 8x)2

x + 1

3 a Differentiate y = (3x + 4)3 using the function of a function rule.

b Expand y = (3x + 4)3 by the binomial theorem and differentiate.

c Show that these results are identical.

4 The position of a particle is given by s = 2t3 ¡ 2t + 1. Find its instantaneous velocity

after 2 seconds.

5 The graph of y = x2 ¡ x ¡ 2 cuts the x-axis at points A and B.

a Find the gradient of the curve at each of these two points.

b Find the equation of the normal at B.

6 The amount of pollution P units in a stream is found to be P = (t3

4 + 1)4 where t is the

time period measured in months. At what rate is the amount of pollution changing after

8 months?




Revisionexercises forchapters

7 to 11

CHAPTER 12

Review exercises are repro-

duced as student worksheets

on the CD and are accessible by clicking on this icon



1 a Express in index notation.

i 8 £ 8 £ 8 £ 8 £ 8 ii 3xxxyyyy

b i Express 273 as prime factors.

ii Express in factor form (5s4t3)2

c Simplify

i 2z5 £ 3z2 ii a2 £ b £ a3 £ b5

d Simplify

i 21d8

7d7 ii

12uv2w3

8v

e Simplify

i (x3)2 ii 6(w4)2

4(w5)2

f Write without brackets, simplifying where possible.

i (ef )3 ii 5ab4(2ab)3

g Simplify

i 3a £ 2b £ c ii (3m)3 £ (¡m)5

h Simplify

i 3a2b £ 5ab2 ¡ (¡ab)3 ii (2 p2q )2 ¡ (¡2 p3) £ (3q 2) £ (2 p)

2 a Simplify without using a calculator.

i 57:8o ii e0

b Evaluate without using a calculator, simplifying where possible. i 16 £ 8¡2 ii 7¡1

c Express with positive indices.

i p¡2q ¡2 ii ( 127 s¡12t6)

1

3

d Evaluate without using a calculator, simplifying where possible.

i 1251

3 ii (18 £ 72)1

4

e Evaluate without using a calculator, simplifying where possible.

i 8¡

2

3 ii 645

3

f Write using radicals. i 16

5

2 ii (3 p)3

5

g Write using index notation. Simplify if possible.

i 1

3p

z ii

9p 49g6

h Simplify, without using a calculator.

i 3 £ 2¡3 ii k7

3 ¥ k1

3

i Simplify

i 3n £ 27n+1

92n¡2

ii 8¡2n

£64n¡2

REVIEW OF CHAPTER 7 A

370 REVISION EXERCISES FOR CHAPTERS 7 T O 11 (Chapter 12)




REVIEW OF CHAPTER 8 B


10 a Find the value of the unknown variable.

A P i n

i $2000 7.4 % p.a. compounded monthly 15 years

ii $54 000 1.5% p.a. compounded annually 12 years

iii $20 000 $15000 ...... compounded annually 8 years

iv $4000 $550 ...... compounded monthly 20 years

v $1 million $300 000 11.5% p.a. compounded monthly

b It is anticipated that an investment will grow to $60000 after 4 years, with an annual

interest rate of 7% p.a. If interest is compounded annually, what initial amount must be

invested?

c Forty thousand dollars was deposited in an account that pays interest compounded monthly.

It has grown to $100000 in 8 years. Calculate the value of i, the annual interest rate.

d How many years will it take for an investment of $1000 earning 7.5% p.a. compounded

annually to grow to $1500?

e The formula for growth due to inflation is F V = P V (1 + i)n where

F V = the future value (cost of a commodity in the future)

P V = present value (cost of a commodity now)

i = the annual percentage increase of the cost, as a decimal (e.g., 4% = 0.04)


i A house is worth $235000 today. If its cost increases by 5% per year, what will it

cost in eight years?

ii A bottle of milk triples in price over a fifteen year period. What was the annual

percentage increase in its cost?

iii How many years will it take an investment that increases in value by 15% per year

to grow in value from $80000 to $300000?

11 Under testing a jet engine is accelerated from 8000 to 10000 revolutions per minute (rpm).

The instrumentation records the following data:

time (seconds) 0:08 0:10 0:12 0:14 0:16 0:18 0:2

speed (rpm) 9590 9720 9810 9870 9910 9940 9960

a Assume that the relationship between the engine speed n and the time t is in the form

m = atk where m = 10000 ¡ n. By first taking logs of m and t, find the values of

a and k .

b Construct a residual plot of the transformed data. Using this residual plot, comment onthe appropriateness of the model.

1 a Which of the following are polynomials? For those that are not, explain why they are not

polynomials.

i f (x) = 3 + x ii g(x) = 4x

x2 + 15 iii k(x) = (x + 1)(x2 + 3x ¡ 1)

b Evaluate each of the following polynomials for the given values of x.

i f (x) = x4

¡ 4x2

+ 7, f (0), f (2) ii h(x) = 7 ¡ x ¡ 2x2

+ 4x3

, h(3), h(¡9)





2 Expand each of these polynomials.

a g(x) = 3(x ¡ 5)(x + 4) b F (t) = (t ¡ 4)(t + 4)(2t ¡ 1)

3 a Expand and simplify, using the binomial expansion for (a + b)n

i (x + 3)3 ii (3 ¡ 2z)4

b Express in the form (x + a)n

i t3 ¡ 3t2 + 3t ¡ 1 ii x4 ¡ 8x3 + 24x2 ¡ 32x + 16

c Write the equations of three polynomials that intersect the x-axis at x = ¡3, 0 and 2.

d Write the equations of three polynomials that intersect the y-axis at y = 7.

e Write the equations of three polynomials that pass through the point (¡1, 3).

4 a Form the composite function y = f (x) if

i y = u + 2 and u = 2x ¡ 12 ii y = t2 ¡ 2t + 3 and t = x2 ¡ 1

b Consider the function f (x) = x ¡ 2x2. Find an expression in simplest form for

i f (

z)

ii f (¡2

c)

iii f (2

t2

+ 1) c f (x) = x2 + 2x ¡ 1 and g(x) = x ¡ 4. Find

i f [g(x)] ii g[g(x)] iii f [g(¡1)] iv f [f (¡1)]]

d f (x) = 3 ¡ x + x3 and g(x) = 2 ¡ x. Find

i f ± g ii g ± f

e Solve the following:

i (y + 2)2 ¡ 3(y + 2) + 2 = 0 ii (x2 ¡ 3x ¡ 1)2 ¡ 12(x2 ¡ 3x ¡ 1) + 27 = 0

5 a Sketch the graphs of these polynomials using a graphics calculator. Adjust the window

to include all points of interest. State the coordinates of the zeros, the y-intercept, and

the local maximum and minimum values.

i y = x(x + 3)(x ¡ 1) ii y = x4 + 2x3 ¡ x2 ¡ 2x

b Sketch the graphs of

i y = x4 ii y = x(x + 3)(x ¡ 2)(x ¡ 1)

iii y = (x2 ¡ x + 4)(x2 ¡ x + 4)

simultaneously in the window (¡3, 3, 1; ¡25, 20, 4). Now change the window to

(¡40, 40, 4; ¡50000, 50000, 10 000). Comment on what the two different views of the

functions show you.

6 Answer the following questions using your knowledge of zeros of a function. Check your

answers using a graphics calculator.

a What are the zeros of the polynomial g(x) = (x + 1)(x ¡ 3)(x + 2)? What happens

to the function for large positive values of x? For large negative values? Draw a sketch

of what you think the graph may look like.

b By hand, sketch both g(x) and h(x) = 2(x +1)(x ¡ 3)(x + 2) on the same set of

axes. What is the effect of the factor of 2? How does the factor of 2 affect the zeros of

g(x)?

c Use a graphics calculator to find the coordinates of the local maximums and local mini-

mums of g(x) and h(x). Comment on your answers.



s s

h

REVIEW OF CHAPTER 9 C


7 a

i y = (x ¡ 1)(x + 1)2 ii y = ¡2(x + 2)(x + 1)2 iii y = (x + 2)2(x ¡ 3)3

b Find the equation of a cubic function for which all of the interesting points lie in the

window (¡1, 1, 1; ¡1, 1, 1).

8 a Describe in words how the graphs of each of these functions is a transformation of thegraph of f (x).

i f (x + 2) ii ¡f (x ¡ 2) + 1

b If f (x) = x4, sketch by hand the graphs of each of these functions above. Check your


9 a The owner of a small business estimates that the profit from producing x items is given

by the function P (x) = 0:002x3 ¡ 1:5x2 + 350x ¡ 1000

i If current production levels cannot exceed 300 items due to limits on space and

resources, how many items should be produced to maximise profit?

ii If, through a re-organisation of the business, production levels can be increased to400 items (with the same profit function), how many items should be produced to

earn the maximum profit?

iii What is the maximum profit that can be earned at this increased production level?

b A business has won a contract to make a large number

of rectangular bins. The bins are to be open on top and

have a square base. These bins must have a volume

of 120 m3. To minimise costs (and therefore maximise

profits) you are given the job of designing a bin of the

required volume that has the minimum surface area.

The formula for the volume V of a bin of side s andheight h is V = s2h while the formula for surface

area is AS = area of vertical sides + area of square

base, or AS = 4sh + s2

i Use the volume formula and the fact that the volume of the bin is 120 m3 to write an

expression for h in terms of s.

ii Hence express the surface area in terms of s only.

iii Use a graphics calculator to sketch the graph of AS (y-axis) against s (x-axis).

iv Hence find the dimensions of the bin to be manufactured if the surface area is to be

a minimum.

1 a If f (x) = 1


You should check your answers with a graphics calculator.

i f (x) + 3 ii ¡f (x ¡ 2) ¡ 7

b Use your knowledge of transformation of functions to sketch the following sets of graphs

by hand. Use a graphics calculator to check your solution.

Draw on your graphics calculator the graphs of the functions below. Set your window toshow all points of interest.




...

...


i y = 1

x, y =

2

x, y =

¡1

4x

ii y = 1

x, y =

1

x ¡ 5, y =

1

x + 3

iii y = 1

x

, y = 1

x + 2 ¡1, y =

1

3x ¡1

c A manufacturer has sheets of metal

to be stamped out as circular discs

for identification tags for a chain of

pet shops. The sheets of metal are

6 metres by 2 metres. Let the discs

be d mm in diameter, and assume

that the discs are stamped out in a

regular rectangular pattern with no

gaps between the discs. The inde-

pendent variable is the diameter of the discs, and the dependent variable

is the number size of the discs.

i Find a function that gives the number of discs stamped from one sheet in terms of

the diameter of the discs (assume a fractional number of discs is possible here).

ii Sketch the graph of the function.

iii Using your function in part i , determine the number of discs with diameter 33 mm

that could be stamped from one sheet.

iv A grade 9 class works out correctly, using mensuration, that the number of discs of

diameter 33 millimetres that can be stamped from one sheet is 10 860. Comment on

the reason why this does not agree with your answer to iii above.

2 On a overland train journey breakfast, lunch and dinner are at 8 am, noon and 6 pm for prices

of $10, $15 and $25. A journey of 1800 km takes 36 hours and costs 30 cents per km. Plot

the cost of a journey leaving at midnight and arriving 36 hours later assuming that meals are

paid for as they are taken. Plot the cost versus the number of elapsed hours of travel time.

3 a A Pennyfarthing bicycle has one large wheel at the front and a small one at the rear. The

front wheel is 150 cm in diameter and the rear wheel is 30 cm in diameter. What is the

rotational speed in revolutions per minute of the rear wheel if the front wheel is rotating

at 30 revolutions per minute?

b The momentum of a body is equal to the mass times the velocity of the body. A collides

with B. During the collision mass A moving at 7 metres per second is stopped and mass

B moves off from rest. Mass A is half that of mass B and the momentum of mass A is

transferred to mass B during the collision. What will be the velocity of mass B after the

collision?

c For a cylinder with a given volume, the square of the radius of the cylinder varies inversely

with the height.

i Explain why this is true.

ii A cylinder with a base radius of 40 cm has a height of 25 cm. If the volume is to

remain constant, what is the height if the base radius is 3 cm?



x

y

64224

4

2

2

4

x

y

10.50.51

2

1.5

1

0.5

0.5

y

x4224

2

2

4

6

8


4 a Sketch the graph of each of these equations:

i x2 + y2 = 4 ii (x + 2)2 + y2 = 16

iii 2(x ¡ 2)2 + 2(y + 4)2 = 50

b Give the equation of each graph below:

i ii iii

c Write each of these equations of a circle in the general form:

i (x + 3)2

+ (y ¡ 4)2

= 9 ii 2(x ¡ 3)2

+ 2y2

= 6d Write each of these equations of a circle in the centre-radius form:

i x2 ¡ 4x + y2 + 3 = 0 ii x2 + y2 ¡ 6x + 4y ¡ 11 = 0

iii x2 + y2 ¡ 3x + 2y = 0

e Sketch the graph of this circle on a graphics calculator: x2 + y2 ¡ 6x + 4y ¡ 11 = 0

5 a Show that the lines 2x + 4y ¡ 2 = 0, x ¡ y ¡ 4 = 0 and x ¡ 7y ¡ 10 = 0 are

concurrent.

b i Find the points at which the line 4x + y ¡ 5 = 0 intersects the parabola

y = x2

¡3x + 3

ii Find the points where the line y + x ¡ 3 = 0 meets the circle x2 + y2 = 5

iii Prove that the line 2y ¡ x ¡ 2 = 0 touches the circle x2 + y2 + 2x ¡ 6y +5 = 0

iv Show that the circles x2 + y2 = 5 and x2 + y2 ¡ 8x + 16y + 35 = 0 touch

each other.

c i Find the points of intersection of the line x + 2y = 11 and the hyperbola

xy = 15

ii Solve algebraically this system of equations: 2x + y ¡ 5 = 0 and xy ¡ 2x2 = 0

d Use a graphics calculator or graphing software to find the points of intersection of the

following curves:the circle x2 + 2x + y2 ¡ 6y ¡ 3 = 0 and the parabola y = x2 + 3x ¡ 3

6 a Classify each of these relations as one-to-one, many-to-one, one-to-many or many-to-

many.

i ii iii iv




b Consider a family consisting of Nanna, her daughter Claire, who is married to Steve and

their two daughters Alicia and Portia. Classify each of these relations as one-to-one,

many-to-one, one-to-many or many-to-many. Justify your answers.

i x ‘is the daughter of’ y ii x ‘is married to’ y

iii x ‘has the granddaughter’ y iv x ‘has the child’ y

1 a i In comparing two silicon chip manufacturing processes, a production manager counted

11 defective chips in a batch of 2000 chips from process A, and 8 defective chips in

1500 chips from process B. Which process was more reliable?

ii The distance from the Moon to the Earth is about 385 000 kilometres. Assuming that

the path of the Moon is a circle, what is the speed of the Moon around the Earth?

iii A stall holder at a local market sold 35 leather wallets at $14 each, and a further 21wallets at the discounted price of $10 each. What was the average price per wallet?

b i Fred received 27 out of 30 marks for the mid-semester exam, 90 out of 100 for assignment work and 52 out of 70 in the final exam. Fred added the marks gained

and the total marks available and calculated the percentage of marks gained.

If assignments count 10%, the mid-semester test counts 20%, and the final exam

70%, did Fred calculate his final percentage correctly? If not, how should he have

calculated it?

ii A car travelled from Kenmore to King George Square at an average speed of 40km/h. What speed must it average on the return journey so the overall average speed

is 50 km/h?

iii A 4 litre can of paint is said to be sufficient to cover 8 sq. metres. What average

thickness of paint does this imply, before evaporation takes place?

c In Queensland before metric measure was adopted, land was often measured in perches,

an abbreviation of square perches in fact. One linear perch is 5.5 yards and one square

perch is 30.25 square yards. An acre is 4840 square yards. How many perches are there

in one acre?

Given that one yard is 36 inches and 1 metre is 39.37 inches, how large in square metres

is a suburban block of 24 perches?

2 a From the table on page 310, determine the average speed for the section from:

i Gympie to Gladstone ii Nambour to Rockhampton

b The table shows the population of Australia over 45 years.

Calculate the annual rate of growth of Australia’s

population over the periods:

i 1950 - 1970

ii 1990 - 1995

Year Population (millions)

1950 82671960 10 3611970 12 6601980 14 6161990 17 0331991 17 2691992 17 4931993 17 7001994 17 8931995 18 079

REVIEW OF CHAPTER 10 D







c

Use the graph to answer the following questions. Show how you arrived at your answers..

i What happened between 3 a.m. and 6 a.m.?

ii When is the rate of evaporation the greatest?

iii Use the graph to estimate the rate of evaporation at 3 p.m.

3 The graph on the right shows the velocity of

a car over a period of sixteen seconds. Recall

that a change in velocity over time is called

acceleration.

a Describe what the car was doing in the

first six seconds.

b What was the average velocity during this

time?

c Describe the motion of the car between

the sixth and tenth seconds.

d What did the car do from the tenth to the fourteenth second?

e What was the acceleration between the fourteenth and sixteenth second?

f What was the total distance travelled in these sixteen seconds?

4 Draw a velocity time graph for this story. You should supply your own distances and times.

Every morning, Jenna walks briskly to the coffee shop. She has her coffee while reading the

Courier Mail . She walks briskly to the corner shop, where she usually buys a few items.

Carrying her purchases, she strolls back home.

5

The graph below shows the depth of water in shallow dish during one hot day inCairns, over hour period.

aa 24

8

2

10

14

4

12

6

03 a.m. 6 a.m. 9 a.m. noon 3 p.m. 6 p.m. 9 p.m. midmid.

mm

15

6

18

24

9

21

12

0

3

82 104 126 14 16

k m

/ h o u r

seconds

200

80

240

320

120

280

160

0

40

4010 5020 6030 70 80

m e

t r e s

minutes

The height of hot air balloon after its launchis shown in the graph. Draw the graph show-ing vertical speed (in metres per minute) for this balloon.

a




6 Three children, aged 8, 5 and 3 had a

bath together. The graph below shows

the water level at various times. Make

up a story to match the graph.

7 Shown below is a height versus time graph for a parachutist doing a parachute jump.

a The jump can be divided into four portions - from P to Q, from Q to R, from R to S and

from S to T. Explain what is happening during each of these portions.

b During which stage of the jump was the parachutist falling fastest?

c Estimate how fast he was falling in metres per second during this section.d How fast was he falling between S and T?

e At what speed did he hit the ground?

f Estimate how fast he was falling between Q and R.

8 a On graph paper, accurately sketch each of the functions given below. Draw a tangent line

to the function at the given point. Estimate the gradient of the tangent line, and hence the

gradient of the function at that point.

Function Coordinates

i h(

x) = 3 ¡

x3 (1

, 2

)

ii i(x) = x2 ¡ 3 (1, ¡2)

b For each function in a, make a table of secants, and hence estimate the gradient of the

function at that point. Compare these answers to those you found by drawing the tangent

line.

c Consider the function y = 1 ¡ x2. Estimate the gradient of the function at each of

these points:

i (0, 1) ii (1, 0) iii (2, ¡3) iv (3, ¡8) v (4, ¡15)


6:00 pm 6:05 pm 6:10 pm 6:15 pm 6:20 pm 6:25 pm

10

20

30

40

depth (cm)

time

4 8 12 16

500

1000

1500

height (m)

time (min)

P Q

R

S

T



REVIEW OF CHAPTER 11 E


1 a Refer to the diagram on page 327. Consider the velocity of the marble rolling down the

incline when it is at B, i.e., after 1 second. By considering the average velocity for small

time intervals just after 1 second, determine the instantaneous velocity of the marble

b Assume the displacement s at time t is given by s = 13 t3

i Where is the marble after 3 seconds?

ii What is a likely value instantaneous velocity at this time?

2 a Evaluate i limx!3

x ¡ 5

x + 2 ii lim

x!¡23

b Using f (x) = 2x2, g(x) = x

3 and a = 3, verify the Limit Laws.

c By first simplifying each of the rational expressions, evaluate each of the following limits.

Use a graphics calculator to check your answers.

i limx!¡1

x + 1

x2 ¡ 1 ii lim

x!2

x2 + 2x ¡ 8

x ¡ 2

d By writing as a single simplified fraction, find lima!0

(x + a)¡1 ¡ x¡1

a

e Evaluate

i limx!1

¡1

x + 2 ii lim

x!18x + 4

2x ¡ 3 iii lim

x!1x2 ¡ 30x ¡ 1000

x ¡ 1

3 a Use your graphics calculator to evaluate 1 + 13 1 + 1

3 + 19 1 + 1

3 + 19 + 1

27

b Continue this pattern for another three terms.

c Make a conjecture about the value of 1 + 13 + 1

9 + 127 + ::::, i.e., if we sum an infinite

number of terms.

4 For each of the following functions, find dy

dx from the definition of a derivative.

a y = x4 b y = 4x3 c y = 3x2 ¡ x ¡ 1

d y = x3 ¡ 6x + 5 e y = 1

(x + 1) f y = (2x + 1)(x ¡ 3)

5 a Differentiate by rule.

i y = x5 ii y = x6 iii y = 5x2

iv y = ¡3 v y = ¡x2 vi y = 13 x4

vii y = t¡3 viii y = 4x4 ¡ 4x + 7x2

b Differentiate by rule.

i s(t) = 3t¡2 ii s(x) = 5x¡1 iii f (x) = x¡2 + x¡4 + x¡6

iv f (x) = 5

x¡1 v g(x) =

¡2

x4 vi y =

1

x2 + 5

vii y = x4

3 viii y =¡

2p

x ix y = 2p

x¡

4x

at B.





c Find

i d

dx

µ 3

4p

x

¶ ii

d

dx

µ34

x2 ¡ 1p

x

¶ iii

d

dp

µ p ¡ 7 p2

p2

¶

iv d

dx

µ 3p

x3

¶ v

d

dx(4x ¡ 2 ¡ 4

p x) vi

d

dt

µ ct

2t2 ¡ b

7t4

¶

d i If s = 3t ¡ 7t3, find dsdt

: ii What is the value of dsdt

when t = 2?

e The formula for the volume of a cone of height h and base radius r is is V = 13 ¼r2h

i Find dV

dr assuming that h is constant.

ii Find dV

dh assuming that r is constant.

6 a The motion of a particle is given by s = 2t3

3 ¡ 7t2

2 + 3t where t is the time in

seconds and s is the displacement in metres. Find i expressions for its velocity and acceleration at any time t

ii when the particle is at rest, i.e., when v = 0

iii when it is travelling with a constant velocity, i.e., when the

acceleration, a , is zero.

b A test rocket’s height in metres above its launching point is given

by h = 90t2 ¡ 2t3 where t is the time in seconds after launch.

i Find the velocity dh

dt of the rocket.

ii At the top of its flight, the velocity of the rocket will

be zero. Find this maximum height.

c The annual inventory cost for a manufacturer is C = 150000

x + 5x, where x is the

order size when the new stock is purchased. Determine the rate of change in inventory

costs when the order size = 50.

7 a Find the equation of the tangent and normal at the points where x has the given value.

i y = x(4 ¡ x), x = 3 ii y = x3 ¡ 10x + 5, x = 3

b Find the point(s) on the curve y = x3 + 6x2 + 12x + 4 where the tangent is parallel

to the x-axis.

c Find the coordinates of the point at which f (x) = ¡x ¡ 1 will touchg(x) = x2 ¡ 5x + 3. Show that f (x) is the tangent to g(x) at that point.

8 a Differentiate each of the following functions.

i y = (x ¡ 8)4 ii y = (3x ¡ 2)2 iii y = (3 ¡ x)6

iv y = (x3 ¡ 3x2 + 1)9 v y = (x ¡ 5)¡3 vi y = (6 ¡ x)¡2

vii y = 2

(x ¡ 5)4 viii y = 3

p 2x + 4 ix y =

1

x2 + 4x ¡ 2

x y = 1p

2x2

¡ 3x + 4



x

d

8 m

B P

A


b i Differentiate y = (x ¡ 1)4 by using the function of a function rule.

ii Expand (x ¡ 1)4 by the binomial theorem, and then differentiate the resulting

expression.

iii Show that these two results are the same.

c The velocity of a tilt train over a section of its journey is given by

v = 2 ¡ 1(t + 1)2

metres per second. Find the instantaneous acceleration at t = 3

seconds.

d If a tank holds 3000 litres of water which takes 20 minutes to drain from the tank, then

the volume of water remaining in the tank after t minutes is given by

V = 3000

µ1 ¡ t

20

¶2

, 0 6 t 6 20

Find the rate at which the water is flowing out of the tank after

i 3 min ii 20 min

9 It is known that the intensity of light at a pointP on the ground, from a light x metres above the

ground, is given by I = x

d3

The diagram shows a point AB which is 8 metres

horizontally from P. The light at A is x metres

above the ground.

a Find I as a function of x.

b On your graphics calculator, draw a graph of I versus x for [¡1, 10, 0:1; ¡0:001, 0:01,

0:001].

c From the graph, find the value of x which gives maximum intensity at P.

10 a Differentiate using the product rule:

i y = (3x + 1)(2x ¡ 2) ii y = 3(2x + 1)(x ¡ 7)

iii y = (x2 ¡ 3x3)(6x4 + 1) iv s = (2t ¡ 3)4(6 ¡ 5t)3

b Find dy

dx if

i y = (x + 1)p

x ¡ 1 ii y = (x + 3)3p

x

iii y =p

xp

3 + x iv y = (x2 + 3)2(2x ¡ 5)¡3

c Students often ask why the product rule is not applied to functions such as y = 2x. Itcan be! Show this, by differentiating using the product rule:

i y = 4x ii y = x3 (Think of this as the product x £ x2)

11 a Differentiate using the product rule:

i y = (3x ¡ 5)

(x + 2) ii y =

(2x + 4)3

(5x ¡ 3)5

b Differentiate using the quotient rule:

i y = 8x + 5

3x

¡7

ii y = (x ¡ 5)2

(4

¡x2)3




c Differentiate, using whichever method you prefer:

i y = 3x ¡ 6

2x + 4 ii y =

(x ¡ 4)2

(2x + 1)3 iii y =

x + 3p 3x ¡ 1

d Differentiate y = 2

(3x ¡ 1)3

i by rewriting as a product, and then using the chain rule

ii by rewriting as a product, and then using the product rule

iii using the quotient rule.

iv Which method of differentiating this function do you prefer? Why?

e Use your knowledge of algebraic fractions and index laws to first simplify these functions,

and then differentiate.

i y = 3x ¡ 4

x ii y =

4x + 5

(4x + 5)2

12 Differentiate these functions, using any appropriate method:

a y = 2(4x ¡ 1)3 b y = 3

x + 35 c y =

(4x + 1)2

2x3

d y = 5x ¡ 5

x3

2

e y = x2(2x ¡ 3)(3x + 2) f y =

p 5 ¡ xp

x3 + 2

13 For each function below:

i For what values of x is the function zero?

ii For what values of x is the function positive?

iii For what values of x is the function negative?

iv For what values of x is the gradient zero?

v For what values of x is the gradient positive?

vi For what values of x is the gradient negative?

a b

c d

14 For each function in question13, for what values of x is the gradient of the function increasing?

y

x

x = 2

(-2' 0)(3' 0)

(1' 2)

y

x

y

x

(-5' 0)(-2'-2)

(-1' 0)

(0' 1) (4' 6)

(6' 0)

y

x(-1' 0) (0'-1)



y

x

(2, 2)

( 2, )

y

x

@=3!+4

y

x

t h o u s a n

d $

year

y

x

d e a

t h s

year


15 Draw neat sketches of functions for which

a the gradient is positive and the gradient is increasing

b the gradient is positive and the gradient is decreasing

c the gradient is negative and the gradient is increasing

d the gradient is negative and the gradient is decreasing.

16 For each of these functions, draw a neat sketch of a gradient function.

a b

c d

17 a A function has a gradient of zero at (¡2, 5) and (0, 3). The gradient is positive for values

of x < ¡2 and x > 0, and negative for ¡2 < x < 0. Draw a sketch of the function.

b A function has a gradient of zero at (3, ¡9) and (0, 0). The gradient is negative for all

other values of x. Draw a sketch of the function.

c A function has zeros at ¡2 and 6, and a zero gradient at (1, 3).

18 a The gradient of a function is zero for all values of x. What does this tell you about the

function?

b The gradient of a function is negative for all values of x. What does this tell you about

the function?

19 a A function has its derivative equal to ¡2 at (¡4, 2), turning points at (¡2, 1) and (0, 4)

and its only zero at (3, 0). Sketch a possible graph of this function.

b A function has a domain that consists of all real numbers, is positive for all values of x,

and has a derivative that is negative for all values of x. It has a horizontal asymptote aty = 1. Sketch a possible graph of this function.

20 For each of the graphs below, make up a news story to match the graph.

a b

y

x

@=4-!S

y

x

x = 2



Maintainingbasicknowledge

andprocedures

SUBJECT MATTER

measurement

percentage

basic algebraic procedures

rational numbers and expressions

linear equations and inequations

formulae

Pythagoras Theorem

distance and mid-point formula

sigma notation

APPENDIX 1



LENGTH AND PERIMETER

The measurement of length, area and volume is of great importance. Constructing a tall building,

building a long bridge across a river, rendezvousing in space to repair a satellite, all require use

of measurement with skill and precision. Builders, architects, engineers and manufacturers need tomeasure the sizes of objects to considerable accuracy. The most common system of measurement

is the International System of Units commonly called the Metric System.

The tables alongside summarise the units of

length and conversion factors. The basic unit

of length in the International System of units

is the metre.

1 kilometre (km) = 1000 metres (m)

1 centimetre (cm) = 1100 m = 0:01 m

1 millimetre (mm) = 11000 m = 0:001 m

Conversions:

1 m = 100 cm

= 1000 mm

= 11000 km

1 km = 1000 m

= 100 000 cm

= 1000000 mm

1 cm = 10 mm

= 1100 m

1 mm = 110 cm

= 11000 m

PERIMETER FORMULAE

The distance around a closed figure is its perimeter. For some shapes we can derive a formula for

perimeter. The formulae for the most common shapes are given below:

Rectangle Circle Arc

P = 4l P = 2(l + b) P = a + b + c C = 2¼r l =¡

µ360

¢2¼r

or C = ¼d

Note: The formula for the length of an arc is relatively simple to derive using the formula for

the circumference of a circle. See if you can derive this for yourself.

MEASUREMENT A

The metric system uses prefixes to indicate an increase or decrease in the size of a unit.

Prefix Symbol Meaning Prefix Symbol Meaning

terra T 1 000 000 000 000giga G 1 000 000 000

mega M 1 000 000kilo k 1 000

hecto h 100deka da 10

deci d 0.1centi c 0.01milli m 0.001

micro 0.000 001

nano n 0.000 000 001 pico p 0.000 000 000 0001

l b

l

ba

c

r

r d

Square Triangle

386 MAINTAINING BASIC KNOWLEDGE AND PROCEDURES (APPENDIX 1)



Find the perimeter of

a b

a Perimeter b Perimeter

= (2 + 3 + 1 + 3 + 2 + 4) cm = +6.4 7.85 +m mm

= 15 cm = 19.2 m

1 a Find the circumference of a circle of radius 13:4 cm. b Find the length of an arc of a circle of radius 8 cm and angle 120o:

2 Find the perimeter of the following shapes:

a b c

3 Solve the following problems:

a Determine the length of fencing around an 80 m by 170 m rectangular playing field, if

the fence is to be 25 m outside the edge of the playing field.

d A cyclist used a bicycle with a ratio of pedal revolutions to wheel revolutions of 1 : 6. If

the diameter of a wheel is 70 cm, how long would it take the cyclist to travel 40 km if

the cyclist averaged 32 pedal revolutions per minute?

2 cm

4 cm

2 cm

3 cm

3 cm

1 cm

5 m 6.4 m

7.8 m

EXAMPLE 13.1

EXERCISE 13A.1

3 cm5 cm

10 cm 4 cm

5 m

2 m

b

c

i

ii

A soccer goal net has the shape shown. If thenetting has cm by cm square gaps, what isthe total length of cord needed to make the

back rectangle of the net?

