College Algebrafd.valenciacollege.edu/file/ashkembi/WorkBook Combined.pdf1.2. QUADRATIC AND...

College Algebra

Armira Shkembic© Draft date January 6, 2014

Contents

1 Algebraic Equations 31.1 Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Quadratic and Quadratic Type Equations . . . . . . . . . . . . . 101.3 Rational Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 191.4 Radical Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 271.5 Absolute Value Equations . . . . . . . . . . . . . . . . . . . . . . 361.6 Overview on Solving Algebraic Equations . . . . . . . . . . . . . 411.7 Systems of Equations . . . . . . . . . . . . . . . . . . . . . . . . . 44

2 Inequalities 592.1 Linear Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . 592.2 Compound Inequalities . . . . . . . . . . . . . . . . . . . . . . . . 662.3 Absolute Value Inequalities . . . . . . . . . . . . . . . . . . . . . 722.4 Nonlinear Inequalities . . . . . . . . . . . . . . . . . . . . . . . . 79

3 Functions 893.1 The Cartesian Coordinate System . . . . . . . . . . . . . . . . . 893.2 Introduction to Functions . . . . . . . . . . . . . . . . . . . . . . 993.3 Representing a Function Algebraically . . . . . . . . . . . . . . . 1013.4 Representing a Function Graphically . . . . . . . . . . . . . . . . 1093.5 Function Operations . . . . . . . . . . . . . . . . . . . . . . . . . 1273.6 One-to-One Functions . . . . . . . . . . . . . . . . . . . . . . . . 1373.7 Inverse of a Function . . . . . . . . . . . . . . . . . . . . . . . . . 1423.8 Even and Odd Functions . . . . . . . . . . . . . . . . . . . . . . . 146

Library of Functions 1513.9 Using Transformations to Graph Functions . . . . . . . . . . . . 153

4 Main Algebraic Functions 1614.1 Linear Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 1614.2 Quadratic Functions . . . . . . . . . . . . . . . . . . . . . . . . . 1814.3 Other Polynomial Functions . . . . . . . . . . . . . . . . . . . . . 1934.4 Rational Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 2204.5 Piecewise-defined Functions . . . . . . . . . . . . . . . . . . . . . 239

CONTENTS 1

5 Exponents and Logarithms 2455.1 Exponential Functions . . . . . . . . . . . . . . . . . . . . . . . . 2455.2 Logarithmic Functions . . . . . . . . . . . . . . . . . . . . . . . . 2525.3 Exponential and Logarithmic Equations . . . . . . . . . . . . . . 2685.4 Exponential Growth and Decay . . . . . . . . . . . . . . . . . . . 275

2 CONTENTS

Chapter 1

Algebraic Equations

1.1 Linear Equations

1.1.1 Definitions and Introduction

Textbook Information. For more information on this topic, moreexamples and more exercises see Section 1.1.

For a video presentation of this section click on the link here.

Definition 1. An equation consists of two algebraic expressions with

an equal sign in between.The solution of an equation consists of the number(s) for which the

equation is true.We say that we are solving an equation or finding the solution to

an equation when we are finding all the possible values of the variablethat change the equation into a true statement (like 0=0).

An equation which turns into the form a constant ·x+ a constant = 0after all the terms are moved on one side of the “=” sign is called a linearequation.

http://youtu.be/Jenc72kgGNE

4 CHAPTER 1. ALGEBRAIC EQUATIONS

Main Concept 1. To solve a linear equation we:

1. isolate the variable on one side of the “=” sign

2. add like terms and

3. divide both sides to isolate the variable by itself.

1.1.2 Linear Equations Involving One Letter

Example 1. Solve the equations:

a. 4x− 7 = 13

b. 12 t−

√3 = 0

Solution 1.

a. 4x− 7 = 13

Process Equation

Terms containing the variable on one side of = 4x = 20

Add like terms and factor the variable

Divide or multiply to find the variable x = 5

Write the solution in set notation {· · · } {5}

1.1. LINEAR EQUATIONS 5

b. 12 t−

√3 = 0

Process Equation

Terms containing the variable on one side of = 12 t =

√3

Add like terms and factor the variable

Divide or multiply to find the variable t = 2√

3

Write the solution in set notation {· · · } {2√

3}

1.1.3 Linear Equations Involving More Than One Letter

Main Concept 2. Some equations involve more then one letter.One of the letters is the variable and the other(s) is (are) considered asconstant(s).

Example 2. Solve the following equations for the indicated variable.

a. 2x + 5y − 2 = 5x− 3y for y.

b. xy + 2x2z + yz + 5 = 13xy3 + z for z

Solution 2.

a. 2x + 5y − 2 = 5x− 3y for y.

Process Equation

Terms containing the variable (y) on one side of = 5y + 3y = 5x− 2x + 2

Add like terms and factor the variable 8y = 3x + 2

Divide or multiply to find the variable y = 3x+28

Write the solution in set notation {· · · } { 3x+28 }


b. xy + 2x2z + yz + 5 = 13xy3 + z for z

Process Equation

Terms containing the variable (z) on one side of = 2x2z + yz − z = 13xy3 − xy − 5

Add like terms and factor the variable z(2x2 + y − 1) = 13xy3 − xy − 5

Divide or multiply to find the variable z = 13xy3−xy−52x2+y−1

Write the solution in set notation {· · · } { 13xy3−xy−5

2x2+y−1 }


Homework Exercises

Name:Please come and see me in 3-245 if you have any questions or difficulties in

solving the problems.

