A guide to the Math Placement Test

© Department of Mathematical Sciences

George Mason University

Revised November 2015

Contents Introduction ............................................................................................ 1 Courses Requiring the Placement Test ................................................ 2 Policies .................................................................................................... 2 Test Format ............................................................................................. 3 Learning Resources ................................................................................ 4 BasicAlgebra ............................................................................................ 5 Fractions .................................................................................... 6 Decimal Notation ....................................................................... 7 Percents ...................................................................................... 8 Ratios…. ..................................................................................... 10 Algebraic Expressions ............................................................... 11 Linear Equations in One Variable ............................................. 11 Linear Equations in Two Variables ........................................... 12 Graph/Slope of a line ................................................................. 13 Algebra I .................................................................................................. 14 Linear Inequalities in 1 Variable ............................................... 14 Absolute Value .......................................................................... 14 Systems of Linear Equations ..................................................... 15 Linear Inequalities in 2 Variables .............................................. 16 Multiplying Algebraic Expressions ........................................... 16 Factoring Algebraic Expressions .............................................. 17 Rational Expressions ................................................................. 17 More on Exponents .................................................................... 18 Radicals ..................................................................................... 19 Quadratic Equations .................................................................. 19 Functions Notation ..................................................................... 20 Equations with Rational Expressions ........................................ 21 Algebra II……………………………………………………...……… 22 More on Functions ..................................................................... 22 Graphs of Functions ................................................................... 24 Transcendentals ...................................................................................... 25

Exponential Functions ................................................................ 25 Logarithmic Functions ............................................................... 26

Angle Measure ........................................................................... 26 Right Triangle and Circular Functions Definitions ................... 27 Graphs of Trigonometric Functions .......................................... 28

Inverse Trigonometric Functions .............................................. 29 Trigonometric Formulas and Identities ..................................... 30 Answers ................................................................................................... 31

1

Dear Student, The Math Placement Test is quite likely the most important exam that you will take as you start your studies at George Mason University. Most majors require a mathematics course. Depending on which mathematics course you intend to take, the placement test will measure your preparedness for that course. Failure to earn the appropriate test score for an intended course will preclude you from registering for that course. Appropriate courses that will prepare you for your intended course will be suggested, so please make all efforts to take this test early and avoid needless frustrations later. This booklet has been prepared by Ellen O’Brien, Director of the Mathematics Learning Center, to give you detailed information concerning the test. The Learning Center supports your learning of mathematics in an ancillary way to the regular classroom lectures. It provides, for example, self paced or instructor lead tutorials for remediation in mathematics. It offers free, limited tutoring for all undergraduates enrolled in mathematics courses. These are staffed by our upper division math undergraduates and graduate students. The booklet also contains sample problems with solutions and a quick review of basic mathematical concepts to help you prepare for the Placement Test. Of course, the best way to prepare is to review the mathematics that you have learned until now. Please take this part of your preparation and the exam itself seriously. I invite you to visit our math website to view services, courses and programs that the department offers. Finally, I wish you a successful and rewarding academic career at George Mason University. Regards, David Walnut Professor and Chair Department of Mathematical Sciences

2

Courses that Require the Placement Test The Math Placement Test is given by the Department of Mathematical Sciences to determine the readiness for the following courses:

• Math 104 or Math 105 Pre-Calculus • Math 110 Introductory Probability • Math 108 Introductory Calculus with Business Applications • Math 113 Analytic Geometry and Calculus I • Math 123 Calculus Algebra/Trigonometry A • Math 125 Discrete Mathematics • CS 112 Computer Science I

Students should talk to an academic advisor to determine which Math course(s) they are required to take. All students admitted to the university are advised to take the Math Placement Test during the orientation process. Policies

For current policies and required minimum scores please visit the webpage: http://math.gmu.edu/placement_test.php

3

Test Format The Math Placement Test is a computer-based test in a multiple choice format. You may need to do some calculations on paper in order to determine the best answer. There is a time limit for completing each section of the test.

The test is divided into four parts: • Basic Algebra • Algebra I • Algebra II • Transcendentals

Students who are attempting to place into Math 110 must take the Basic Algebra section. All other students must start with the Algebra I section. The sections of the test are sequential. For example, students must pass Algebra I in order to proceed to the Algebra II section. If you are unsure of which math course you are required to take, please consult your academic advisor.