‘Ideal athletics tracks’ are m long with two“straights” and semi-circular ends of diameter

m. Find

the length of each straight to the nearest cm

the distance the

5 5

400

80

athlete in the next toinnermost lane must start in front of theathlete inside him so they both run m.Each lane is m wide.

4001

80 m

MAINTAINING BASIC KNOWLEDGE AND PROCEDURES (APPENDIX 1) 387



AREA

The area of a closed figure is the number of square units it contains.

As for the length measurements, we will use the International System of Units. The relationships

between the various units of area can be obtained using the conversions for length.

For example, since 1 cm = 10 mm

then 1 cm £1 cm = 10 mm £10 mm

i.e., 1 cm2 = 100 mm2

AREA UNITS & CONVERSIONS

1 cm2 = 10 mm £ 10 mm = 100 mm2 fsquare centimetreg1 m2 = 100 cm £ 100 cm = 10000 cm2 fsquare metreg1 ha = 100 m £ 100 m = 10000 m2 fhectareg

1 km2 = 1000 m £

1000 m = 1000000 m2 or 100 ha.

The formulae for calculating the area of the most commonly occurring shapes should be familiar

to you and are given below:

Area = length £ breadth Area = 1

2(base £ height)

Area = base £ height Area =·

average length of parallel sides¸£·distance betweenparallel sides

¸

Area = ¼r2 Area =

µ

360£ ¼r2

Rectangle:

length

breadth

Triangle:

base base

height

1 cm

10 mm

Trapezium:

Circle:

r

Sector:

r

Parallelogram:

height

base

height




Find the area of each of the following figures:

a b

a Area = 12 (base £ height)

= 12 £ 10 £ 7 cm2

= 35 cm2

b Area = µ360 £ ¼r2

= 60360 £ ¼ £ 82

+ 33:51 cm2

Find the shaded area of the following figures:

a b

a Area = trapezium area¡ triangle area

=

µ12 + 18

2

¶£ 12 ¡ 1

2 £ 10 £ 6

= 15 £ 12 ¡ 5 £ 6

= 150 cm2

b Area = area of large semi-circle¡ area of small semi-circle

= 12 £ (¼ £ 82) ¡ 1

2 £ (¼ £ 52)

= 61:26 cm2

A sector has area 25 cm2 and radius 6 cm. Find the angle subtended at the centre.

Area =¡

µ360

¢¼r2

) 25 = µ360 £ ¼ £ 62

) 25 = µ¼

10) µ¼ = 250

) µ = 250

¼ + 79:58

) the angle measures 79:58o:

EXAMPLE 13.2

EXAMPLE 13.3

EXAMPLE 13.4

8 cm

60°

7 cm

10 cm

6 cm

10 cm

12 cm

18 cm 3 cm5 cm

25 cm2

6 cm°




1 Converta 12 cm2 into mm2 b 7 m2 into cm2 c 2 km2 into m2

d 9 m2 into mm2 e 200 mm2 into cm2 f 13 ha into m2

2 Find the area of the following figures:a b c


a Find the angle of a sector with area 30 cm2 and radius 12 cm.

b What is the cost of laying ‘instant lawn’ (turf) over an 80 m by 120 m rectangular playing

field if the turf comes in 12 m wide strips and costs $15 for 10 m?

c A cropduster can carry 240 kg of fertiliser

at a time. If it is necessary to spread 50 kg

of fertiliser per hectare, how many flights are

necessary to fertilise a 1:2 km by 450 m rect-

angular property?

SURFACE AREA

² Solids with plane faces

The surface area of a three dimensional solid with plane faces is the sum of the areas of

the faces.

Find the surface area of the rectangular prism.

The figure has 2 faces of 6 cm £ 4 cm

2 faces of 6 cm £ 3 cm

and 2 faces of 3 cm £ 4 cm.

Total surface area = [2 £ 6 £ 4 + 2 £ 6 £ 3 + 2 £ 3 £ 4] cm2

= [48 + 36 + 24] cm2

= 108 cm2:

EXERCISE 13A.2

6 cm

14 cm

5 cm 4 cm 10 cm

100 m

50 m

EXAMPLE 13.5

6 cm4 cm

3 cm




² Solids with curved surfaces

There are three shapes for which a simple formula can be derived:

Cylinder: Cone: Sphere:

Area = area of ends

+ area of curved surface

Area =2¼r2+2¼rh

Area = area of base

+ area of curved surface

Area =¼r2+¼rs

(s is called the slant height)

Area=4¼r2

The formulae for a cone and a cylinder can be proved relatively easily as shown below, but the proof for a sphere is beyond the level of this course.

Cylinder: The net for a cylinder is used to derive the formula for the surface area.

Notice that the curved part of a cylinder is made from a rectangle of length 2¼r and width h.

Surface area of closed cylinder = 2 £ (area of circle) + area of rectangle

= 2(¼r2) + 2¼rh

= 2¼r2 + 2¼rh

Cone: The curved surface of a cone is made from a sector of a circle with radius equal to theslant height of the cone. Also, the circumference of the base must be the same as the arc

length of the sector.

r

s

r

h

r

h

cut

h

r

r

2 r

r

r

s

cut

2 r r

s

B

A°




arc AB =

µ µ

360

¶2¼s

But arc AB = 2¼r

)

µ µ

360¶2¼s = 2¼r

) µ

360 =

r

s

Hence, the total area

A = area of circle + area of sector

= ¼r2 +

µ µ

360

¶¼s2

= ¼r2

+³r

s´

¼s2

= ¼r2 + ¼rs

Find the surface area of a solid cone of base radius 5 cm and height 12 cm.

Let the slant height be s cm.

s2 = 52 + 122

fPythagoras

g) s2 = 169

) s =p

169 = 13 fas s > 0g Now A = ¼r2 + ¼rs

) A = ¼ £ 52 + ¼ £ 5 £ 13

) A + 282:7

Calculator: 5 5 13

Thus the surface area is approximately 282:7 cm2:

1 Find the total surface area of

a a box 20 cm by 15 cm by 8 cm

b a sphere of radius 6 cm

c a hemisphere of base radius 10 m

d a cylinder of base radius 9 cm and height 20 cm

e a cone of base radius and perpendicular height both 10 cm

f a cone of base radius 8 cm and vertical angle 60o


a The cost of manufacturing a hemispherical glass dome is given by $(5200 + 35A)where A is its surface area in square metres. Find the cost of making a glass hemispherical

dome of diameter 10 m.

b Determine the cost of painting the exterior walls and top of a cylindrical wheat silo which

is 40 m high and 20 m in diameter, given that each litre of paint costs $7:25 and covers

8 m2:

c Determine the total area of leather in 20 dozen cricket balls each with diameter 7 cm.

+× × × = x 2

EXAMPLE 13.6

s cm

12 cm

5 cm

EXERCISE 13A.3

8 cm

60°




VOLUME AND CAPACITY

The volume of a solid is the amount of space it occupies, i.e., the number of cubic

units of space it occupies.

The capacity of a container is the volume of fluid, (liquid or gas), that it may contain.

The units of volume and capacity are very closely related and are summarised below:

Volume and capacity - units and conversion

1 cm3 = 10 mm £ 10 mm £ 10 mm = 1000 mm3 fcubic centimetresg1 m3 = 100 cm £ 100 cm £ 100 cm = 1000000 cm3 fcubic metresg1 mL = 1 cm3 fmillilitreg

1 L = 1000 cm3 = 1000 mL flitreg1 kL = 1000 L = 1 m3 fkilolitreg

1 ML = 1000 kL fmegalitreg

The following volume formulae are very common and have been considered in previous years.

Volume of prism = area of base £ height

Volume of a pyramid or cone = 1

3(area of base £ height)

Spheres

Volume of a sphere = 4

3¼r3

Prisms (solids of uniform cross-section)

Pyramids and cones

Rectangular prism Cylinder Uniform solid

height height

height base base

base

base

Square-based pyramid

height

base

Triangular-based pyramidor tetrahedron

height

Cone

base

height

r




Find the volume of the following:

a b

a Volume = area of base £ height

= length £ breadth £ height

= 16:8 £ 8 £ 4:5 cm3

= 604:8 cm3

b Volume = area of base £ height

= ¼ £ 72 £ 10 cm3

+ 1539 cm3

1 Find the volume of

a a rectangular box 12 cm by 15 cm by 10 cm

b a cone of radius 10 cm and slant height 18 cm

c a cylinder of height 3 m and base diameter 80 cm

d an equilateral triangular prism with height 12 cm and triangles of side 2 cm

2 a A sphere has surface area 100 cm2. Find its volume.

b A sphere has volume 27 m3. Find its surface area.


a The contents of a rectangular tank 4 m by 3 m by 2 m are emptied into a cylindrical tank

of base diameter 5 m. How high will the water level rise?

b Determine the volume of water required to fill a

swimming pool with dimensions shown alongside.

c A concrete path 1 m wide and 10 cm deep is placed around a circular lawn of diameter

20 m. Find the volume of concrete required to lay the path.

d How long would it take to fill a rainwater tank which is cylindrical with base diameter

2:4 m and height 4 m, if the water enters it at 80 litres per minute?

e A cubic metre of brass is melted down and cast

into solid door handles with shape as shown.

How many handles can be cast?

EXAMPLE 13.7

4.5 cm

16.8 cm8 cm

10 cm

14 cm

EXERCISE 13A.4

1.5 m

3 m

25 m

8 m

2 cm




Percentages are comparisons with the whole amount (which we call 100%).

% reads per cent which is short for per centum.

Loosely translated from Latin, per cent means out of every hundred.

If an object is divided into one hundred parts then each part is called 1 per cent, written 1%:

Thus, 1100 = 1% and 100

100 = 100%

From this it may be seen that a percentage is really a special case of a fraction where the denominator

is 100. Consequently all fractions and decimals may be easily converted into percentages.

Convert the following into percentages by multiplying by 100%:

a 35 b 0:042

a 35

= 35 £ 100% f100% = 1g

= 60%

b 0:042

= 0:042 £ 100% f100% = 1g= 4:2% fshift dec. point 2 places rightg

Express as decimals: a 88% b 116%

a 88%

= 0:88

= 0:88

b 116%

= 1:16

= 1:16 fshift decimal point 2 positions leftg

A TV set has a sale price of $550 after it has been marked up by 25%.

At what price did the shopkeeper buy the TV set?

cost price £ multiplier = selling price

) C £ 125% = $550

) C £ 1:25 = $550

) C = $550

1:25

) C = $440

i.e., it cost the shopkeeper $440:

PERCENTAGE B

EXAMPLE 13.8

EXAMPLE 13.9

EXAMPLE 13.10




1 Convert the following into percentages:

a 1425 b 3

4 c 0:92 d 1:16

e 1:024 f 4 g 1 h 1120

2 Express as fractions in lowest terms:

a 85% b 42% c 105% d 15%

e 33 13 % f 7 1

2 % g 6 14 % h 132%

i 16 23 % j 48% k 160% l 0:25%

3 Express as decimals:

a 92% b 106% c 112:4% d 88:2%

e 7:5% f 1% g 256% h 0:05%

i 1150% j 0:0037% k 342:8% l 63:7%

4 A bicycle costs $160 and is sold for $280. Calculate the profit as a percentage of the

a cost price b selling price.

5 A greengrocer buys fruit and vegetables from the market and sells them at a 25% mark up.

On one particular morning, her fruit and vegetables cost her $500.

a What will be her profit if she should sell all of her produce?

b Express this profit as a percentage of her sales.

6 A 30 m roll of wire mesh was bought wholesale

for $216. If it is sold for $8:50 per metre, find

the profit and express it as a percentage of thewholesale (cost) price.

7 Find the cost to a retailer who sells a television

set at $640 for a 25% profit on the cost price.

8 An electrical firm sells a washing machine for

$585, having made a 30% profit on the wholesale

(cost) price. Find the wholesale price.

SIMPLE INTEREST

I = P £ r£ t

where P is called the principal

r is called the rate of interest, (p.a.)

t is the time in years

EXERCISE 13B.1

If is borrowed for years at p.a. (per annum) simple interest, then thecan be calculated by using the formula

$ , ,P t r I% simple interest




Find the simple interest payable on an investment of $20000 at 12% p.a. over a period of 4years.

P = $20000 r = 12% = 0:12 t = 4 years

I = P £ r £ t

) I = $20000 £ 0:12 £ 4

) I = $9600

) simple interest is $9600

Find the simple interest payable on an investment of $10000 at 8% p.a. over a period of 9months.

P = $10000 r = 8% = 0:08 t = 0:75 fas 9 months = 34 year = 0:75 yearsg

I = P £ r £ t

) I = $10000 £ 0:08 £ 0:75

) I = $600

) simple interest is $600

1 Find the simple interest payable on an investment of

a $4000 at 8% p.a. for 5 years b $1500 at 11% p.a. for 6 months

c $2500 at 10 12 % p.a. for 2 years d $20000 at 12 1

4 % p.a. for 9 months

2 How long will it take for a $10000 investment to earn $10000 interest if it is invested at 10% p.a. simple interest?

3 $8000 becomes $9800 when invested over a 10 year term where simple interest is paid annually.

What rate of interest is being paid to the investor?

COMPOUND INTEREST

We can write a formula for Compound Growth:

A= P £ (1+ i)n

where A = final amount , (i.e., principal + interest)

P = principal (the original amount)

i = interest rate per annum (expressed in decimal form)

(1+ i) = the multiplier


EXAMPLE 13.11

EXAMPLE 13.12

EXERCISE 13B.2




Notice that the formula for A above gives the total final amount (i.e., principal plus interest).

To find the interest only we use:

Compound Interest = final amount ¡ principal = A¡ P

What will $5000 invested at 8% p.a. compound interest amount to after 2 years?

How much interest is earned?

An interest rate of 8% indicates a multiplier of 1:08.

After 2 years it amounts to $5000 £ (1:08)2

= $5832

) interest earned = $5832 ¡ $5000

= $832

1 a What will an investment of $3000 at 10% p.a. compound interest amount to after 3 years?

b What part of this is interest?

2 $5000 is invested for 2 years at 10% p.a. What will this investment amount to if the interest

is calculated as

a simple interest b compound interest?

3 You have $8000 to invest for 3 years and there are 2 possible options you have been offered:Option 1: Invest at 9% p.a. simple interest.

Option 2: Invest at 8% p.a. compound interest.

a Calculate the amount accumulated at the end of the 3 years for both options and decide

which option to take.

b Would you change your decision if you were investing for 5 years?

4 What percentage increase will occur if I invest any amount over a 4 year period at 10% p.a.

compound interest? [Hint: Let the principal be $1000.]

Consider the number crunching machine alongside:

EXAMPLE 13.13

EXERCISE 13B.3

BASIC ALGEBRAIC PROCEDURES C

ALGEBRAIC SUBSTITUTION

5 – 7 calculator x

input, x

output




If we place any number x, into the machine, it calculates 5x ¡ 7, i.e., x is multiplied by 5 and

then 7 is subtracted.

For example, if x = 2, 5x ¡ 7

= 5 £ 2 ¡ 7

= 10 ¡ 7

= 3

and if x = ¡2, 5x ¡ 7

= 5 £ (¡2) ¡ 7

= ¡10 ¡ 7

=¡

17

If p = 4, q = ¡2 and r = 3, find the value of

a 2 pq ¡ r b p

4r + pq c p ¡ q ¡ r

r ¡ p

a 2 pq ¡ r

= 2(4)(¡2) ¡ 3

=

¡16

¡3

= ¡19

b p

4r + pq

=

p 4(3) + (4)(¡2)

= p 12 ¡ 8=

p 4

= 2

c p ¡ q ¡ r

r ¡ p

= 4 ¡ (¡2) ¡ 3

3 ¡ 4

= 4 + 2 ¡ 3

¡1

= 3

¡1

= ¡3

1 If p = 4, q = ¡3, r = ¡2 and s = ¡1, evaluate

a 2q ¡ 3 p b 6 pq

s c (2r ¡ s)2 d s p

ep

p2 + q 2 f 3

p ¡ r g 2q 3 h

p + q ¡ r

p ¡ q + r

2 For u = ¡5, v = 4, w = ¡2, x = 1, y = 6 and z = 3, evaluate

a (2u)2 b 2u2 c 2u ¡ v

x d

4y

3z

e u

w ¡ x f

¡v

w2 g

uv + w

y

¡u

h x + z2

2u

PRODUCTS AND EXPANSIONS

A product of two (or more) quantities is the result obtained when multiplying them together.

To make products easy to perform, follow the procedure below:

Step 1: multiply the signs

Step 2: multiply the numerals (numbers)

Step 3: multiply the pronumerals (letters)

EXAMPLE 13.14

EXERCISE 13C.1

¡3xsign coefficient variable




So,

Simplify the following products:

a ¡3 £ 4x b 2x £ (¡x2) c ¡4x £ (¡2x2)

a ¡3 £ 4x b 2x £ (¡x2) c ¡4x £ (¡2x2)

= ¡12x = ¡2x3 = 8x3

Other simplifications are possible using the expansion (or distributive) rules:

a(b + c) = ab + ac

a(b ¡ c) = ab ¡ ac

Geometric Demonstration:

Expand the following:

a 3(4x + 1) b 2x(5 ¡ 2x) c ¡2x(x ¡ 3)

a 3(4x + 1) b 2x(5 ¡ 2x) c ¡2x(x ¡ 3)

= 3 £ 4x + 3 £ 1 = 2x £ 5 ¡ 2x £ 2x = (¡2x) £ x ¡ (¡2x) £ 3

= 12x + 3 = 10x ¡ 4x2 = ¡2x2 + 6x

Note:

² In practice we do not usually include the second line of working.

² In c , observe the placing of a bracket around ¡2x.

Expand and simplify a 2(3x ¡ 1) + 3(5 ¡ x) b x(2x ¡ 1) ¡ 2x(5 ¡ x)

a 2(3x ¡ 1) + 3(5 ¡ x)

= 6x ¡ 2 + 15 ¡ 3x

= 3x + 13

b x(2x ¡ 1) ¡ 2x(5 ¡ x)

= 2x2 ¡ x ¡ 10x + 2x2

= 4x2 ¡ 11x

¡ ¡ ¡

33

x £ £££ +

22

xxx

2

2

== ==

¡66

xx

3

3

EXAMPLE 13.15

b c

a

b c+

EXAMPLE 13.16

EXAMPLE 13.17

The overall area is and could also be found by adding the areasof the two small rectangles, i.e.,

Hence, equating areas

a b c

a b c

( + )+

( + ) = +

,.

.

ab ac

ab ac f g




1 Simplify the following:

a 2 £ 3x b 4x £ 5 c ¡2 £ 7x d ¡3 £ 4x

e 2x £ x f 3x £ 2x g ¡2x £ x h ¡3x £ x2

i ¡2x £ (¡x) j ¡3x £ 4 k ¡x2

£ (¡2x) l 3d £ (¡2d) m (¡a)2 n (¡2a)2 o 2a2 £ a2 p a2 £ (¡3a)


a 3(x + 2) b 2(5 ¡ x) c ¡(x + 2) d ¡(3 ¡ x)

e ¡2(x + 4) f ¡3(2x ¡ 1) g x(x + 3) h 2x(x ¡ 5)

i a(a + b) j ¡a(a ¡ b) k x(2x ¡ 1) l 2x(x2 ¡ x ¡ 2)


a 3(x ¡ 4) + 2(5 + x) b 2a + (a ¡ 2b) c 2a ¡ (a ¡ 2b)

d 3(y + 1) + 6(2 ¡ y) e 2(y ¡ 3) ¡ 4(2y + 1) f 3x ¡ 4(2 ¡ 3x) g 2(b ¡ a) + 3(a + b) h x(x + 4) + 2(x ¡ 3) i x(x + 4) ¡ 2(x ¡ 3)

j x2 + x(x ¡ 1) k ¡x2 ¡ x(x ¡ 2) l x(x + y) ¡ y(x + y)

m ¡4(x ¡ 2) ¡ (3 ¡ x) n 5(2x ¡ 1) ¡ (2x + 3) o 4x(x ¡ 3) ¡ 2x(5 ¡ x)

Rational numbers can be written as a ratio of two integers in the form

a

b :

Rational numbers appear in many forms.

For example, 4, ¡2, 0, 10%, ¡47 , 5 : 3, 1:3, 0:6 are all rational numbers.

i.e., they can be written as ratios: 41 , ¡2

1 , 01 , 1

10 , ¡47 , 5

3 , 1310 , 2

3

Note that 4, ¡2 and 0 are integers, and 10%, ¡47 , 5 : 3, 1:3 and 0:6 are fractions.

We can extend our classification

of numbers as follows:

A common fraction consists of two whole numbers, a numerator and a denominator, separated

by a bar symbol. For example,

RATIONAL NUMBERS (FRACTIONS) D

EXERCISE 13C.2

RATIONAL NUMBERS

INTEGERS

ZERO POSITIVE INTEGERS NEGATIVE INTEGERS

FRACTIONS

2

3

numerator

denominator

bar (which also means ‘ )divide’




TYPES OF FRACTIONS

45 is a proper fraction fas the numerator is less than the denominator g76 is an improper fraction fas the numerator is greater than the denominator g2 3

4 is a mixed number

fas it is really 2 + 3

4

g12 , 3

6 fas both fractions represent equivalent portionsg

RULES FOR OPERATING WITH FRACTIONS

The rules for adding, subtracting, multiplying and dividing number fractions are essential to learn,

as they are important when we operate with algebraic fractions.

Addition and subtraction:

i.e., a

c +

b

c =

a + b

c or

a

c ¡ b

c =

a ¡ b

c

Find 34 + 5

6

34 + 5

6 fLCD = 12g= 3

4 £ 33 + 5

6 £ 22 fto achieve a common denominator of 12g

= 912 + 10

12

= 1912

= 1 712

Find 1 23 ¡ 1 4

5

1 23 ¡ 1 4

5

= 53 ¡ 9

5 fwrite as improper fractionsg= 5

3 £ 55 ¡ 9

5 £ 33 fto achieve a common denominator of 15g

= 2515 ¡ 27

15

= ¡215 or ¡ 2

15

To aadd subtractor two fractions we convert each one to an equivalent fraction with commondenominator and add or subtract the new numerators.

EXAMPLE 13.18

EXAMPLE 13.19

are equivalent fractions




Multiplication:

a

b £ c

d =

ac

bd

Find a 14 £ 1 2

3 b (3 12 )2

a 14 £ 1 2

3

= 14 £ 5

3 fconvert 1 23 to

an improper fractiong= 512

b (3 12 )2

= 3 12 £ 3 1

2

= 72 £ 7

2

fconvert 3 12 to

an improper fractiong= 49

4 or 12 14

fconvert to a mixed number

g

Division:

To divide by a number, we multiply by its reciprocal, i.e.,

a

b ¥ c

d =

a

b £ d

c

Find a 3 ¥ 23 b 2 1

3 ¥ 23

a 3 ¥ 23

b 2 13 ¥ 2

3

= 31 ¥ 2

3 = 73 ¥ 2

3

= 31 £ 3

2 fTo divide by 23 , multiply by 3

2 :g = 73 £ 3

2

= 92 = 7

2

= 4 12 = 3 1

2

1 Find

a 314 + 6

14 b 79 ¡ 2

9 c 716 + 2

16 d 34 ¡ 5

6

e 58 + 3

4 f 47 ¡ 1

2 g 1 12 + 5

8 h 1 12 ¡ 5

8

i 2 12 + 4 1

6 j 2 12 ¡ 4 1

6 k 23 + 4 l 2

3 ¡ 4

m 5 2

3 ¡3 1

8

n 2 1

3

+ 3 3

4

o 2 1

3 ¡3 3

4

p 2 4

5 ¡1 1

3

To multiply two fractions, we multiply the numerators together and multiply the denominatorstogether and then simplify if possible, i.e.,

EXERCISE 13D.1

EXAMPLE 13.20

EXAMPLE 13.21




2 Calculate

a 34 £ 5

6 b 47 £ 1

2 c 2 £ 34 d 2

3 £ 4

e 1 12 £ 5

8 f (2 13 )2 g (1 1

2 )3 h 1 38 £ 4

11

3 Evaluate

a 34 ¥ 56 b 34 ¥ 23 c 712 ¥ 34 d 25 ¥ 34

e 1 12 ¥ 5

8 f 2 13 ¥ 3 3

4 g 34 ¥ 2 1

2 h 4 23 ¥ 1 1

7

i 23 ¥ 4 j 2 ¥ 3

4 k 3 12 ¥ 4 1

3 l 1

23

4 Calculate

a 4 38 + 2 2

5 b 712 ¡ 5 3

8 c ( 23 )4 d (1 1

3 )3

e 34 £ 1 1

2 ¥ 2 f 34 ¥ 1

4 ¥ 12 g

6 £ 3 £ 12

34

h 4 ¡ 1

2

3 £ 23

i 1 ¥ ( 12 + 35 ) j 1 ¥ 12 + 35 k 6 ¡ 3 £ 34 l ( 34 )2 ¥ 2 12

RATIONAL EXPRESSIONS IN ALGEBRA

² Cancellation

Simplification of algebraic fractions is sometimes (but not always) possible.

If the numerator and denominator of an algebraic fraction are both written in factored

form and common factors are found, we can simplify by cancelling the common factors.

This process was observed when we cancelled number fractions such as 1228 ,

i.e., 1228 = 4£3

4£7 = 37 where the common factor 4 was cancelled.

Similarly, algebraic fractions can be simplified by cancelling common factors if they exist.

For example, 4ab

2a

= 2 £ 2 £ a £ b

2 £ a ffully factorisedg

= 2b

1 fafter cancellationg= 2b

It is a good idea to check both numerator and denominator to see if they can be expressed as the

product of factors then look for common factors which can be cancelled.

Algebraic fraction

Express bothnumerator anddenominator as a product of factors.

If there are common factors,simplify by cancelling.

If there are no commonfactors, you cannot simplify.




² Illegal cancellation

Take care with fractions such as a + 3

3 :

The expression in the numerator, a + 3 cannot be written as the product of factors other than

1 £ (a + 3) as a and 3 are terms of the expression not factors.

A typical error in illegal cancellation is: a + 3

3 =

a + 11

12

11

1 1

= a + 1.

You can check that this cancellation of terms is not correct by substituting a value for a.

For example, if a = 3, LHS = a + 3

3 =

3 + 3

3 = 2, whereas

RHS = a + 1 = 4.

Simplify if possible a a2

2a b

6a2b

3b

a a2

2a

= a £ a

2 £ a

= a

2

b 6a2b

3b

= 6 £ a £ a £ b

3 £ b

= 2 £ a £ a

1= 2a2

1 Simplify

a a2

¡2a b

¡a

2a2 c

¡a3

¡4a d

4a2

¡a2

e ¡6t

¡3t2 f

4d2

¡2d g

¡ab2

2ab h

¡4ab2

6a2b

2 Simplify if possible

a (2a)2

a2 b

(4n)2

8n c

(¡a)2

a d

a2

(¡a)2

e (¡2a)2

4 f

(¡3n)2

6n g

(x + y)2

x + y h

2(x + 2)

(x + 2)2

3 Simplify if possible

a 2(a + 3)

2

b 4(x + 2)

2

c 6(c + 3)

3

EXERCISE 13D.2

EXAMPLE 13.22




d 2(d ¡ 3)

6 e

4

2(x + 1) f

12

4(2 ¡ x)

g 3(x + 4)

9(x + 4) h

12(a ¡ 3)

6(a ¡ 3) i

(x + y)(x ¡ y)

3(x ¡ y)

j

2xy(x

¡y)

6x(x ¡ y) k

5(y + 2)(y

¡3)

15(y + 2) l

x(x + 1)(x + 2)

3x(x + 2)

ADDING AND SUBTRACTING ALGEBRAIC FRACTIONS

Pronumerals are used in algebraic fractions to represent unknown numbers. Because of this we

can treat algebraic fractions in the same way that we treat numerical fractions as they are in fact

representing numerical fractions.