1. (8 points) Solve each equation:a. (4 pts) 5x− 15 = 4

b. (4 pts) 3x + 4 = 2x− 4


2. (8 points) Solve

3xy + 5x− 2 = 3x2 − 7y for y.

3. (8 points) Solve

xy2 − 5x + 7y2 − 3 = 3y for x.


4. (4 points)(a) I feel comfortable solving linear equations.

Strongly Disagree-1 2 3 4 5-Strongly Agree

b. (4 pts) Write down and solve a linear equation.

Comments about what you learned in this section.

Was any particular exercise, type of problem or concept hard to solve/understand?


1.2 Quadratic and Quadratic Type Equations

1.2.1 Quadratic Equations in the General Form

Textbook Information. For more information on this topic, moreexamples and more exercises see Section 1.2 and 1.4.

For a video presentation of this material on Quadratic Equations click onthe link here.

Definition 1. A quadratic equation is one that, after placing all

terms on one side and simplifying, is changed to the form a constant ·x2 +a constant · x + a constant = 0, where the constant multiplying x2 is notzero.

Usually we write this as ax2 + bx + c = 0, where a 6= 0.

Main Concept 1. There are different ways to solve quadratic equa-tions. The main ones are:

1. Writing it in the form ax2+bx+c = 0, then factoring it, if it is factorable,and setting each factor equal to zero

2. Using the quadratic formula:

x =−b±

√b2 − 4ac

2a.

3. Completing the square.

Here we will focus on the first two ways.

Example 1. Solve each equation:

a. −x2 + 5x = 4

b. 4x2 + 3x = 7x + 15

c. 5x2 + 2x = 3x2 − x− 5

http://youtu.be/TWgfFobbU3Y

http://youtu.be/TWgfFobbU3Y

1.2. QUADRATIC AND QUADRATIC TYPE EQUATIONS 11

Solution 1.

a. −x2 + 5x = 4

Process Equation

Write the equation in the form ax2 + bx + c = 0 −x2 + 5x− 4 = 0

Factor and set each factor equal to zero or use the Quadratic Formula (−x + 1)(x− 4) = 0

−x + 1 = 0 x− 4 = 0

x = 1 x = 4

Write the solution in set notation {· · · } {1, 4}

b. 4x2 + 3x = 7x + 15

Process Equation

Write the equation in the form ax2 + bx + c = 0 4x2 − 4x− 15 = 0

Factor and set each factor equal to zero or use the Quadratic Formula (2x + 3)(2x− 5) = 0

2x + 3 = 0, 2x− 5 = 0

x = − 32 x = 5

2

Write the solution in set notation {· · · } {− 32 ,

52}


c. 5x2 + 2x = 3x2 − x− 5

Process Equation

Write the equation in the form ax2 + bx + c = 0 2x2 + 3x + 5 = 0

Factor and set each factor equal to zero

or use the Quadratic Formula The equation does not factor so

x = −3±√32−4·2·52·2

x = −3±√−31

4

x = −3±√31i

4

Write the solution in set notation {· · · } {−3+√31i

4 , −3−√31i

4 }

1.2.2 Quadratic Type Equations

For a video presentation of this material on Quadratic Type Equations clickon the link here.

Definition 2. An equation of quadratic type is one where an

expression is raised to two different powers and one power is double theother.

http://youtu.be/apTJg82nObc

http://youtu.be/apTJg82nObc


Main Concept 2. To solve an equation of quadratic type

1. First make the appropriate substitution by making t equal to the repeat-ing expression raised to the power whose double is the other power.

2. At this point the equation changes to the form at2 + bt + c = 0 whichcan be solved for t.

3. After solving for t use the substitution and the solution for t to solve forthe original variable.

Example 2. Solve the equation.

a. (3x− 5)2 + 7(3x− 5) + 12 = 0

b. 4(4x + 1)16 + (4x + 1)

13 + 3 = 0


Solution 2.

a. (3x− 5)2 + 7(3x− 5) + 12 = 0

Process Equation

Substitute and change the equation to quadratic t = 3x− 5

t2 + 7t + 12 = 0

Solve for t.

Since 4 and 3 multiply to 12 and add to 7

the equation factors: (t + 4)(t + 3) = 0

Set factors equal to zero: t = −4 or t = −3

Solve for the original variable x. Use t = 3x− 5, t = −4, t = −3

to solve for x

3x− 5 = −4 or 3x− 5 = −3

Terms containing the variable on one side of =

3x = 1 or 3x = 2

Divide or multiply to find the variable

x = 13 or x = 2

3

Write the solution in set notation {· · · } { 13 ,23}


b. 4(4x + 1)16 + (4x + 1)

13 + 3 = 0

Process Equation

Substitute and change the equation to quadratic t = (4x + 1)16

4t + t2 + 3 = 0

Solve for t. t2 + 4t + 3 = 0

Since 1 and 3 multiply to 3 and add to 4

the equation factors: (t + 1)(t + 3) = 0

Set factors equal to zero: t = −1 or t = −3

Solve for the original variable x. Use t = (4x + 1)16 , t = −1, t = −3

to solve for x

(4x + 1)16 = −1 or (4x + 1)

16 = −3

Recall that something raised

to the 16

thpower

is the same as the 6th root.