4

Learning Resources The Math Learning Center offers two programs designed to prepare students for their required math courses. The Basic Math Program refreshes skills needed for success in Math 110. The Algebra Program prepares students for Math 105, Math 108 or Math 125. Both are self-paced programs offered in an online format. http://math.gmu.edu/math-learning-center.php The center also manages the Undergraduate Mathematics Tutoring Center which specializes in general help for freshman/sophomore math classes. It is staffed by upper division mathematics majors and graduate students. http://math.gmu.edu/tutor-center.php Both the Mathematics Tutoring Center and the Math Learning Center are located in Johnson Center Room 344. Hours are posted on the webpage above. We are closed on holidays. More information about the Math Placement Test is available at http://math.gmu.edu/placement_test.php

5

Basic Algebra The Basic Algebra section of the test measures your ability to perform basic arithmetic and Algebraic operations and to solve problems that involve these operations. The following exercises and examples provide a sample of the type of material that is tested in the Basic Algebra section. Only those students attempting to place into Math 110 will be given questions from the Basic Algebra section of the test. Basic Operations Example: John’s car insurance premium is $1515.00. He has the option of making four equal payments but will be charged an additional $25 processing fee. If he chooses the four-payment option, what is the amount of each payment?

Solution: The processing fee is added to the premium

00.1540$

00.2500.1515$

+

This amount is then divided by 4 38515404

Each payment will be $385.00

Exercise 1: A shipment of 1344 CD’s is to be packed into cartons containing 24 CD’s each. How many cartons are necessary to ship these CD’s? Exercise 2: A loan of $10,020.00 is to be paid off in 60 equal monthly payments. How much is each payment (excluding interest)? Exercise 3: Susan’s Cell phone provider charges her $35.00 per month for the first 300 minutes of calls and $0.70 for each minute over 300. If she uses 340 minutes of calls in a given month, how much will she be charged?

6

Fractions

A fraction or rational number is of the form qp where p and q are integers, and

0≠q p is called the numerator and q is called the denominator. The fraction is said to be reduced to lowest terms if p and q have no factors (other than 1) in common. The same number can be expressed as a fraction in many ways. These fractions

are called equivalent fractions: 41 ,

123 ,

205 ,

4010 ,

4812

Only the first fraction is in reduced form. ► To change from a mixed number to an improper fraction, multiply the whole number by the denominator of the fraction and add the result to the numerator of the fraction. This gives the numerator of the improper fraction.

Example: 35

321 = ( ( ) 2315 +×= )

843

835 = ( ( ) 38543 +×= )

► The product of two fractions dc

ba× is

bdac . That is, multiply numerators to

get the numerator of the product and multiply denominators to get the denominator of the product.

Examples: 998

94

112

=×

512

52

16

526 =×=×

► The quotient of two fraction dc

ba÷ is equal to

cd

ba× or

cbad

Example: 49 ounces of a solution is to be poured into test tubes with capacity

213 ounces. How many test tubes will be filled?

Solution: 147249

2749

21349 =×=÷=÷ 14 test tubes will be filled.

7

Exercise 4: Perform the indicated operation and give the result in reduced form.

a. 51

31÷ b. 6

83÷ c.

432

521 × d.

85

326 ÷

► The sum of two fractions with a common denominator is cba

cb

ca +

=+

If fractions do not have the same denominator then they must be changed to an equivalent form.

Example: Find the sum 85

311 +

Change each fraction to an equivalent form

2432

88

34

311 =⋅=

2415

33

85

85

=⋅=

Then add the equivalent fractions to get 12447

2415

2432

85

31

=+=+ or 24231

Exercise 5: Perform the indicated operation and give the result in reduced form

a. 87

416 + b.

43

107+ c.

61

85− d.

127

412 −

Decimal Notation Rational numbers or fractions can also be expressed in decimal notation.

100137 can be written as 7.13

.125 is equivalent to 1000125 or

81

► When adding or subtracting numbers in decimal notation, make sure to align the decimal point. Example: Find the sum 78.31.23 +

8

Solution:

88.26

78.31.23

+

► When multiplying two numbers in decimal notation, first multiply the numbers disregarding the decimal point(s). Next put the decimal point in the product. The number of places to the right of the decimal point in the product should be the SUM of the places to the right of the decimal point in each of the factors. Example: Find the product 1.235.6 × Solution:

13335

12700635

1.235.6

×

The product should have three places to the right of the decimal point. The result will be 13.335 Example: Carlos is getting paid $10.80 per hour and time and a half for hours over 40 per week. If he works 43.5 hours this week, how much will he earn? Solution: Carlos is paid 20.16$5.180.10$ =× per hour of overtime He will earn: 70.488)20.16$5.3()80.10$40( =×+× Percents

The symbol % is used to indicate the fraction 1001 or the ratio: 1001 to

23% is equivalent to 23.10023 or

100% is equivalent to 00.1100100 or

Exercise 6: Change the decimal to percent

a. .25 b. .085 c. 1.52

9

Exercise 7: Change the percent to a decimal a. 2% b. %

215 c. 0.3%

Example: A class contains 12 female and 8 male students. What percent of the class is female? Solution: There are a total of 20 students in the class.