Simplify a x

2 +

3x

4 b

a

3 ¡ 2a

5

a x

2 +

3x

4 fhas LCD of 4g

= x

2 £ 2

2 +

3x

4

= 2x

4 +

3x

4

= 2x + 3x

4

= 5x

4

b a

3 ¡ 2a

5 fhas LCD of 15g

= a

3 £ 5

5 ¡ 2a

5 £ 3

3

= 5a

15 ¡ 6a

15

= ¡ a

15

Simplify a 4

a +

3

b b

5

x ¡ 4

3x

a 4

a +

3

b fhas LCD of abg

= 4

a £ b

b +

3

b £ a

a

= 4b

ab +

3a

ab

= 4b + 3a

ab

b 5

x ¡ 4

3x fhas LCD of 3xg

= 5

x £ 3

3 ¡ 4

3x

= 15

3x ¡ 4

3x

= 11

3x

EXAMPLE 13.23

EXAMPLE 13.24




Simplify a b

3 + 1 b

a

4 ¡ a

a b

3 + 1

= b

3 +

3

3

= b + 3

3

b a

4 ¡ a

= a

4 ¡ a

1 £ 4

4

= a

4 ¡ 4a

4

= ¡3a

4

1 Simplify, i.e., write as a single fraction.

a a

2 +

a

3 b

b

5 ¡ b

10 c

c

4 +

3c

2

d x

7 ¡ x

2 e

a

3 +

b

4 f

t

3 ¡ 5t

9

g m

7 +

2m

21 h

5d

6 ¡ d

3 i

3 p

5 ¡ 2 p

7

j m

2 +

m

3 +

m

6 k

a

2 ¡ a

3 +

a

4 l

x

4 ¡ x

3 +

x

6

2 Simplify

a 7

a +

3

b b

3

a +

2

c c

4

a +

5

d

d 2a

m ¡ a

m e

a

x +

b

2x f

3

a ¡ 1

2a

g 4

x ¡ 1

xy h

a

b +

c

d i

a

b ¡ x

y

j 23

+ a2

k x3

+ 34

l xy

+ 23

3 Simplify

a x

2 + 1 b

y

3 ¡ 1 c

a

2 + a

d b

4 ¡ 3 e

x

2 ¡ 4 f 2 +

a

3

g 3

¡

x

5

h 2 + 1

x

i 5

¡

2

x

EXAMPLE 13.25

EXERCISE 13D.3




j a + 2

a k

3

b + b l

x

3 ¡ 2x

4 Simplify (miscellaneous)

a x

2 +

2x

5 b

4x

5 ¡ 3x

2 c

3

a +

2

3a

d 4

y ¡ 3

2y e

5

a +

3

b f

4

3a ¡ 5

b

g x

7 + 2 h 3 ¡ x

4 i 2a ¡ a

3

MULTIPLYING AND DIVIDING ALGEBRAIC FRACTIONS

Simplify a 3m

£ m6

b 3m

£ m2

a 3

m £ m

6

= 3 £ m

m £ 6

= 1

2

b 3

m £ m2

= 3

m £ m2

1

= 3m2

m

=

3

1

1

21

11

£m

£m

m

= 3m

1= 3m

Simplify a 4

n ¥ 2

n2 b

3

a ¥ 2

a 4

n ¥ 2

n2

= 4

n £ n2

2

= 4 £ n2

n £ 2

= 2 £ 2 £ n £ n

n £ 2= 2n

b 3

a ¥ 2

= 3

a £ 1

2

= 3 £ 1

a £ 2

= 3

2a

EXAMPLE 13.26

EXAMPLE 13.27




1 Simplify

a x

2 £ y

5 b

a

2 £ 3

a c

a

2 £ a d

a

4 £ 2

3a

e c5 £ 1c f c5 £ c2 g ab £ cd h ab £ ba

i 1

m2 £ m

2 j

m

2 £ 4

m k

a

x £ x

b l m £ 4

m

m 3

m2 £ m n

³a

b

´2

o

µ2

x

¶2

p 1

a £ a

b £ b

c

2 Simplify

a a

2 ¥ a

3 b

2

a ¥ 2

3 c

3

4 ¥ 4

x d

3

x ¥ 4

x

e 2

n ¥ 1

n f

c

5 ¥ 5 g

c

5 ¥ c h m ¥ 2

m

i m ¥ m

2 j 1 ¥ m

n k

3

g ¥ 4 l

3

g ¥ 9

g2

3 Simplify (miscellaneous)

a 2a

3 £ 6

4a b

2

9 ¥ b

3 c

2c

5 +

3c

4 d

a

c ¡ b

2

e 7b £ 3a2b f 83a ¡ 2a g abcd ¥ abcd h a3b + 52b

Solve for x a 5 ¡ 3x = 6 b 3x ¡ 4 = 2x + 6

a 5 ¡ 3x = 6

) ¡3x = 6 ¡ 5 fsubtracting 5 from both sidesg) ¡3x = 1

) x = 1

¡3 fdividing both sides by ¡3g

) x =¡

1

3

LINEAR EQUATIONS E

EXERCISE 13D.4

EXAMPLE 13.28

Linear equations

unknown variable constants

are equations which are in the form (or can be converted to the form)where is the and are .ax + = 0b x a b, ( ) ,




b 3x ¡ 4 = 2x + 6

) 3x ¡ 2x = 6 + 4 ftake 2x from both sides, add 4 to both sidesg) x = 10

1 Solve for x

a 2x + 3 = 7 b 3 ¡ 4x = ¡11 c 6 = 11 ¡ 2x

d x

3 + 2 = ¡1 e

x + 3

4 = ¡2 f 1

2 (5 ¡ x) = ¡3

2 Solve for x

a 2x ¡ 3 = x + 6 b 4x ¡ 2 = 5 ¡ x c 4 ¡ 5x = 3x ¡ 7

d ¡x = x + 4 e 12 ¡ 7x = 3x + 8 f 5x ¡ 9 = 1 + 6x

3 Solve for x

a 2(x + 8) + 5(x ¡ 1) = 6 b 2(x ¡ 3) + 3(x + 2) = 0

c 5(x + 2) = 2(3 ¡ 2x) d 3(x + 3) ¡ 2(x + 1) = ¡2

e 4 ¡ x ¡ (2 ¡ x) = 6 f 5 ¡ 3(1 ¡ x) = 2 ¡ x

g 3 ¡ 2x ¡ (2x + 1) = ¡1 h 3(4x + 2) ¡ 2x = ¡7 + x

FRACTIONAL EQUATIONS

Fractional equations are simplified by multiplying both sides by the least common denominator

(LCD). Consider the following fractional equations:

The equation x

2 = x

3 has LCD is 6

The equation 5

2x =

3x

5 has LCD is 10x

The equation x ¡ 7

3 =

x

2x ¡ 1 has LCD is 3(2x ¡ 1)

Solve for x: x

2 =

3 + x

5

x

2 =

3 + x

5 has LCD = 10

) x

2 £ 10 =

µ3 + x

5

¶£ 10 fmultiplying both sides by 10g

) 5x = 2(3 + x) fcancelling common factorsg) 5x = 6 + 2x fexpandingg) 3x = 6 fsimplifyingg) x = 2 fdividing both sides by 3g

EXERCISE 13E.1

EXAMPLE 13.29




EXAMPLE 13.30

EXAMPLE 13.31

EXAMPLE 13.32

Solve for x: 4

x =

3

4

4

x =

3

4 has LCD = 4x

) 4

x £ 4x =

3

4 £ 4x fmultiplying both sides by 4xg

) 16 = 3x fcancelling common factorsg) 16

3 = x fdividing both sides by 3g

Solve for x: 2x + 1

3 ¡ x =

3

4

2x + 1

3 ¡ x =

3

4 has LCD 4(3 ¡ x)

)

µ2x + 1

3 ¡ x

¶£ 4(3 ¡ x) =

3

4 £ 4(3 ¡ x) f£ both sides by LCDg

) 4(2x + 1) = 3(3 ¡ x) fcancelling common factorsg) 8x + 4 = 9 ¡ 3x fexpandingg) 8x + 3x = 9 ¡ 4 fadd 3x, take 4 both sidesg

) 11x = 5) x = 5

11

Solve for x: x

3 ¡ 1 ¡ 2x

6 = ¡4

x

3 ¡

1 ¡ 2x

6

=

¡4 has LCD of 6

) 6 £ x

3 ¡ 6

µ1 ¡ 2x

6

¶ = 6 £ ¡4 fmultiplying both sides by 6g

) 2x ¡ (1 ¡ 2x) = ¡24 fcancelling common factorsg) 2x ¡ 1 + 2x = ¡24 fexpandingg

) 4x ¡ 1 = ¡24

) 4x = ¡24 + 1 fadding 1 to both sidesg) 4x = ¡23

) x = ¡234




1 Solve for x

a x

2 =

3

7 b

3

5 =

x

6 c

x

5 =

x ¡ 2

3

d x + 13

= 2x ¡ 18

e 2x3

= 5 ¡ x4

f 3x + 23

= 2x ¡ 52

2 Solve for x

a 5

x =

3

5 b

6

x =

2

7 c

4

3 =

7

x d

3

2x =

5

4

3 Solve for x

a 2x + 3

x + 1 =

4

3 b

x + 1

1 ¡ 2x =

3

5 c

2x ¡ 1

4 ¡ 3x = ¡2

3

d x + 3

2x ¡ 1 =

1

2 e

4x + 3

2x ¡ 1 = 6 f

3x ¡ 2

x + 4 = ¡5

4 Solve for x

a x

2 ¡ x

6 = 4 b

x

4 ¡ 3 =

2x

3

c x

8 +

x + 2

2 = ¡1 d

x + 2

3 +

x ¡ 3

4 = 1

e 2x ¡ 1

3 ¡ 5x ¡ 6

6 = ¡2 f

x

4 = 4 ¡ x + 2

3

LINEAR INEQUALITIES

Rules for handling linear inequalities

Examples such as, if 5 > 3 then 3 < 5, and if 2 > ¡3 then ¡3 < 2, indicate that if we

interchange the LHS and the RHS then we must reverse the inequality sign.

Note: > and < are reverses; > and 6 are reverses.

Recall also that

² if we add or subtract the same number to both sides the inequality sign is maintained,

² if we multiply or divide both sides by a positive number the inequality sign is maintained,

² if we multiply or divide both sides by a negative number the inequality sign is reversed.

Consequently, we can say that the method of solution of linear inequalities is identical to that of

linear equations with the exceptions that

² interchanging the sides reverses the inequality sign,

² multiplying (or dividing) both sides by a negative, reverses the inequality sign.

EXERCISE 13E.2




Graphical solutions

If we solve an inequality and finish with x > 4, this means that every number which is 4 or greater

than 4 is a solution to the original inequality. We could represent this on a number line by

Likewise if we finish with x < 5 our representation would be

Solve for x and graph the solution: 3 ¡ 2x < 7

3 ¡ 2x < 7

) ¡2x < 7 ¡ 3 fsubtract 3 from both sidesg) ¡2x <

< <

4

) )

)

¡2x

¡2 >

4

¡2 f

f

divide

divide

reverse

reverse

both

both

sides

sides

by

by

,

,

to

to

¡

¡

2

2

g

g

) x > ¡2

Solve for x: 3 ¡ 5x > 2x + 7

3 ¡ 5x > 2x + 7

) ¡5x ¡ 2x > 7 ¡ 3 fsubtract 2x, subtract 3 both sidesg) ¡7x >

>

4

) ¡7x¡7 6 4¡7

) x 6

6

¡47

1 Solve for x and graph the solution.

a 3x + 2 < 0 b 5x ¡ 7 > 2 c 2 ¡ 3x > 1

d 5¡

2x 6 11 e 2(3x¡

1) < 4 f 5(1¡

3x) > 8

4

The filled-in circleindicates that

4 included.is

The arrowhead indicatesthat all numbers on thenumber line in that directionare included.

x

5

The hollow circle indicatesthat 5 included.is not

x

EXAMPLE 13.33

EXAMPLE 13.34

EXERCISE 13E.3

x

xu R _




2 Solve for x

a 3x + 2 > x ¡ 5 b 2x ¡ 3 < 5x ¡ 7

c 5 ¡ 2x > x + 4 d 7 ¡ 3x 6 5 ¡ x

e x

2 ¡ 2x

5 > ¡2 f 5 ¡ 2x < 15 ¡ 2(1 + x)

A formula is an equation which connects two or more variables.

The plural of formula is formulae or formulas.

In a formula it is usual for one of the variables to be on one side of the equation and the other

variable(s) and constants to be on the other side.

The variable on its own is called the subject of the formula.

For example, C = 2

¼r is the formula for finding the circumference of a circle. C has been

written in terms of the radius, r, and the constants 2 and ¼: C is the subject of the formula.

In this chapter we deal with ² substitution into formulae,

² formulae rearrangement, and

² formula construction.

SUBSTITUTION

If a formula contains two or more variables and we know the value of all but one of them, we can

use the formula to find the value of the unknown variable.

² Write down the formula.

² State the values of the known variables.² Substitute into the formula to form a one variable equation.

² Solve the equation for the unknown variable.

When a stone is dropped down a well the total distance fallen, D metres,

is given by the formula D = 12 gt2 where t is the time of fall (in seconds)

and g is the gravitational constant of 9:8. Find

a the distance fallen after 5 seconds

b the time (to the nearest

1

100 th second) taken for the stone to fall 100 m.

a D = 12 gt2 where g = 9:8 and t = 5

) D = 12 £ 9:8 £ 52 Calculator:

) D = 122:5 0:5 9:8 5

) the stone has fallen 122:5 metres.

b D = 12 gt2 where D = 100, g = 9:8

)

)

100 = 12 £ 9:8 £ t2

100 = 4:9t2

FORMULAE F

EXAMPLE 13.35

× × = x 2

D

The Method:




) 100

4:9 = t2

) t =

r 100

4:9 fas t > 0g Calculator: 100 49

) t + 4:5175:::::

) time taken is approx. 4:52 seconds.

In this exercise, unless it is otherwise stated, give your answer correct to 1 decimal place where

necessary.

1 The formula for finding the circumference, C , of a circle of diameter d, is

C = ¼d where ¼ is the constant with value approximately 3:14159. Find

a the circumference of a circle of diameter 11:4 cm

b the diameter of a circle with circumference 250 cm.

2 When a cricket ball is dropped from the top of a building the

total distance fallen is given by the formula D = 12 gt2

where D is the distance in metres and t is the time taken in

seconds. Given that g = 9:8, find

a the total distance fallen in the first 3 seconds of fall

b the height of the building, to the nearest metre, when

the time of fall to hit the ground is 5:13 seconds.

3 The formula for calculating the circumference C , of a circle of radius r , is C = 2¼r.

Finda the circumference of a circle of radius 8:6 cm

b the radius of a circle of circumference 100 metres.

4 When a car travels a distance d kilometres in time t hours,

the average speed s km/h for the journey is given by

the formula s = d

t. Find

a the average speed of a car which travels 200 km in

2 12 hours

b the distance travelled by a car in 3 14 hours if its average speed is 80 km/h

c the time taken, to the nearest minute, for a car to travel 865 km at an average speed of

110 km/h.

5 A cylinder of radius r , and height h, has volume given by V = ¼r2h:

Find

a the volume of a cylindrical tin can of radius 12 cm and height

17:5 cm

b the height of a cylinder of radius 4 cm given that its volume is

80 cm3

c the radius, in mm, of copper wire of volume 100 cm3 and length

2 km.

÷ =( )

EXERCISE 13F.1

d

D

r

h




6 The formula D = 3:56p

h gives the approxi-

mate distance (D km) to the horizon which can

be seen by a person with eye level h metres

above the level of the sea. Find

a the distance to the horizon when a

person’s eye level is 10 m above sea level

b how far above sea level a person’s eyemust be if the person wishes to see the

horizon at a distance of 30 km.

7 The following formula was discovered around 100AD by the Greek mathematician, Heron. It

now bears his name, Heron’s formula:

To find the area of a triangle with sides a, b and c units long we

find s, its semi-perimeter, using the formula

s = a + b + c

2 , and then use A =

p s(s ¡ a)(s ¡ b)(s ¡ c).

If a triangle has sides of length 5 cm, 6 cm and 7 cm, finda its semi-perimeter and hence b its area.

FORMULA REARRANGEMENT

For the formula a = bx + c we say that a is the subject. This is because a is expressed in terms

of the other variables, b, x and c.

In formula rearrangement we require one of the other variables to be the subject.

For example, if a = bx + c, we might wish to write x in terms of a, b and c, i.e., to make x

the subject of the formula.

To rearrange a formula we use the same processes as used for solving an equation for the

variable we wish to be the subject.

Make x the subject of a = bx + c:

If a = bx + c

then bx + c = a

) bx = a ¡ c fsubtract c from both sidesg

) x = a ¡ c

b fdivide both sides by bg

Make x the subject of ax + 3 = bx + d:

a

b

c

Earth

h D

EXAMPLE 13.36

EXAMPLE 13.37




ax + 3 = bx + d

) ax ¡ bx = d ¡ 3 fterms containing x on LHS, all others on RHSg) x(a ¡ b) = d ¡ 3 fx is a common factor on LHSg

) x = d ¡ 3

a ¡ b fdivide both sides by a ¡ bg

Make z the subject of a c = m

z b

a

z =

z

b

a c = m

z b

a

z =

z

b

) cz = m ) ab = z2 fmultiply both sides by bzg) z =

m

c ) z2 = ab

) z = §p ab

Make t the subject of s = 12 gt2 where t > 0.

s = 12 gt2

) 2s = gt

2

fmultiply both sides by 2g)

2s

g = t2 fdivide both sides by gg

) t2 = 2s

g

) t =

r 2s

g fas t > 0g

1 Make x the subject of

a a + x = b b ax = b c 2x + a = d

d y = mx + c e ax + by = c f 3 ¡ tx = p

g a ¡ bx = m h c ¡ x = t i 8 = p ¡ dx

2 Make x the subject of

a 3x + a = bx + c b ax = c ¡ bx c mx + a = nx ¡ 2

d 8x + a =¡

bx e a¡

x = b¡

cx f rx + d = e¡

sx

EXAMPLE 13.38

EXAMPLE 13.39

EXERCISE 13F.2




3 Make z the subject of

a az = b

c b

a

z = d c

3

d =

2

z

d z

2 =

a

z e

b

z =

z

n f

m

z =

z

a ¡ b

4 Make

a r the subject of A = ¼r2, (r > 0) b x the subject of N = x5

a

c r the subject of V = 43 ¼r3 d x the subject of D =

n

x3

e x the subject of y = 4x2 ¡ 7 f Q the subject of P 2 = Q2 + R2

THE RULE OF PYTHAGORAS

In geometric form, the Rule of Pythagoras is:

PYTHAGOREAN TRIPLES

The simplest right-angled

triangle with sides of integer

length is the 3-4-5 triangle.

The numbers 3, 4, and 5 satisfy the rule 32 + 42 = 52.

THE RULE OF PYTHAGORAS G

h y p o t e

n u s e

legs

In a right-angled triangle, with

hypotenuse c and legs a and b,

c2 = a2 + b2.

c

b

a

c

c

c2

a

aa2

b2b

b

53

4

A a aright-angled triangle

hypotenuse

legs

is triangle which has right-angle as one of its angles.

The side opposite the right-angle is called theand is the longest side of the triangle.

The other two sides are called the of the triangle.

In any right-angled triangle, the area of the square on the hypotenuse is equalto the sum of the areas of the squareson the other two sides.




The set of numbers f3, 4, 5g is called a Pythagorean triple since it obeys the rule a2 + b2 = c2

where a, b and c are integers.

Other examples are: f5, 12, 13g, f7, 24, 25g, f8, 15, 17g.

Find x, given

x2 + 52 = 62 fPythagorasg) x2 + 25 = 36

) x2 = 11

) x = §p 11

Note: If x2 = k,

then x = §p k

and we then reject

¡p k. Why?

But x > 0, ) x = p 11.

Solve for x:

x2 + ¡12¢

2= 12 fPythagorasg

) x2 + 14 = 1

) x2 = 34

) x = §q

34

) x =p

32 fas x > 0g

Find the value of x:

(2x)2 = x2 + 62 fPythagorasg) 4x2 = x2 + 36

) 3x2 = 36

) x2 = 12

) x =p

12 fas x > 0g

EXAMPLE 13.40

EXAMPLE 13.41

EXAMPLE 13.42

6 cm

5 cm

x cm

1 cm

x cm

Qw _ cm

2 m x x m

6 m




PROBLEM SOLVING USING PYTHAGORAS’ RULE

² Draw a neat, clear diagram of the situation.

² Mark on known lengths and right-angles.

² Use a symbol, such as x, to represent the unknown length.

² Write down Pythagoras’ rule for the given information.² Solve the equation.

² Write your answer in sentence form (where necessary).

Right-angled triangles occur frequently in problem solving and often the presence of such triangles

indicates that the rule of Pythagoras is likely to be used.

A rectangular gate is 3 m wide and has a 3:5 m diagonal. How high is the gate?

Let x m be the height of the gate.

Now (3:5)2 = x2 + 32 fPythagorasg) 12:25 = x2 + 9

) 3:25 = x2

) x =p

3:25 + 1:803 m

So, the gate is approximately 1:803 m high.

A man travels due east by bicycle at 16 kmph. His son travels due south on his bicycle at

20 kmph. How far apart are they after 4 hours, if they both leave point A at the same time?

After 4 hours the man has travelled 4 £ 16 = 64 km

and his son has travelled 4 £ 20 = 80 km.

Thus x2 = 642 + 802 fPythagorasgi.e., x2 = 4096 + 6400

) x2 = 10496

) x =p

10496 + 102:4

) they are 102:4 km apart after 4 hours.

SPECIAL GEOMETRICAL FIGURES

All of these figures contain right-angled triangles where Pythagoras’ rule applies.

Things to ememberr

3 m

3.5 m x m

N

S

W E

ManA 64 km

80 km x km

Son

EXAMPLE 13.43

EXAMPLE 13.44

rectangle square rhombus isosceles triangle equilateral triangle

d i a g o n a

l




1

2 A square has diagonals of length 8 cm. Find the length of its sides.

3

4 An isosceles triangle has equal sides of length 8 cm and a base of length 6 cm. Find the area

of the triangle.

5 and a base radius of 5 cm.

How high is the cone?

DISTANCE BETWEEN TWO POINTS

Consider the points A(1, 3) and B(4, 1). The distance

from A to B can be found by joining A to B using a

straight line then drawing a right-angled triangle with

hypotenuse AB and with sides parallel to the axes.

It is clear that d 2 = 22 + 32 fPythagoras’ ruleg) d 2 = 13

) d =p

13 fas d > 0g) the distance from A to B is

p 13 units.

To avoid drawing a diagram each time we wish to find a distance, a distance formula can be

developed.

If A(xA, yA) and B(xB , yB) are two points in a plane, then the distance, d,

between these points is given by dAB =p

(xB ¡ xA)2 + (yB ¡ yA)2

or dAB =q

(x-step)2 + (y-step)2

:

[Note: We use the symbol dAB

to represent the distance between points A and B.]

EXERCISE 13G.2

COORDINATE GEOMETRY H

5 cm

13 cmA acone has slant height of cm13

T A B CA C A C B a

own is km south of town and town is km east of town B. Is it traveldirectly from to by car at kmph or from to via in trainkmph?

80 15090 130

quicker totravelling at

y

x3

2

B (4, 1)

A (1, 3)

d

A yacht sails km due west and then km due south.

How far is it from its starting point?

9 7




Find the distance from A(¡2, 1) to B(3, 4).

x-step = 3 ¡ (¡2) = 5

y-step = 4

¡1 = 3

) AB =p

52 + 32

) AB =p

34 units

Use the distance formula to determine if the triangle ABC, where A is (¡2, 0), B is (2, 1)

and C is (1, ¡3), is equilateral, isosceles or scalene.

A to B x-step = 2 ¡ (¡2) = 4

y-step = 1 ¡ 0 = 1

) AB =

p 4

2

+ 12

=p

17 units

B to C x-step = 1 ¡ 2 = ¡1

y-step = ¡3 ¡ 1 = ¡4

) BC = p (¡1)

2

+ (¡4)2

=p

1 + 16

=p

17 unitsA to C x-step = 1 ¡ (¡2) = 3

y-step = ¡3 ¡ 0 = ¡3

) AC =p

32 + (¡3)2

=p

18 units

As AB = BC, triangle

ABC is isosceles.

1 Find the distance between the following pairs of points.

a A(3, 1) and B(4, 2) b C(¡1, 2) and D(5, 2)

c O(0, 0) and P(3, ¡4) d E(3, 0) and F(7, 0)

e G(0, ¡2) and H(0, 3) f I(2, 0) and J(0, ¡4)

g R(1, 2) and S(¡1, 1) h W(3, ¡2) and Z(¡1, ¡4)

2

a A(5, ¡3), B(1, 8), C(¡6, 1) b A(1, 0), B(3, 1), C(7, 3)

c A(¡2, ¡1), B(0, 3), C(4, 1) d A(p 2, 0), B(¡p 2, 0), C(0, p 6)e A(¡2, 0), B(1,

p 3), C(1, ¡p

3) f A(a, b), B(a, ¡b), C(1, 0)

MIDPOINT FORMULA

In general,

if A(xA

, yA

) and B(xB

, yB

) are two points then the midpoint M of AB has coordinatesµxA

+ xB

2 ,

yA

+ yB

2

¶:

Use the distance formula to classify triangle ABC, as either equilateral, isosceles or scalene.

EXAMPLE 13.45

EXAMPLE 13.46

EXERCISE 13H.1




Find the coordinates of the midpoint of AB for A(¡1, 3) and B(4, 7).

x-coordinate of midpoint = ¡1 + 4

2

= 32

= 1 12

y-coordinate of midpoint = 3 + 7

2

= 5

) midpoint of AB is (1 12 , 5).

1 Find the coordinates of the midpoint of the line segment joining the following pairs of points:

a (6, 3) and (2, 5) b (4,

¡1) and (0, 1)

c (3, 0) and (0, 5) d (¡1, 6) and (1, 6)

e (3, ¡1) and (¡1, 0) f (¡1, 3) and (3, ¡1)

g (6, 8) and (¡3, ¡4) h (¡2, 3) and (¡5, 1)

Sigma notation is a mathematical shorthand way of writing sums of numbers having a definite

pattern.

P is an upper case Greek letter which for our purposes stands for “the sum of”.

Instead of writing the sum 1 + 2 + 3 + 4 + 5 + :::::: + 50, that is the sum of the first 50 wholenumbers, we write

50Xi=1

i

This reads

“the sum of all numbers of the form i where i runs through the integers ..... etc. up to1 2 3, , , ”50

If we actually add the terms we say we have found the sum.

Evaluate a

10Xi=1

i b

10Xx=5

x2

a10X

i=1

i

= 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10

= 55

b10X

x=5

x2

= 52 + 62 + 72 + 82 + 92 + 102

= 25 + 36 + 49 + 64 + 81 + 100

= 355

EXERCISE 13H.2

EXAMPLE 13.47

EXAMPLE 13.48

SIGMA NOTATION I




1 Expand and find the sum of

a

8

Xi=1

i b

8

Xi=1

(i + 1) c

5

Xx=1

µ1

x

¶

d

5Xi=1

i3 e

10Xn=0

(2n + 1) f

5Xn=0

(3 £ 2n)

2 Sigma notation is used in statistics.

For example, the mean (x) is given by the formula

x =

nXi=1

f ixi

nXi=1

f i

where xi is a score and f i is its frequency.

ax f

x1 f 1

x2 f 2

x3 f 3

x4 f 4

is a frequency table.

Write the mean x, in sigma notation.

b i Write f 1x1 + f 2x2 + f 3x3 + :::::: + f 20x20

f 1 + f 2 + f 3 + :::::: + f 20in sigma notation.

ii Describe what this fraction represents in statistical terms.

EXERCISE 13I




Chapter 3survey onstatistics

APPENDIX 2



Conducting Student Surveys: INSTRUCTIONS TO TEACHERS

Materials: one survey per student, 9 plastic cups, scissors.

The students circle Male or Female for each question, and then answer each survey question as best

they can. While they are doing this, you should label each cup with a survey topic (for example,

Hours of T.V.) and place them in a row at the front of the room.

Once the students have completed the survey, they cut the questions apart, and place each completed

survey question in the appropriate cup.

The class now needs to form nine groups, one for each survey topic. Each group will collate the

data for their question, as shown in this example:

Hours of TV watched in a typical week

Male 20 18 14 16 10 16 9 ::::

Female 15 15 23 35 16 20 0 ::::

Make enough copies of the survey form so each student has one copy The page contains fivesurveys, each consisting of nine questions. Distribute the surveys to the students. ou will get

better data if the survey questions are discussed. For example, does typical week include theweekend? The desired outcome is for each student to interpret the questions in the same way

.Y

a.

CHAPTER 3 SURVEY ON STATISTICS (APPENDIX 2) 427

The data should then be entered into a graphics calculator in three lists - a Male list, a Female

list, and a Combined list. If your calculator has the functionality, use descriptive list names, for

example, TVM, TVF, TVC.

Once the students have checked and double-checked the data to ensure it is accurate, they should

transfer their lists to the teacher’s calculator.

Visit the mathematics-for-queensland.com website for a method of storing these datasets in pro-grams in your graphics calculator. This is our recommended method of storing and sharing datasets,

because:

² it is the only way to store the lists for future use in low-end calculators such as the TI-82

² the students and the teacher only have to store 9 programs rather than 27 lists

² the data for each question is always copied into List 1, List 2 and List 3 when the

program is run. This simplifies the teaching, as exactly the same keypresses will

work for each dataset.

The survey form is also available in Word format from the website, so it is easy to modify if you

wish to change the survey questions.

Just remember that this section of the chapter is dealing with quantitative data, so questions such

as, “How did you get to school today?” will not be able to be analysed using the methods outlined

in the chapter.





428 CHAPTER 3 SURVEY ON STATISTICS (APPENDIX 2)

Male Female Male Female Male Female Male Female Male FemaleI guess the width I guess the width I guess the width I guess the width I guess the widthof the room to be of the room to be of the room to be of the room to be of the room to be _______ metres _______ metres _______ metres _______ metres _______ metres

Male Female Male Female Male Female Male Female Male FemaleThe numbers of The numbers of The numbers of The numbers of The numbers of

hours I watch TV hours I watch TV hours I watch TV hours I watch TV hours I watch TVin a typical week is in a typical week is in a typical week is in a typical week is in a typical week is

__________ __________ __________ __________ __________

Male Female Male Female Male Female Male Female Male FemaleThe number of The number of The number of The number of The number of

hours I am on the hours I am on the hours I am on the hours I am on the hours I am on thecomputer in a computer in a computer in a computer in a computer in atyp typ typ typ typical week ical week ical week ical week ical week

is _________ is _________ is _________ is _________ is _________


hours I spend on hours I spend on hours I spend on hours I spend on hours I spend on

the telephone in a the telephone in a the telephone in a the telephone in a the telephone in atypical week typical week typical week typical week typical weekis _______. is _______. is _______. is _______. is _______.

Male Female Male Female Male Female Male Female Male FemaleThe number of The number of The number of The number of The number ofhours I spend hours I spend hours I spend hours I spend hours I spend

doing homework doing homework doing homework doing homework doing homeworkand assignments in and assignments in and assignments in and assignments in and assignments in

a typical week a typical week a typical week a typical week a typical weekis _______. is _______. is _______. is _______. is _______.

Male Female Male Female Male Female Male Female Male FemaleThe number of The number of The number of The number of The number ofhour hour hour hour hours I do paid s I do paid s I do paid s I do paid s I do paid

work in a typical work in a typical work in a typical work in a typical work in a typicalweek is week is week is week is week is

________ hrs. ________ hrs. ________ hrs. ________ hrs. ________ hrs.

Male Female Male Female Male Female Male Female Male FemaleThe amount of The amount of The amount of The amount of The amount of

money I earn/am money I earn/am money I earn/am money I earn/am money I earn/amgiven in a week is given in a week is given in a week is given in a week is given in a week is

$_________ $_________ $_________ $_________ $_________


hours I spend hours I spend hours I spend hours I spend hours I spendplaying sport in a playing sport in a playing sport in a playing sport in a playing sport in a

typical week typical week typical week typical week typical weekis ________, is ________, is ________, is ________, is ________,


CDs I ow CDs I ow CDs I ow CDs I ow CDs I own n n n nis ________ is ________ is ________ is ________ is ________



Glossary

APPENDIX 3



430 GLOSSARY (APPENDIX 3)

acceleration (Ch 10) Acceleration is a measure of how the velocity of a moving

object changes over time. This is the time rate of change in

velocity.

arc (of a circle) (Ch 5) An arc of a circle is a part of the curved boundary of a circle.

asymptotes (Ch 9) An asymptote is a line (or curve) to which a graph of a curve

continually gets closer to but never touches it.average (Ch 3) An average of a data set is any measure of its centre. Usually

the average which is used is the mean.

axis of symmetry (Ch 2) An axis of symmetry of any shape is a straight line about

which the shape may be reflected so that it leaves the shape

unchanged.

base number (Ch 7) In index notation, the base is the quantity which is raised to

the power. For example, in a3 and 5¡4 the base numbers are

a and 5.

bearing (Ch 5) The bearing of point P from point Q is the direction in which a

person would travel in a straight line from Q to P. It is usually

given as a compass angle.

binomial expansion (Ch 8) In (x + 2)3 = x3 + 6x2 + 12x + 8, x + 2 is a binomial and

x3 + 6x2 + 12x + 8 is a binomial expansion.

bivariate data (Ch 3) Bivariate data is data which contains two variables that are

under consideration. For example, data consisting of height

and weight of kangaroos.

boxplot (Ch 3)

Cartesian plane (Ch 1) The Cartesian plane is the number plane which contains a set

of coordinate axes and points in it can be described by using

ordered pairs of numbers.

centre (Ch 3) The centre of a circle is the point which is inside it and always

the same distance from any point on the circle.

chain rule (Ch 11) The chain rule is a rule for differentiating composite functions.

For example,

dy

dx =

dy

du

du

dx .

chord (Ch 5) A chord of a circle is a straight line which joins any two points

on the circle.

circle (Ch 9) A circle is the set of points which are a fixed distance from a

given point.

circular functions (Ch 5) Circular functions are the trigonometric functions:

S (x) = sin x, C (x) = cos x and T (x) = tan x.

class intervals (Ch 3) The class intervals of a dataset are the intervals (usually of

equal length) into which the data is subdivided. The end points

of each interval are called class limits.

A simple boxplot is a representation of a set of data with a

box between the lower quartile and the upper quartile with the

median shown between, and with whiskers out to the lowestand highest values on either side of the box. Other forms of a

boxplot show outliers.

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GLOSSARY (APPENDIX 3) 431

coefficient (Ch 2) A coefficient is a constant which is multiplied by a variable

quantity.

For example, 3x, 4x2, 5xy have coefficients 3, 4 and 5.

coefficient of

determination

(Ch 4) The coefficient of determination is r2 where r is the correlation

coefficient. It is used to interpret the strength or weakness of

correlation between two variables.

common logarithm (Ch 7) The common logarithm of a number is its power of 10.

For example, as 2 = 100:3010:::: then log10 2 = 0:3010::::

composite function (Ch 8) A composite function is a function which is made up of a

function within a function.

For example (2x + 3)4 is u4 where u = 2x + 3.

constant term (Ch 2) A constant term is a term in a polynomial which does not

contain the variable.

