mathematics chapter 2f5

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Mathematics F5 Chapter 2 :Graphs of Functions II

Chapter 2: Graphs of Functions II2.1 Basic skills

A. Insert brackets to x for each of the following.

x=1x

=

x2 = 1 x2

=

5x3 = 5 x3

=

-2

1x=

3x2 = 3 x2

=

-6x = 6 x =

B. Simplify the following using a calculator.

12(2)=4(1.5)=

5(-1)=

3(-0.5)=

-12(3)=-1(0.5)=

-3(-4)=

-8(-1.5)=

1(4)2=2(1.5)2=

6(-1)2=

4(-1.5)2=

C. Complete the following tables.

y =2x2 + 5x 1 y =x3 -12x + 10

x 1 3 4

y

y =x

6y = 2x -x2

x -3 -1 4

y

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x -1 -2 3

y

x y=2x2 + 5x 1=2(x)2 + 5(x) 1

1

3

4

x y= x3 -12x + 10= (x)3 -12(x) + 10

-1

-2

3

1


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2.2 Curve Sketching

Steps involved in curve sketching:a) Determine the shape of the graph

b) Find the y-intercept by substituting x = 0 into the given function

c) Find the x intercept by substituting y = 0 into the given function.d) Plot the y-intercept and x intercept on y-axis and x- axis.

e) Draw a straight line(using ruler) or a curve through the intercepts.

Act 1:Sketch the graph for each of the following functions

a) y = 2x - 6 b) y = 4+ x

x 0

y 0

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x 1 2.5 3

y

xy=

x

6=

)(

6

x

-3

-1

4

x y=2x -x2=2(x) (x)2

1

2.5

3

2


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c) y =x2 - 16 d)y = -x2 + 9 e)y =x3 - 8

f)y =x

8 g)y =

x

20

2.3 Solving Equation Graphically

A) Intersection of two graphs

Ex 1:

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x 0

y 0

3


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Ex 2:

Ex 3:Complete the tables below.

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B) Solving equation graphically

Ex 4:

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Ex: 5

Find the solution of the equationx2 - 6x + 3 = 7 by graphical method.

Solution:

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Mathematics F5 Chapter 2 :Graphs of Functions IIDraw the graphs of functionsy =x2 - 6x + 3 for -1 x 7 andy = 7 on the same axes.

y =x2 - 6x + 3

From the graph,x =________ and _________

Ex 6:

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x -1 0 1 2 3 4 5 6 7

y

7


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2.4 Region Representing Inequalities in Two Variables

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A Determining whether a given point satisfiesy = ax + b , y > ax + b ory < ax + b

Example 71.Determine whether the following points satisfy y = 2x -7 or y > 2x -7 or y < 2x -7

a) (4, 2) b) (-3, -13) c) (2, -6)

2. Determine whether each of the following points satisfies y = 2x -5 or y > 2x -5 or y < 2x -5a) (4, 2) b) (6, 7) c) (-3, 12)

B Position of a point relative to the graph of y = ax+ b

Mission 6 :To investigate the position of given point relative to the graph y= ax + b

1) Plot the graph ofy = 2x + 3 using Geometers Sketchpad below.

y

2) Based on the graph drawn, copy and complete the following table

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9

x


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3) From the activity, we can conclude that:

i)

ii)

iii)

C Identify the region satisfying satisfiesy = ax + b , y > ax + b ory < ax + b

Example 8

1.For each of the following graph, determine whether the shaded region satisfies

a) y >3x + 4 ory 2

1 x + 2 ory 3x 2

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Point y 2x + 3 y 2x + 3 Below the

line?

On the line

?

Above

the line?

(1,5)

(1,0)

(0,4)

(-4,1)

(3,9)

(-1,4)

(3,2)

(-1,1)

(0,-2)

10


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b) y < 3x - 2

D Shading the region representing the given inequality

Linear inequality in one variable

Type of line to be drawn on graph depends on the inequality symbol.Symbol Type of line

< or > Dashed line --------------

or Solid line

To shade the region representing an inequality, follow the steps below.a) Draw the graphy = ax + b

i) For y > ax + b ory < ax + b draw a dashed line fory = ax + b

ii) For y ax + b ory ax + b draw a dashed line fory = ax + b

b) Choose any point not on the line. You may use (0,0) .

c) Determine whether the point satisfies the given inequality.i) if yes, shade the region that contain the pointii) if not, shade the region on the other side of the line.

Example 9a) Draw the liney = 2x + 4 for domain -4 x 2. Then shade the region that satisfiesy > 2x + 4.