12 out of 20 or 2012 are female

%6010060

2012

== of the class is female

Exercise 8: Find the following

a. 30% of 75 b. 4% of 196 c. 120% of 300 Example: Suppose 60% of a class is female. Of these females, 20% are freshman. What percent of the class is female upper class students? Solution: If 20% of the females are freshman, then 80% of females are upper class students. 80% of 60%= 48.60.80. =× so 48% of the class are female upper class students Example: Suppose we know that 40% of a class or 16 students are male. How many students are in the class? Solution: We know 40% of (the number) is 16 That is ( ) 1640. =× numberthe

So ( ) 4040.16

==numberthe students

Example: Before graduating, Joe worked in a convenience store and earned $5.85/hr. He started a new job that is paying him $18.95/hr. What is the percent increase in his hourly rate?

Solution: originalincreaseactualincrease =% so

85.585.595.18 − or 223%

10

Ratio

The ratio of two numbers a and b is written a to b, a:b or ba

We can think of this as a fraction in order to simplify it. The ratio 10:5 is equivalent to 2:1.

The ratio 32.5 to 5 is equivalent to 65 to 10 since 1065

22

55.32

55.32

=⋅=

► To show two ratios are equal use the following:

DC

BA= if and only if BCAD =

Example: Show that 12.5 : 5 is equivalent to the ratio 5:2

Solution: 25

55.12= if and only if 5525.12 ×=×

or 25 = 25 Example: How much water should be mixed with 7 cups of salt to make a 12% saline solution?

Solution: We have the proportion 100127

=x

⎟⎠

⎞⎜⎝

⎛totalpart

Multiply to get 70012 =x

3158=x

So we need 31517

3158 =− cups of water.

Exponents ►When working with algebraic expressions involving exponents, the following properties of exponents are used:

1. baba xxx += 4. ( ) aaa yxxy =

2. bab

a

xxx −= 5.

a

aa

yx

yx

=⎟⎟⎠

⎞⎜⎜⎝

⎛

3. ( ) abba xx = 6. 10 =x for any 0≠x

11

Exercise 9: Simplify each of the following expressions using the rules of exponents

a. ( )243x b. 5

10

24aba c.

( )322

25

2516

xyxyx

Exercise 10: Find the product ( )26073 125 zxyzyx Algebraic Expressions Most of Algebra involves manipulating expressions that contain variables. Example: Evaluate the expression using the indicated value of the variable

( )( ) 2513 −=+− kforkk Solution: ( )( ) ( )( )52123 +−−− substitute -2 in the expression for k ( )( )5216 +−−−

( )( )37− = -21 Exercise 11: Evaluate the expression using the indicated value of the variable

a. 2362 2 −=+− xforxx b. ( )

414

15=

+

+xfor

xx

Exercise 12: Combine the terms below and simplify the result. a. ( ) 2834 −++− xx

b. ( )baba 34216 ++−

c. 1013

54 +

−xx

Linear Equations in One Variable A linear equation in one variable can have a unique solution, no solution or infinitely many solutions. When it has no solution, the equation is called inconsistent.

12

Example: Solve the following equations by isolating the variable. 214152 +−=− xx

Solution: xxxx 42144152 ++−=+− Add 4x to both sides 21156 =−x 152115156 +=+−x Add 15 to both sides 366 =x

36616

61

⎟⎠

⎞⎜⎝

⎛=⎟⎠

⎞⎜⎝

⎛ x Multiply both sides by 61

6=x The equation has a unique solution Example: Solve the equation ( )34416 −= xx Solution: 121616 −= xx Distribute the 4 xxxx 1612161616 −−=− Add -16x to both sides

120 −= The equation is inconsistent. Example: Solve the equation ( ) ( ) 510625310 ++−=−+ xxxx Solution:

5201215310 ++−=−+ xxxx Clear the parentheses by distributing

15131513 −=− xx Combine like terms 0 = 0

Since the last statement is always true, every real number will be a solution.