For example, in x3 + 5x2 ¡ 4x + 6, the 6 is the constant

term.

continuous variable (Ch 1) A continuous variable is a quantitative variable describing acharacteristic of a population which takes values over a certain

range. It can have any value within the range and is often a

result of measuring.

correlation coefficient (Ch 4) The correlation coefficient is a measure of the degree of corre-

lation (linear association) between two variables in a bivariate

dataset.

corresponding sides (Ch 5) When two triangles are similar, their corresponding sides are

opposite their equal angles.

cosecant (Ch 5) The cosecant of angle µ (cosec µ) is the reciprocal of sin µ

i.e., cosecµ = 1

sin µ.

cosine axis (Ch 5) The cosine axis is the horizontal axis (x-axis) when doing unit

circle trigonometry.

cosine of an angle (Ch 5) In a right-angled triangle the cosine (cos) of an angle is the

ratio of the side adjacent to the angle to the hypotenuse.

cosine rule (Ch 5) a2 = b2 + c2 ¡ 2bc cos A or

cos A = b2

+ c2

¡ a2

2bc

cotangent (Ch 5) The cotangent of angle µ (cot µ) is the reciprical of tan µ

i.e., cot µ = 1

tan µ.

coterminal (Ch 5) Angles are coterminal if they have the same position on the

unit circle diagram, for example, 42o and 402o.

counter-example (Ch 2) A counter-example to a statement is an example which shows

that the statement is not true.

B

A

Ca

bc

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cubic function (Ch 8) A cubic function is any function of the form ax3+bx2 +cx+d

where a 6= 0 and a, b, c, d are constants.

data (Ch 3) Data are pieces of information collected about a situation.

degree of a

polynomial

(Ch 8) The degree of any polynomial is the highest power that the

variable can have.

For example, f (x) = x

4

¡ 3x

2

+ x ¡ 5 has degree 4.dependent variable (Ch 1) If y is a function of x, then y is the dependent variable as its

values depend on the values of x.

derivative (Ch 11) The derivative (or derived function) is the function which gives

the slope of the curve for all values in its domain.

derived function (Ch 11) See derivative.

difference of squares (Ch 2) The difference of squares is one perfect square subtracted from

another, for example, x2 ¡ 9.

differentiation (Ch 11) Differentiation is the process by which we differentiate a func-

tion. It is the fundamental process of differential calculus.

direction of a point (Ch 5) The direction of a point from another is the number of degrees

East or West of either the North or South direction.

discrete variable (Ch 1) A discrete variable is one describing a characteristic of a popu-

lation in numerical terms where only certain values are possible

(usually whole numbers), for example, the number of children

in family. A discrete variable is often derived from counting.

discriminant (Ch 2) For the quadratic ax2 +bx+c, the discriminant ¢ is defined

as ¢ = b2 ¡ 4ac.

displacement (Ch 10) Displacement is the distance between the initial position of anobject and its final position.

domain (Ch 1) The domain of the function y = f (x) is the set of all num-

bers that x may have which produce real values of y .

ellipse (Ch 9) An ellipse is the oval shape traced out by all points for which

the sum of the distances to two fixed points is a constant.

expanded form (Ch 8) The expanded form of a polynomial consists of terms connected

by + and ¡ signs and contains no brackets. For example,

(2x ¡ 3)(x + 4) has expanded form 2x2 + 5x ¡ 12:

experiment (Ch 3) A mathematical experiment is the process of examining a par-ticular series of events and gathering the information obtained.

exponent (Ch 7) The exponent (power or index) is the superscript which shows

how many times the base number is multiplied.

For example, in x3 the number 3 is the exponent.

extrapolation (Ch 4) An extrapolation for a dataset is the estimation of the possible

value of the variable which is outside the known dataset. This

estimation is however based on the known dataset.

factorial notation (Ch 8) n! is the product of the first n whole numbers.

For example, 5! = 5£

4£

3£

2£

1 = 120.

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factorisation (Ch 8) Factorisation is the process of writing a polynomial expression

into a product of linear factors (or linear and quadratic factors).

factorised form (Ch 8) Example: The factorised form of

² x2 ¡ 3x + 2 is (x ¡ 1)(x ¡ 2)

² x3 ¡ 8 is (x ¡ 2)(x2 + 2x + 4)

family of functions (Ch 2) A family of functions is a collection of functions which have a particular characteristic, for example, the functions of the form

f (x) = x2 + c where c is a constant.

fitted values (fit) (Ch 4) Fitted values are values which seem to fit a given functional

relationship.

five-number summary (Ch 3) A five-number summary of a set of data consists of the lowest

value, the lower quartile, the median, the upper quartile, and

the highest value.

frequency table (Ch 3) A frequency table is a table which summarises the count of the

number of times a category occurs in the data. function (Ch 1) A function is a relation in which no vertical line can cut its

graph more than once.

gradient (Ch 1) The gradient (or slope) of a curve at a point is the gradient of

the tangent at that point.

histogram (Ch 3) A histogram is a frequency diagram which consists of vertical

rectangles whose heights are proportional to the frequency for

each class interval.

identity (algebraic) (Ch 2) An identity (algebraic) is an algebraic equation which is true

for all values of the variable(s).For example, x2 + 2x = x(x + 2) and xy + 4y = y(x + 4)are identities.

independent variable (Ch 1)

index (Ch 7) See exponent .

instantaneous velocity (Ch 11) Instantaneous velocity is the velocity of an object at a particular

instant in time. If s = f (t) then ds

dt is its instantaneous

velocity.

interpolation (Ch 4)

interquartile range (Ch 3) The interquartile range is the difference between the upper

quartile and the lower quartile of a set of data.

inverse variation (Ch 9)

irrational number (Ch 2)

If y is a function of x then x is the independent variable, as y

depends on it.

An interpolation for a dataset is the estimation of the possiblevalue of the variable which falls within the range of the dataset.

The line of best fit is usually used.

Two variables have inverse variation if an increase in one of

them causes a proportional decrease in the other.

An irrational number is a real number which is not rational.

It has a decimal expansion which is non-recurring and non-

terminating.

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knot (Ch 5) A knot is a speed measure in navigation. 1 knot is one nautical

mile per hour.

least squares

regression line

(Ch 4) The least squares regression line gives a method for finding the

‘line of best fit’. It is the line which gives the smallest sum of

the squares of the residuals.

limiting value (Ch 11) A limiting value of a function is a number which is beingapproached as x approaches some particular value.

line of best fit (Ch 4) It is a line that ‘best fits’ a set of points. It is also called a

‘trendline’.

linear relation (Ch 1) A linear relation is a function of the form f (x) = mx + c

where m and c are constants. Its graph is a straight line with

slope m and y-intercept c

local maximum (Ch 8) A local maximum is a point of a graph for which the y-value is

greater than the y-values for all other points in the neighbour-

hood.

local minimum (Ch 8) A local minimum is a point of a graph for which the y-value is

less than the y-values for all other points in the neighbourhood.

logarithm (Ch 7) If a = bn then we write n = logb a and say that n is the

logarithm of a in base b.

many-to-many

mapping

(Ch 9) A many-to-many mapping is a mapping where at least one value

in the domain maps to two or more values in the range and at

least one value in the range is mapped onto two or more values

in the domain.

many-to-onemapping

(Ch 9) A many to one mapping happens when more than one member of the domain is mapped onto one member of the range.

mapping (Ch 1)

mathematical model (Ch 1) A mathematical model is a way of representing a real-life sit-

uation mathematically.

maximum (Ch 2) The maximum value of a function is the greatest value it may

take for all values of x in the domain.

mean (Ch 3) The mean is the average of a set of data values found by sum-ming all the values and dividing by the number of values.

mean deviation (Ch 3) The mean deviation of a dataset is the average of the differences

between each member of the set and the mean of the set.

median (Ch 3) The median is the middle point of a set of data values. It may

be one of the data values (for an odd number of values) or

the average of the two middle values (for an even number of

values).

minimum (Ch 2) The minimum value of a function is the least value it may take

for all values of x in the domain.

A amapping is matching of members of the domain to themembers of the range.

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modal class (Ch 3) The modal class is the class which has the greatest frequency

of data in it.

mode (Ch 3) The mode of a dataset is the score which occurs with the great-

est frequency.

multimodal (Ch 3) If two or more modes exist for a dataset we say that the dataset

is multimodal.natural log (Ch 7) The natural logarithm of a number is its power of e, that is, if

a = eb then b = loge a:

nautical mile (Ch 5) A nautical mile is a distance measure based on travelling

through 1 minute of arc on a great circle on the earth’s surface.

1 nautical mile = 1.852 km.

normal (Ch 11) The normal to a curve at any point is the line which is perpen-

dicular to the tangent at that point.

(Ch 9) A one-to-many mapping is a mapping in which one member of

the domain maps onto more than one element of the range.

one-to-one mapping (Ch 9) A one-to-one mapping occurs when each member of the domain

is mapped onto one member of the range without using any

member twice.

ordered pairs (Ch 1) Ordered pairs are used to represent points in the number plane.

For example, (2, 3) is found by starting at the origin, moving

2 units x-wards and then 3 units y-wards.

outlier (Ch 3) An outlier is a data value significantly different from the ma-

jority of data values.

parabola (Ch 2) A parabola is a curve which results when a quadratic function

is graphed.

Pascals triangle (Ch 8) 1 11 2 1

1 3 3 11 4 6 4 1

1 5 10 10 5 1etc.

Each new row is

constructed by adding

the two numbers

directly above it.

perfect square (Ch 2) As 16 = 42 where 4 is a whole number we say that 16 is a

perfect square.

piecewise function (Ch 2) A piecewise function is a function which can be described al-gebraically by giving two or more different rules, each defined

over a given domain.

polynomial (Ch 8) A polynomial is a function made up of two or more terms of the

form §axn where a is a constant and n is a positive integer..

power (Ch 7) See exponent .

product rule (Ch 11) If u and v are functions of x and y = uv then the product

rule of differentiation is dy

dx = u

dv

dx + v

du

dx

one-to-many mapping

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quadratic equation (Ch 2) A quadratic equation is an equation which can be simplified to

ax2 + bx + c = 0 where a 6= 0.

quadratic formula (Ch 2) If ax2 + bx + c = 0 then x = ¡b § p

b2 ¡ 4ac

2a which

is called the quadratic formula.

quadratic function (Ch 2) A quadratic function has form f (x) = ax2 + bx + c where

a, b and c are constants and a 6= 0.

quartic function (Ch 8) A quartic function has form f (x) = ax4 + bx3 + cx2 +dx +e

where a, b, c, d and e are constants and a 6= 0.

quartile (Ch 3) If a data set is placed in ascending order then Q2, the second

quartile, is the score which divides the set into two equal groups

(the bottom half and the top half). Q2 is the median, Q1 is the

median of the bottom half. Q3 is the median of the top half.

quotient rule (Ch 11) If u and v are functions of x and y = u

v then the quotient

rule of differentiation is dy

dx =

du

dx v ¡ udv

dxv2

.

radian (Ch 5) A radian is the angle subtended at the centre of a circle by an

arc of equal length to the radius.

radical sign (Ch 2) The radical sign is thep

symbol used to represent the square

root of a number.

range (statistics) (Ch 3) The range of a dataset is the difference between the highest and

lowest values in the dataset.

range (function) (Ch 1) The range of a function y = f (x) is the set of all values

that y may have for all values of x in the domain.

rate (Ch 10) A rate is an expression of how much one quantity changes

in comparison with changes in another quantity, for example,

speed of 60 km/h, distance changes by 60 km as time changes

by 1 hour.

rational function (Ch 8) A rational function is of the form f (x) = g(x)

h(x).

rational number (Ch 2) A rational number is one which can be written in the form p

q where p and q are integers which have no common factors,

q 6= 0. Rational numbers have terminating or recurring decimal

expansions.

reciprocal (Ch 7) The reciprocal of a non-zero real number a is 1

a.

reciprocal function (Ch 8) The reciprocal function of f (x) is g(x) = 1

f (x).

rectangular hyperbola (Ch 9) A rectangular hyperbola is a function of the form f (x) = a

x

or f (x) = k + a

x ¡ h

where a, h and k are constants.

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regression line (Ch 4) Same meaning as ‘line of best fit’.

relation (Ch 1) A relation is any set (or collection) of points of the Cartesian

plane.

repeated roots (Ch 8) If an equation has two or more identical roots we say they are

repeated roots.

For example if an equation has roots 2,

¡1,

¡1, 3 and 1

2

we

say that ¡1 is a repeated root.

residual (Ch 4) Residual is the difference between the actual and the predicted

values.

residual plot (Ch 4)

revenue (Ch 2) Revenue is the amount of income received from the sale of

goods.

robust measure (Ch 3) A robust measure is one which is not greatly affected by out-

liers.root of an equation (Ch 2) A root of an equation is a solution of the equation, i.e., a value

of the variable which satisfies the equation.

For example, x = 2 is a root of x2 ¡ 3x + 2 = 0.

sector

segment

A residual plot is a scatterplot of the values of the residuals

against those of the independent variable. It shows the differ-

ences between the data points and their predicted values.

sample (Ch 3) A sample is a subset of a population.

scatterplot (Ch 3) A scatterplot is the plotting of points when bivariate data is

shown in the Cartesian plane.

secant of a curve (Ch 10) A secant of a curve is a straight line which joins any two points

on the curve.

secant of an angle (Ch 5) The secant of the angle µ (sec µ) is the reciprocal of cos µ, that

is, sec µ = 1

cos µ.

sector (Ch 5) A sector is a region of the interior of a

circle between an arc and the two radii

to the end points of the arc.

segment (Ch 5) A segment of a circle is the region be-

tween a chord of a circle and the arc

to its end points.

sexagesimal form (Ch 5) Sexagesimal form is expressing angles in degrees, minutes and

seconds.

similar triangles (Ch 5) Two triangles are similar if they have the same shape (look the

same) but are different in size. Corresponding angles are equal

and corresponding lengths are in the same ratio.

simulation (Ch 3) A simulation is the imitation of a particular random process by

use of a random number generator, usually a computer.

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sine axis (Ch 5) The sine axis is the vertical axis (y-axis) when considering the

unit circle.

sine of an angle (Ch 5) In a right-angled triangle, the sine (sin) of an angle is the ratio

of the side opposite the angle to the hypotenuse.

sine rule (Ch 5) In any triangle ABC

sin A

a =

sin B

b =

sin C

c

skewed distribution (Ch 3) A skewed distribution has a frequency diagram which is not

symmetrical about the mean.

spread (dispersion) (Ch 3) The spread of a dataset is the measure of how widely it is

spread out (or dispersed).

standard deviation (Ch 3) The standard deviation is a measure of the spread of a set of

data. It is a type of average of all the distances of data valuesfrom the mean.

stemplot (Ch 3) A stemplot (also called a stem-and-leaf plot) is a statistical

representation of a data set. The stem represents intervals and

the leaves are individual data values.

subscript notation (Ch 8) A subscript is a number or letter written in small size and placed

to the right and at the bottom of the letter or symbol written in

normal size.

summation notation (Ch 8) Summation notation (or sigma notation) is the writing of a sum

of numbers which have a particular form, in a shorter form.

For example, 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 =9P

i=1i

surd (Ch 2) Square roots, cube roots, etc., of numbers which do not produce

exact results are called surds. For example,p

2, 3p

4 are surds.

symmetric distribution (Ch 3) A symmetric distribution is a distribution of a dataset which

shows symmetry about the mean.

tangent of an angle (Ch 5) In a right-angled triangle, the tangent (tan) of an angle is the

ratio of the side opposite the angle to the side adjacent to the

angle.

tangent to a curve (Ch 10) A tangent to a curve is a straight line which touches the curve

at a point.

trial (Ch 3) A trial is one repetition of a particular mathematical experiment,

for example, rolling a die once.

turning point (Ch 2) A turning point on a function is either a local maximum or a

local minimum.

univariate data (Ch 4) Data which deal with one variable only is univariate data.

variable (Ch 1) A variable is a pronumeral that may have any value from a

given set of values.

B

A

Ca

bc

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variance (Ch 3) The variance of a dataset is the square of its standard deviation

i.e., s2.

velocity (Ch 10) The velocity of an object is its speed in a particular direction,

i.e., both the speed and direction have to be given at a given

point in time.

vertex (Ch 2) The maximum or minimum turning point of a quadratic functionis called its vertex.

whole numbers (Ch 1) A whole number is a number which has no decimal or fractional

part attached to it. For example, 7, 39, 1036 are whole numbers.

y-intercept (Ch 1) If a function cuts the y-axis at point (0, a) then a is the y-

intercept.

zero of a function (Ch 2) If y = f (x) then a zero of f (x) is a value of x which makes

f (x) have value 0.

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ANSWERS



442 ANSWERS

Exercise 1A

1 a i independent - no. of hours (h),

dependent - total charge (c)

ii No. of hours ( h ) Charge ( c )

0 601 100

2 1403 180...

...

iii f(0, 60), (1, 100), (2, 140), (3, 180), ......g

iv

v

vi C = 40h + 60

b i independent - no. of people (n),

dependent - amount received/person (A)

ii No. of people ( n ) Amount received ( A )

1 6 000 0002 3 000 000

3 2 000 000...

...

6 1 000 000...

...

iii f(1, 6000000), (2, 3 000 000), (3, 2 000 000),

...., (6, 1000000), ....giv

v A = 6000000

n

c i independent - diameter of table (d),

dependent - area of table top (A)

ii Diameter ( d ) Area ( A )

0 02 3:144 12:566 28:27...

...

iii f(0, 0), (2, 3:14), (4, 12:56), .....g

iv

v

vi A = ¼

³d

2

´2d table, graph, equation

Exercise 1B.1

1 a i f (¡1) = ¡7 ii f (3) = 9 iii f (0) = ¡3iv f (b) = 4b¡ 3

b i g(1) = ¡9 ii g(¡2) = ¡6

iii g(11

2) = ¡6

iv g(1+a) = 2(1+a)2+1+a¡12 = 2a2+5a¡9

c p(¡2) = 0 p(¡3) = 0 ) p(¡2) = p(¡3)

d h(a) = a3 h(¡a) = ¡a3

¡h(¡a) = ¡(¡a3) = a3

) h(a) = ¡h(¡a)

2 a i y = ¡11 ii x = 4

b i y = ¡5 ii x = 1

3

c i y = 3 1

4 ii x = 7

3 a f (2) = 2 b f (¡3) = 1

3 c f (1

2) = 5

d f (0) is undefined

4 a i g(3) = ¡12 ii g(¡3) = ¡60iii g(0) = 0

b i h(0) = 100 ii h(¡2) = 84 iii h(2) = 84

5 a yes b no

6 a no - one value of the independent variable has more

than one corresponding dependent value

b yes - each point in the domain maps to one point

in the range

c yes - each point in the domain maps to one point

in the range

7 a yes - each point in the domain maps to one point

in the range

60 100 140 180

0 1 2 3

c

h

0 5 10 15 20 25 30

0 2 4 6

A

d

50

100

150

200

1 2 3 4

c

h

charge

number of hours

1

1 106 106 106 106 106 106

4

4

2

2

5

5

3

3

6

6 A

n

1 2 3 4 5 6

A

n1 106

2 106

3 106

4 106

5 106

6 106

5

10

15

20

25

30

1 2 3 4 5 6

A

d

a r e a

diameter



ANSWERS 443

b no - domain points map to more than one point in

the range

Exercise 1B.2

2 a discrete b continuous c discrete

d continuous e continuous f continuous

g discrete

3 class discussion would be valuable

Exercise 1B.3

1 a i dependent - rent paid (R),

independent - no. of weeks rented (n)

ii R discrete n discrete

iii n R

1 1202 2403 360...

...

52 6240

f(1, 120), (2, 240), (3, 360), ...., (52, 6240)gR = 120n

iv 0

6 n 652 0

6 R 66240

b i independent - no. of cards (n)

dependent - cost of cards (C )

ii n discrete, C discrete

iii n C

0 460...

10 462:50...

100 485...

200 510

f(0, 460), ...., (10, 462:50), ...., (100, 485),

...., (200, 510)gC = 0:25n + 460

iv domain: 0 6 n 6 200, range: 460 6 C 6 510

c i independent - no. of km (n)

dependent - depth of tread (T )

ii n continuous, T continuous

iii n (km) T (mm)

0 15

3000 146000 139000 12

12 000 11...

33 000 4

f(0, 15), (3000, 14), (6000, 13), ....,

(33000, 4)gT = 15 ¡

n

3000

iv domain: 0 6 n 6 33 000range: 4 6 T 6 15

d i independent - date (D)

dependent - closing price (P )

ii D discrete, P discrete

iii f(5, 3126:7), (6, 3178:2), (7, 3167:0),(8, 3104:8), (9, 3137:9)g no equation

iv domain: 5 6 D 6 9range: 3104:8 6 P 6 3178:2

e i independent - length of metacarpal (l)

dependent - height of person (h)

ii l continuous, h continuous

iii l h

39 15741 16343 175

45 17146 17347 17348 17249 18351 178

iv domain: 39 6 l 6 51v range: 157 6 h 6 183

Exercise 1D.1

1 a mAB = ¡4

7 b mAC = ¡ 4

11 c mBC = 0

d mFI = ¡ 2

3 e mEH = 4:5

f mBF is undefined g mHE = 4:5

h mEG = ¡ 2

11

T (mm)

n (km × 10 )3

0

5

10

15

20

0 10 20 30 40

2000

4000

6000

10 20 30 40 50 60

n

R

number of cards

C

c o s

t

100 200 300

460

480

500

520



444 ANSWERS

2 a 1

2 b 1

5 c ¡1

8 d 1 e ¡11

6 f 1

2 g 4

h 0 i undefined j ¡0:95 k ¡0:75 l 1

12

m b ¡ d

a ¡ c n ¡

n

m3 c perpendicular

4 a horizontal b vertical

Exercise 1D.21 a mAB = ¡0:5 mCD = 2 ) perpendicular

b mAB = 1

3 mCD = 3 ) neither

c mAB = 1 mCD = 1 ) parallel

d mAB = ¡2 mCD = ¡2 ) parallel

e mAB = 0 mCD = 0 ) parallel

f AB is a vertical line, CD is the x-axis

) perpendicular

2 mAB = 1

2 mCD = 1

2 ) parallel

3 mAB = ¡2 mCD = 1

2

) perpendicular

4 1st segment 2nd segment

a m = 1 1

3 m = 11

3 collinear

b m = ¡1

3 m = ¡1

3 collinear

c m = 1 1

4 m = 2 not collinear

d m = 1

3 m = 1

3 collinear

e m = 1 1

3 m = 11

3 collinear

f m = 1:2 m = 1:2 collinear

g m = 1

9 m = ¡1:5 not collinear

h m =

b

a m =

b

a collinear

5 mHI = 2

3 mJK = 2

3 mIJ = ¡0:5

mKH = ¡0:5 ) parallelogram

6 mAB = 1, mBC = 1

3, mAC = ¡3

) BC?AC ) right-angled triangle

7 any point satisfying y = ¡x + 2 for example,

x = ¡3, y = 58 a mAB = 0 mCD = 0 mBC = 1 1

3 mAD = 1 1

3

AB = 5 CD = 5 BC = 5 AD = 5) ABCD is a rhombus

b mAC = ¡2, mBD = 1

2 ) diagonals are

perpendicular

Exercise 1E

1 a i volume of ink (v) ii length of line L

iii 0 iv metres v 300 vi metres/mL

b i minutes typed ii words typed iii 0iv words v 32 vi words/min

c i length of cable ii cost of connection

iii $800 iv $ v 1500 vi $/km

d i weekly sales ii salary iii $500 iv $

v 0:1 vi $/$

e i Celsius ii Fahrenheit iii 30 iv degrees

v 2 vi deg F/deg C

f i no. of pages ii printing time iii 15iv seconds v 6 vi pages/second

g i call time ii cost iii 5 iv dollars

v 0:8 vi dollars/minute

Exercise 1F.11 a m = 3, c = 4

b y = ¡5x + 2; m = ¡5, c = 2

c y = 1

2x + 0; m = 1

2, c = 0

d y = 1x ¡ 5

4; m = 1, c = ¡5

4

e y = 0x ¡ 3, m = 0, c = ¡3

f y = 2

3x + 0; m = 2

3, c = 0

g y = ¡3x + 4; m = ¡3, c = 4

h y = 1

2x + 1

2; m = 1

2, c = 1

2

i y = ¡2x + 8; m = ¡2, c = 8

j m undefined, c indeterminantk y = 5

3x + 23

3 , m = 5

3, c = 23

3

l m = 1200, c = ¡4500

m y = 3x + 2, m = 3, c = 2

n y = ¡x + 3; m = ¡1, c = 3

o y = ¡ 5

8x + 3

8; m = ¡5

8, c = 3

8

p y = 0x + 0; m = 0, c = 0

3 a y = 2x + 5 b y = ¡x + 6 c y = 2

3x ¡ 3 2

3

d y = 1

2x e y = 4 f x = ¡2

g y = ax ¡ a2 h y = ¡3x ¡ 8

i y = 2x ¡ 3 1

2 j y = 1

3x + 5

6

4 a y = x + 1 b y = 2x ¡ 5 c y = 2x + 1

d y = ¡3x ¡ 5 e y = x

8 + 2 3

4

f y = ¡7

6x + 2 2

3

5 m = ¡3, y = ¡3x 6 m = 3, y = 3x + 3

Exercise 1F.2

1 a 2x ¡ y + 3 = 0 b x + y ¡ 2 = 0c 3x ¡ y = 0 d y ¡ 2 = 0e 3x ¡ 4y + 4 = 0 f 2x ¡ 4y + 3 = 0

g x = 0 h ax ¡ y + b = 02 a y ¡ 8 = 0 b x + 1 = 0

c 3x ¡ 2y ¡ 16 = 0 d 8x + 6y ¡ 13 = 0

e bx ¡ ay = 0 f gx ¡ f y = 0

3 a y = 0 b x = 0 c y = 3 d x = ¡2e y = d f x = c

4 a m = 2; 2x ¡ y ¡ 11 = 0

b m = ¡1

2; x + 2y ¡ 8 = 0

5 a 2x + 3y ¡ 11 = 0 (m = ¡2

3)

b m = 3

2; 3x ¡ 2y + 3 = 0



ANSWERS 445

6 AB: y = 1

2x + 2 1

2 AC: y = ¡2x + 10

mAB = 1

2, mAC = ¡2

7 AB: 2x ¡ 5y + 24 = 0 AC: 9x ¡ 4y ¡ 3 = 0BC: 7x + y + 10 = 0

8 a C = 50n + 2000 b $2000 c $7000

9 a D = ¡5 p + 100 b 87:5 c p = $20

Exercise 1H

1 a x y

¡1 ¡20 11 4

b x y

¡1 80 51 2

c x y

¡1 ¡400 601 160

d x y

¡1 ¡70000000 ¡20000001 3 000 000

e x y

¡1 0:970 11 1:03

f x y

¡1 00 01 0

2 a m = 1,

c = ¡3b m = 1

2,

c = ¡1

c m = ¡2,

c = 1000d m = ¡5000,

c = 40000

e m = 0:006,

c = 0:002f m = 0:007,

c = 0:002

3 a (6, 0) (0, 4) b (¡6, 0) (0, 2)

c (3, 0) (0, ¡3

5) d (4, 0) (0, 8)

e (¡2, 0) (0, 12

) f (12, 0) (0, ¡16)

Exercise 1I.1

1 a i n - no. sold ii earnings E iii discrete

iv 1:5 v E / n vi E = 1:5n

b i n - no. of sleepers ii length of edging L

iii discrete iv 1:8 v L / n vi L = 1:8n

c i n - mL of brandy ii A amt of alcohol

iii continuous iv 0:17 v A / n

vi A = 0:17n

2 a C / D b ¼ c C = ¼D

3 a no b no c yes d yes e no f yes

4 (600, 108) (950, 171) (700, 126)

5 R = kL k = 172

7 a T b F c F d T e T f F

8

tyre tread decreases etc. Rate of change is negative.

Exercise 1I.2

1 a d - continuous, C - continuous

b independent - distance driven, dependent - cost

c Distance Cost

0 20100 45200 70300 95480 140

d domain: 0 6 d 6 480e range: 20 6 C 6 140

f

g 20 h fixed cost of hiring $20 i 0:25 $/km

j charge for each km driven k C = 0:25d + 20

2 a both discrete

b no. of instruments: independent cost : dependent

c domain: 06

I 6

120 range: 3006

C 6

2200

As the no. of kilometres driven increases, the depth of

y

x

y

x

y

x

y

x

y

x

y

x

y

x

y

x

-Et _

y

x

x

y

x y

Qw _

x

y

0

50

100

150

0 200 400 600



446 ANSWERS

d $300 fixed manufacturing costs e ¼ 16f cost to make each instrument g C = 16I + 300

3 a independent variable: time taken

dependent variable: oven temperature

b T = 20:5t + 29:7 c 29:7d Class discussion would be valuable.

4 Q = ¡4n + 248 max. capacity 248 million kL5 (¡3, ¡3) (¡1, ¡1), (0, 0), (1, 1), (3, 3);

appears linear from the points.