So we have 6√

4x + 1 = −1 and 6√

4x + 1 = −3.

But the 6th root of any number

is always positive.

Therefore the equation has no solution.

Write the solution in set notation {· · · } {}


Homework Exercises



1. (19 points) Solve the equations:a. (5 pts)

3

5x2 − x =

1

5

b. (6 pts)

(2x− 1)4 − 5 (2x− 1)

2+ 4 = 0


c. (8 pts)

3

(x + 3

2

) 14

+

(x + 3

2

) 12

+ 2 = 0

2. (4 points) (a) I feel comfortable solving quadratic equations.


b. (4 pts) Write down and solve a quadratic equation.


3. (4 points)(a) I feel comfortable solving quadratic type equations.


b. (4 pts) Write down and solve a quadratic type equation.



1.3. RATIONAL EQUATIONS 19

1.3 Rational Equations

Textbook Information. For more information on this topic, moreexamples and more exercises see Section 1.1 and 1.2.


Definition 1. An equation containing the variable in (a) denomina-

tor(s) is called a rational equation.

Main Concept 1. To solve a rational equation multiply both sides ofthe “=” sign by the least common denominator, then solve the equation byfactoring, using the quadratic equation or as a linear equation. At the endcheck that the solution does not make the denominator zero.


a. 3x+1 + 1

x+2 = 7x+1

b. 2x−1 − 3 = 2

x−1

c. 2x−1x+2 = 2x+1

3x−1 −13

3x2+5x−2 + 1

http://youtu.be/fb1HxTom-y0


Solution 1.

a. 3x+1 + 1

x+2 = 7x+1

Process Equation

3x+1 + 1

x+2 = 7x+1

Multiply by the common denominator: 3(x+1)(x+2)x+1 + (x+1)(x+2)

x+2 = 7(x+1)(x+2)x+1

3(x + 2) + (x + 1) = 7(x + 2)

3x + 6 + x + 1 = 7x + 14

Solve the equation obtained: −3x− 7 = 0

x = − 73

Check the answer since the original equation

has denominators. − 73 checks as a solution.

Write the solution in set notation {· · · } {− 73}


b. 2x−1 − 3 = 2

x−1

Process Equation

Multiply by the common denominator 2(x−1)x−1 − 3(x− 1) = 2(x−1)

x−1

2− 3x + 3 = 2

Solve the equation obtained: −3x + 3 = 0

−3x = −3

Check the answer since the original equation 1 makes the denominator zero.

Therefore the equation has no solution.



c. 2x−1x+2 = 2x+1

3x−1 −13

3x2+5x−2 + 1

Process Equation

Multiply by the The LCD can be found after

common denominator factoring all the denominators.

and simplify: To factor 3x2 + 5x− 2 we need to find two

numbers that multiply to −6 and add to 5.

The numbers will be 6 and −1.

To get 3x2 we need 3x and x as the first terms

(3x )(x ).

To get 6x and −x as the inside and outside terms

we need to factor as:

(3x− 1)(x + 2).

So that is the LCD.

Multiplying by it gives

(2x− 1)(3x− 1) = (2x + 1)(x + 2)− 13 + (3x− 1)(x + 2)

6x2 − 5x + 1 = 2x2 + 5x + 2− 13 + 3x2 + 5x− 2

x2 − 15x + 14 = 0.

Solve the equation obtained: Factor:

(x− 1)(x− 14) = 0

x = 1 and x = 14.


Check the answer Neither 1 nor 14 make

since the original equation the denominator zero,

has denominators. so they check as solutions.

Use set notation {· · · } {1, 14}


Homework Exercises



1. (19 points) Solve the equations:a. (3 pts)

2x

x + 3= − 6

x + 3− 2.

b. (4 pts)6t + 7

4t− 1=

3t + 8

2t− 4


c. (7 pts)x + 2

2x− 1=

x

x + 1− x− 8

2x2 + x− 1− 1

d. (5 pts)3x

x− 2+

1

x= 4


2. (4 points) (a) I feel comfortable solving rational equations.


b. (4 pts) Write down and solve a rational equation.



1.4. RADICAL EQUATIONS 27

1.4 Radical Equations



Definition 1. An equation where the variable is inside a root (or a

non-integer, rational exponent) is called a radical equation.

Main Concept 1. To solve a radical equation, isolate (one of) the rad-ical(s) and raise to powers to cancel the root. Solve the obtained equationand check the equations that had even root(s) originally.


a.√

2x− 3 + 1 = x

b. (6x + 1)13 + 4 = 6

c.√x + 1 +

√3x + 1 = 2

http://youtu.be/BpX6zjZM_j4


Solution 1.

a.√

2x− 3 + 1 = x

Process Equation

Isolate (one of) the radical(s)√

2x− 3 = x− 1

Raise to powers to cancel the isolated radical 2x− 3 = (x− 1)2

Simplify 2x− 3 = (x− 1)(x− 1)

2x− 3 = x2 − 2x + 1

Repeat this process until there are no radicals left

Solve the equation obtained 0 = x2 − 4x + 4

Since −2 and −2 multiply to 4 and add to −4

we factor.