Linear Equations in Two Variables A solution for an equation in two variables is a replacement for each of the variables so that when substituted, the resulting statement is true. We generally write the solution as an ordered pair. For example, ( )5,2 is a solution of the equation 113 =+ yx . When the variable x is replaced by 2 in the equation and the variable y is replaced by 5, the resulting statement is true. The linear equation in two variables has an infinite number of solutions. The graph of these solutions is a line in the x-y plane.

13

Example: The formula in two variables gives the conversion between Fahrenheit and Celsius temperatures. ( )FytoCx !!

3259

+= xy

Find the temperature in degrees Celsius that corresponds to F!212

Solution: 3259212 += x Substitute the value 212 for y

x59180 = Add -32 to both sides

x=100 Multiply both sides by 95

A solution of the equation is ( )212,100 The Graph of a Line We can graph the solutions of any linear equation on a set of axes called the Euclidean Plane. In the case of the linear equation, the solutions form a line.\We can measure how much and in which direction the line tips using the slope. The slope-intercept form of the equation is : bmxy += If the equation is in this form, the line has slope m and y-intercept (0, b) The y-intercept is the point where the line intersects the y-axis. Example: In the graph below, find the x and y intercepts and the slope.

Example: Find the slope and y-intercept of the line with equation: 25105 =− yx

Solution: Write the equation in slope-intercept form: 25

21

−= xy

The slope is 21 and the y-intercept ⎟

⎠

⎞⎜⎝

⎛−25,0

Parallel lines have identical slopes. Lines are perpendicular if their slopes have

the relationship 2

11m

m −=

Solution: x intercept: (-12, 0) y intercept (0, 6) Slope is 1/2

14

Example: Find the slope of a line that is perpendicular to the line 25105 =− yx

Solution: The line 25105 =− yx has slope 21 .

Any line perpendicular to it will have slope m= - 2. Algebra I The Algebra I section of the test measures your ability to perform basic algebraic operations and to solve problems that involve algebraic concepts. Only those students attempting to place into Math 108, Math 112, Math 125, Math 113, CS 112 will be given questions from the Algebra I section. In addition to all material in the Basic Algebra, the following exercises and examples provide a sample of the type of material that is tested in the Algebra I section. Linear Inequalities in One Variable Unlike linear equations, linear inequalities have an infinite number of solutions. We can use a number line or interval notation to indicate the solutions. Any real number can be added to or subtracted from both sides of the inequality. We can multiply or divide both sides by any nonzero real number but, if we multiply or divide by a negative number, the inequality sign must be reversed.

Example: Solve the inequality 3≤ −95x + 2

Subtract 2 from both sides: 1≤ −95x

Multiply both sides by -5/9: −59≥ x

The solution is the interval: ⎥⎦

⎤⎢⎣

⎡ −∞−95,

Absolute Value Equations and Inequalities The absolute value of a number is defined to be its distance from 0 on the number line.

⎩⎨⎧

<−

≥=

00xforx

xforxx

Example: Solve the equation for x 75315 =− x

Solution: 75315 =− x or 75315 −=− x 20−=x or 30=x

15

Example: Solve the inequality 1552 ≥+x

Solution: 1552 −≤+x or 1552 ≥+x 20−≤x or 5≥x Distance Formula ► The distance between two points ( )11, yx and ( )22, yx is given by

the formula: ( ) ( )2122

12 yyxxd −+−= The formula uses the Pythagorean Theorem Exercise 1: Find the distance between the points ( )5,2− and ( )8,1 Systems of Linear Equations We sometimes group equations together and investigate whether they have a common solution. This is called a system of equations. The system of linear equations has a solution if there is an ordered pair ( )yx, that satisfies each equation in the system. A system of linear equations can have exactly one solution, no solution or infinitely many solutions. There are several methods available to solve systems of linear equations. The next example illustrates the Elimination Method.