6 a Weight of chicken ( W ) cooking time ( T )

0 201 452 703 95

T = 25W + 20

b 65 min c 1:6 kg

1 a i discrete ii discrete iii continuous

iv continuous

b i independent ii dependent

2 a i y = 1

4x + 2 ii y = 3 b i collinear ii not

3 a ¡ 2

3 b ¡ 1

2

4 s = 30¡n

2

¢+ 100, discrete, range: 100 6 s 6 1660

5 a (0, ¡8)

b i m = 3

2, c = 3 ii m = 1:5, c = 4

iii m = 1, c = ¡1; no common point as i and

ii are parallel

6 a i and ii b all lines are parallel

7 a no b v = d

t v / d t / d

Exercise 2B.1

1 a 2p

7 b 2p

2 c 3p

6 d 3p

3 e 4p

2

f 2p

10 g 3p

5 h 10p

7 i 7p

2 j 8

k 6p

2 l 12

2 a 7p

3 b 9 c 4p

2 d 9p

6 e ¡25p

2

f 2a g ¡3x h 4000p

10 i 32p

2 j ¡6

k f 3 l 2cp

2c

3 a 1

3 b 4

5 c 9

2 d 5 e 0:1 f 0:2 g 4

3

h 1:5

4 ap

21 bp

6 c 6 d 24 e 8 f 6p

6

g 72 h 20

p 14 i 6 j 36 k 240 l 72

5 a 2 bp

2 c 1 d 3p

3p 5

e 5

2 f

p 6

gp

2 h 2 i 3

2 j

1p a

k

p 3

2 l 9

4

6 a x = §10 b x = 10 c x = ¡10

d x = §10 e x = §1 f x = 0

g x = §p ¡4 which is not a real number h x = ¡2Exercise 2B.2

1 a 7p

3 b 10p

6 c ¡2 d ¡2p

2e 3

p 2 f ¡7

p 3

2 a ¡9p

6 b ¡3p

5 c ¡5p

2 d 3p

7e ¡9

p 2 f ¡2

p 6

3 a 4p

3 b 9p

3 c ¡2p

2 d 7p

17 e 5p

3f 7

p 2 g ¡

p 3 h 5

p 2 i 1 +

p 2 ¡

p 3

j 5p

3 k 16p

3 l ¡p

6

4 a 3p

5 b 24p

2 cp

a d a(p

2 +p

3)e 7

p 2 f 4a g

p c(c + 4) h d(

p e ¡ e)

5 a 5p a b 2p a c 2b d 9a e a2 f ¡4p bg

m

5 h

c

8 i abc j 6xy k a l b2 m 2

n 1 op

b

Exercise 2B.3

1 a

p 6

2 b

5p

2

2 c

p 2

2 d 2 e

p 21

3

f

p 2

2 g

p 6 h

p 3 i

p 3

3 j

p 3

2

k ¡2p

3 l ¡

p 15

52 a

2p

15

3 b 4

5

p 30 c

p 6 d 2

9

p 3 e 2

f 3p

2

5 g

p 6(1 +

p 2)

6 h

p 10(3 + 2

p 5)

10

i 6 j ¡ 5

3 k 6 l

p 6

2

3 a

p a

a b

2p

c

c c

p a d

p ap

b

b e ¡2

p b

f p

3c gp 6

9 h ¡

p b

a

Exercise 2C.1

1 a x2 + 3x + 2 b y2 + 11y + 24 c t2 + t ¡ 6

d r2 + 3r ¡ 18 e c2 + 3c ¡ 10 f x2 ¡ 6x + 8

g a2 ¡ 6a + 9 h z2 ¡ 25 i k2 + 2k + 1

2 a x2 + 5x + 6 b x2¡ 5x¡ 6 c x2¡ 20x + 99

d 2x2¡x¡6 e 6x2¡33x+45 f 2x2+15x¡27

g 27¡21x+2x2 h x2¡a2 i 6x2¡14ax+8a2

j 18x2 ¡ 12x + 2 k x2 + 10x + 25

l x3 ¡ x2 ¡ 5x ¡ 3

3 a x2+4x+4 b 9x2¡12x +4 c 81¡18x+ x2

d 4a2 ¡ 12ax + 9x2 e 25x2 ¡ 70x + 49

f 9¡12x+4x

2

g x

2

+2x+1 h 1¡10x+25x

2

0

25

50

75

100

0 1 2 3 4

W (kg)

T (min)

Chapter 1 Revision Set



ANSWERS 447

i 9y2 ¡ 12xy + 4x2 j 36x2 ¡ 6x + 1

4

k 104x2 + 200x + 1 l 9 ¡ 24x + 16x2

4 a x2¡ 81 b x2¡ 36 c 81¡x2 d 4x2¡ a2

e 4x2 ¡ 4a2 f 18x2 ¡ 2 g 1

4x2 ¡ 1 h 1¡ x2

i 100x2 ¡ 9

5 a 9x2 ¡ 1 b 9r2 + 30r + 25 c 4x2 ¡ 2x ¡ 12

d 1 ¡ 2t + t2

e r2

¡ 2

3r + 1

9 f 15 + 2r ¡ r2

g 3x2 ¡ 75 h ¡r2 ¡ 6r ¡ 9 i ¡t2 ¡ 2t + 8

j d3 ¡ d k 12x2 ¡ 25x + 12 l t4 ¡ 2t2 + 1

Exercise 2C.2

1 a (x + 2)(x + 3) b (x + 2)(x + 5)c (x + 3)(x + 5) d (x + 2)(x + 9)e (x + 5)(x ¡ 3) f (x + 2)(x ¡ 11)g (x + 6)(x ¡ 4) h (x + 3)(x ¡ 4)

i (x + 7)(x ¡ 3) j (x ¡ 3)2 k (x ¡ 6)(x ¡ 1)

l (x +3)(x + 6) m (x¡2)2 n (x¡14)(x¡1)o (x + 6)(x ¡ 7) p (x ¡ 4)(x + 9)

q (t ¡ 3)(t ¡ 5) r (t ¡ 4)(t ¡ 6)s (r¡16)(r + 2) t (h + 6)2 u (s¡15)(s + 4)

2 a (x + a)2 b (x ¡ a)2 c (x + 2)(x ¡ 2)

d (x + 5)(x ¡ 5)

3 a (2x + 1)(x ¡ 5) b (3x ¡ 1)(x + 3)

c (2x+1)2 d (3x+5)(x¡3) e (2x¡3)(x¡4)f (3x + 4)(2x ¡ 1) g (2x + 3)(5x ¡ 1)h 2(2x + 1)(3x ¡ 1) i 6(2x ¡ 1)(x ¡ 1)

j (8x + 1)(2x + 1) k (2x ¡ 3)(x ¡ 2)l (3x ¡ 1)(7x + 1) m x(2x + 1)(x ¡ 2)n 2(x + 4)(x ¡ 3) o 3(2x ¡ 1)(3x + 4)p 2x(2x ¡ 5)(3x + 1) q 3(r ¡ 5)(2r + 3)r 2(2r ¡ 1)(3r + 1)

Exercise 2C.3

1 a (a + b)(a ¡ b) b (x + 4)(x ¡ 4)c (x + 1)(x ¡ 1) d (x + 4)(x ¡ 4)e (c + 5)(c ¡ 5) f (y + 40)(y ¡ 40)g (x + 10)(x ¡ 10) h (x + 1

2)(x ¡ 1

2)

i (2x + 5)(2x ¡ 5) j (3x + 4)(3x ¡ 4)k 4(x + 2)(x ¡ 2) l (2w + 3)(2w ¡ 3)

m (x + 1

3)(x ¡ 1

3) n 16(x + 1)(x ¡ 1)

o (2x + c)(2x ¡ c) p (p

5x +p

7)(p

5x ¡p

7)

q (p

3x + 4)(p

3x ¡ 4) r ( 12

t + 1)( 12

t ¡ 1)

2 a 2(a + 4)(a ¡ 4) b (a2 + 4)(a ¡ 2)(a + 2)

c (x + 5)(x ¡ 1) d (x2 + y2)(x + y)(x ¡ y)

e x2 f 3(2a + 5b)(2a ¡ 5b)

g (1 + x2)(1 + x)(1 ¡ x) h (x2 + 2)(x2 + 3)

i (a ¡ 2)(4a ¡ 1) j (d + 2)(d + 6)k 2(n ¡ 1)2 l 2(x ¡ 2)

3 a 5(y ¡ 2) b 4(4 ¡ t) c 2(2t + r ¡ 3s)d (x + 11)(x ¡ 11) e (x ¡ 4)(x ¡ 7)f (t + 9)(t ¡ 3) g 3( p ¡ 3)( p ¡ 1)h 2(c + 1)(c ¡ 1) i 4(3 + x)(3 ¡ x)

j x(x + 5)(x ¡ 5) k 3(x ¡ 1)2

l (a2 + b)(a2 ¡ b) m (2d ¡ 1)(d + 7)n (1 + 10x)(1 ¡ 10x) o ¡1(x + 2)(x + 3)

5 a i a3 + b3 ii a3 ¡ b3

b i (x+3)(x2¡3x+9) ii (x+4)(x2¡4x+16)

iii (2x + 1)(4x2 ¡ 2x + 1)

iv 2(x ¡ 2)(x2 + 2x + 4)

v (a + b)(a ¡ b)(a

4

+ a

2

b

2

+ b

4

)Exercise 2C.4

1 a i 0 ii 0 iii ¡3 iv a2 + 2a ¡ 3

b i 0 ii 0 iii ¡10 iv a2 ¡ 3a ¡ 10

c i 0 ii 0 iii ¡6 iv 8a2 ¡ 22a ¡ 6

2 a yes b no 3 a above b below

4 (0, 3) (1, 4) etc.

Exercise 2D.1

1 a x = 3, x = ¡2 b x = 2, x = 0:5c x = 1:79, x = 2:79

2 a x = ¡3 or x = ¡1 b x = 3 or x = 8c x = 6 or x = ¡2 d x = ¡3 e x = 4f x = 1

2, or x = ¡4 g x = 3 or x = ¡1

3

h x = ¡1

2 or x = ¡1 1

2 i x = 2

3

3 a x = 1 or ¡1 b x = §5 c x = §2d x = §10 e x = §5 f x = §3

g x = §5

2 h x = §4 i x = §1

9

4 a x = ¡3 or x = 5 b x = ¡2 or x = ¡5

c x = ¡1

3 or x = 2 d x = 4 or x = ¡6

e x = ¡2 or x = ¡3 f x = ¡5 or x = 3g x = §6 h x = §4 i x = §1

j x = ¡a k x = 13 or x = ¡1

2

l x = 1

2 or x = ¡11

3

Exercise 2D.2

1 a (x + 1)2 b (g + 3)2 c (x ¡ 7)2

d (a + 1

2)2 e (m ¡ 5)2 f (x + 100)2

2 a 4 b 9 c 16 d 25

e x2 + 10x + 52 = (x + 5)2

f x2 ¡ 6x + 32 = (x ¡ 3)2

g x2 + 12x + 36 = (x + 6)2

h x2 + 3x + 9

4 = (x + 3

2)2

i x2 ¡ 5x + 25

4 = (x ¡ 5

2)2

3 a 9, 3 b 25, 5 c 12y, 6 d 81, 9 e 25

4 , 5

2

f 494

, 72

4 a 1, (x + 1)2 b 9, (x + 3)2 c 25

4 , (x ¡ 5

2)2

d 1

4, (x + 1

2)2 e 25

4 , 2(x + 5

2)2 f 1, 3(x ¡ 1)2

g 1

16, 2(x¡ 1

4)2 h 4

9, 3(x¡ 2

3)2 i 4, 4(x¡2)2

5 a x = 1 §p

3 b x = 3 c x = 5, x = 2

d x = 4, x = 1 e x = 3, x = ¡1

2

f x = 3

2, x = 1

3



448 ANSWERS

6 a x = 2 §p

3 b x = ¡1 §p

6

c x = ¡5, x = ¡1 d x = ¡ 3

2 §

p 21

2

e 5

2 §

p 21

2 f x =

1 §p

13

2 g x =

3 §p

33

4h x = 4, x = ¡1 i x = 1

3, x = ¡5

7 a x = 3 §p

13 b x = 1 §p

2 c x = 4 §p

6

d x = 3

2§

p 52

e x = ¡5

2§

p 372

f t = 1

2§

p 212

g x = 3

2§

p 3

2 h x = 1

6§

p 13

6 i x = 1

4§

p 33

4

8 x = ¡1 §p ¡2

Exercise 2D.3

1 a x = 3 or x = ¡2 b x = 1 or ¡5

c x = 3 or x = ¡1

2 d x = 2 §

p 3

e x = ¡1 §p

6 f x = ¡1 or ¡5

g x = ¡3 §

p 21

2 h x =

5 §p

21

2

i x = 1 §

p 13

2 j x = 3 §

p 33

4k x = 4, ¡1 l x = 1

3, x = ¡5

m x = 1

3 n x = 2

3 or x = ¡4

5

o x = ¡1 §

p ¡11

6

2 a x2 + 9x + 8 = 0, x = ¡1 or ¡8

b x2 ¡ 6x + 9 = 0, x = 3

c 3x2 ¡ x ¡ 5 = 0; x = ¡1 §

p 61

6d x2 + x + 1

4 = 0; x = ¡1

2

e x2

¡ 9x + 18 = 0; x = 3 or x = 6

f x2 + 5x + 1 = 0; x = ¡5 §

p 21

4

3 a x = ¡3 or x = ¡1 b x = +2 or x = ¡3

c x = ¡1

2 or ¡3

Exercise 2D.4

1 a ¢ = 60 b ¢ = ¡27 c ¢ = 181d ¢ = 57 e ¢ = 145 f ¢ = 0

g ¢ = 16 h ¢ = 0 i ¢ = ¡11

2 a one b 0 c 2 irrational d 0 e one

f 2 irrational g 2 irrational h 2 irrational

i 2 irrational

Exercise 2E.1

1 a x ¡3 ¡2 ¡1 0 1 2 3

y 0 ¡6 ¡10 ¡12 ¡12 ¡10 ¡6

b x ¡3 ¡2 ¡1 0 1 2 3

y 25 12 3 ¡2 ¡3 0 7

c x ¡3 ¡2 ¡1 0 1 2 3y 6 0 ¡4 ¡6 ¡6 ¡4 0

2 a x ¡3 ¡2 ¡1 0 1 2 3

y 9 4 1 0 1 4 9

b x ¡3 ¡2 ¡1 0 1 2 3

y 1 0 1 4 9 16 25

c x ¡3 ¡2 ¡1 0 1 2 3y 11 6 3 2 3 6 11

d x ¡3 ¡2 ¡1 0 1 2 3

y 18 8 2 0 2 8 18

Exercise 2E.2

1 a x = ¡4, x = 2; (0, ¡8); (¡1, ¡9)

b x = 7, x = 5; (0, 35); (6, ¡1)

c x = ¡5, x = 2; (0, ¡10); (¡3

2, ¡ 49

4 )

d x = ¡5, x = 5; (0, ¡25); (0, ¡25)

e x = ¡1

2, x = 1

2; (0, ¡1); (0, ¡1)

f x = ¡1

2, x = ¡1; (0, 1); (¡3

4, ¡ 1

8)

2 a (0, 5); (¡1, 4); no real zeros

b (0, ¡2); ( 56

, ¡ 49

12); x = 2, x = ¡1

3

c (0, 6); (

3

4 ,

39

8 ); no real zeros

3 a x = 4, x = ¡2; (0, ¡8); (1, ¡9)

b x = ¡1, x = 3; (0, ¡9); (1, ¡12)

c x = 1, x = ¡4; (0, ¡4); (¡3

2, ¡ 25

4 )

d x = 2, x = 5

2; (0, ¡20); ( 9

4, 14

)

e x = d §p

d2 ¡ 4; (0, 4); (d, 4 ¡ d2)

f x = ¡b §

p b2 ¡ 4ac

2a ; (0, c); (

¡b

2a, 4ac ¡ b2

4a )

y

x

y

x

y

x



ANSWERS 449

Exercise 2F.1

1 a b

c d

e f

2 a b

c d

e f

3 a b

c d

e f

Exercise 2F.2

1 a y = (x ¡ 1)2 ¡ 3 b y = (x + 3)2 ¡ 1

c y = 2(x ¡ 1)2 + 4 d y = 2(x + 1)2 + 3

e y = 2(x ¡ 11

2)2 + 3 1

2

f y = ¡3(x + 2)2 + 4

2 a y = 2(x ¡ 13

4 )2 ¡ 225

8

b (31

4, ¡281

8) c ¡7 d x = 7, x = ¡1

2

e

3 a i x = 3 ii (0, 18) iii (3, 0)

b i no zeros ii (0, 6) iii (1, 5)

c i x = ¡5 §

p 29

2 ii (0, ¡1) iii (¡ 5

2,¡29

4 )

d i (0, 0) ii (0, 0) iii (0, 0)

e i x = § 2p

3ii (0, ¡4) iii (0, ¡4)

f i x = §2 ii (0, 12) iii (0, 12)

4 a y = (x ¡ 2)2 (2, 0); (0, 4)

b y = (x + 1

2

)2 + 3

4

; (¡1

2

, 3

4

); (0, 1)

y

y y

y

y

x

x

x

x

x

y

x

y

x

y

x

y

x

y

x

y

x

y

x

y

x

y

x

y

x

y

x

y

x

y

x

y

x

-\Qw _ 7

-7

(3\Q r _\'-28\Qi _\)



450 ANSWERS

c y = 3(x ¡ 1

6)2 ¡ 3 1

12; ( 1

6, ¡3 1

12); (0, ¡3)

d y = ¡1

2(x + 2)2 + 6; (¡2, 6); (0, 4)

5 a g(x) = (x ¡ 2)2 b g(x) = (x + 1)2 ¡ 2

c g(x) = ¡(x + 3)2 d g(x) = 2x2

e g(x) = 2(x ¡ 1)2 ¡ 2 f g(x) = ¡x2 + 2

6 All functions have roots x = ¡2, x = 1

7, 8, 9, 10 class discussion would be valuable.

Exercise 2G

2 a b

c

3 a b

c

4 a y = ¡ jx ¡ 1j b y = 1

2jxj c y = jxj + 2

d y = jx ¡ 1j + 1

2 e y = ¡ jx ¡ 1j+ 2

f y = 3 jx + 3j5

x ¡6 ¡5 ¡4 ¡3 ¡2 ¡1 0y ¡1

6 ¡1

5 ¡1

4 ¡1

3 ¡1

2 ¡1 undef.

x 1 2 3 4 5 6

y 1 1

2

1

3

1

4

1

5

1

6

6 a b

c d

7 a b

c

Exercise 2H

1 a no wind resistance, hit at ground level

b t = 0 or t = 4, t = 0 is start

) hit ground at 4 sec.

c t h

0 01 152 203 154 0

d 20 m e 18:75 m g ii t = 1 or 3 sec

h t = 2 §p

2, or t = 0:6, t = 3:4

2 a class discussion b $1600 c 38

d 0 6 x 6 N , 200 6 C 6 RN and R depend upon demand

3 a x = 1

2 mm b 0:3419 mm 6 x 6 0:6581 mm

4 10:4 secs

5 a $140 b n = 25 c $4800d R = ¡2n2 + 200n

e

y

x

y

x

y

x

y

x

y

x

y

x

y

x

x y

1

y

x

x

y

x x

y 2

41 x

y y

x

x y 3

x y 1

y

x

y

x

y

x

4321

25

20

15

10

5

h

t

100

5000

4000

3000

2000

1000

R

n

y

x

2 x

2 y

32

1

y

x



ANSWERS 451

f $5000 g 50

6 a A = 2xy b L = 4x + 3y

c 1200 = 4x + 3y ) y = 1200 ¡ 4x

3

d A = 2x(1200 ¡ 4x

3 ) = 800x ¡

8x2

3e

f x = 150, y = 200

7 a y = ¡2:041x2 + 4:49x ¡ 1:47

b As a quadratic model has three parameters a,

b and c, three points will determine a, b and c

uniquely.


1 a i 7p

2 ii 16 iii ¡18p

2 iv 4b2

b i 2p

3

3 ii 3

p a

2 a 7p

3 bp

7 c 48p

2 d 85 + 28p

3

e 207 f 7 + 5p 23 a ¡20x2 + 6x + 2 b 9x2¡ 4 c x2¡ 15x¡ 28

d ¡3x2 + 9x ¡ 9

4 (x + 1)(x ¡ 3) = x2 ¡ 2x ¡ 3

5 a (2x ¡ 3)(2x + 1) = 0 x = 3

2 or x = ¡1

2

b x = 3

5 or ¡ 1

2 c x = 3

2 or ¡4

6 a ¢ = ¡20 ) zero b ¢ = 49 ) 2c ¢ = 0, ) 1

7 y = 6(x + 1

12)2 ¡ 2 1

24

8 g(x) = ¡2(x + 2)2 + 2

9 (0, 6), (1, 7) 10 a = ¡3

11 L = 150 m, B = 225 m

12 a y = jxj¡ 3 b y = jx ¡ 3j

c y = jxj+ 2 d y = jx + 2j

e y = 2 jx + 1j¡ 3 f y = ¡0:5 jx ¡ 1j+ 2

Note: Statistics is quite different to other areas of Math-

ematics - many questions in the field of statistics do not

have definite answers. Answers in this section are often

suggested, rather than definitive, which means that you can

disagree with them, and quite possibly be right!

Exercise 3A

1 a Here are some of the patterns. There are others.

Many more males were exposed to the occurrence.

There are a large number of people whose classif-

ication of Economic Status was ‘Other’.

For females, more so than males, economic status

was a major factor in the death rate.

For children, more so than adults, economic status

was a major factor in the death rate.b The incident was the sinking of the Titanic.

2 A - Rockhampton B- Stanthorpe C - Cairns

D - Lady Elliot Island E - Longreach

F - Thursday Island G - Toowoomba

H - Coolangatta

Exercise 3B

1 a i teenagers in the local community

ii the variable is the frequency of car accidents

b for example, mean number of accidents involving

teens on a Saturday night

c It depends on what the researcher wants to know, but a census may be possible since the police

would have a record of each automobile accident.

2 a i students at my school

ii student’s opinion on the number of school dances

next year

b for example, the percentage of students that want

six dances next year

c a sample should be adequate.

3 a i for example, age of Labor voters

ii for example, gender of Labor voters

b i for example, distance from the Sun

ii colour of the planet, as seen through a telescope

Exploring data solutions

300

60000

40000

20000

A

x

x

y

0.4 1.8

(1"1' 1)

y

x

y

x

(-1'-3)

-2\Qw _ -1 Qw _

y

x

y

x

(1' 2)1\Qw _

y

x

y

x



452 ANSWERS

c i for example, length

ii for example, gender

d i for example, weekly income from paid work

ii for example, preferred career

4 a D b C c D d D

e C (in our opinion - even though money is discrete,

practically speaking large sums of money are con-

tinuous)f C

g D (while foot length is continuous, shoes come in

discrete sizes)

h D i C

j trick question! - neither, since this is a categorical

variable

5 a population, since the tuckshop manager has this data

available

b sample - unless data on all golf holes on all golf

courses in Australia is readily available

c sample - it is impossible to ensure that every fish

has been captured and identifiedd population - this data is readily available

e sample - it is not possible or even practical to de-

termine the voting intentions of all electors in a

Queensland seat

6 a Not practical - it is difficult to determine the lifes-

pan of even one giraffe in the wild, much less all

of them.

b Probably not practical. The number of zoos and

wild animal parks in the world is very large.

c Depends on the size of the school. For most high

schools, it is more practical to survey rather than

conduct a census.d Practical, since the number of staff in secondary

schools is generally less than 100.

e Not practical - a restaurant owner cannot gather

data from all customers, as some would refuse to

participate.

f Practical - the Australian government does conduct

a census every five years, to answer such questions

as this.

Exercise 3C

1 a observation b survey c experiment

d survey (but the survey must be anonymous so

students feel free to answer truthfully)

e available data f survey or observation

2 These answers are examples. There are other equally

valid methods.

a Number each customer. Choose 50 numbers using

a random number generator.

b She must not only sample oranges from the top of

each crate (guess why!). The wholesaler needs to

select fruit from throughout a crate. This might be

easiest to do as the oranges are removed from the

crate for resale.

c If the population of koalas that will be affected

by the proposed highway is known, number each

koala (or identify each in some other way), and

choose a sample using a random number generator.

3 a This is a self-selecting sample. A phone survey

would be preferable.

b This is a self-selecting sample. Most magazines are

bought over the counter, so it is not possible to se-

lect randomly from the population of readers, since

the population isn’t known. To increase reader par-

ticipation in the survey, the magazine might offer a

computer system as a prize, with all of those who

complete the survey being in the draw.

c Weekend users of boat ramps are excluded. The

DoT researcher should survey on the weekend as

well.

4 Only listeners of Triple J had a chance to participate,

and of those, only those with a strong opinion would

choose to participate. Despite its impressive sample size,

the sample tells us nothing about the opinions of the

Australian population on this issue.

5 A class discussion of student solutions would be

valuable. See website for development of model.

Exercise 3D

1 a 19 b 21.12 a It will disadvantage schools that had a large number

of absent students on the day.

b For example, the Education Dept thought that

schools might otherwise only give the test to stu-

dents who were expected to do well.

3 10.7682, 8.7311, 9.5314.

4 A class discussion of student solutions would be

valuable.

Exercise 3E.1

1 a time series

b pie graph, to show how the budget is apportioned

c scatterplot d column graph

e pie graph is probably best f scatterplot.

2 This time series graph shows, over a year: distance be-

low ground, size of larvae, when it is above ground, the

time when the cycle begins anew, and length of each

stage.

3 a battle radius = 1 cm, non battle radius = 3:4 cm

) proportion = 8:6%

b The coxcomb graph shows month by month, the

relative number of deaths in the Crimean War.

More striking is the clear picture that a large ma-

jority of the deaths did not occur on the battlefield

and hence may have been preventable.

Exercise 3E.2

1 Lie Factor = ( 5:3

2:4)2

( 80236212

) = 3:8



ANSWERS 453

2 a The area of the large doll is 3:8 times the area of

the small doll, while 5% is 1.099 times as large as

4.55%. The Lie Factor = 3:8 ¥ 1:099 = 3:46

b The visual impression is that the rate increase is

much bigger than it actually is.

3 a b

c The changes in the value of the Aussie dollar aren’t

visible in the second graph.

4 a The university’s ranking is improving faster than tu-

ition costs are rising.

b A picture of a university where the quality of edu-

cation is decreasing and the costs are rising.

c The initial message that is opposite to what was in-

tended is its worst feature. Other problems with the

graph: the two line graphs measure different things,

so there can be no vertical scale. Because of this, the

lines could have been placed anywhere, for exam-

ple, the ranking graph could have been placed above

the tuition costs graph. The graph implies that these

changes took place over the same time period, but

the dates show differently.

d Put a vertical scale on the left showing rankings, but

with #1 at the top of the scale. A second vertical

scale on the right could show tuition costs. Only use

data from 1989 to 1999.

5 a

b The rate of improvement in the fuel econ-

omy standards increases gradually from 1981to 1999, but slows between 1999 and 2001.

Exer

1 a b

c d

e, f They all show a slight right skew. c exaggerates

the skew. a, with bin size 2, seems to show the

right amount of detail.

2 b One useful display has these bins - 0-99, 100-

199, ...1900-1999. It shows the basic shape of

the dataset, along with two low outliers and two

high outliers. Smaller bin sizes show extraneous

detail, while larger ones (for example, 250 or 400mm wide) don’t show all of the outliers.

Exer

1 a i 0 685578588585345921 0633501826

ii 0 3420 685578588585591 0330121 6586

iii 0 32

0 55555450 670 8888891 0011 3321 51 661 8

iv 0 2

0 30 40 5555550 60 70 888880 91 001 11 21 3311 5

1 6611 8

b median = 8.

c either iii or iv. iii shows the general shape - some-

what right-skewed, while iv shows that the dataset

contains a large number of 5s and 8s.

3 a 0 000035991 25572 13773 4

45 67

Units: 2 j 1 represents 210 cents

e 59601000784840988950335199 0 000035991535213 1 2557

05 2 13773 4

2 45 67

Units: 2 j 1 represents 210 cents

A class discussion on the differences would be

valuable.

cise 3G

cise 3H



454 ANSWERS

5 a Oxford Cambridge

49:8 low 49:59 7 538441

38334 8 53844122 9 72

b Cambridge rowers’ weights are slightly greater, on

average; both datasets have an outlier (the cox).

6 116 minutes

7 a 53: 0 j 0; 506: 0 j 5; 5157: 5 j 1b 900 hectares

8 Tool: Our machine is much more consistent in

the distance it throws the ball.

Machine: Our machine had an average length of

throw much closer to the specified 55metres.

Exer

1 An outlier affects the mean more than the median

because the mean is calculated from the data val-

ues. The median is the middle value of a dataset,and doesn’t use the actual data values.

2 a mean = 8.48, median = 9;

b mean = 83, median = 22.5;

c mean = 3.43, median = 0:95 Your dataset should have a significant skew, or an

outlier.

6 47 a 0

b It is not a good measure of the centre, as it

is the minimum value.

8 mean = 114.6, median = 116.

median, because there is an outlier 9 mean = 758 mm, median = 673 mm.

median, as there are outliers

10 201 cm

Exercise 3J.1

1 a 1.73 b 17.06 c 0 d 1.41 e 35.36f 8

Exercise 3J.2

1 a 3 b 3 c 0 d 3 e 752 a min = 1 Q1 = 1.5 median = 2.5 Q3 = 4.5

max = 6b min = 1 Q1 = 1.5 median = 2.5 Q3 = 4.5

max = 54c min = 12 Q1 = 12 median = 12 Q3 = 12

max = 12d min = 48 Q1 = 48.5 median = 50 Q3 = 51.5

max = 52e min = 0 Q1 = 12.5 median = 50 Q3 = 87.5

max = 100f min = 2 Q1 = 3 median = 6 Q3 = 11 max = 12

3 a Cambridge: ¾ = 12.7 Oxford: ¾ = 12.1b Cambridge: ¾ = 5.0 Oxford: ¾ = 4.4c It has a large effect.

d Cambridge: IQR = 8.3 Oxford: IQR = 9:3e IQR without: Cambridge 6.8; Oxford 6:9f The outlier had much less effect.

g With outlier:

Cambridge f49:5, 82:15, 84:6, 90:45, 97:3gOxford f49:8, 81:15, 83:9, 90:45, 92:7gWithout outlier:

Cambridge f81:1, 83:65, 85:15, 90:45, 97:3gOxford f79:1, 83:55, 84:25, 90:45, 92:7g

6 a ¾ = 447 mm, IQR = 305.5b min = 25 Q1 = 571 med = 673 Q3 = 876.5

max = 18297 For the standard deviation to be zero, all the data values

must be equal.

8 median = 1.5; Q1 = 0.59 a D b A c B d E e C

Exercise 3K

1 a b

c

2 a b

c

3

6 Class discussion would be valuable.


cise 3I



ANSWERS 455

8 a

b yes


10 The distribution is right-skewed, with 4 outliers from

$61 to $69. The amounts range from $2 to $69, with

a median = $20:74, with an IQR = $19:80.

11 a

b Data is skewed for Circ, with 1 outlier, so median

and IQR are more appropriate.

c Circ: median = 136, IQR = 121.4;

Csrc: median = 217.3, IQR = 110.3

d

e The Circ data has one extreme outlier. The box-

plots show that maximum concentration is signifi-

cantly greater with sustained release codeine. There

is an indication that Circ has greater variation.

12 a D b A c C d E e B


1 a i all battery hens

ii amount of calcium in diet, thickness of egg shells

b mean weekly intake of calciumc sample, as the number of battery hens is very large

d continuous

e discuss f experiment

2 a discuss b discuss c column

3 A graph with integrity gives an honest portrayal

of the relative differences in the data values or

frequencies. Graphical excellence implies that a

graph clearly shows the relationships between mul-

tiple variables.

4 Discuss. Some features are the proportion of winds

from each direction, and the relative wind strength

in winter and summer.

5 a i 2 13 0837604 523405 85889286 337 518 656

9 910 2211121314 115 316 1

ii Data is right

skewed, with

3 outliers.

iii 58

c discuss

d Median, as the

data is skewedright, and has

outliers.

e i refinery: mean = 66.4, ¾ = 34.8,

DoE: mean = 33.5, ¾ = 48.6ii refinery: median = 58, IQR = 44.5,

DoE: median = 20, IQR = 8.75

f

g The DoE measurements have much less variation

than the refinery measurements, though there is

one outlier. The DoE average reading is much less

than the refinery’s.

h It is to the refinery’s benefit to have high initial

levels. The higher these levels are, the less theapparent impact of the increased future production.

Exercise 4A

1 a straight b straight

c curved

2 a

0.0

2.0

4.0

6.0

8.0

0.0 125.0 250.0 375.0 500.0

Histogram

count

0.0

1.3

2.5

3.8

5.0

100.0 175.0 250.0 325.0 400.0

Histogram

count

0.0

125.0

250.0

375.0

500.0

Circ Csrc

Boxplots

variables

amount



456 ANSWERS

b There is no obvious model for this dataset.

c The 3 books with the largest cover area tended to

be heaviest.

Exercise 4C

1 b

c Y 1 = ¡5:6x + 123:4 d 0:994e It appears to be a very close fit, except for possibly

the last few data values.

2 b

c Y 1 = ¡0:15x + 4:69 d 0:888e The data is roughly linear, but seems to be clustered

into 3 or 4 groups, with little change within a group,

but a significant drop from one group to the next.

3 a B b F c E d D e C f A

4 actual values are a 0:71 b 1:0 c 0:82 d 0e 0:35 f 0:58 g 0:96 h 0:92

5 g ¼ (about 3:1)

Exercise 4D

1 a

b Y 1 = 0:73x + 0:13 c 0:999d It appears to fit very well.

e

f The residual plot shows that the dataset is not

modelled well by a linear function.

g Y 1 = ¡0:34x2 + 1:24x ¡ 0:05h

i The residual plot shows no pattern, so it is an

appropriate model.

2 a

b Y 1 = 2:1x ¡ 54:3 c 0:81d A linear model looks appropriate.

e

f Nopattern inresiduals, therefore linear model is

appropriate.

g Y 1 = ¡0:02x2 + 7:2x ¡ 412h

i Since a linear function is appropriate, a quadratic

will be as well. (why?)

3 b

c for y1 vs x1 only

4 a Y 1 = 3695x ¡ 259b The gradient is the cost per carat; y-intercept is

cost of a ring with no diamond. As it is negative,

this implies that it is not appropriate to extrapolate

using this model.

c d



ANSWERS 457

e There are no strong underlying patterns in the data,

so a linear model is appropriate.

5 a i ii

b i taking 1982 to be year 0, Y 1 = 0:002x +0:87ii Y 1 = ¡0:158x + 8:1

c i

There is a curved pattern to the data implying

that a linear model is not suitable.

ii

There is a curved pattern to the data implying

that a linear model is not suitable.

d 4:47 goals per game e save percent: 92%

6 For 82-89, the gradient is ¡0:08, implying the number

of goals per game was gradually decreasing. For 90-

97, the gradient is ¡0:243, implying that the number of goals per game decreased at a faster rate.

Exercise 4E.1

1 Y 1 = 2x + 1:33 2 Y 1 = ¡1:42x + 14:81

Exercise 4E.2

1 a i y = 7:09x ii

iii No pattern in residual plot, therefore a linear

model is appropriate.

b i Y 1 = 3:73x ii

iii No strong pattern, therefore a linear model is

appropriate.