0 = (x− 2)(x− 2)

Setting the factors equal to zero

x = 2

Check your answer(s) in the original equation x = 2 gives√

4− 3 + 1 = 2

which is true.

Write the solution in set notation {2}


b. (6x + 1)13 + 4 = 6 is the same as 3

√6x + 1 + 4 = 6

Process Equation

Isolate (one of) the radical(s) 3√

6x + 1 = 2

Raise to powers to cancel the isolated radical 6x+1=8

Simplify 6x + 1 = 8

Repeat this process until there are no radicals left

Solve the equation obtained 6x = 7

x = 76

Check your answer(s) in the original equation Since the original equation

does not have an even root

there is no need to check the equation.

Write the solution in set notation { 76}


c.√x + 1 +

√3x + 1 = 2

Process Equation

Isolate (one of) the radical(s)√

3x + 1 = 2−√x + 1

Raise to powers to cancel the isolated radical(√

3x + 1)2

=(2−√x + 1

)2Simplify 3x + 1 =

(2−√x + 1

) (2−√x + 1

)3x + 1 = 4− 2

√x + 1− 2

√x + 1 + x + 1

3x + 1 = 4− 4√x + 1 + x + 1

3x + 1 = 5− 4√x + 1 + x

Repeat this process until there are no radicals left Have the radical by itself:

2x− 4 = −4√x + 1

2x−4−4 =

√x + 1

x−2−2 =

√x + 1

Square both sides

(x−2)2(−2)2 = x + 1

x2−4x+44 = x + 1

Solve the equation obtained

x2 − 4x + 4 = 4x + 4

x2 − 8x = 0

x(x− 8) = 0

x = 0, x = 8


Check your answer(s) in the original equation x = 0 gives√

1 +√

1 = 2,

which is true.

But x = 8 gives√

9 +√

25 = 2

which is not true.

Write the solution in set notation {0}


Homework Exercises



1. (20 points) Solve the equations using the template as a guide. Foreach process, rate your comfort level using the scale:

Don’t know where to start-1 2 3 4 5-I am confident with this part

a. (6 pts)x = 2

√x− 1

Process Problem given ComfortLevel

Isolate (one of) theradical(s)

Raise to powers tocancel the isolatedradical

Simplify

Repeat this processuntil there are noradicals left

Solve the equationobtained

Check your an-swer(s) in theoriginal equation

Write the solutionin set notation


b. (7 pts)2x =

√2x + 3− 1




Simplify






c. (7 pts) √x2 − x− 4 = x + 2




Simplify






2. (4 points) Write down and solve a radical equation involving thevariable inside at least two radicals.




1.5 Absolute Value Equations


Definition 1. The absolute value of a real number r, denoted by

|r|, is its distance from zero.

Definition 2. An equation where the variable shows inside an absolute

value is called an absolute value equation.


Main Concept 1. To solve an absolute value equation

1. isolate the absolute value,

2. set the inside of the absolute value equal to the number(s) that haveabsolute value equal to the constant that is on one side of “=”,

3. solve the equation(s) obtained.

Example 1. Solve the equations.

a. 2|2x− 1| − 2 = 4

b. 4 + |7− 2x| = 4

c. 2|5x + 1|+ 3 = 2

http://youtu.be/XjHnNVluJfk

1.5. ABSOLUTE VALUE EQUATIONS 37

Solution 1.

a. 2|2x− 1| − 2 = 4

Process Equation

Isolate the absolute value 2|2x− 1| = 6

|2x− 1| = 3

Make the expression inside the absolute value

equal to the numbers that have absolute value = 3 2x− 1 = 3 or 2x− 1 = −3

Solve the two equations 2x = 4 or 2x = −2

x = 2 or x = −1

Write the solution in set notation {· · · } {−1, 2}

b. 4 + |7− 2x| = 4

Process Equation

Isolate the absolute value |7− 2x| = 0

Make the expression inside the absolute value

equal to the numbers that have absolute value = 0 7− 2x = 0

Solve the equation 7 = 2x

x = 72

Write the solution in set notation {· · · } { 72}


c. 2|5x + 1|+ 3 = 2

Process Equation

Isolate the absolute value 2|5x + 1| = −1

|5x− 1| = − 12

Make the expression inside the absolute value There are no numbers

equal to the numbers that have absolute value = − 12 that have absolute value of − 1

2

So the equation has no solution.


1.5. ABSOLUTE VALUE EQUATIONS 39

Homework Exercises



1. (14 points)a. (7 pts)

4|2x− 9|+ 3 = 11

b. (7 pts)|4x− 7|+ 2 = 10


2. (4 points)(a) I feel comfortable solving equations involving absolute value.


b. (4 pts) Write down and solve an equation involving absolutevalue.



1.6. OVERVIEW ON SOLVING ALGEBRAIC EQUATIONS 41

1.6 Overview on Solving Algebraic Equations

Main Concept 1. You may think of the process of solving an equationas separated into two main parts. The first part is reducing the process ofsolving the given equation to solving a linear or quadratic equation. Thisis described by the first chart below.

The second part is solving the linear or quadratic equation. This partis described in the second chart.

Make sure you check the answers make sense and give real numbers whensubstituted in the equation. So when solving algebraic equations involvingdenominators, make sure the answer does not make the denominator zero,and when solving an equation containing an even root substitute the an-swer in the equation to make sure it checks.