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11


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b) Draw the liney =2

5 x + 3 for the domain -4

x 2. Then shade the region that satisfies

y 2

5

x + 3

E Determining region which satisfies two or more simultaneous linear inequalities

Example 10

1. Shade the region that satisfies the inequalities

a)y


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c) x < y 3, y 5 x andx0 d)x + y 3, y < 3and x 2

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13


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Example 11

1. Write down the inequalities which satisfy the shaded region in each diagram.

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End of chapter

Example 9

Draw and shade each of the inequalities using y > ax + b ory < ax + b

a graph.

`

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2.5 Drawing graphs

5.2.2 Calculate the gradient of a straight line

Ex 3:Calculate the gradient of each of the following lines

Ex 4 : Find p, given that

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a) A(1, 0) , B (5,p) and the gradient of AB =4

3

(p=3)

b) G (p, 2), H (1,5) and the gradient of GH =2

1

(p=7)

Exercise 5.2A pg 122 textbook Question 2 and 4

5.2.3 Relating value of the gradient with steepness and direction of inclination of a straight line

y

A line that is inclined upward to the right has a positive gradient.

A line that is inclined downwards to the right has a negative gradient.

Ex 5 Ex 6

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xo

17


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a) i)

ii)b)

Exercise 5.2B pg 124-125 textbook Question 3 and 4

Ex 7: Ex 8:

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Line A B C D

Direction of

inclination

Gradient

18


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1.1. Intercepts

The intercept of a straight line is the x-coordinate of the point where the line cuts the x-axis

The intercept of a straight line is the y-coordinate of the point where the line cuts the y-axis

Thex-interceptof the line is 4

They-interceptof the line is -3

5.3.1 Determine the x-intercept and the y-intercept and the y-intercept of a straight line.

Ex 9:

State the x-intercept and y-intercept of the lines below.

Exercise 5.3A pg 127 textbook Question 1 and 2

5.3.2 Formula of gradient using intercepts

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a) y

(0,3)

x

(3,0)

b) y

(0,6)

x(-5,0)

c) y

(4,0) x

(0,-2)

x-intercept= y-intercept= x-intercept= y-intercept= x-intercept= y-intercept=

19

yl3l1Oxl2

LineGradient=x-intercepty-interceptl1l2l3


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Gradient of a line, m= )int

int(

erceptx

ercepty

Ex 10: Calculate the gradient of the following lines.

a) y

(2, 0) x

(0,-5)

b) y(0,9)

x(-12,0)

c)(0,4) , (6,0)

m = m =

Exercise 5.3A pg 127 textbook Question 3

5.3.3 Calculations involving gradient and intercepts

From gradient, m= )int

int(

erceptx

ercepty

Then, y-intercept= -(m x x-intercept)

x-intercept= -m

ercepty int

Ex 11 :

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a) Find thex-interceptor they-

intercept, given

i)gradient =2

5 ,x-intercept= 4

ii)gradient = 2, y-intercept =8

iii)gradient =-2, y-intercept=-6

iv) gradient =-4 x-intercept=-3

b)Calculatex-interceptory-

intercept, given the gradient,m. y

i)

4 m=5

2

O x

yii)

x -8 O

m= -2

1

c) y

A

B(5,2)

xC 0Based on the graph given,

determine

a) gradient of the straight lineBOC

b)y-interceptof the straight line

AB, given the gradient of AB is

-3

Exercise 5.3B pg 128 textbook Question 1,2 and 31.2. Equation of a Straight Line

5.4.1 Drawing the graph of equation y=mx + c

Ex 12:a)Draw a graph fory = x + 4 b)Draw a graph fory =-x 2 c)Draw agraph y = 3x - 6

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Steps to draw graph of equation y=mx + cOn your graph paper, draw x-axis and y-axis.

Spot two points.Try to use intercepts/ You may draw a table tofind the points.

Now, plot the points.

Join the points with a straight line


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Exercise 5.4A pg 132 textbook Question 1 b , d and f

5.4.2 Determine whether a point lies on a straight line

To determine whether apoint ( a, b ) lies on the liney = mx + c :

Substitutex-coordinate and y-coordinate of the point into the equation

If the equation is satisfied, LHS = RHS, then the point lies on the line.If the equation is not satisfied, LHSRHS, then the point does not lie on the line.