2 a Y 1 = 0:239x b

c The extension for the 200g weight appears to be

too large. He should measure it again, if possible,or delete this point from the dataset and re-calculate

the regression line.

3 a Y 1 = 8:19x2 b

c No strong pattern in the residuals, so the quadratic

model is appropriate.


1 Class discussion would be valuable here.

2 b Data appears to be

curved, and convex

down.

c i Y 1 = 1:1x + 35

ii 0:985

d Residuals show a

pattern, implying

that a linear model

is not appropriate.

e i Y 1 = ¡0:003x2 + 1:45x + 30 ii 0:992f i

ii While there are still patterns in the residual

plot, the residuals are small, and the r2 valueis higher than that of the linear model.

g Class discussion would be valuable.

h linear model: 158 km/hr

quadratic model: 149 km/hr

i linear model: 107 metres

quadratic model: 119 metres

Exercise 5B.1

1 a 0.8290 b 0.9511 c 3.0777 d 0.5000e 0.5000 f 1.0000 g 1.2062 h 3.2709i 1.2868 j 1.000 k 2.000 l 1.0002



458 ANSWERS

2 a 40 b 77 c 80 d 30 e 45 f 60g 48 h 55 i 6

3 a ]BAC = 70o AB = 3.4202 cm BC = 9.3969 cm

b ]DFE = 40o DE = 5.0346 cm DF = 7.8324 cm

c BC = 13 cm B = 22:6199o C = 67:3801o

d FG = 3 cm H = 36:85990 G = 53:1301o

4 a 0 6 y 6 1 b 0 6 y 6 1 c 0 6 y < infinity

5 sin 39o

cos39o = 0:8098; tan39o = 0:8098

sin 30o

cos30o = 1

2

p 3

2 = 0:5774; tan30o = 0:5774

sin 45o

cos45o =

1p 2

1p 2

= 1; tan45o = 1

6 µ = 38:7o

7

A sin A cos(90 ¡ A) cos A sin(90 ¡A)

10 0:1736 0:1736 0:9848 0:9848

20 0:3420 0:3420 0:9397 0:939730 0:5000 0:5000 0:8660 0:866040 0:6428 0:6428 0:7660 0:766050 0:1736 0:7660 0:6428 0:642860 0:8660 0:8660 0:5000 0:500070 0:9397 0:9397 0:3420 0:342080 0:9848 0:9848 0:1736 0:1736

c complementary d cot A is the reciprocal of tan A

8 a 16.20 m/s b 11.5o

9 2.5981 m2

Exercise 5B.2

1 a 25o360 b 105o2204800 c ¡63o703000

d 235o2105800 e 14o202400 f 546o120 g 60

h 300

2 a 15.4100 b 125.1467 c ¡44.2567 d 83.7153e 123.7517 f ¡90.2033 g 0.83333 h 0:0014

3 a 0.3180 b 0.0950 c 1.0632 d 0.2178e 0.3874 f 1047.0

4 a FH= 8 cm; G = 53o704800; H = 36o5201200

b PQ = 25 cm; P = 16o1503700; Q = 73o4402300

c AB = 15; A = 28o402100; C = 61o5503900

d XY = 9; Z = 12o4004900, X = 77o1901100

Exercise 5B.31 119.0353 metres

2 a 21.2557 metres b 22:8557 metres

3 1425.3 metres 4 72.04 metres

5 9.9066 metres 6 7.03 metres

7 2:099 km

8p

24 = 4:899

9 a AB=p

2 bp 2

2 ,

p 2

2 ,1

10 a ]P = 60o, ]PQT = 30o QT =p

3

b 1

2,p 3

2 , 1p

3

cp 3

2 , 1

2,p

3

Exercise 5C

1 a b

c d

e f

g h

i j

k l

2 a 1st b 3rd c 4th d 2nd e 2nd f 3rd

g 3rd h 1st i none j none k 4th l 1st

3 a 1 minute = 1

360£ 60 of equator =

12800£ ¼

21 600b 53 minutes assuming that the plane flies

along a great circle or the earth is flat.

4 a 1080 b 30000c 180+360n, where n is any whole number, positive

or negative

d See part c

Exercise 5D

1 Angle sin µ cos µ tan µ

0o 0 1 0

90o 1 0 not defined

180o 0 ¡1 0

270o ¡1 0 not defined

360o 0 1 0

240°

60°

450°

0°

330°

720°

90°

270°

180°

240° 540°

180°



ANSWERS 459

2 a 0:87 b 0:50 c 1:73 d ¡0:71 e 0:71f ¡1 g 0:50 h ¡0:87 i ¡0:58 j ¡0:50k 0:87 l ¡0:58 m ¡0:5 n ¡0:87o 0:58

3 a ¡ b + c ¡ d + e ¡ f ¡ g + h 04 a between 0 and 180 b between 90 and 270

c between 0 and 90 and 180 and 270

d between 180 and 360 e between 90 and 180f between 0 and 90 g between 180 and 270

5 a 20o b 45o c 80o d 27.2o e 40o

f 64o2701500

6 a ¡660o, ¡300o, 420o, 780o, 1140o, 1500o

b, c, d, all the sines (cosines, or tangents) have the

same value

7 297.1268 or 242.8732

8 ap

1 ¡ a2 b ap

1 ¡ a2

9 a 45o and 225o b 180o c 38.17o and 141.83o

Exercise 5E

1 a 0.5 b 1.5 c 3 d 4 e x f 2¼2 a ¼

6 b ¼

2 c ¼

4 d ¼

3 e ¼

12 f ¼

3

g ¼

8 h 4¼

3 i 5¼

3 j 5¼

4 k ¼ l 3¼

2

3 a 0.8343 b 2.6786 c 3.8572 d 7.2955e 1.1222 f 4.4986 g 1.7149 h 5.9073i 0.0548

4degrees 30 45 60 90 120 180

radians ¼

6

¼

4

¼

3

¼

2

2¼

3 ¼

degrees 225 270 360

radians 5¼

4

3¼

2 2¼

5 a 45o b 540o c 60o d 240o e 30o

f 210o g 495o h 252o i 800o

6 a 57o1704500 b 43o3204100 c 79o3802800

d 114o3503000 e 208o3302400 f 15o2801100

g 277o530400 h 155o1601800 i 286o2804400

7 ap 3

2 b ¡1p

2 c 1 d 0 e

p 3

2 f

p 3

g 0.9967 h ¡0:8968 i 1.5574

8 a ¼

6 or 5¼

6 b ¼

3 or 5¼

3 c ¼

4 or 5¼

4

d 0.9973 or 2.1443 e 2.1688 or 4.1144f 1.2278 or 5.0554

Exercise 5F

1 a 30.0702 cm2 b 11.8177 cm2 c 27.5804 cm2

d 19.0568 cm2 e 16.9706 cm2 f 24 cm2

2 a 66.7572 cm2 b 13:649 m2

3 No such triangle exists. With side of 4 and 5,

10 cm2 is the largest possible area.

4 a 16.277 cm b 4 cm 5 35.47 cm2

6 30o or 150o 7 7.88 cm

8 There are 6 equilateral triangles with side s. An equi-

lateral triangle has an angle of 60 degrees so the area

of an equilateral triangle is1

2s2 sin60o =

p 3

4 s2 and 6 times this is 3s2

p 3

2 .

9 Each dodecagon is made up of 12 triangles each isosce-

les with an apex angle of 30o. If s is the side of the

triangle the triangle has area 1

2 sin 30os2 = s

2

4 . So

the dodecagon has area 3s2. Now the square which en-

closes the dodecagon has side 2s so has area 4s2 . Thus

1 dodecagon is 34

the area of the surrounding square and

this is also the ratio of the area of the 8 dodecagons to

the note, as it consists of 8 dodecagons with 8 surround-ing squares.

Exercise 5G.1

1 a 85.1915 b 4.9742 c 5.4873 d 13.8412e 3.0196

2 a angle C = 65o; AB = 23.6620 cm = AC

b angle P = 80o; PQ = 7.5445 m, QR = 9.9979 m

c angle X = 39o; XY = 121:7040 km,

YZ = 130.3041 km

d angle E = 55o; DE = 0.1381 m, EF = 0.1924 m

3 angle ADB = 15o; BD = 163.2872 m,

CD = 104

.9590

m

4 222.9 metres

Exercise 5G.2

1 a A = 54:68o, B = 80:32o, c = 18:1230 cm or

A = 125:32o, B = 9:68o, c = 3:0913 cm

b P = 108:19o, R = 41:81o, p = 8.55 cm or

P = 11:81o, R = 138:19o, p = 1.84 cm

2 Either C = 62:11o, then A = 77:89o and so

a = 12:17 cm or C = 117:89o, then A = 22:11o and

so a = 4:6844 cm

3 a i C = 67:48o, B = 52:52o b = 68:7248 cm

ii C = 112:52o, B = 7:48o b = 11:2739 cm

b C = 27:73o

, B = 112:27o

b = 37:7856 cmOnly one solution possible.

c The data requires sin C = 8£ sin 100

3 which is

greater than 1, so no triangle exists.

d B = 25:66o, C = 34:34o c = 19:5411 cm

Only one solution is possible.

e A = 26:9250o, C = 118:1o c = 116.8835 cm

Only one solution.

4 No, the third angle is 27o and 9:8

sin 27o does not

equal 11:4

sin 85o:

Exercise 5H

1 a cos A = b2 + c2 ¡ a2

2bc b cos B =

a2 + c2 ¡ b2

2ac

c cos C = a2 + b2 ¡ c2

2ab2 a a = 5:1006 cm b c = 9:6437 cm

c c = 298:3002 m d a = 26:0249 m

e b = 18:8076 cm

3 a P = 125:0996o b Q = 27:5527o

c R = 27:2660o

4 a x = 5:5678 b x = 16:64335

smallest angle = 41.4096

o



460 ANSWERS

6 a C = 22:3316o, B = 130.5416o, A = 27:1268o

b YZ = 12.4788 km, Z = 48.0799o, Y = 75:9203o

c largest angle = 100.2970o, smallest angle = 29.4687o,

third angle = 50.2344o

d third side = 11.7419 m, largest angle = 83o400,third angle = 55o550

e smallest angle = 26.6275o

, largest angle = 99.5940o

,third angle 53.7785o

f third side = 7.2 cm, largest angle = 94.57o,

third angle 48o210

7 The largest angle is 100.2866o.

8 This is the third side of a triangle with sides 15 and 9enclosing an angle of 140o. The length is 22.6458 cm.

9 a = 6:6532, B = 38.5382o, C = 85:4574o

10 Third side is 27.5608 cm. Other two angles are 62.0919o

and 48.3955o:

11 2 hrs 17 minutes (the distance between the cyclists in-

creases at 35.1158 km/hr using the cosine rule on a

triangle with sides 20 and 20, and included angle 135o

.)Exercise 5I

1 20.77 m 2 height of tree = 4.62 m

distance from the building = 26.20 m

3 47.51 m 4 height of the hill is 125.1 m

5 S71.5oE, 14.10 km from his starting point

6 2.06 km 7 2028 6.234 km on a bearing of 166o

9 within an angle of 14.6 degrees

10 a 296 b 560:6 km

11 angle CAB = 30:8o 12 20:3 cm2

13 26.832 cm

2

14 114.64 metres 15 29.16 m16 11.59 hectares


1 b i AB = 3.3166 cm, C = 33.5573o, B = 56:4427o

ii PQ = 8:0944 cm, K = 35:6149o, P = 54:3851o

iii W = V = 66:4218o, U = 47.1564o

c height of Nelson = 9.5999 metres

2 angle of pyramid = 45:6o

3 a i 28.5600 ii 135.1700 b i 36o420 ii 137o120

c i 0.6405 ii 2.39464 2 hrs 42 mins

5 a 3rd angle = 83o, side opposite 58o = 16.1707 cm,

side opposite 83o = 18.9261 cmb angle opposite 11 m = 41.6144o or 138.3856o

third angle = 113.3865o or 16.6144o

side opposite the 3rd angle = 15.2028 cm or

4.7360 cm

c 3rd side = 8.8012 km

angle opposite the side length 13 km = 81.2483o

angle opposite the side length 11 km = 56.7517o

6 a area = 38.7314 cm2 b area = 74.9310 metres2

7 a 132.7734o b 27.8527o or 152:1473o

c 52:019o d 51.0505o or 128.9495o

8 b r = a

2sin A

Chapter 6 Section A - Review of Chapter 1

1 Sales 0 50 100 150 1000Wage 800 807:50 815 822:00 950

a sales; total wage

b i as above ii (0, 800), (50, 807:50) ... (1000, 950)

iii iv

v w = 0:15s + 800 where w = total wage

and s = sales2 a ¡2 b 18 c 03 a discrete b continuous c discrete

4 a i wages ii hours worked b discrete

c i Hours worked 8 9 10 11 12

Wage 96 114 132 150 168

ii

iii w = 12n iii w = 96 + 18(n ¡ 8) where

n = number of hours worked

and w = wage

d domain 0 6 n 6 12; range 0 6 w 6 1685 a 1

6 b 0

6 mHI

= 2

3, m

JK = 2

3, m

HJ = ¡5, m

IK = ¡5

7 a no. of metres b total cost c 125d dollars e 16 f dollars per metre

8 a m = ¡2, c = 11 b m = 1

2, c = 7

2

9 a y = 5x¡15 b y = ¡2x¡1 c x = 1 d y = 210 a

b

0

1000

800

9501000800600400200

1500

1000

500

W

s

1412108642

160

120

80

40

(8' 96)

(12' 168)W (wages)

n

y

x(-1' 2)

@=-2!+3

y

x-5

2\Qw _

1\Qw _

3

32 x y

25

21

x y



ANSWERS 461

11 a x ¡3 ¡2 ¡1 0 1 2 3y 16 13 10 7 4 1 ¡2

b

c

12 a number of kilograms b cost in dollars

c if measured in whole number of kg, discrete

d 3:29 e C

/n f C = 3:29n

13 a Number of pages can be a positive whole number

only. Therefore it is discrete.

As time in this case is measured in whole seconds,

the variable is discrete.

b i number of pages printed ii total time

c

Pages 1 2 3 4 5 6 7 13

Time 15 20 25 30 35 40 45 75

d n 2 positive whole numbers; n > 1e T 2 positive whole numbers which are multiples

of 5; T > 15

f

g gradient = 5; seconds per page

h Time to print one page.

i T = 5n + 10 n > 1

14 a Not a function as 2 maps onto 3 and onto 5.

b Not a function as 2 maps onto ¡2 and onto ¡3.

15 The longer the trench, the longer it will take to dig it,

given that the number of workers remains the same.

16 a Size Width (mm) Length (mm)

A0 840 1188A1 594 840A2 420 594A3 297 420A4 210 297A5 148:5 210A6 105 148:5

A7 74:25 105A8 52:5 74:25

b

c Length =p

2 width

d and e This is of the form y = kx with

the constant of proportionality = p 2.

Chapter 6 Section B - Review of Chapter 2

1 a 2p

14 b 11 c x d 90 e 2cp

c

f 6p

2 g 24 h ¡18 i ¡36 j 3k

p 2 l 3

2 a §6 b x = 6 c ¡6 d §6

3 a ¡3p

13 b 8p

5 c 8p

2 d ¡12p

6e ¡3

p x f ¡

p a g 8d h ¡2f 2 i 6n

j a

4 k w

p w l

p bc

bc

4 a p 102

b p 7 c 2p 3 d 2 + 3p 2 e p 32

f

p b

b g

7p

2

2 h 2

p c

5 a x2 + x¡ 30 b 6x2 + x¡ 2 c x2¡ 18x + 816 a x2 ¡ 16x + 64 b 4x2 + 20x + 25

c 25 ¡ 20x + 4x2

7 a x2 ¡ 49 b 9x2 ¡ 1 c 16 ¡ 4x2

8 a (x ¡ 2)(x ¡ 8) b (3x + 1)(2x ¡ 3)c (x + 6)(x ¡ 6) d 3(x + 3)(x ¡ 3)e 2(x ¡ 8)(x + 2) f 5(x + 1)(x + 3)

g 2(x

2

+y

2

)(x+y)(x¡y) h (x+2+ y)(x+2¡y)

y

x

y

x

y

x

T

n

(13' 75)

(1' 15)

number of hours

length of trench

length

width

@=~2\!



462 ANSWERS

i (x + 2)(2x ¡ 3) j (x ¡ 4)(3x + 1)k 2(x ¡ 1)(2x + 3) l 3(x ¡ 1)(2x ¡ 5)

9 a ¡18 b 0 c 0 d 8

10 a (x + 1)(x2 ¡ x + 1) b (x ¡ 5)(x2 + 5x + 25)

11 a x = ¡1

3, x = 2

b x = ¡2, x = 2

12 a x = 11 or x = ¡2 b x = 3 or x = 3i.e., a repeated root c x = 4

3 or x = ¡ 3

2

13 a x = ¡12 or x = 12 b x = ¡5 or x = 5c x = ¡2 or x = 2

14 a (x + 6)2 b (2g ¡ 3)2 c (2x ¡ 5)2

15 a x = 1 §

p 5

2 b x =

¡4 §p

14

2

c x = ¡1 or x = ¡416 as above

17 a ¡71 b 137 c 0

18 x ¡3 ¡2 ¡1 0 1 2 3

y 0 ¡4 ¡6 ¡6 ¡4 0 6

19 x = ¡6 or x = 4 y = 24

appropriate

v-window:

¡6:3 6 x 6 6:3

¡1 6 y 6 26

20 x = 1, x = 2 y = 6 x = 1:5, y = ¡0:7521 a

b

c

22 y = (x ¡ 2)2 + 6

23 a y = 3(x ¡ 4

3)2 ¡ 41

3 b (1 1

3, ¡4 1

3) c +1

d x = 4 §

p 13

3e

24 a

y

x

2

-2

-\Qe _

y

x-2

-8

2

y

x

-3

-6

2

62 x x y

y

x

(-1' 25)

-6 4

2-2

10

-20

y

x

2

2

3 x y

x y

2-2

4

y

x

2

22 2 x y

x y

10(2' 6)

y

x

6)2( 2 x y

y

2 x y

4-4 -3

x

y x ( 3) 62

y

x

(1\Qe _ '-4\Qe _\)

3

134

3

134

1 +

+

0.13

2.54

y

x

y x



ANSWERS 463

b

25 a b

26 a

b

27 a A and d are measured continuously, projection is

perpendicular to screen.

b d A

0 0:4

1 0:55

2 1

3 1:75

4 2:8

5 4:156 5:8

7 7:75

8 10

c 1:06 m2

d 3:27 metres

(2 = 0:15d2 + 0:4)

28 Examples are:

a (¡1, ¡1), (0, ¡6)

b (¡1, ¡2), (5, ¡5)

c (0, 2), (1, 3)

29 a x =

1 +p

5

2

b 1:6180

30 Square root may be + or ¡

31 k = ¡15

32

x 3 5 7 9 11 13 15 17

y 4 12 24 40 60 84 112 144distance 5 13 25 41 61 85 113 145

x, y and the distance from the origin form

Pythagorean triads.

33 x = 2:27 x = ¡5:27

Chapter 6 Section C - Review of Chapter 3

1 a i all people who have the Melbourne flu

ii the number of days a person with the flu is

infectious

b for example, the maximum number of days a

person is infectious

c a sample, as it would be very difficult to gather

such data on all people who have the flu2 a for example, number of possums born in a litter

b for example, the type of food the possum eats

3 a discrete, as it tends to be measured in days

b continuous

c discrete, since memory tends to come in multiples

of 16 Mbytes

4 a population, as the data is available, and the

population is not too large

b sample, as gathering data for the entire issue would

be too time-consuming and expensive

5 a using available data b observation c survey

6 a randomly choose page numbers and then an articleon that page

b randomly choose a number of location/time com-

binations, and count the number of each species

seen from each of these

7 People living in the same street tend to have a simi-

lar socio-economic background and similar views. A

better alternative is a phone poll (though this excludes

folks without phones) with people selected from the

electoral roll.

8 a time series graph, as absenteeism depends partly

upon the day of the week

b column graph9 For example, in 1999: more females than males over

80 years old; large jump in both males and females

in 1947 (due to end of WWII and soldiers returning

home); same in 1920 (end of WWI); declining birth

rates in recent years; 1901: very few people over 80years old; significantly more males in age range from

35 to 50

10 comparing Yr 8 and Yr 12, lie factor = ( 3819

)2

( 3519

) = 2:2

11 a for example,

c 0 4444333322222221110 999988887766651 4332201 972 20 High 47

d i 8:56 ii 6:9 iii 8:0 iv 7:2e median and IQR because the dataset is skewed

right, and has an outlier

f min = 1:7, Q1 = 3:1, median = 6:9,

y

x

y

x

y

x

)1(21)(

x xh

y

x

!=-1

y

x

x y 1

x y 1

x y 1

x y 1



464 ANSWERS

Q3 = 10:3, max = 47:2g

h class discussion would be valuable

13 when the dataset is skewed, or contains one or more

outliers (or both)

14 for example, 0 0 0 0 0 1 1 2 2 415 1010111111111214152116 3 4 5 6 6 6 7 7 1 517 class discussion would be valuable

18 when the dataset contains no outliers

19 a i increase by 3 ii no change

iii increase by 3 iv no change

v if constructed on a graphics calculator, there is

no apparent change (why is this?)b i increase by a factor of 10

ii increase by a factor of 10iii increase by a factor of 10iv increase by a factor of 10v if constructed on a graphics calculator, there is

no apparent change

20 97:1 kg

21 for example, 3 7 10 13 1722 a 1:5 b 0.5

Chapter 6 Section D - Review of Chapter 4

1 a i

ii Y 1 = ¡6:2x + 77 iii 0:911iv no apparent pattern in residuals, so the model

is appropriate

b i

ii Y 1 = 5:2x ¡ 3:6 iii 0:959iv the residuals show a curved pattern, so another

model may be more appropriate

2 a Y 1 = 0:38x + 6:5 b m = 0:38, c = 6:5c gradient: increase in number of chirps/min for a

one degree increase in temperature, y-intercept:

number of chirps per minute when temperature is

0o

C

d

e class discussion would be valuable.

3 a The quadratic mode1 Y 1 = 0:007x2+0:1x+3:7fits the data well, with r2 = 0:9998 and no pattern

in the residuals. Class should discuss if a model

of the form Y = ax2 would be appropriate here.

b 95 m c using predicted values, 7 m

5 a Y 1 = 6:45x b

c No apparent pattern in residual plot so the model

is appropriate.

Chapter 6 Section E - Review of Chapter 5

1 a 0:3907 b 0:9703 c 0:6009 d 1:15472 a 36o520 b 59o560 c 73o520 d 48o460

3 a AB = 5:69, BC = 12:79, ]ABC = 66o

b CB = 20, ]ABC = 36:87o, ]ACB = 53:13o

4 18:25 m

6 a 15:41 b 125:147 c ¡44:257 d 83:7157 0:29088 a b

c d

9 a 0 b ¡1 c 1 d not defined

10 a 180 < µ < 360 b ¡90 < µ < 90c 0 < µ < 90; 180 < µ < 270d µ = 0o, 180o, 360o

11 510o, 870o In general, 150o + n£ 360o,

12 µ = 72o 320 3300 µ = 287o 270 2700

13 a 1

4

Rb 5R c 2¼R

14 a ¼

3

R b 7¼

6

Rc 7¼

4

Rd ¼R

15 a 0:4485R b 1:7149R c 4:4986R

d 17:4533R

16 a 60o b 300o c 210o d 220o

17 a 85o5603700 b 13o1004100

150°

120° 0°

330°

where n

is an integer



ANSWERS 465

18 ap 3

2 b 0 c 1:557

19 a ¼

3, 5¼

3 b 3¼

4 , 7¼

4 c 1:737, 4:457

20 a 82:6 cm2 b 612:84 cm2

21 A = 2:34o or A = 177:66o

22 a = 19:62 cm

23 A = 53o,

a = 6:41 cm,c = 5:37 cm

24

25 2 hours 24 minutes 26 51:74; 24:9

27 61:56 cm 28 38:94 m

29 For angles from 1 to 6 degrees.

Hint: Try doing this on your graphics calculator by

generating Lists.

List 1 = angle in degrees

List 2 = List 1 ¥ 60List 3 = tan List 1

30 a ]OCM = 62o

b MC = r

tan 62o = 0:5317rc r = 32:6435

Exercise 7A

1 a 105 b 69 c 2x9 d p4q 4 or ( pq )4 e a7b5

f y1

2 a 2, 4, 8, 16, 32, ...., 1024

b 8, 27, 216; 216; 63 = 23 £ 33

c 1

3, 19

, 127

, 181

, 1

243

3 a 2 £ 2£ 2£ 3£ 3b 2

£2

£2

£2

£3

£3

£3

£3

c 2£ 2£ 2£ 2£ 3£ 3£ 5£ 7d 2£ 3£ 5£ 7£ 11 e aaabbbbb f 2yyyyyy

g 2£ 2£ 2£ 2£ 2£ 2£ y £ y £ y £ y £ y £ y

h aaabbbccc i cccddd j cccddd

k 2£ 2£ 2£ ddd l 5£ 5£ 5£ aaaaaabbb

m a n 5 pppqq o ¡3£¡3£¡3£ aaa

p 3£ 3£ aaaa£ bbbbbb::::

q 3£ 2£ 2yyyy

r 3£¡2£¡2£¡2£ aaaaaa£ bbbbbb

Exercise 7B

1 a y7 b r16 c 2y7 d 6x13 e ¡15d6

f h12

g b10

h a5

b6

i x7

y5

j 24a4

b6

k ¡30g8n9 l 12a2bc2d3

2 a 6x5 b ¡48x3 c ¡5d19 d ¡48y6 e r15

f ¡720t3 g ¡x4y4 h 6f 3g6 i ¡15k5t5

3 a a5 b b3 c 4c2 d 2a4 e 2

3y2 f 1

g 5a6 h 4 i a2b2 j 3a2 k 12

a7b8c9

l

4

3 p

3

qr4 a y6 b x20 c 3c10 d 4a8 e 5c8 f d12

g c22 h 8a18 i ¡10c19 j c3 k 23 p2

l a6b4

5 a a2b2 b 27c3 c 32x5y5 d x4y4z4

e 64 p6q 6r6 f a3b6 g 256d8c12 h a16b6

i 12s3t7 j a2b3 k 4a2b2 l a4b4

6 a d7 b b4 c 2a6 d 6a8 e 2f 4 f 12

x

g 3 h ¡10a5 i 12 p7 j ¡20m4n4 k y10

l 6abc m a2b2 n 4g o 8 p4s4 p ¡ x4y5

q 20b4

r ¡4k5

s 108y5

t ¡8q 6

u 4

2

t14

v ¡32a7 w ¡72r5 x 12a4b3 y c3d

z 8ab8

7 a 8a4b4 b 28 p2q 2 c 44a2b2

Exercise 7C

1 a 1 b 1 c 1 d 1 e 1 f 1

2 a 1

16 b 1

64 c 1

12 d 4 e 1

108 f 1

108

g 2

3 h 1

5 i 50 j 1

12 k 1

36 l ¡72

m 1

64 n 1

144 o 1

36 p 27 q 16 r 20

s 1

16

t 16

3 a 1

a2 b

1

b c

5

c3 d

3

x4 e

7

z2 f

6

c4

g 1

y1

2

h ¡2

z1

3

i 1

(ab)3 j

x6

a k

2y2

3

x

l a3

2b2 m

2y

z2

3

n 1

abc o 3x2 p

2b2

a

4 a 5 b 12 c 2 d 3 e 2 f 15

g 1

3

h 6 i 1

6 j 1

3 k 1

2 l 0

5 a 4 b 16 c 1

17 d 8 e 9 f 1

81 g 8

h 1

125 i 1

8 j 1

3 k 6 l 125

6 a 3p

82 b 2p

43 c 3p

b d 4 3p

b e 1p

6

f 1

3p

dg

2p

163 h 3p

a2 i 1p

g j

4p

8a3

k 13p

a2l

5p b5

7 a a1

2 b a3

2 c a3 d t1

3 e h2

3 f a

gp

3d h (3d)3

2 i 2x4

3 j h¡1

2

k 3£ (2y)¡1

3 l 2a¡2

8 a 1 b 1 c 1 d 4 e 3 f 2 g

1

8

85°

42°A

8 C

B

93.69°

46.31°

8

12.42

40°

9

C

A

B

6.31°40°

8

9

1.37

133.69°



466 ANSWERS

h 1

9 i 1

125 j 2

9 k 4 l 8 m 9

n 16 o 1

2 p 1

3 q 1

32 r a

1

2 s 2x

t a u a v a3

2 w 4 x 3 y 1

5

z 1

9 a 6p

x5 b 4p

a7 c 6p

b7 d 12p

a7

10 a 33n

b 35n

c 214n

d 25n

e 64n or 26n f 28 g 53n¡4 h 222n

11 16¡1

4 , 1

161

4

, 4

r 1

16,

14p

16

Exercise 7D

1 a log2

8 = 3 b log5

25 = 2 c log2

64 = 6

d log7

343 = 3 e log2

16 = 4 f log2

1

8 = ¡3

g log6

1 = 0 h log144

1

12 = ¡1

2

i log8

1

2 = ¡1

3 j log

10 10 = 1

2 a log2

32 = 5 b log5

125 = 3

c log2

256 = 8 d log4

64 = 3

e log2

1

4 = ¡2 f log

2

1

32 = ¡5

g log5

1

5 = ¡1 h log

9 9 = 1 i log

2

1

2 = ¡1

j logn1

n = ¡1 k logn

1

n3 = ¡3

l loga1

ab = ¡b

3 a 128 = 27 b 6 = 361

2 c 25 = 52

d 10 = 10001

3 e 1000 = 103 f 49 = 72

g 2 = 81

3 h 1

2 = 2¡1 i 1

27 = 3¡3 j 1 = 20

4 a 1024 = 210 b 81 = 34 c 16 = 42

d 729 = 93 e 1

32 = 2¡5 f 1

125 = 5¡3

g 1

64 = 4¡3 h 1

6 = 6¡1 i 2 = 16

1

4

j 2 = 641

6 k 3 = 271

3 l 8 = 163

4

5 a 6 b 4 c 0 d 2 e ¡¡

12

f 32

g x

h x i

Exercise 7E

1 a 2 b 2 c 3 d 2 e 1

2 f 3 g 4

h 2 i 0 j 2

2 a log4 b log 20 c log5 d log72

e log 24 f log6 g log 20 h log 30i log 25

8 j log 16

3 a log 16 b log5 c log 24 d log 1 = 0

e log 3 f log 1

4 = ¡ log4 g log2 h log 72

i log 20 j log 200

4 a 3 b 3

2 = 1 1

2 c 5

4 = 1 1

4 d 1

4 e 1

2

f 13

g ¡1 h ¡3 i ¡3

2 = ¡11

2

5 a 3 b ¡3 c 1

2 d ¡2

Exercise 7F

1 a x = 500 b x = ¡1 c x = 10 d x = 11

3

e x = 11

2 f x = 1

3 g x = 2

9 h no solution

i x = 1 j x = 8 k x = 4 l x = 1

201

2 a x = 25 b x = 1 c x = 100 d x = ¡9

2

e x = 1 f x = 2

Exercise 7G

1 a 1:129 b 2:807 c 0:849 d 2:322e 3:322 f 0:431 g 1:771 h 2:585i 0:774

2 a 1 b 6 c 2 d 1 3 a 3 b y = a

4 a i 1 + 2a

3 ii

b + 2

a + b b

2 p + q

1 + p + q

c i 9 ii 8

Exercise 7H.1

1 a 2:585 b 2:262 c 0:631 d 0e ¡1:904 f 3:322 g 0 h not possible

2 a 2:096 b 1:861 c 0:792 d 1:002e

1:070

f 0

:926

g 1

:431

h 1

:1453 a 2:322 b 2:771 c 3:210 d 2:322

e 6:644 f no value g 2:151 h 1:107

Exercise 7H.2

1 a $35917:13 b $164553:40 c $319:87d $44 766:80 e 4:2% p.a. f 40:34% p.a.

g 15:725 years h 10:52 years

2 $108 893:31 3 5:2% p.a. 4 11:6% p.a.