Note 1. In later sections you will see that we also check the solution does notlead to taking the logarithm of zero or a negative number, but this sectiondoes not include these types of equations.


QuadraticType Equations

RationalEquations

Absolute ValueEquations

RadicalEquations

The equation can bechanged to the form

at2 + bt + c = 0

by substitution.

Solve for t by fac-toring or using thequadratic formula:

t =−b±

√b2 − 4ac

2a.

Use the value(s) of tand the expressionsubstituted for t tofind the value of the

original variable.

The variableshows in the

denominator(s).

Multiply bythe commondenominator.

The variableshows inside

absolute value(s).

Isolate the absolutevalue: Expres-sion involving

absolute value=Expression without

absolute value

Make the expressioninside the absolutevalue equal to the

number(s) that havethe same absolute

value as theExpression without

absolute value

The variable showsinside root(s).

Isolate the morecomplicated root.

Raise to the indexof the radical

isolated, simplify.

Repeat the pre-vious two steps

until there are noradicals and the

equation is a poly-nomial or rational.

Figure 1.1: Reducing to Solving a Polynomial Equation

1.6. OVERVIEW ON SOLVING ALGEBRAIC EQUATIONS 43

Simplify theoriginal equationso that it writtenas a polynomial.

Move all termson one side ofthe = and zeroon the other.

The equationis of the form

ax2 + bx + c = 0.

The equation isa polynomial ofdegree at least 3

Make sure all theterms are one

one side of = andfactor the equation.

Make each factorequal to zero and

solve using the pro-cess shown on the

columns on the left.

The equation is ofthe form ax + b = 0.

Place all the termscontaining the

variable on one sideof the = sign andthe other terms

on the other side.

If possible factorthe expression onthe left of = andset each factor

equal to zero. Iffactoring is not

easily seen use thequadratic formula

x =−b±

√b2 − 4ac

2a.

Check the answerof the original

equation containsdenominator(s) or

even root(s). Writethe solution in set

notation {· · · }.

Add like termsand factor outthe variable.

Divide or mul-tiply to find x.

Figure 1.2: Solving a Polynomial Equation


1.7 Systems of Equations

We have already seen in applications problems how different variables can beconnected through an equation. In this section we will look at situations whentwo or more equations are used for the same 2 or more variables.

Example 1. A movie theater sells tickets for $8.00 each, with seniorsreceiving a discount of $2.00. One evening the theater took in $3580 in revenue.Let x represents the number of tickets sold at $8.00 and y represents the ticketssold at the discounted price of $6.00.

a. Write an equation that relates these variables.

b. Suppose we also know that 525 tickets we sold. Write another equationsrelating the variables x and y.

Solution 1.

a. The revenue from selling x tickets at a price of $8.00 each is 8x. The revenuefrom selling y tickets at a price of $6.00 each is 6y. Since the total revenueis $3580, 8x + 6y = 3580.

b. Since x is the number of $8.00 tickets sold and y is the number of $6.00tickets sold, the total number of tickets sold is x + y. Therefore x + y = 525is another equation relating the variables.

Definition 1. A system of equations is a collection of two or more

equations each containing one or more variables. A system of equation isdenoted by:

First equation

Second equation

More equations may follow

A solution to a system of equations is written as an ordered list ofnumbers (x0, y0, · · · ) consisting of the values for each of the variables usedthat satisfy all equations of the system.

1.7.1 Solving Systems of Equations By Substitution

Main Concept 1.One way to solve equations is by substitution. In this method we use

one of the equations given to solve for one of the variables, and substitutethe expression for the variable in the other equations. We continue thisprocess until we have an equation of one variable left. Then we solve for

1.7. SYSTEMS OF EQUATIONS 45

this variable and substitute its value in the other equations. Continue thisprocess until the values for all the variables are found.

At the end write the solution in the coordinate form( value of x, value of y, value of z)

Example 2. Solve each system using the substitution method:

a.

{2x + y = 5

−4x + 6y = −2

b.

{x + y2 = 5

2x + y = 4

c.

x + y + z = 6

3x− 2y + 4 = 9

x− y − z = 0

Solution 2.

a. Use 2x + y = 5 to solve for y:

y = 5− 2x.

Substitute this expression for y in the equation

−4x + 6y = −2

−4x + 6(5− 2x) = −2

−4x + 30− 12x = −2

−16x = −32

x = 2

The solution to the system will give the values of all variables used, so weneed to find the value of y. Since we have y = 5− 2x, substituting the valuefor x here will give us the value for y:

y = 5− 2(2)

y = 1

Therefore the solution in coordinates form is (2, 1).


b. Use x + y2 = 5 to solve for x:

x = 5− y2.

Substitute this expression for x is the second equation:

2x + y = 4

2(5− y2) + y = 4

10− 2y2 + y = 4.

Solve for y:10− 2y2 + y − 4 = 0

−2y2 + y + 6 = 0

2y2 − y − 6 = 0

(2y + 3)(y − 2) = 0

y = −3

2, y = 2.

Substituting these values of y in the equation x = 5 − y2 gives the corre-sponding values of x:

y = 2: x = 5− 4 = 1.

Therefore (1, 2) is a solution.

y = −3

2: x = 5− 9

4

=20

4− 9

4

=11

4.