Example: Solve the system ⎩⎨⎧

−=−

=+

24894823

yxyx

Solution: 4823 =+ yx multiply by -3: 14469 −=−− yx 2489 −=− yx 2489 −=− yx Add: 16814 −=− y Solve for y: 12=y The value 12 is substituted for y in either of the original equations. ( ) 481223 =+x Solve the resulting equation for x 48243 =+x 243 =x 8=x The solution to the system is (8, 12) Exercise 2: Solve the systems

a. ⎩⎨⎧

−=+−

=−

102553yx

yx b. ⎩⎨⎧

=−

−=−

102623

yxyx

16

Linear Inequalities in Two Variables Linear inequalities in two variables can be solved by graphing. The solution is an infinite set of ordered pairs in the plane. We can indicate the solution by shading the region that contains these ordered pairs. Example: Solve the inequality 5y ≥ 3x −15 Solution: Graph the line that results from replacing the inequality sign with an equal sign. Here, graph the line 5y = 3x −15 This line partitions the plane into three sets: The solutions of 5y > 3x −15 5y < 3x −15 5y = 3x −15 In the inequality, test an ordered pair from the plane that does not lie on the line. Using (0,0), we get: 5(0) ≥ 3(0)−15 which is a true statement. Shade the half-plane of points that satisfy the inequality. See the graph below.

In a similar way we can solve a system of inequalities. The graph below shows

the solutions to the system 5y ≥ 3x −15y ≤ 0x ≥ 0

$

%&

'&

Multiplying Algebraic Expressions We use the Distributive Law to multiply algebraic expressions. Example: Find the product ( )1026 −xx

Solution: ( )1026 −xx = ( ) ( )10626 −+ xxx = xx 6012 2 −

17

To multiply 2 binomial factors, we apply the Distributive Law twice and combine like terms. Example: Find the product ( )( )127 −+ xx

Solution: ( )( )127 −+ xx = ( ) ( )12712 −+− xxx

= ( )172712 −⋅+⋅+⋅−⋅ xxxx

= 7142 2 −+− xxx = 7132 2 −+ xx The result of this multiplication is called a quadratic expression. We “undo” the multiplication in a process called factoring. Factoring Algebraic Expressions The next two examples show the factoring process. In the first, we factor out the highest common factor among the terms. In the second, we write the quadratic expression as a product of 2 binomial factors. Example: Factor the expression xx 2416 2 + Solution:

xx 2416 2 + = ( )2416 +xx The terms have the factor x in common

= ( )328 +xx They also have a factor of 8 in common

Example: Factor the expression 40182 −+ tt Solution: 40182 −+ tt = ( )( )220 −+ tt Special Factorings: ))((22 bababa −+=−

))(( 2233 babababa ++−=−

))(( 2233 babababa +−+=+ Rational Expressions Rational expressions are fractions where the numerator and/or denominator involve variables. We can reduce, add, subtract, multiply and divide these rational expressions using the rules of fractions. You may want to review the rules of fractions in the Basic Algebra section on page 6.

18

Example: Reduce the rational expression 103

7352 −−

−

xxx

Solution: In order to reduce the expression we must factor both the numerator and the denominator, then cancel any factors common to both.

103

7352 −−

−

xxx = ( )

( )( )5257

−+

−

xxx

= ( )( )( )52

57−+

−−

xxx Factor the numerator

=( )27+

−

x Cancel the common factor

Example: Find the sum 31

934

2 −

+−

−

+

xx

xx

Solution:

31

934

2 −

+−

−

+

xx

xx =

( )( )( )( )( )( )33

1333

34−+

++−

−+

+

xxxx

xxx Use a common denominator

= ( )( )( )33

3434 2

−+

++−+

xxxxx

= 92

2

−

−

xx

Example: Simplify the expression

ba

x

+

15 2

Solution: Using the rules of dividing fractions we get:

( )baxbax +=+

⋅ 22 51

5

More Exponents ►The definition of exponents is extended to include all rational numbers:

a

a

xx 1

=− b aba

xx =

Exercise 3 : If x=

32 find the value of x-3

Exercise 4: Find the value of 21

32

927 ⋅

19

Radicals When simplifying algebraic expressions involving radicals, the following properties of radicals are used:

1. nnn xyyx =⋅

2. n

n

nyx

yx= where 0≠y

3. ( )nmm n xx = Example: Find the product ( ) ( )5156 −⋅+

Solution: ( ) ( )5156 −⋅+ = 55515616 −⋅+⋅−⋅

= 551− Exercise 5: Find the sum: 7328 + Exercise 6: Simplify each of the following expressions

a. 568 ba b. 2

37

728uvvu c. 3 1558 yx

Exercise 7: Rationalize the denominator to simplify: 23

22+

Quadratic Equations An equation of the form 02 =++ cbxax is called a quadratic equation. The equation will have two, one or no real number solutions. There are several methods available to find the solutions of a quadratic equation. The following examples illustrate these methods. Example: Solve the equation 202 2 =x Solution: 202 2 =x 102 =x Isolate the term involving the radical

10=x Take the square root of both side of the equation

20

Example: Solve the equation 25442 =++ xx Solution: 25442 =++ xx 02142 =−+ xx bring all terms to one side ( )( ) 037 =−+ xx factor the left side 07 =+x or 03 =−x set each factor equal to zero 7−=x or 3=x the equation has 2 solutions Example: Solve the equation 0169 2 =++ xx

Solution: In the quadratic formula: a

acbbx2

42 −±−= we use

9=a , 6=b and 1=c

92

19466 2

⋅

⋅⋅−±−=x =

186− since 042 =− acb

we get only one solution 31

−=x to the equation.