5 4:693 years 6 4:463 years

7 a 8:04; approx. 0:5% b 17:67 approx. 1:87%c 3:22; approx. 6:88%

8 a $222 036:64 b 16:24% p.a. c 81:71%d 9% p.a. e 7:213 years

Exercise 7I

1 d = 0:8t1:2 2 d = 14:8t0:2


1 a b12 b b12 c b1 d b1

5

2 a 1

a3 b

1

b1

2

c 2

c1

4

d 4e3

d2 e

e2

4 f

3h2k4

5 j2g3

3 a 4p

a3 b 1p

bc

2p c

d 2£ 4p

d3

4 a

1

a5

6b

1

2x1

2 y1

2c

4

9a2x2

5 a 2

5 b 7

8

6 a log 10 = 1 b log2 c log 10 = 1

8 a x = 3:465 b x = 4:29 c x = 1:21d x = 33 e x = 2 f x = 10

9 a 4:9% p.a. b 5 years

10 V = 41:78t0:912; $2785:92

Exercise 8A

1 a f (x) is a polynomial. b f (x) is a polynomial.

c g(

x)

is a polynomial.



ANSWERS 467

d h(x) is not a polynomial since the denominator has

a term in x.

e f (x) is a polynomial.

f m(x) is not a polynomial since it has a negative

power of x

g F (x) is not a polynomial as it has a term which is

a fractional power of x.

h g(x) is not as polynomial as its denominator is aquadratic function.

i d(x) is not a sum of powers of x so it is not a

polynomial.

j g(x) is a polynomial.

k h(x) has a negative power of x and so is not a

polynomial.

l k(x) is a polynomial. After multiplying out the

terms a cubic polynomial is obtained.

2 a 9 b 1 c 2 d 0 e 2f Not a polynomial since there is an x term in the

denominator.

3 a x6+2x¡3 b 4x¡7 c ( 13

)x4+x3 d 29x29

ep

2x3 + 7p

5x f x + x4 ¡ x5

4 a a0 + a1x + a2x2

b a0 + a1x + a2x2 + a3x3 + a4x4

c a0 + a1x + a2x2 + a3x3 + a4x4 + a5a5

5 Because it only works for polynomials of degree

less than or equal to 26 and it cannot represent a

polynomial whose degree is not specifed.

6 a a0 + a1x + a2x2 + a3x3 + a4x4

b x2 + x3 + x4 + x5 + x6

c a1(x ¡ 2) + a2(x ¡ 2)2

d (x + 1) + (x + 1)2 + (x + 1)3

Exercise 8B

1 a i ¡109 ii 5 iii 515

b i 4 ii 7 iii 52

c i 5 ii 5 iii ¡1831

Exercise 8C.1

1 a i 3x ¡ 1 ii ¡x + 11

b i x2 + 5x + 1 ii ¡x2 ¡ x + 3

c i 8x2 ¡ 10x + 4 ii ¡2x2 + 2x + 6

d i 4x3¡ 6x2¡ 2x + 9 ii ¡4x3 + 4x2¡ 6x¡3

e i ¡3x3+7x2¡2x¡12 ii ¡5x3¡7x2+2x¡2

f i x4+x3¡3x2¡3x+2 ii x4¡x3¡3x2+3x+2

2 a 4x ¡ 12 b ¡10 + 8x c 3t2 ¡ 2t ¡ 5

d 2x2 + 16x + 14 e 2x3 + 3x2 ¡ 2x ¡ 3

f x4 + 2x3 ¡ 5x2 ¡ 6x g t3 ¡ 2t2 + t

h x4 ¡ 6x3 + 12x2 ¡ 8x i a2 + 2ab + b2

j 12t3 ¡ 8t2 ¡ 3t + 2 k a3 + 3a2b + 3ab2 + b3

l a4 + 4a3b + 6a2b2 + 4ab3 + b4

3 a i 2 ii 15 b i 2 ii ¡4

c i 3 ii ¡4 d i 4 ii 4

e i 7 ii no term f i 7 ii no term

g i 4 ii no term h i 5 ii ¡1

4 a x + 3, for x 6= ¡2 b x ¡ 5, for x 6= ¡3c x + 3 for x 6= ¡4 d x ¡ 8 for x 6= 3

e 3x ¡ 2 for x 6= ¡2

3

f (x + 5)(2x ¡ 1) for x 6= ¡5

g 2x + 1 for x 6= 1h (x ¡ 3)(x + 5) for x 6= 3 and x 6= ¡5

Exercise 8C.2

1 a x3 ¡ 3x2 + 3x ¡ 1 b x3 ¡ 12x2 + 48x ¡ 64

c x5 ¡ 10x4 + 40x3 ¡ 80x2 + 80x ¡ 32

d 32x5 + 80x4 + 80x3 + 40x2 + 10x + 1

e 27a3 ¡ 81a2 + 27a ¡ 27

f 8c3 ¡ 60c2 + 150c ¡ 125

g 32x5 ¡ 240x4 + 720x3 ¡ 1080x2 + 810x ¡ 243

h a7 + 7a6 + 21a5 + 35a4 + 35a3 + 21a2 + 7a + 1

i 25x2 ¡ 50x + 25 j 64 ¡ 48y + 12y2 ¡ y3

k 27 ¡ 54x + 36x2 ¡ 8x3

l 16 ¡ 32x + 24x2 ¡ 8x3 + x4

m x3 ¡ 6 + 12x¡3 ¡ 8x¡6

n a4 ¡ 4a2 + 6 ¡ 4a¡2 + a¡4

o x3 ¡ 3

2x2 + 3

4x ¡ 1

8

p x8

81 ¡

2x6

27 +

x4

6 ¡

x2

6 ¡ 1

16

2 a (x+1)3 b (x¡1)4 c (2x+2)3 d (x+5)3

Exercise 8D

1 a 0, ¡ 1

2, 2 b 2, ¡1, ¡5, ¡9 c ¡1, ¡5

d 13

e ¡1, 3, 2, 4 f 0, ¡3, 2, 11

2 x(x +5)(x¡ 1), 2x(x +5)(x¡ 1), ¡x(x +5)(x¡ 1)

3 a 2, ¡3 b 0, ¡ 1

2, 43

c 4 d 2, ¡3

e 0, ¡ 1

2, 43

f ¡3, 4

Exercise 8E.1

1 a

b y

x

(0"6' 0"03)

1

( )23 1 x x y

y

x-1

(-0"53' 1"13)

(2"53'-13"13)

4

( )( )14 x x x y



468 ANSWERS

c

d

e

f

2 an example is: y = x3 ¡ x2

3 a jxj < 3 roots are ¡2, 1b jxj < 12:75 roots are ¡3, 25, 4c jxj < 3:1 roots are ¡2, 0:1, 1

Exercise 8E.2

1 They show that on a large scale the graphs appear identical.

2 The shape of these curves is identical if the scale

is chosen appropriately for each graph.

Exercise 8F

1 a y = 6x ¡ 10 b y = (8x + 2)1

3

c y = 3(25 ¡ 10x + 4x2) d y = x2 ¡ 1

x2 + 4

e y = (t + 2)3 f y = t4 ¡ t2 + 2

g y = 2jxj h y = x2 + 16x + 64

2 a y = x2 ¡ 4x + 4 b y = 1

(x ¡ 3)2

3 a y =p

u where u = 2x + 3

b y = 1

u where u = 7 ¡ x2

c y = u3 where u = x2 + 5x ¡ 3

d y = 3p u where u = 4 ¡ x2

e y = u¡4 where u = 2x

f y = 2

u where u = x4

4 a f (y) = 2y2 + y ¡ 1

b f (2a) = 8a2 + 2a ¡ 1

c f (3x ¡ 1) = 18x2 ¡ 9x

d f (2 ¡ c) = 2c2 ¡ 9c + 9

e f (t2) = 2t4 + t2 ¡ 1

f f ([t ¡ 1]2) = 2t4 ¡ 8t3 + 13t2 ¡ 10t + 2

5 a i x2 ¡ 13x + 41 ii x2 ¡ 3x ¡ 4

ii x4 ¡ 6x3 + 8x2 + 3x ¡ 1 iv x ¡ 10

b i 27¡ 27x + 9x2 ¡x3 ii ¡x3 + 4x + 2 iii x

iv x9 ¡ 12x7 + 3x6 + 48x5 ¡ 24x4 ¡ 65x3

+48x2 + 4x ¡ 2

6 a x = 0 or ¡4 b x = 1 or 9

c x = 1 +

p 5

2 or

1 ¡p

5

2 d x = §2

Exercise 8G

1 a Inverted with y values increased by a factor of 3.

b Moved to the right by a fixed amount of 3 units.c Displaced downward by 1 unit.

d Moved to the left by a fixed amount of 3 units

followed by a scaling up of all vertical lengths by

a factor of 2.e Moved to the right by 1 unit, inverted and then a

fixed value of 3 added to all y values.

4 a

b

y

x

(-1"22'-1"25) (1"22'-1"25)

-1"62

-0"62 0"62

1"621

13 24 x x y

y

x

1

-1 x 133 23 x x y

y

x

-10

2

1094 23 x x x y

y

x

(0"53'-0"13)

(-2"53' 14"13)

0"36

4

0"69

143 23 x x x y

y

x-1 2

1\Et _

4

( )( )( )41251

x x x y

y

x

-2 1

-2\Wt _

3

( )( )( )32152

x x x y



ANSWERS 469

c

d

5 a The zeros are at x = ¡1 and x = 5.

b At the same values as in part a.

c The zeros are those of f (x) shifted 3 units to theright, so they are at x = 2 and 8.

6 a 0 and 1 b The same as those in part a.

c ¡5, ¡1 and 0 d 0

7 a The graph of g is moved up 3 units relative to that

of f .

b The graph of h is scaled up, i.e. magnified, by a

factor of 4 and inverted.

c The graph of j is the graph of f shifted 2 units to

the right.

Exercise 8H

1 a 7.75 cubic km b 7.44 cubic km c 77:7%d 5418

2 a 0 6 x 6 36 b 24

3 a total profit = 2:7x ¡ 50 ¡ 0:001x2 b 1350

4 350

5 a a is about 700 milliseconds, the time for the crash

barrier to return to its normal shape.

b t = 0 up to 700 milliseconds

c k = 2:36£ 10¡6

d Maximum amount of depression, 120 mm, occurs

after approximately 230 milliseconds ( 7003

, to be

exact).

6 a h = 1

20¼r2 b SA =

1

10r + ¼r2

d r = h = 25.15 cm


1 a f has a negative power of x.

b g has a fractional power of x.

c h is not a sum of powers of x.

2 a x + x2

2 +

x3

3 +

x4

4 +

x5

5b (x + 1) + 2(x + 1)2 + 3(x + 1)3

3 a i 2x4 + 6x3 + 2x2 ¡ 5x + 15

ii x4 + 3x3 + x2 ¡ 8x ¡ 20

iii x5 + 8x4 + 16x3 + x2 ¡ 20x

iii x3 + 3x2 + x ¡ 4

b P (x) = 2x3 ¡ 7x + 16, P (¡3) = ¡17

4 a f (g(x)) = x6 + 1

b g(f (x)) = x6 + 3x4 + 3x2 + 1

5 a 32x5 ¡ 240x4 + 720x3 ¡ 1080x2 + 810x ¡ 243

b (3x + 1)4 = 81x4 + 108x3 + 54x2 + 12x + 1

6 a i ¡3, 2 ii ¡1, 1, 2b i Zeros at ¡1, 1 and 2. y-intercept y = 2.

Graphing you see a greatest value of + 2:1 at

x + ¡0:2, and a least value of + ¡0:6 at

x + 1:5

ii Graphing you see only 1 zero, at x = 1, the y-

intercept is 3, and it seems that y has a least value

of zero at x = 1. You can verify this by express-

ing y = (x¡1)2(x2¡x +3). The second factor has no real factors and is thus always positive.

iii Graphing you see the function is zero at 0 and 1and the graph touches the x-axis there. There is

a local maximum value at x = 0:5. The function

can be analysed completely by expressing it in

factored form

x4 ¡ 2x3 + x2 = x2(x ¡ 1)2

7 225 cm2

8 Volume is x(32 ¡ x)(40 ¡ 2x) and the maximum

value is at x = 8 when the volume is 4608 cm3

9 16.2121 feet

Exercise 9A

1 a

b

c

x

y

3

y x

4

4

4

4

x

y

3 x

31

y x

Qe _

x y

4 y

41 y

x

Q r _

x

y

1)3)(2)(1(51

x x x y

0.1871.58 3.390.2

x

y

)3)(2)(1(251

x x x y

3.660.8



470 ANSWERS

d

2 a

b

c

d

e

f

g

3 a A = 12000000N

A = area of each plot N = number of plots

b

c + 17142 (note variation if trace function is used)

Exercise 9B1

2 Day Rate ( $ )

1 19:702 153 154 125 126 127 128 99 9

10 9

4

4

4

4

x

y

31

x y

31 x

y

1

x y

y

x

x 1 y 3

1

Qe _

3

1 x y

11 x

y

1

x y

1

wQ _ x

y

y 1

y 221

x y

11 x

y

1 x

y

Qy _ x

y

321 x

y

3 y

1

x y

x 1 x 3

y 1

y 2

x

y

1

We _

2

Qw _

13

1 x

y

21

1 x

y

1 x

y

x

y

4

4

4

4

1

x y 3

x

y

1

x y

N

A

N A

61012

0 100 200 300 400

4

8

12

16 r a t e s

power (KWh)

0 1 2 3 4 5 6 7 8 9 10

4

8

12

16

20

daily rate ($)

days

4

1.33 x

y

y 3

x 1

31

1

x y

x

y

y 1

12

x y

1

x y



ANSWERS 471

3 a

b

Exercise 9C

1 a The belt is travelling at a certain speed. If the roller

is small, a point on the belt will take a shorter time

to move round the roller than for a large roller, so

the smaller the roller the faster it must rotate to

“keep up ” with the belt.

b 1013:33 revolutions per minute

2 400 metres 3 833:14 rpm 4 1:92 kg/cm2

5 5:3333 m/sec2

6 a Since area equals width times length, we have

A = wl and if A is fixed then l = Aw

.

b l = 1000

4 = 250 cm

7 24 days. There is 40 days supply left with 3 horses.

With 5 horses the food is eaten at a rate 5

3 faster,

so the supply will last for 40 £ 3

5 days.

Exercise 9D.1

1 a

b

c

d

f

e

g

h

i

j

0 10 20 30 40 50 60 70 8 0 90 100

1

2

3

4

5

i n t e r e s

t r a

t e

( % )

sum invested 10 ($) 3

0 10 20 30 40 50 60 70 8 0 90 100

1

2

3

4

5

sum invested 10 ($) 3

i n t

e r e s

t g a

i n e

d

1 0

( $ )

3

x y Qe _

-Qe _ ( )313

1 , ( ) ( ) 912

312

31

y x

x

y

2.83

2.83

22 8 y x

x

y3

4.584.58

7

(0, 2)

( ) 22252 x y

x

y10

1010

10

22 100 x y

x y

1

5

(0, 3)

( )22 12333 y x

x

y5.36

5.36

3.36

3.36( 1, 1)

( ) ( )222011 y x

x

y

8

4

4

2

(3, 0)

( ) 22253 y x

x y

4

2

(0, 3)( )22 13 y x

x

y

2.17

7.83

(1, 5)

( ) ( )22951 y x

x

y

4

4

4

4

22 16 y x



472 ANSWERS

2 a

b

c

d

e

f

g

h

i

j (x ¡ a)2 + (y + b)2 = c2

3 a (x ¡ 1)2 + y2 = 4 b x2 + (y + 2)2 = 16

c (x + 3)2 + (y ¡ 2)2 = 1

4 (x ¡ 2)2 + (y ¡ 1)2 = 100

5 a y = §4 b y = §3 c y = §5d y = 0 e y = §

p 21

6 a y = ¡11; y = 5 b y = ¡9; y = 3c y = ¡13; y = 7 d y = ¡3 § 4

p 6

e y = ¡12; y = 6

7 b no point of intersection

8 a e.g. (x + 2)2 + (y ¡ 2)2 = 4b (x ¡ a)2 + (y ¡ a)2 = a2

9 a b

c y = ¡p

36 ¡ x2 d

x

y

1

1

22 1 y x

x

y

( , )a b

c

x

y

( 1, 1)

2

2

( ) ( ) 211 22 y x

x

y

( 3 , 4)

( ) ( ) 543 22

y x

x

y

(1, 7)

7( ) ( ) 171

22 y x

x

y

(1, 7)

1

0.07

13.93 ( ) ( ) 4971 22

y x

x

y

( 3 , 4)

6

8

( ) ( ) 2543 22

y x

x

y

(5, 6)

( ) ( ) 465 22 y x

x

y

2.241

5

2.24

(0, 2)

( ) 92 22 y x

x

y

7

2.65

2.65

(3, 0)

( ) 163 22 y x

WINDOW: [ 10, 10, 1; 10, 10, 1] WINDOW: [ 10, 10, 1; 7, 7, 1]

WINDOW: [ 10, 10, 1; 7, 7, 1] WINDOW: [ 15, 1 5, 1; , 10 , 1]

x

y

5 6

5 x 6

522 y x



ANSWERS 473

Exercise 9D.2

1 a x2 + y2 + 6x ¡ 7 = 0

b x2 + y2 ¡ 2x ¡ 10y + 17 = 0

c x2 + y2 + 6y + 6 = 0

d x2 + y2 ¡ 6x ¡ 16 = 0

e x2 + y2 + 2x ¡ 2y ¡ 8 = 0

f x2

+ y2

+ 6y + 5 = 0

2 a x2 + y2 ¡ 4x = 0 b x2 + y2 ¡ 10y + 16 = 0

c x2 + y2 + 4x + 6y + 12 = 0

d x2 + y2 + 6x + 6y + 12 = 0

e x2 + y2¡10y + 24 = 0 f x2 + y2¡4x¡8 = 0

3 a (x ¡ 2)2 + (y + 1)2 = 16

b (x + 3)2 + (y¡8)2 = 100 c x2 + (y¡2)2 = 1

d x2 + (y + 4)2 = 18 e (x¡ 3

2)2 + (y + 1)2 = 13

4

f (x ¡ 1

2)2 + (y ¡ 1

2)2 = 3

2

4 a (x ¡ 3)2 + (y + 5)2 = 1 b x2 + (y ¡ 2)2 = 9

c (x ¡ 6)2 + y2 = 7 d (x ¡ 1

2)2 + (y + 1

3)2 = 1

Exercise 9D.3

1 a

b

c

d

e

f

Exercise 9E

1 a x = ¡3, y = ¡1 is a point common to all 3 lines.

b x = 1

10, y = ¡1

5 fits all three equations.

2 a (3, 0) and (¡1, 12) b (1, 0) and ( 32

, 14

)

c (¡6

5, ¡ 8

5) and (2, 0)

d (1, ¡2) is the only point so the straight line is a

tangent.

e (2, 4) is the only solution of the two equations so

the straight line is a tangent.

f The radii are eachp

13 and the distance between

the centres is 2p

13 so the circles must touch,

also point of intersection is (3, 2).

g Here the radii arep

20 and p 54 and the sum of

these is p 5(2 + 12 ) = 5

2p 5.

The distance between the centres isp

125

4 and this

equalsp

25£p 5£p

1

4 = 5

2,

also point of intersection is (¡1, ¡2).

3 a (1, 0) and (¡ 15

13, ¡ 42

13)

b (x, y) = (8, 3) and (3, 8)

c (x, y ) = (¡4, ¡6) and (12, 2)

d (x, y) = (0, ¡5) and (¡15

4 , 5

2)

4 a fx = 1:278, y = 0:5471g

fx = 2:137, y = 7:272gfx = ¡2:050, y = 2:358gfx = ¡2:365, y = 4:822g

b fx = ¡5:078, y = 5:554gfx = ¡4:119, y = ¡0:3929gfx = 1:336, y = 0:7930gfx = 1:861, y = 4:046g

Exercise 9F

1 many to many; many to one; one to one; one to many;

one to one; many to many

2 a many to many b one to one c many to many

WINDOW: [ 7, 7, 1; 5, 5, 1]

WINDOW: [0, 33, 1; 10, 10, 1]

WINDOW: [ 7, 7, 1; 5, 5, 1]

WINDOW: [ 3.5, 1, 1; 2, 1, 1]

WINDOW: [ 1, 5, 1; 4, 0.5, 1]

WINDOW: [ 1, 20, 1; 4, 10, 1]



474 ANSWERS

d many to one e many to many f many to one


2

3 3000 rpm

4 centre is at (3, ¡4), the radius is 1

5 a x2 + y2 + 2x + 2y + 1 = 0b x2 + y2 ¡ 4y ¡ 5 = 0

6 x = 1, y = ¡2, and x = ¡11

2 , y = 50

7 a one to many b many to many c many to one

8 The intersection points will be seen to be x = 0,

y = ¡1, and x = ¡9

5, y = ¡2

5.

9 x = 2 and y = ¡3 fits all three equations 10 p 1811 k = 3 or k = ¡2 12 x = ¡1 or x = 5

13 x = 0, y = 1, and x = 4

5, y = ¡3

5

Exercise 10A

1 a Trixie b 94:4 km per hour

c 1:58 mm per 10 000 km

d approx. 107 589 km per hour

e 9424777961 km f $1:32g approx. 0:26 error/100 words

h assuming a life span of 75 years 2:3652£ 10¡9

2 c 63:3% 3 not possible

4 a 1000 cm2/cm3 or 110

m2/cm3 c 50 m2

6 a 16:67 m/sec b 360 km/hr c 120 kg/ha

d $800/tonne

7 a 9:4 L/100 km b 22 L/100 km

c 43:4 mpg d 23:3 mpg

8 a 104:6 km per hr b 35450 L/1000 km

c 2254:4 m/sec d 1:7£ 10¡8 km per hour

10 men (km/h) 36:6; 33:3; 28:5; 26:2women (km/hr) 34:3; 30:3; 25:4; 23:4

11 a Ndeti 12:36 mph; 19:90 km per hr

Pippig 11:1 mph; 17:88 km per hr

b 20:09 km per hour c 2 hr 6 min 38 sec

d 21:11 km per hr

Exercise 10C

1 a 98:57 km per hour b 75:53 km per hour

2 a 107800 (Q), 61 400 (NSW)

b 67429 (Q), 60933 (NSW)

3 a 1910-1926 b population decline 29553/year

4 a 4 million kL/day b y = ¡4x + 248c approx. 248 million kL

5 132:5 km2 per day

6 a 1:3 mm per hour b approx. 1:7 mm per hour

10 a approx. 5:00 am

b approx. 6:35 am; 190th day

c about 33 seconds per day

11 a ¡21 points per month

b approx. 25 points per month

12 a accelerating b 15 km per hour

c Accelerates from 0 to 30 km per hour for 1 minute;

then travels with constant velocity of 30 km per

hour for 4 minutes.

d Brakes from 30 km per hour to 15 km per hour inabout 1

4 min; comes to rest in the next 3

4 minute;

at rest for one minute and then accelerates to 10km per hour for 1 minute.

e zero f 2 km g 83 metres

13 a

b

c

14

15 a

b

30

0 1 2 3 4 5 6

6

9

12

weight

cost

displacement

time

displacement

time

displacement

time

0

25

50

75

100

125

0 4 8 12 16 20 24 28 32

m/min

time

m/sec

time012345

6

0 1 2 3 4 5 6 7

m/sec

time0

1

2

3

0 2 4 6 8 10



ANSWERS 475

Exercise 10D

1 a ¡2 b 5 c 3 d 52 a 0 b approx. 20 metres/sec c approx. 3:8 secs

d h = 0 e approx. 40 metres/sec

3 a 8:25 kL b 3 kL c 8 kL per hour

d approx. 3 kL per hour

e approx. 0:5 kL per hour

6 a 0 b 2 c 4 d 6 e 8gradient = 2£ x-coordinate

7 a 12 b 3 c 0 d 3 e 12gradient = 3 £ (x-coordinate)2


1 approx. 0:12 cm per min

2 553 846 cubits per day assuming a 24 hour day

3 a 20 litres per min b 40 litres per min

c 40000 cm3 d 50 litres per min

4 a 15 seconds b approx. 10 secs

c approx. 12 metres/sec d about 2 1

2 metres/sec

e i from zero to about 8 seconds

ii from about 8 seconds to about 13 seconds

iii from about 13 seconds

5

Note: 2 mins for metres would equal the800Australian record

6 12:50 pm7 x-coord y-coord gradient of secant

¡3 31 ¡16¡2:9 29:13 ¡15:7¡2:8 27:32 ¡15:4¡2:7 25:57 ¡15:1¡2:6 23:88 ¡14:8¡2:5 22:25 ¡14:5¡2:4 20:68 ¡14:2¡2:3 19:17 ¡13:9¡2:2 17:72 ¡13:6¡2:1 16:33 ¡13:3

¡2:01 15:13 ¡13¡2:001 15:013 ¡13

From the table the gradient of the curve at (¡2, 15)

appears to be ¡13:

Exercise 11A

1 6 metres/sec

2 a 8 metres b 8 metres/sec

3 6 metres/sec

4 a at 8 metres b 4 metres/sec

5 a 3 metres/sec b 6:75 m/sec c 12 metres/sec

6 a 0

:354

metres/sec

b depends upon the accuracy of each arithmetical

calculation

Exercise 11B

1 a 20 b 2:5 c 5 d 0 e ¡37 f ¡32 a 1

2 b 0 c 0 d 8 e 4 f 7

4 a 5 b ¡2 c ¡1 d 1:75 e 1 f 1g 0 h 0

5 3x2

6 a 1

2 + 1

4 = 3

4 b 1

2 + 1

4 + 1

8 + 1

16 + 1

32

c 1

2 + 1

22 + 1

23 + 1

24 + 1

25 + :::: + 1

2n d 1

e Note1Pn=1

1

2n ! 1 but lim

n!1(sum) = 1

Exercise 11C.1

1 a 8 b 12 c 5 d 4 e 7 f 4 g 1h 0

Exercise 11C.2

1 a 4

x b 3

x2 c 2

x+ 3

d ¡4

x e 3 ¡ 2

x

f 12x g 2x + 2 h ¡6x

Exercise 11C.3

1 a 6 b 18x c ¡6x2 d 8x ¡ 2 e 0f 4x ¡ 5 g 6x2 h ¡4x i 3x2 ¡ 2x ¡ 2

Exercise 11D

1 a dy

dx = 3x2 b

dy

dx = 8x c

dy

dx = 2x

d dy

dx = 2x ¡ 6 e

dy

dx = 3x2 + 1 f

dy

dx = 5x4

g dy

dx

= 6x2 h dy

dx

= 6x5 i dy

dx

= ¡3x2

2 dy

dx = 2(x ¡ 3) = 2x ¡ 6 3

dy

dx = 8x7

Exercise 11E

1 a dy

dx = 4x3 b

dy

dx = 7x6 c

dy

dx = ¡5x4

d dy

dx = 6x e

dy

dx = 5 f

dy

dx = ¡1

2

g dy

dx = 0 h

dy

dx = 0 i

dy

dx = ¡1

j dy

dx = x2 k

dy

dx = 4x

l dy

dx = 12x2 ¡ 8x + 7 m

dy

dx = 3

2x2 + 2

3x ¡ 1

4

n dy

dx = 3ax2 + 2bx + c o

ds

dt = ¡2t¡3

2 a s(t) = ¡9t¡4 b s0(x) = 30x¡3

c y0(x) = ¡3x¡3

2 d f 0(x) = ¡3x¡4 + 4x¡5

e f 0(x) = ¡x¡2¡ 2x¡3¡ 3x¡4 f f 0(x) = ¡ 5

x2

g f 0(x) = ¡2

x3 h f 0(x) = ¡

12

x5 i

dy

dx =

6

x4

j dy

dx

= 1¡ 1

x2

k dy

dx

= 1

3x¡

2

3 l dy

dx

= 1

2x¡

1

2

10

5

20

30 60 90

110

120 time (sec)

metres per sec



476 ANSWERS

m dy

dx = ¡1p

x n

dy

dx = 5

3x¡

2

3 = 5

3 3p

x2

o ds

dt = 2 ¡

2p x

3 a ¡ 1

2p

x3b ¡5

4x3

2

c ¡1

x2 ¡

4

x3 + 1

d ¡12p x ¡ 3x2 e 2 p ¡ 2 p3

f 6x ¡ 5

g 2x + 3

x2 h

1

3x2

3

i ¡6

x5

2

j u + gt

k ¡1

t2 +

2

t3 l

¡1

3x2

3

¡ 1 m ¡1

2t3

2

+ 4

t3

n ¡ 1

t4

3

o ¡a

2t2 +

2b

3t3

4 a ds

dt = 10 ¡ 10t b ¡10 c

dv

dt = 5

5 a dA

dr

= 2¼r; dA

dr , circumference of circle

c dV

dr = 4¼r2;

dV

dr , surface area of sphere

Exercise 11F

1 a dy

dx = 5x4 ¡ 2;

d2y

dx2 = 20x3

b dy

dx = 1 1

2x

1

2 ; d2y

dx2 =

3

4p

x

c dy

dx =

¡2

5x3;

d2y

dx2 =

6

5x4

d dy

dx

= 1

4 +

1

2x2

; d2y

dx2

= ¡1

x3

e dy

dx = ¡

1

10x3

2

; d2y

dx2 =

3

20x5

2

f dy

dx = 6x5 ¡ 4x3 ¡ 3x2 + 6x + 1;

d2y

dx2 = 30x4 ¡ 12x2 ¡ 6x + 6

g dy

dx = 3

2x

1

2 ¡ 1

x3

2

; d2y

dx2 = 3

4x¡

1

2 + 3

2x¡

5

2

h dy

dx = 14x(x2 ¡ 1)6;

d2y

dx2 = 14(x2

¡ 1)5

(13x2

¡ 1)

Exercise 11G

1 a v = 3t2 ¡ 12t + 9, a = 6t ¡ 12b v = 0 when t = 1 and t = 3c a = 0 when t = 2

2 a s = 0 b v = 2t ¡ 3, a = 2c The particle is travelling with constant acceleration.

d v = ¡3 when t = 0e When v = 0, t = 1 1

2 seconds and s = ¡21

4 metres

i.e., 21

4 metres to the left of its starting point.

f s = 0, when t = 0 and t = 3 seconds

g

3 v = 2t ¡ 6 a = 2 = constant

) moving with constant acceleration

4 a v = dh

dt = 90t ¡ 3t2 b 13500 metres

5 a t = 0, h = 2 metres Therefore the stone is thrown

from a point 2 metres above the ground.

b t = 1:5 secs

c v = 6 ¡ 10t; t = 0, v = 6 m/sec

d t = 3

5 secs e h = 3:8 metres

f v = ¡9 m/sec i.e., downwards at 9 m/sec

6 a h = 200 metres b when t = 0, v = 0c the stone is moving downwards

d when h = 0, t =p

250 + 15:8 secs

e ¡25:3 metres/sec f a = 1:6 m/sec/sec

g 16 m/sec down; h = 120 metres

h t = 12 secs; h = 84:8 metres above the crater

floor.