And(114 ,− 3

2

)is another solution.

c. To solve this system you have to use one of the equations to solve for onevariable

x = 6− y − z

and substitute it in the other two equations:{3(6− y − z)− 2y + 4 = 9

6− y − z − y − z = 0{18− 3y − 3z − 2y + 4 = 9

6− 2y − 2z = 0


{22− 5y − 3z = 9

6− 2y − 2z = 0.

Use one of these equations to solve for one of the remaining variables:

22− 5y − 3z = 9

−5y = −13 + 3z

y =−13 + 3z

−5

y =13− 3z

5.

Substitute this expression for y in the other equation and solve for z:

6− 2y − 2z = 0

6− 213− 3z

5− 2z = 0

6− 26− 6z

5− 2z = 0

30− 26 + 6z − 10z = 0

4− 4z = 0

−4z = −4

z = 1

Substitute this value of z in y = 13−3z5 to find y:

y =13− 3

5

y =10

5

y = 2.

Substitute the values of y and z to find x:

x = 6− y − z

= 6− 2− 1

= 3.

Therefore the solution is(3, 2, 1) .


1.7.2 Solving Systems of Equations By Elimination

Main Concept 2. Another way of solving systems of equations is byelimination. Here we first rewrite (if needed) the equations so that eachvariable shows in the same column, then the “=” sign and the constant.Multiply (or divide) equations by constants to get it to the form wherewhen adding two equations one of the variables add to zero. We continuethis process until there is only one variable left, solve for that variable andsubstitute its value in the other equations to find the other variable(s).

Example 3. Solve each system using the elimination method:

a.

{2x + y = 5

−4x + 6y = −2

b.

{x + y2 = 5

2x + y = 4

c.

x + y + z = 6

3x− 2y + 4 = 9

x− y − z = 0

Solution 3.

a. If we multiply the first equation by 2

{2(2x + y) = 2 · 5−4x + 6y = −2{

4x + 2y = 10

−4x + 6y = −2

and add the two equations,8y = 8

the terms containing x will cancel and give

y = 1.

Substitute this in one of the original equations and solve for x:

2x + y = 5

2x + 1 = 5

2x = 4

x = 2.

Therefore the solution is (2, 1).


b. If we multiply the first equation by −2:{−2(x + y2) = −2 · 52x + y = 4{−2x− 2y2 = −10

2x + y = 4

and add the two equations:

−2y2 + y = −6

−2y2 + y + 6 = 0

2y2 − y − 6 = 0

(2y + 3)(y − 2) = 0.

So y = − 32 and y = 2. Substituting these values of y into any of the original

equations and solving for x:

y = −3

2: x +

9

4= 5

x = 5− 9

4

x =11

4

(11

4,−3

2

)

y = 2: x + 4 = 5

x = 1 (1, 2).

Therefore the solutions are {(114 ,− 3

2

), (1, 2)}.

c. First make sure all three equations are written in the form ax+by+cz+d = 0.x + y + z = 6

3x− 2y = 5

x− y − z = 0

Use one pair of the equations and multiply to cancel one of the variables, useanother pair to cancel the same variable:

Using the first and second equations and canceling x:{−3(x + y + z) = −3 · 63x− 2y = 5{−3x− 3y − 3z = −18

3x− 2y = 5.


Adding the two gives: −5y − 3z = −13.

Use another pair (second and third) to cancel x also:{3x− 2y = 5

−3(x− y − z) = −3 · 0{3x− 2y = 5

−3x + 3y + 3z = 0

Adding the two: y + 3z = 5.

Now the process reduces to solving the system:{−5y − 3z = −13

y + 3z = 5 · 0.

Adding the two equations gives: −4y = −8, y = 2.

Substituting the value of y in one of the above equations: −5y − 3z = −13.

−10− 3z = −13

−3z = −3

z = 1

Substituting the values y = 2 and z = 1 into any of the original equationsand solving for x:

x + y + z = 6

x + 2 + 1 = 6

x = 3.

Therefore the solution is (3, 2, 1).

1.7.3 Inconsistent, Dependent, Independent Systems

If we take a geometrical look at systems of equations, we would see that eachequation is represented by a graph and the solution to the system is the inter-section of these graphs.

Definition 2. A system of equations is called consistent if it has

a solution, and it is called inconsistent if it does not have a solution. Asystem of equations with infinitely many solutions or no solution is calleda system of dependent equations and a system of equations with finitenumber of solutions is called a system of independent equations.

Example 4. Solve each system of equations. State whether the system isinconsistent, dependent or independent.


a.

{2x + y = 5

4x + 2y = 8

b.

{2x + y = 4

−6x− 3y = −12

c.

2x− 3y − z = 0

−x + 2y + z = 5

3x− 4y − z = 1

Solution 4.

a. Solving the system by elimination gives:{−4x− 2y = −10

4x + 2y = 8

0 = −2

which is not true, therefore the system is inconsistent.

b. Solving the equation using the elimination method:{6x + 3y = 12

−6x− 3y = −12

0 = 0

which is always true and the system is dependent.

c. Solving by elimination: {2x− 3y − z = 0

−2x + 4y + 2z = 10

y + z = 10

and {−3x + 6y + 3z = 15

3x− 4y − z = 1

2y + 2z = 14

Solving the system {y + z = 10

2y + 2z = 14


gives {−2y − 2z = −20

2y + 2z = 14

0 = −7.