Ex 13:Determine whether the given point lies on liney = 3x 2

a)P (1,1) b)Q(-2,-3) c)R(-4,-14)

Exercise 5.4A pg 132 textbook Question 2 a, c and e

5.4.3 Writing the equation of a straight line

Substitute the given gradient and y-intercept into the equation y = mx + cThe equation of a straight line can be written in the form of:

y = mx + c

gradient y-intercept

Ex 14:

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d)6x + 3y = -2c) 4x + 3y = 12

Mathematics F5 Chapter 2 :Graphs of Functions IIWrite equation of the straight line, given

a) m = 3, c=2 b) m=-3, c=-4 c) m=2

1 , c = -2 d) m=2,

c=0

Ex 15:Write the equation of the straight line that passes through the given pairs of pointsbelow.a) (0,3) and (2, 9) b)(2,7) and (4, -3) c) y

(0,6)

X(2,0)

Exercise 5.4A pg 132 textbook Question 3 a, b and c , Question 4

5.4.4 Determining the gradient and y-intercept of a straight line The equation of a straight line can be written in the form of

y = mx + c, m is the gradient and c is the y-intercept

ax+ by= c a, b and c are constants

When the equation is written in the form of ax + by= c, change it to the form of y= mx + c , then identify the gradient and y-intercept.

Ex 16:Find the gradient and y-intercept of each of the following straight lines.a)y=1-4x b)y=2x + 6

Exercise 5.4B pg 135 textbook Question 1 a, d, g and j

5.4.5 Finding the equation of a straight linea)Equation that is parallel to thex-axis,For a line parallel to the x-axis where the y-interceptis k, the equation is y =k.

y

M N

b)Equation that is paralel to y-axisFor a line parallel to the y-axis where the y-interceptis h, the equation is y =h.

K y R

c)Equation that passes through agiven point and has a specific

gradient to find c, substitute the

coordinate of given point into

equation .

Write the equation as y=mx + c

Ex 17: y

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3

x

PQ

-5

i) y-intercept of MN is 3,

equation MN is y=3

ii)y-intercept of PQ is -5,

equation PQ is y=-5

x

-2 8

L S

i) y-intercept of KL is -2,

equation KL is x=-2

ii)y-intercept of RS is 8,

equation RS is x=8

Q

x 2

P

In the diagram, the gradient of PQ is

2

3. Find the equation of PQ.

d) Equation that passes through two given points

Find the gradient of the straight lineby using the two given points

To find c,Substitute coordinates of anyone of the two points and the gradientinto y=mx + c

Write the equation in y=mx + c

Ex 18:Find the equation of a straight line whichpasses through (4,-1) and (6,-9).

Ex 19:Find the equation of the line which passes

through (-4, 7 ) and gradient = -2

Ex 20: Find the equation of the line which passes

through (-1,3) and (-3,9)

Exercise 5.4B pg 135 textbook Question3 a, b; 4a, b; 5d, f5.4.6 The point of intersection of two straight lines

y point of The point of intersection of two straight lines can beintersection determined by:

a) drawing the two lines on the same axesb) solving simultaneous equations.

x

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Ex:21

Find the point of intersection of the given two straight lines by solving the simultaneous equations.

a) y = 2x + 1x + y = 7

Solution:

b) 2y = 4x- 7, 4y = 12x = -9

Solution:

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x

y

x

y

x

y

x

y


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Exercise 5.4C pg 137 textbook Question 2 a, c and e1.3. Parallel Lines

5.5.1 Verifying two parallel lines5.5.2 Determining whether two straight lines are parallel

Two parallel lines have the same gradient same astwo lines that have the same gradient are parallel.

To determine whether two straight lines are parallel,

Arrange the two given equations, if necessary, in the formy=mx + c

If both m are same, conclude that the two straight lines are parallel.

If both m are not same, conclude that the two straight lines are not parallel.

Ex 22:

Determine whether the given lines are parallel.

a)y=3x + 5

y=3x 10

b) 3x-4y =12

8y-6x = 5

c)2y = 6x + 7

y + 5x = 8

Ex 23:Find the value ofm if the lines given are parallel.

a) 3x +y = 6 is parallel tox + my = 12 b) 2x +y = 5 is parallel toy = mx + 8

Exercise 5.5A pg 140 textbook Question 2a, c and 3 a, c

5.5.3 Finding the equation of the parallel lineTo find the equation of a straight line,lwhich passes through a point (h,k) and is parallel to theother line,

Find the gradient of the other parallel line

Substitute (h,k) and gradient intoy = mx + c and find c.

Write the equation of the line lasy = mx + c withgradientand c as step above .

Ex 24:Find the equation of the straight line which passes through

a) (1, 4) and is parallel to the line b)(0,-3) and is parallel to the line c)(3, 4) and is parallel to the line

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y= 3x -11 2x +y 5 =0 4y 2x = 3

Exercise 5.5A pg 140 textbook Question 4a and b5.5.4 Solving problems

Ex 25 : Ex 26 :

Ex 27 :

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Exercise 5.5B pg 143 textbook Question 1 and 3Assessment Corner :Objective questions 1 to 5

Subjective questions 3 and 4

End of chapter 5

mathematics chapter 2f5

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