Exercise 11H

1 a y = 4x ¡ 4; y = ¡1

4x + 4 1

2 b y = 9; x = 3

c y = 3

4x ¡ 1

4; y = ¡4

3x + 19

24

d y = 10x ¡ 22; y = ¡ 1

10x + 8 3

10

e y = 4x ¡ 3; y = ¡1

4x ¡ 1 1

4

f y = ¡2

3x + 4; y = 3

2x ¡ 21

2

g y = 1

4x + 1; y = ¡4x + 18

h assuming positive root only y = ¡ x

16 + 3

4,

y = 16x ¡ 127

2

2 (1, 8) and (3, 4) 3 (1, ¡1) 4 x = 4, y = 85 x = 4, y = 13 solving f (x) and g(x) simultaneously

gives one root only viz x = 4

Exercise 11I.1

1 a 4(x + 7)3

b 6(x ¡ 6)2

c 15(5x ¡ 2)2

d ¡6(3 ¡ x)5 or 6(x ¡ 3)5

e 27x(x ¡ 2)(x3 ¡ 3x2 + 1)8

f 7(2ax + b)(ax2 + bx + c)6

g ¡3(8x + 5)(4x2 + 5x ¡ 3)2

h ¡3(x ¡ 5)¡4 or ¡3

(x ¡ 5)4 i

2

(6 ¡ x)3

j ¡3

(x ¡ 7)2 k

¡8

(x ¡ 5)5 l

5

2p

5xor

p 5

2p

x

m 2

3

3p (2x + 4)2

n 3

2

p x + 3 o

¡3

(3x + 1)

4

3

s t t 2 3 s t 2 3

s 2

s

t 1 2 3 4

23

2

4

6



ANSWERS 477

p ¡1

2(x + 3)3

2

q ¡2(x + 3)

(x2 + 3x + 1)2

r ¡3x + 1

(3x2 ¡ 2x + 1)3

2

2 a 5

(15x ¡ 23)2

3

b ¡3

2p

2 ¡ 3x+

5

x2

c3p 3

3(x ¡ 2)2

3

d 2x

(3 ¡ 2x2)3

2

e 4x

3(x2 ¡ 1)43

f 3x

(x2 ¡ 9)3

2

3 a 6(2x ¡ 1)2

b 8x3 ¡ 12x2 + 6x ¡ 1; dy

dx = 24x2 ¡ 24x + 6

4 2

27 metres per sec per sec

5 ¡1500

(d + 30)2; ¡0:6 mg per m2 per day

6 dV

dt = 8(t ¡ 50) a ¡360 L/min b ¡320 L/min

c ¡240 L/min

7

¡m0

¯1

c

¯vp

¡(v2 ¡ c2)rate of change of mass wrt velocity

8 15

2 x¡

1

2 (3p

x ¡ 1)4

Exercise 11I.2

1 a dy

dx = 6(3x + 4) b

dy

dx = 14(2x + 1)6

c dy

dx = ¡8(3 ¡ 4x) d

dy

dx = 6x(x2 + 2)2

e dy

dx = 12(x ¡ 1)(x2 ¡ 2x + 1)5 f

¡3

(3x ¡ 4)2

g dy

dx =

¡6

(3x ¡ 5)2 h

dy

dx =

¡6x

(x2 + 3)2

i dy

dx =

3(3x2 + 1)

(x3 + x ¡ 5)2

2 a dy

dx =

p 3

2p

x b

dy

dx =

5

2p

5x + 7

c dydx

= ¡15x2

2p

4 ¡ 5x3d dy

dx = ¡1

2x3

2

e dy

dx =

2(x ¡ 2)

(x2 ¡ 4x + 1)3

2

f dy

dx =

8x

3(2x2 ¡ 3)4

3

Exercise 11J

1 a 24x ¡ 5 b ¡8x ¡ 4 c 12x ¡ 8

d 12x2 + 20x ¡ 5 e 10x4 ¡ 3x2 ¡ 2x ¡ 15

f x(16x6 ¡ 42x5 + 15x ¡ 30)

g 6(2x ¡ 3)2(3x + 1)3(7x ¡ 5)

h 2(x ¡ 4)7(x + 3)9(9x ¡ 8)

i ¡6(3t ¡ 4)(6t ¡ 5)

2

(15t ¡ 17)

j (3x ¡ 2)2(5x + 3)3(105x ¡ 13)

2 ap

x ¡ 5 + x

2p

x ¡ 5b

p 3x ¡ 1 +

3x

2p

3x ¡ 1

c ¡2(3x + 1)1

3 ¡ 2x

(3x + 1)2

3

d (x + 2)2(7x + 2)

2p xe ¡3

p x + 1 ¡

3x

2p

x + 1+

1p x + 1

f

p ¡(x ¡ 4)

2p

x ¡

p x

2p ¡(x ¡ 4)

g ¡(3x ¡ 1)

(3x + 1)3

h ¡2(x + 3)(x + 14)

(2x ¡ 5)4 i

¡(5x + 1)

27(x ¡ 1)4(x + 1)3

j 2(5x ¡ 7)(40x2 ¡ 84x + 15)

(4x2 ¡ 3)4

4 a f 0(x) =

p x + 2

2p x + 1 +

p x + 1

2p x + 2

b f 0(x) = 10x2 + 1

2p

x c f 0(x) =

2

x2

d f 0(x) = ¡2(24x2 ¡ 27x ¡ 4)

(2x ¡ 3)3(3x2 ¡ 2)4

e f 0(x) = ¡(2x + 3)

(x + 1)2(x + 2)2

f f 0(x) = ¡4x

(1 ¡ x2)3

2

5 The derivative is 16x3 ¡ 2x

Exercise 11K

1 a 11

(2x + 1)2 b

¡3(2x2 ¡ 1)

(2x2 + 1)2 c

¡(2x3 + 3)

(x3 ¡ 3)2

d ¡24(2x + 1)2(4x + 7)

(3x ¡ 1)6

2 a ¡71

(3x ¡ 7)2 b

(x ¡ 3)2(2x + 15)

(2x + 1)3

c x4 + 5x2 ¡ 2

(x2 + 1)2 d

¡(x ¡ 4)2(x2 ¡ 16x + 15)

(x2 ¡ 5)3

3 a ¡83(x ¡ 2)2

b 2(x ¡ 13)(2x + 1)2

(x ¡ 4)3

c (2x + 1)2(4x ¡ 25)

(x ¡ 4)2 d

3p x ¡ 3

¡ 3x + 1

2(x ¡ 3)3

2

e 1

2p

x + 1(x + 2)3

2

4 ¡12

(2x ¡ 1)3

5 a 3

x2 b 1 c

3

2p

x d

¡10

(5x + 4)3



478 ANSWERS

6 a 12(2x¡1) b 2(3x¡2)(18x¡1) c 2

(x ¡ 5)2

d 3p

x ¡ 3 + 3x

2p

x ¡ 3e ¡(4x + 1)(4x + 3)

2x4

f 3x + 5

2x3

2

g 12x2 ¡ 4

h ¡1p ¡(2x ¡ 3)(4x2 + 1)

¡ 4xp ¡(2x ¡ 3)(4x2 + 1)

3

2

7 a dy

dx = ¡2

x2 b

dy

dx =

¡(2x + 1)

2p

1 ¡ x ¡ x2

c dy

dx =

4xp 4x2 ¡ 1

d dy

dx = 0 e

dy

dx = 3

f dy

dx =

(3x ¡ 1)(3x + 1)

3x2 g

dy

dx = 3x2 + 2x¡ 6

h dy

dx =

2x + 1

2x3

2

i dy

dx =

¡6

(3x + 4)3

j dy

dx = ¡7

2(x ¡ 3)3

2 :(2x + 1)1

2

k dy

dx =

2(2x + 3)

3(x + 1)4

3

l ¡1

6(5 ¡ x)5

6

Exercise 11L

1 a i ¡4 and 3 ii x < ¡4; x > 3iii ¡4 < x < 3 iv ¡1 v x > ¡1vi x < ¡1

b i ¡2; ¡1; 2 ii ¡2 < x < 1; x > 2

iii x < ¡2; ¡1 < x < 2 iv ¡11

2, 1

v x < ¡11

2; x > 1 vi ¡1 1

2 < x < 1

c i 0 ii x > 0 iii x < 0 iv no values

v for all real x vi no values

d i no values ii x > 3 iii x < 3iv no values v no values

vi for all x except x = 3

2 a x > ¡1 b x > ¡1

2 c gradient is constant

d x > 33 a

b

c

d

4 a

b

c

d

5 a

x

y

x

y

x

y

x

y

x

y

x

y y x x ( )( )3 2

x

y

y 0

x

y x 2

x

y( 1, 4)

1



ANSWERS 479

b

c

d

e

6 a

b

c

d

e

f

7 a nothing

b function is increasing for all values of x

8 a

b

c


1 dy

dx = 3x2 ¡ 2

2 a dy

dx = 1 ¡

1

x2 ¡

4

x3 b

dy

dx = 60x3(3x4 ¡ 5)4

c

dy

dx =

23

(2x + 5)2

x

y

(2, 16)

(0, 0)

x

y

( 2, 16)

(0, 0)

x

y

x

y

(0, 2)

( 2, 3)

(2, 3) x

y

x

y

y x k

x

y

x

y y x wQ _ 2

x

y

x

y

1

x

y

x

y

( 2, 3) (2, 3)

x

y 2

43 x y



480 ANSWERS

d dy

dx = (x3 ¡ 2)(2x3 ¡ 3x ¡ 1)

3 a 10 metres/sec b ¡10 metres/sec/sec

c t = 3 secs d 45 metres e t = 6 secs

4 a i y = 3x ¡ 9 ii y = ¡1

3x + 11 b 20

5 a $120000 (increasing)

b P = 5:27 + 0:18(t ¡ 3) or P = 0:18t + 4:73;

$5:81 million

8 a = ¡2

1 a i 85 ii 3x3y4

b i 3£ 7£ 13 ii (5ssssttt)£ (5ssssttt)

c i 6z7 ii a5b6 d i 3d ii 3uvw3

2

e x6 ii 6w8

4w10 =

3

2w2 f i e3f 3 ii 40a4b7

g i 6abc ii ¡27m8 h i 16a3b3 ii 16 p4q 2

2 a i 1 ii 1 b i 1

4 ii 1

7

c i 1

p2q 2 ii

t6

3s4 d i 5 ii 6

e i 1

4 ii 1024 f i

2p

165 ii 5

p 27 p3

g i 1

z1

3

ii 9

7g3 h i 3

8 ii k2

i i 37 ii 1

212 j A = D, B = C = E = F

3 a i log3

81 = 4 ii log25

1

5 = ¡1

2

b i 32 = 25 ii 1

100 = 10¡2

c i log5 125 = 3 ii loga ax = x

4 a i 0 ii 1 ¡ 1

2 log

5 2

b i log 20 ii log 49

125

5 a x = 4

9 b x = 5 6 a 3 b 1

3 c ¡2

7 a 5 b ¡1

8 a i 1:0913 ii 0:3705 b i 9 ii logp r

c i 1

3 ii z = d d i 1 + p ii

2 p + pq

1 + pq e i 1

6 ii 1

2

9 a i x = 1:4037 ii x = ¡2

b i x = 2:3652 ii ¡4:4205

10 a i $6048:07 ii $4516:49 iii 3:66% p.a

iv + 10% p.a. v + 10:52 years

b $45 773:71 c 11:51% p.a. d 5:61 years

e i $347202 ii 7:6% iii 9:4572 years

11 a m = 0:8184x¡2:52

b

There is an underlying pattern in the residuals, so

a power function is not an appropriate model for

this data.

Chapter 12 Section B – Review of Chapter 8

1 a i and iii are polynomials; ii is not a polynomial as

it is a rational function

b i f (0) = 7; f (2) = 7ii h(3) = 94; h(¡9) = ¡3062

2 a g(x) = 3x2 ¡ 3x ¡ 60

b F (t) = 2t3 ¡ t2 ¡ 32t + 16

3 a i x3 + 9x2 + 27x + 27

ii 16z4 ¡ 96z3 + 216z2 + 81

b i (t ¡ 1)3 ii (x ¡ 2)4

c i f (x) = kx(x + 3)(x ¡ 2)for example, k = 1, 2, 3

d f (x) = mx + 7for example with m = ¡1, 2, 3

e f (x) = ¡4x + 7, g(x) = 2x

2

+ 4x + 5,h(x) = x3 ¡ x2 + x + 6

4 a i y = 2x ¡ 10 ii x4 ¡ 4x2 + 6b i f (z) = z ¡ 2z2 ii f (¡2c) = ¡2c ¡ 8c2

iii f (2t2 + 1) = ¡8t4 ¡ 6t2 ¡ 1

c i f [g(x)] = x2 ¡ 6x + 7 ii g[g(x)] = x ¡ 8

iii f [g(¡1)] = 14 iv f [f (¡1)] = ¡1

d i f og = ¡x3 + 6x2¡11x + 9 ii ¡1 + x¡x3

e i y = 0; y = ¡1ii x = 5; x = 4; x = ¡1; x = ¡2

5 a i

[

WINDOW:

WINDOW:

¡4, 2, 1; ¡10, 10, 1]

ii

[¡2:5, 2:5, 0:2;

¡3, 3, 0:2]

b

local min at ,(0.54 0.88)

local mins at , and ,( 1.62 1) (0.62 1)

y x-int at ; -int at , and0 3 0 1

y x-int at ; -ints at , , and0 2 1 0 1

local max at ,( 1.87 6.06)

local max at ,( 0.5 0.56)

Chapter 12 Section A – Review of Chapter 7



ANSWERS 481

The first window shows the graphs are quite dif-

ferent near the origin, but for large negative and

large positive values of x, the graphs are nearly

identical.

6 a

For f (x) x !1, f (x) !1x ! ¡1, f (x) ! ¡1

b See graphs above. The 2 produces a vertical shift

of 2 (note y-coordinates of local maximum and

local minimum). The zeros are not affected.

ii ¡f (x ¡ 2) + 1 is a reflection in the x-axis is

followed by a translation to the right of 2 units

and a vertical translation of 1 up.

b

9 a i + 185-186 ii 400 iii + $27000

b i h = 1

20s2 ii As =

1

5s + s2

iii Note s, h > 0

f x x x x ( ) ( )( )( )1 2 3

f x x x x ( ) ( )( )( )2 1 2 3

123

10

20

1 2 3

( )1.53, 1.13

( )1.53, 13.13

( )1.53, 26.26

( )1.53, 2.26 x

y y x 4

y x ( )2 4

x

y

y x 4

y x ( )2 14( )2, 1

[¡3, 1, 1; ¡6, 1, 1 ]

ii

iii

[¡3, 6, 1; ¡120, 30, 4]

b simplest function is y = x3 or y = 6x(x ¡ 1)2

8 a i f (x + 2) is a translation to the left of 2 units.

1 a i

ii

x

y

y 3

31 x

y

y 7

x 2

72

1

x

y

y

x

c see above

7 a i

[¡2, 2, 1; ¡3, 3, 1 ]

iv s = 0:464 metres h = 0:232 metres

Chapter 12 Section C – Review of Chapter 9



482 ANSWERS

b i

ii

iii

c i N = 12000 000

d2

ii

iii 11 019

2

3 a 150 revs/min b 3:5 metres/sec

c i As V = ¼r2h, r2 = V

¼h =

k

h as

V

¼ is a constant

ii 4444:4 cm

4 a i ii

iii

b i (x ¡ 1)2 + y2 = 16 ii x2 + (y + 3)2 = 9iii x2 + (y ¡ 1)2 = 1

4

c i x2 + y2 + 6x ¡ 8y ¡ 16 = 0

ii x

2

+ y

2

¡ 6x + 6 = 0d i (x¡2)2+y2 = 1 ii (x¡3)2+(y +2)2 = 24

iii (x ¡ 1 1

2)2 + (y + 1)2 = 31

4

e

5 a Point of concurrency is (3, ¡1)

b i x = ¡2, y = 13; x = 1, y = 1

x

y

x y 1

x y 2

x y

41

d

N

x

y

x y 1

x

y

x y 1

x

y

x y2 2 4

2

2

2

2

x

y

( ) x y 2 162 2

26

( )2, 0

x

y

( ) ( ) x y 2 42 2

( )2, 4

x

y

31 x

y

y 3

x

y

51 x

y

x 5

x

y

12

1 x

y

x 2

y 1

131 x

y

x

y

y 1

100

200

300

400

500

600

$

midnight noon midnight noon

120130

190

205295

320

530

540

600

8:00 6:00 8:00



ANSWERS 483

ii x = 1, y = 2; x = 2, y = 1iii point of tangency is (0, 1)

iv radii arep

5 and 3p

5 respectively

distance between centres is 4p

5c i (5, 3) and (6, 2:5) ii (0, 5) and (1 1

4, 2 1

2)

d

6 a i one to many ii one to one iii one to many

iv one to one

b i many to one ii one to one iii one to many

iv one to many

Chapter 12 Section D – Review of Chapter 10

1 a i 8

1500 ii 3600 km/hr iii $12:50

b i No. 2730£0:1 + 90

100£0:2 + 52

70£ 0:7 = 0:79

= 79%

c 160 perches/acre, 607 m2

ii 200

3 = 662

3 km/hr iii 0:5 mm

2 a i 72:7 km/hr ii 75:8 km/hr

b i 220000 people per year ii 209 000 per year

c i depth of water increased, possibly due to rain

ii between 1 p.m. and 2 p.m.

iii about 1:1 mm/hr

3 a accelerating at 4 km/hr/sec b 12 km/hr

c constant velocity of 24 km/hr

d constant deceleration of 6 km/hr/sec e 0f 60 m

5

7 a QR parachutist jumps and accelerates

RS parachute opens

ST parachutist slows descent rate

b between Q and R

c average speed about 300 metres per minute

d approx. 60 m per minute

e 60 m/min or about 3:6 km/hr

f 350 metres/min

8 a i m = ¡3 ii m = 2

b i m = ¡3 ii m = 2c i m = 0 ii m = ¡2 iii m = ¡4

iv m = ¡6 v m = ¡8, m = ¡2x

Chapter 12 Section E – Review of Chapter 11

1 a 2 metres/sec

b i s = 9 metres ii v = 9 metres/sec

2 a i ¡2

5 ii 3 c i ¡1

2 ii 6 d ¡ 1

x2e i 0 ii 4 iii 1

3 a 11

3; 1 4

9; 1 13

27 b 1 40

81; 1 121

243; 1 364

729 c 11

2

4 a dy

dx = 4x3 b y = 12x2 c y = 6x ¡ 1

d 3x2 ¡ 6 e ¡1

(x + 1)2 f 4x ¡ 5

5 a i dy

dx = 5x4 ii

dy

dx = 6x5 iii

dy

dx = 10x

iv dy

dx = 0 v

dy

dx = ¡2x vi

dy

dx = 4

3x3

vii dydt

= ¡3t¡4 viii dydx

= 16x3 ¡ 4 + 14x

b i s0(t) = ¡6t¡3 ii s0(x) = ¡5x¡2

iii f 0(x) = ¡2x¡3 ¡ 4x¡5 ¡ 6x¡7

iv f 0(x) = ¡ 5

x2 v g0(x) =

8

x5

vi dy

dx = ¡

2

x3 vii

dy

dx = 4

3x1

3

viii dy

dx = ¡

1p x

ix dy

dx =

1p x ¡ 4

c i ¡ 3

8x3

2

ii 1

2x

3

2

¡ 68

x3 iii ¡

1

p2 iv

¡9

x5

2

v 4 ¡ 1

4x3

4

vi ¡ c

2t2 +

4b

7t5

d i ds

dt = 3 ¡ 21t2 ii ¡81

e i dV

dr = 2

3¼rh ii

dV

dh = 1

3¼r2

6 a i v = 2t2 ¡ 7t + 3, a = 4t ¡ 7

ii t = 1

2 sec, t = 3 sec iii t = 13

4 secs

b dh

dt = 180t ¡ 6t2 hmax = 27 000 m

c

dC

dx = ¡557 a i y = ¡2x + 9, y = 1

2x + 1 1

2

ii y = 17x ¡ 49, x + 17y ¡ 37 = 0

b (¡2, ¡4)

c (2, ¡3); solving f (x) and g(x) simultaneously gives

rise to one point of intersection only and

f 0(2) = g 0(2)

8 a i dy

dx = 4(x ¡ 8)3 ii

dy

dx = 6(3x ¡ 2)

iii dy

dx = ¡6(3 ¡ x)5

iv 9(3x2 ¡ 6x)(x3 ¡ 3x2 + 1)8

y x x 2 3 3

x x y y2 2 2 6 3 0

( )4, 1

3

( )4.51, 3.81

( )1.75, 5.32

( )0.76, 0.15 x

y

10

4

12

6

14

8

0

2

4010 5020 6030 70 80

m e

t r e s

/ s e c

minutes



484 ANSWERS

v dy

dx = ¡3(x ¡ 5)¡4 vi

dy

dx = 2(6 ¡ x)¡3

vii dy

dx =

¡8

(x ¡ 5)5 viii

dy

dx = 2

3(2x + 4)¡

2

3

ix dy

dx =

¡(2x + 4)

(x2 + 4x ¡ 2)2

x

dy

dx =

¡(4x ¡ 3)

2(2x2 ¡ 3x + 4) 32

b i dy

dx = 4(x ¡ 1)3

ii y = x4 ¡ 4x3 + 6x2 ¡ 4x + 1dy

dx = 4x3 ¡ 12x2 + 12x ¡ 4

= 4(x3 ¡ 3x2 + 3x ¡ 1)

iii ) dy

dx = 4(x ¡ 1)3

c a = 72

(t + 1)3; 1:125 m/sec/sec

d dv

dt = 15(t¡ 20)

; ¡255

L/min; 0

9 a I = x

(x2 + 82)3

2

b

c 5:65 metres

10 a i dy

dx = 12x ¡ 4 ii

dy

dx = 12x ¡ 39

iii dy

dx = ¡x(126x5 ¡ 36x4 + 9x ¡ 2)

iv ds

dt = ¡(2t ¡ 3)3(5t ¡ 6)2(70t ¡ 93)

b i dy

dx =p

x ¡ 1 + x

2p

x ¡ 1+

1

2p

x ¡ 1

ii dy

dx =

(x + 3)2(7x + 3)

2p

x

iii

p x + 3

2p x +

p x

2p x + 3

iv 2(x2 + 3)(x2 ¡ 10x ¡ 9)

(2x ¡ 5)4

11 a i dy

dx =

11

(x + 2)2 ii

dy

dx = ¡8(x + 2)2(10x + 59)

(5x ¡ 3)6

b i dy

dx =

¡71

(3x ¡ 7)2

ii dy

dx =

2(x ¡ 5)(2x2 ¡ 15x + 4)

(x2 ¡ 4)4

c i dy

dx =

6

(x + 2)2 ii

dy

dx = ¡2(x ¡ 13)(x ¡ 4)

(2x + 1)4

iii dy

dx =

3x ¡ 11

2(3x ¡ 1)3

2

d dy

dx =

¡18

(3x ¡ 1)4

e i dy

dx =

4

x2 ii

¡4

(4x + 5)2

12 a dydx

= 24(4x ¡ 1)2 b dydx

= ¡3(x + 35)2

c dy

dx = ¡

(4x + 1)(4x + 3)

2x4 d

dy

dx = ¡5(x ¡ 3)

2x5

2

e dy

dx = 3x(8x2 ¡ 5x ¡ 4)

f dy

dx = ¡3x2

p 5 ¡ x

2(x3 + 2)3

2

¡ 1

2p

x3 + 2p

5 ¡ x

13 a b

i x = ¡2; x = 3 x = ¡5; x = ¡1; x = 6

ii ¡2 6 x 6 3 x < ¡5; ¡1 < x < 6

iii x < ¡2; x > 3 ¡5 < x < ¡1; x > 6iv x = 1 x = ¡2; x = 4

v x < 1 ¡2 < x < 4

vi x > 1 x < ¡2; x > 4

c d

i x = ¡1 zero

ii x < ¡1 x < 2

iii x > ¡1 x > 2

iv none none

v none all real x x 6= 2

vi real none

14 a x < 1 b x < 0 c noned for all x, except x = 2

15 a

b

c

x

y

x

y

x

y



ANSWERS 485

d

16 a

b

c

d

17 a

b

c

18 a function is constant

b function is decreasing for all values of x

19 a

b

Exercise 13A.1

1 a 84:2 cm b 16:8 cm

2 a 94:2 cm b 13:7 cm c 37:7 cm

3 a 700 m b 407 m c i 74:34 m ii 6:28 m

d 1 h 35 min

Exercise 13A.2

1 a 1200 mm2 b 70000 cm2 c 2000000 m2

d 9000000 mm2 e 2 cm2 f 130 000 m2

2 a 30 cm2 b 65:97 cm2 c 6963 m2

3 a 23:87o b $28 800 c 12 trips

Exercise 13A.3

1 a 1160 cm2 b 452:4 cm2 c 942:5 m2

d 1640 cm2 e 758:4 cm2 f 603:2 cm2

2 a $13446:68 b $2562:36 c 3:695 m2

Exercise 13A.4

1 a 1800 cm3 b 1567 cm3 c 1:508 m3

d 20:78 cm3

2 a 94:03 cm3 b 43:52 m2

3 a 1:222 m b 450 m3 c 6:597 m3

d 3 h 46 min e 15915 handles

Exercise 13B.1

1 a 56% b 75% c 92% d 116% e 102:4%f 400% g 100% h 55%

2 a 17

20 b 21

50 c 21

20 d 3

20 e 1

3 f 3

40

g 1

16 h 33

25 i 1

6 j 12

25 k 8

5 l 1

400

3 a 0:92 b 1:06 c 1:124 d 0:882 e 0:075

x

y

x

y

y x= 2

x

y y = 3

x

y

y x x ( )( )2 2

2 2

x y x 2

x

y( )2, 5

( )0, 3

x

y

( )3, 9

(0, 0)

x

y

2 6

(1, 3)

x

y

y 1

x

y

( )2, 1

( )0, 4

3



486 ANSWERS

f 0:01 g 2:56 h 0:0005 i 11:5 j 0:000037 k 3:428 l 0:637

4 a 75% b 42:9% 5 a $125 b 20%6 $39, 18:1% 7 $512 8 $450

Exercise 13B.2

1 a $1600 b $82:50 c $525 d $1837:502

10 years 3

2:25% p.a.

Exercise 13B.3

1 a $3993 b $993 2 a $6000 b $60503 a option 1, $10160; option 2, $10 077:70

) choose option 1

b option 1, $11600; option 2, $11 754:62) choose option 2

4 46:41%

Exercise 13C.1

1 a ¡18 b 72 c 9 d 1 e 5 f 12

g ¡54 h 3

5

2 a 100 b 50 c ¡14 d 2 2

3 e 11

2 f ¡1

g ¡2 h ¡1

Exercise 13C.2

1 a 6x b 20x c ¡14x d ¡12x e 2x2

f 6x2 g ¡2x2 h ¡3x3 i 2x2 j ¡12x

k 2x3 l ¡6d2 m a2 n 4a2 o 2a4

p ¡3a3

2 a 3x + 6 b 10 ¡ 2x c ¡x ¡ 2 d x ¡ 3

e ¡2x¡8 f 3¡6x g x2 + 3x h 2x2¡10x

i a2 + ab j ab ¡ a2 k 2x2 ¡ x

l 2x3 ¡ 2x2 ¡ 4x3 a 5x ¡ 2 b 3a ¡ 2b c a + 2b d 15 ¡ 3y

e ¡6y¡10 f 15x¡8 g a+5b h x2+6x¡6

i x2 + 2x + 6 j 2x2 ¡ x k ¡2x2 + 2x

l x2 ¡ y2 m 5¡ 3x n 8x¡ 8 o 6x2 ¡ 22x

Exercise 13D.1

1 a 9

14 b 5

9 c 9

16 d ¡ 1

12 e 13

8 f 1

14

g 2 1

8 h 7

8 i 6 2

3 j ¡1 2

3 k 42

3 l ¡3 1

3

m 2 13

24 n 6 1

12 o ¡1 5

12 p 1 7

15

2 a 5

8 b 2

7 c 11

2 d 22

3 e 15

16 f 54

9

g 3 3

8 h 1

2

3 a 9

10 b 11

8 c 7

9 d 8

15 e 22

5 f 28

45

g 3

10 h 4 1

12 i 1

6 j 22

3 k 21

26 l 1 1

2

4 a 6 31

40 b ¡4 19

24 c 16

81 d 2 10

27 e 9

16 f 6

g 12 h 1 3

4 i 10

11 j 23

5 k 3 3

4 l 9

40

Exercise 13D.2

1 a ¡a

2 b ¡

1

2a c

a2

4 d ¡4 e

2

t

f ¡2d g ¡b

2 h ¡

2b

3a

2 a 4 b 2n c a d 1 e a2 f 3n

2

g x + y h 2

x + 2

3 a a + 3 b 2(x + 2) c 2(c + 3) d d ¡ 3

3

e 2

x + 1 f

3

2 ¡ x g 1

3 h 2 i

x + y

3

j y

3 k

y ¡ 3

3 l

x + 1

3

Exercise 13D.3

1 a 5a

6 b

b

10 c

7c

4 d ¡

5x

14 e

4a + 3b

12

f ¡2t

9 g

5m

21 h

d

2 i

11 p

35 j m

k 5a

12 l

x

12

2 a 7b + 3a

ab b

3c + 2a

ac c

4d + 5a

ad d

a

m

e 2a + b

2x f

5

2a g

4y ¡ 1

xy h

ad + bc

bd

i ay ¡ bx

by j

4 + 3a

6 k

4x + 9

12 l

3x + 2y

3y

3 a x + 2

2 b

y ¡ 3

3 c

3a

2 d

b ¡ 12

4

e x ¡ 8

2

f 6 + a

3

g 15 ¡ x

5

h 2x + 1

x

i 5x ¡ 2

x j

a2 + 2

a k

3 + b2

b l ¡

5x

3

4 a 9x

10 b ¡

7x

10 c

11

3a d

5

2y e

5b + 3a

ab

f 4b ¡ 15a

3ab g

x + 14

7 h

12 ¡ x

4 i

5a

3

Exercise 13D.4

1 axy

10b 3

2 c

a2

2 d 1

6 e 1

5 f

c2

10

g acbd

h 1 i 12m

j 2 k ab

l 4

m 3

m n

a2

b2 o

4

x2 p

1

c

2 a 3

2 b

3

a c

3x

16 d 3

4 e 2 f

c

25

g 1

5 h

m2

2 i 2 j

n

m k

3

4g l

g

3

3 a 1 b 2

3b c

23c

20 d

2a ¡ bc

2c e

21a

2b2

f

2

3a g 1 h

2a + 15

6b



ANSWERS 487

Exercise 13E.1

1 a x = 2 b x = 31

2 c x = 21

2 d x = ¡9

e x = ¡11 f x = 11

2 a x = 9 b x = 12

5 c x = 13

8 d x = ¡2

e x = 2

5 f x = ¡10

3 a x = ¡5

7 b x = 0 c x = ¡4

9 d x = ¡9

e no solution f x = 0 g x = 3

4 h x = ¡ 13

9

Exercise 13E.2

1 a x = 6

7 b x = 18

5 c x = 5 d x = ¡11

2

e x = 15

11 f no solution

2 a x = 25

3 b x = 21 c x = 21

4 d x = 6

5

3 a x = ¡5

2 b x = ¡ 2

11 c no solution

d no solution e x = 9

8 f x = ¡9

4

4 a x = 12 b x = ¡36

5 c x = ¡16

5

d x =

13

7

e x = 16

f x =

40

7

Exercise 13E.3

1 a x < ¡2

3 b x > 9

5

c x 6 1

3 d x > ¡3

3 a z = b

ac b z =

a

d c z =

2d

3 d z = §

p 2a

e z = §p

bn f z = §p

m(a ¡ b)

4 a r =

r A

¼ b x = § 5

p aN c r = 3

r 3V

4¼

d x = 3q nD e x = §r y + 74

f Q = §p

P 2 ¡ R2

Exercise 13G.1

1 a 8 cm bp

15 cm cp

72 cm dp

8 m

ep

24 km f p

24 cm

2 a x =p 10

3 b x =

p 5

2 c x =

p 7

2

3 a x =p

48 b x =p

13 c x =p

3

d x =p

7 e x =p

3 f x = 1

4 a x = p 3 b x = p 5 c h = p 13

Exercise 13G.2

1 11:40 km 2 5:66 cm 3 by train 4 22:25 cm2

5 12 cm

Exercise 13H.1

1 ap

2 units b 6 units c 5 units d 4 units

eW _ x tO _ x

mathematics for queensland year 11b

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