Therefore the system is inconsistent.

Example 5. Find real numbers a, b, c so that the graph of the quadraticfunction y = ax2 + bx + c contains the points (−1,−4), (1, 6) and (3, 0).

Solution 5. Since the point (−1,−4) is on the graph we will have

−4 = a(−1)2 + b(−1) + c

−4 = a− b + c

Also, since the point (1, 6) is on the graph we will have

6 = a(1)1 + b(1) + c

6 = a + b + c

And the point (3, 0) gives

0 = a(3)2 + b(3) + c

0 = 9a + 3b + c

The values of a, b and c will be given by the solution to the system:a− b + c = −4

a + b + c = 6

9a + 3b + c = 0

.

{a− b + c = −4

a + b + c = 6

2a + 2c = 2

a + c = 1

{−3a− 3b− 3c = −18

9a + 3b + c = 0

6a− 2c = −18

3a− c = −9

{a + c = 1

3a− c = −9


4a = −8

a = −2

Substituting the value of a in a + c = 1 gives −2 + c = 1 so c = 3, andsubstituting a and c in a + b + c = 6 gives (−2) + b + 3 = 6, 1 + b = 6, b = 5.Therefore the function will be y = −2x2 + 5x + 3.


Homework Exercises



1. (16 points) Solve each system by the elimination method.a. (5 pts) {

2x− 3y = 5

3x + 2y = 14


b. (5 pts) {x + y = 5

2x + 2y = 10

c. (6 pts) 4x + 2y + 3z = 6

x + 2y + 2z = 1

2x− y + z = −1


2. (6 points) Solve the system using the substitution method.{2x + y = 2

3x− y = −7


3. (4 points) (a) I feel comfortable solving systems ofequations using the elimination method.


b. (4 pts) Write down a system of equations and solve it usingelimination.

4. (4 points) (a) I feel comfortable solving systems ofequations using the substitution method.


b. (4 pts) Solve the system of equations in the previous exerciseusing substitution.



Chapter 2

Inequalities

2.1 Linear Inequalities



2.1.1 Interval Notation

Main Concept 1. The set of all numbers (whole or decimal) is calledthe set of real numbers and is denoted by R.

The real numbers can be placed in increasing order on a straight line.Smaller numbers are to the left of larger numbers. Each point of the linecorresponds to a number and each number corresponds to a point on theline.

−5 −4 −3 −2 −1 1 2 3 40

0

Negative Numbers Positive Numbers

Definition 1. A set of numbers lying in an uninterrupted range of

real numbers is called an interval.

http://youtu.be/Qhd01abEnSA

60 CHAPTER 2. INEQUALITIES

Intervals can be categorized into finite (length) intervals, and infinite (length)intervals. And the main ways to express an interval include verbal expression,in set notation, in interval notation, and graphically.

Verbal Description Set Notation IntervalNotation

Graphical Description

FiniteIntervals

The set of numbersgreater than a andless than b.

{x ∈ R|a < x < b} (a, b) ba)(

The set of num-bers greater than orequal to a and lessthan b

{x ∈ R|a ≤ x < b} [a, b) ba)[

The set of numbersgreater than a andless than or equal tob

{x ∈ R|a < x ≤ b} (a, b] ba](

The set of num-bers greater than orequal to a and lessthan or equal to b

{x ∈ R|a ≤ x ≤ b} [a, b] ba][

InfiniteIntervals

The set of numbersless than a

{x ∈ R|x < a} (−∞, a) a)

The set of numbersless than or equal toa

{x ∈ R|x ≤ a} (−∞, a] a]

The set of numbersgreater than a

{x ∈ R|x > a} (a,∞) a(

The set of num-bers greater than orequal to a

{x ∈ R|x ≥ a} [a,∞) a[

2.1. LINEAR INEQUALITIES 61

Example 1. For each statement, write the inequality, the interval notationand graph the numbers that satisfy it on the number line:

a. x is less than 3.

b. x is less than or equal to −1.

c. x is less than or equal to 5 and greater than 2.

d. x is greater than 4 and less than 9.

Solution 1.

a. x < 3, (−∞, 3),

3)

b. x ≤ −1, (−∞,−1],

−1]

c. 2 < x ≤ 5, (2, 5],

52](

d. 4 < x < 9, (4, 9)

94)(

2.1.2 Linear Inequalities

Definition 2. An inequality consists of two algebraic expressions

with an inequality sign in between.An inequality is called linear if the two algebraic expressions separated

by the inequality sign are of the form ax + b.We solve an inequality when we find all the values that satisfy it.


Main Concept 2. A linear inequality is solved the same way as alinear equation:

1. Terms that involve the variable should be on one side.

2. Multiply or divide to find x.

with the exception that when multiplying or dividing by a negativenumber switch the sign of the inequality. At the end the solutionis written as an interval notation and graphed.