►The expression acb 42 − is called the discriminant. It tells us how many solutions a quadratic equation will have. If 042 >− acb the equation has 2 real number solution If 042 =− acb the equation has 1 real number solution If 042 <− acb the equation has no real number solution Exercise 8: Solve the equations a. 0494 2 =−x

b. 0592 2 =−+ xx c. 0642 =−+ xx

Function Notation Functions can be defined using a list of ordered pairs, a graph, a table, or an equation. The following equation defines a function: ( ) 12 −= xxf . Each value of x is paired with the value ( )xf .

( ) 8133 2 =−=f so the pair ( )8,3 is in the function f.

( ) 9911010 2 =−=f so the pair ( )99,10 is in the function f.

( ) ( ) 9911010 2 =−−=−f so the pair ( )99,10− is in the function f.

21

Example: For the function ( )312

+

−=xxxf find ( )2−f and ( )1+bf

Solution: ( ) 3

13

321)2(2

2

==+−

−−=−f

( )

42

3)1(1)1(1

22

+

+=

++

−+=+

bbb

bbbf

Complex Rational Expression A complex rational expression is one in which a rational expression appears in the numerator and/or denominator. Simplifying the complex rational expression is equivalent to finding the quotient of the expression in the numerator with the expression in the denominator.

Example: Simplify the complex rational expression

yx

xy11

1

−

Solution: Combine the terms in the denominator first.

xyxy

yx−

=−11

Next, find the quotient of the numerator with the denominator

xyxy

xy−

÷1 =

xyxy

xy −⋅

1

= xy −

1

Equations with Rational Expressions Given an equation involving rational expressions we can find an equivalent equation with no rational expressions. To find the equivalent equation we multiply the entire equation by the least common denominator. After solving the equivalent equation, we must check the solution in the original to see if it makes any denominator equal to zero. If so, we discard that solution.

Example: Solve the equation xx 23

511+=

Solution: The LCD of the rational expressions is 10x

x

xxx

x2310

5110110 ⋅+⋅=⋅ Multiply equation by 10x

22

15210 += x Solve the resulting equation 52 =− x

25

−=x

Example: Solve the equation xx

=−31

34

Solution: The LCD of the rational expressions is 3x

xxxx

x ⋅=⋅−⋅ 3313

343 Multiply equation by 3x

234 xx =− Solve the resulting equation 043 2 =−+ xx Use the quadratic formula

32

)4(3411 2

⋅

−⋅⋅−±−=x

32491

⋅

±−=x

341 −= andx

Exercise 9: Solve the equation 32

56132

32

2 −+

−=

−

+−

+ xxx

xx

x

Algebra II The Algebra II section of the test measures your ability to solve problems that involve more advanced Algebra than the previous section. Only those students attempting to place into Math 113 and CS 112 will be given questions from the Algebra II section. The following exercises and examples provide a sample of the type of material that is tested in the Algebra II section. More on Functions A function is a set of ordered pairs where each first component is paired with exactly one second component. The set of first components is called the domain and the set of second components is called the range. Unless otherwise stated, the domain of the function ( ) 12 −= xxf , is all real numbers. Any real number a can be paired with another real number using the equation ( ) 12 −= xxf . The pair can be written ( )1, 2 −aa .

23

Suppose we have the function ( )t

tg 3= . We say ( )tg is undefined at 0=t

since the expression 03 is undefined. On the other hand, ( )tg is defined at any

real number other than 0. The domain of ( )tg is all real numbers except 0=t .

Suppose we have the function ( ) 5+= xxf . In order for the expression

5+x to represent a real number, the expression 5+x must not be negative. For this function the domain is all real numbers 5−≥x . ►The average rate of change for a function f(x) is the change in f(x) over the change in x.

xxfchangeofrateaverage

Δ

Δ=

)(

Alternatively, it is the slope of the secant line passing through the two points (x1, f(x1)) and (x2, f(x2))

Exercise 1: Let ( )62

1+

=x

xf ; Find each of the following:

a. the domain of ( )xf b. the range of f(x) c. the average rate of change from x=-1 to x=5

Exercise 2: Let ( )x

xg−

=31 ; Find each of the following:

a. the domain of ( )xg b. the range of g(x) c. the average rate of change from x=-2 to x=2 ►We can compose two functions by applying one right after the other. Example: For the functions f and g defined in the exercises 1 and 2, find a. ( )( )6−gf b. ( )( )2−fg

Solution: ( )( )6−gf =⎟⎟⎠

⎞⎜⎜⎝

⎛

−− )6(31f = ⎟

⎠

⎞⎜⎝

⎛31f =

203

( )( )2−fg = ⎟⎟⎠

⎞⎜⎜⎝

⎛

+−⋅ 6)2(21g = ⎟

⎠

⎞⎜⎝

⎛21g =

2/51 or

52

24

Example: The function values of f and g are given in the table below. Use them to find a. ( )( )0gf b. ( )( )3fg

x f(x) g(x) 0 -3 3 1 0 2 2 -2 1 3 1 0

Solution: a. ( )( )0gf = ( )3f =1 b. ( )( )3fg = ( )1g =2 ►Some functions have an inverse. The inverse of a function will interchange the real numbers in the ordered pair. If ( )ba, is in the function f then ( )ab, is in

1−f , the inverse of f . Keep in mind, that not all functions have inverses.

The functions ( ) 53 += xxf and ( )351 −

=− xxf are inverses.

Notice f contains the pair (3,14) and f-1 contains the pair (14, 3). This is true for all ordered pairs of the function f. Graphs of Functions We can graph the set of ordered pairs of a function on a set of axes just as we do with linear equations in 2 variables. The graph of a quadratic function cbxaxxf ++= 2)( is a parabola.

The vertex of the parabola is located at ⎟⎠

⎞⎜⎝

⎛ −− )2(,

2 abf

ab

This point on the graph is either a minimum or a maximum for the function. The leading coefficient a, determines which. For a<0, the function takes on a

maximum at abx2−

= and the range of the function is ⎭⎬⎫

⎩⎨⎧

⎟⎠

⎞⎜⎝

⎛ −≤

abfyy2

.

For a>0, the function takes on a minimum at abx2−

= and the range of the

function will be ⎭⎬⎫

⎩⎨⎧

⎟⎠

⎞⎜⎝

⎛ −≥

abfyy2

.

25

Example: In the graph of 74)( 2 +−= xxxf below, find the x and y intercepts and the vertex.

► To graph the function baxfcy +−⋅= )( we shift the graph of f(x) right a units (for a>0), left a units for a<0) up b units for b>0, down b units for b<0 vertically stretch the graph of f(x) by a factor of c, c>1 vertically shrink the graph of f(x) by a factor of c, for 0<c<1 Example: The graph above can be viewed as a shift of 2xy =

(2 units right and 3 units up). The function can be written in the form: 3)2()( 2 +−= xxf

Transcendentals The test measures your ability to work with exponential, logarithmic and trigonometric functions of the right triangle and the unit circle. The graphs of these functions and some common trigonometric formulas are given in the next few pages. Exponential Functions An exponential function is one in which the variable occurs in the exponent. The base determines whether the function is increasing or decreasing.

1)( >= aforaxf x 1)( <= aforaxf x Domain: all real numbers Domain: all real numbers Range: y>0 Range: y>0

Solution: y intercept: (0, 7) x intercept: none vertex: (2, 3) note: (2, 3) is the minimum of f(x)

26

Logarithmic Functions A logarithmic function is the inverse of an exponential function.

Example: Evaluate: a. 8log2 b. 1log5 c. 001.log10

Solution: a. 38log2 = since 823 = b. 01log5 = since 150 =

c. 3001.log10 −= since 001.10 3 =−

Example: Solve the equation for x: 103 =+xe Solution: Isolate x by taking the log of both sides. Here use ln the natural log. 10lnln 3 =+xe 10ln3 =+x 10ln3+−=x 6974.−≈

Example: Solve the equation for x: 6)13(log4 =+x Solution: Use the inverse function to isolate x

6)13(log 44 4 =+x 6413 =+x

4083=x Angle Measure An angle can be measured in degrees or in radians. A full revolution is !360 or π2 radians. We use this information to convert from one measure to the other.

The following proportion shows the relationship between x degrees and y

radians. π2360

=yx

Example: Find the degree measure of an angle whose radian measure is 32π

Solution: ( ) ππ 2

3603/2=

x

( )3/23602 ππ =x so 120=x

This is the natural log function, (the inverse of y=ex) f(x)= ln x Domain: x>0 Range: all real numbers

27

Right Triangle Definitions

hypopp

=θsin opphyp

=θcsc

hypadj

=θcos adjhyp

=θsec

adjopp

=θtan oppadj

=θcot

Definitions of Circular Functions

Special Angles Degree Radian sin θ cos θ tan θ csc θ sec θ cot θ

0 0 0 1 0 -- 1 -- 30 π/6 2/1 2/3 3/3

2 3/32

3

45 π/4 2/2

2/2

1 2 2 1

60 π/3 2/3

2/1 3 3/32

2 3/3

90 π/2 1 0 -- 1 -- 0 Sum and Difference Formulas

( ) ( ) ( ) ( ) ( )vuvuvu sincoscossinsin ±=±

( ) ( ) ( ) ( ) ( )vuvuvu sinsincoscoscos ±=±

28

Graphs of the Trigonometric Functions f(x)=sin(x) period: 2π

f(x)=cos(x) period: 2π

f(x)=tan(x) period: π

Exercise 1: Find the period of

a. ( )x2sin b. ⎟⎠

⎞⎜⎝

⎛4

cos x c. ⎟⎠

⎞⎜⎝

⎛3

tan x

Exercise 2: Which has the shortest period?

a. xy cos= b. xy cos21

= c. )2cos( xy = d. xy21cos=

29

Inverse Trigonometric Functions The inverse sine function , written )(sin 1 x− or )arcsin(x , has

domain [-1,1] and range ⎥⎦

⎤⎢⎣

⎡−

2,

2ππ . It is defined to be the number in the

interval ⎥⎦

⎤⎢⎣

⎡−

2,

2ππ whose sin is x.

In a similar way, )(cos 1 x− or )arccos(x is defined to be the number

in the interval [ ]π,0 whose cosine is x. The inverse tangent function, written

)(tan 1 x− or )arctan(x , is defined to be the number is the interval ⎥⎦

⎤⎢⎣

⎡−

2,

2ππ

whose tangent is x.

Exercise 3: Find each of the following:

a. ⎟⎟⎠

⎞⎜⎜⎝

⎛

33arctan b. )

3arccos(cos ⎟

⎠

⎞⎜⎝

⎛π c. )0arcsin(

30

Double Angle Formulas ( ) ( ) ( )uuu cossin22sin =

( ) ( ) ( ) ( ) ( ) 1cos2sin21sincos2cos 2222 −=−=−= uuuuu Law of Sines

cC

bB

aA sinsinsin

==

Law of Cosines Cabbac cos2222 −+= Pythagorean Identity 1cossin 22 =+ xx Exercise 4: In the triangle below, find the length of the side marked with x.

Exercise 5: In the triangle below, find the length of the side marked with x.

Exercise 6: If 135sin =A and

125tan =A , find Acos

31

Answers to Exercises Basic Algebra 1. 56 2. 167 3. $63.00

4. (a) (b) 161 (c)

20173 (d)

3210

5. (a) 817 (b)

2091 (c)

2411 (d)

321

6. (a) 25 % (b) %218 (c) 152 %

7. (a) .02 (b) .055 (c) .003 8. (a) 22.5 (b) 7.84 (c) 360

9. (a) 89x (b) 5

92ba

(c) 452y

10. 213460 zyx 11. (a) 23 (b) 21/80

12. (a) 144 −x (b) ba 1020 + (c) 1015 −x

Algebra I

1. 18 or 23 2. (a) (0, -5) (b) inconsistent system

3. 827

4. 33 or 27 5. 75

6. (a) bba 22 23 (b) vu 32 (c) 3 252 xxy

7. 624 +−

8. (a) 27,

27− (b)

21,5− (c) 102,102 +−−−

9. 6,21

−−

Algebra II 1. (a) all real numbers except 3−=x (b) all real numbers except y=0 (c) -1/32 2. (a) all real numbers (b) all real numbers y>0

(c) or

321

3<x

415

−− ( )51

41

−

32

Transcendentals 1. (a) (b) (c) π3 2. c

3. (a) !30 (b) 3π

(c) 0

4. !70sin2

1 5. 12 6.

1312

π π8