Example 2. Solve the following inequalities. Graph the solution on thenumber line and write it using interval notation.

a. 4x− 1 ≤ 7x

b. x−53 > −3− x

Solution 2.

a. 4x− 1 ≤ 7x

Process Inequality

Terms containing the variable on one side of ≤ −3x ≤ 1

Divide or multiply to find the variable. Since we divide by −3 we switch the sign

When dividing or multiplying by x ≥ − 13

a negative number, switch the sign

Write the answer in interval notation.[− 1

3 ,∞)

Graph the answer on the number line.− 1

3

[


b. x−53 > −3− x

Process Inequality

Terms containing the variable on one side of > x3 −

53 > −3− x

x3 + x > −3 + 5

3

4x3 > − 4

3

Divide or multiply to find the variable. Since we multiply by 34 we do not switch the sign

When dividing or multiplying by x > −1


Write the answer in interval notation. (−1,∞)

Graph the answer on the number line.−1

(


Homework Exercises



1. (10 points) Solve the inequalities.a. (5 pts)

4x− 3 > 5x + 7

b. (5 pts)2x− 1

3≤ 7x− 5


2. (4 points)(a) I feel comfortable solving linear inequalities.


b. (4 pts) Write down and solve a linear inequality.




2.2 Compound Inequalities



Definition 1. A compound inequality consists of two inequalities

combined by an and or or conjunction.

Note 1. An inequality written in one combined line as:

Expression©1 Expression©2 Expression,

where ©1 and ©2 stand for a < or ≤ inequality sign, is the same as:

Expression©1 Expression and Expression©2 Expression

Main Concept 1. When solving a compound inequality

1. Solve both inequalities.

2. Graph both solutions on the same number line.

3. If the conjunction between the inequalities is and, the solution isthe part of the number line where both inequalities are true. If theconjunction between the inequalities is or, the solution is the part ofthe number line where either of the inequalities is true.

4. Write the solution in interval notation.

Example 1. Solve each inequality.

a. 3x + 4 ≤ 2x or 1− 7x < 15

b. 5x− 4 ≥ 2x and 3x− 2 > x + 2

c. −2 < 3x + 4 ≤ 2x

http://youtu.be/Xcm2ttXHDvA

2.2. COMPOUND INEQUALITIES 67

Solution 1.

a. 3x + 4 ≤ 2x or 1− 7x < 15

Process Inequality Inequality

Terms containing the variable on one side 3x + 4 ≤ 2x 1− 7x < 15

x ≤ −4 −7x < 14

Divide or multiply to find the variable. x > −2

When dividing or multiplying by


Write the answer for each in interval notation. (−∞,−4] (−2,∞)

Graph the answer on the number line.−4 −2

] (

Write the answer in interval notation, or-Solution will include the numbers

taking in consideration the conjunction. where either inequality is true:

(−∞,−4] ∪ (−2,∞)


b. 5x− 4 ≥ 2x and 3x− 2 > x + 2


Terms containing the variable on one side 5x− 4 ≥ 2x 3x− 2 > x + 2

3x ≥ 4 2x ≥ 4

Divide or multiply to find the variable. x ≥ 43 x > 2



Write the answer for each in interval notation.[43 ,∞) (2,∞)

Graph the answer on the number line.

43 2

[ (

Write the answer in interval notation, and-Solution will include the numbers

taking in consideration the conjunction. where both inequalities are true:

(2,∞)


c. −2 < 3x + 4 ≤ 2x, which is the same as −2 < 3x + 4 and 3x + 4 ≤ 2x.


Terms containing the variable on one side −2 < 3x + 4 3x + 4 ≤ 2x

−6 < 3x x ≤ −4

Divide or multiply to find the variable. −2 < x

When dividing or multiplying by x > −2


Write the answer for each in interval notation. (−2,∞) (−∞,−4]

Graph the answer on the number line.−4 −2

] (

Write the answer in interval notation, and-Solution will include the numbers

taking in consideration the conjunction. where both inequalities are true.

The inequalities have no common parts

in their solution, so the compound

inequality has no solution.

{}


Homework Exercises



1. (16 points) Solve the inequalities.a. (8 pts) 3x + 2 ≤ 12 or 5x− 1 > 3

b. (8 pts) x− 2 + x+13 ≥ 3 and 2x− 4 ≥ 3x− 2


2. (4 points)(a) I feel comfortable solving compound inequalities.


b. (4 pts) Write down and solve a compound inequality.




2.3 Absolute Value Inequalities



Main Concept 1. When solving absolute value inequalities firstisolate the absolute value so that the inequality is of the form

|Expression| © a,

where © stands for the inequality sign and a is a real number. Thenthink about the numbers in the number line that have absolute value © a.Write a compound inequality describing these numbers and solve the in-equality.

Example 1. Solve the inequalities.

a. 3|2x− 4|+ 2 ≤ 5

b. |x− 5| − 3 > 1

c. |2x− 4|+ 4 ≥ 2

d. |5− 3x|+ 9 < 4

http://youtu.be/AKT8b1fH5_8

2.3. ABSOLUTE VALUE INEQUALITIES 73

Solution 1.

a. 3|2x− 4|+ 2 ≤ 5

Process Inequality

Isolate the absolute value 3|2x− 4|+ 2 ≤ 5

3|2x− 4| ≤ 3, |2x− 4| ≤ 1

What numbers have absolute value ≤ 1?−1 1

[ ]

Describe the above interval using compound inequality −1 ≤ u ≤ 1

Write the inequality using the expression −1 ≤ 2x− 4 ≤ 1

inside absolute value

Terms containing the variable on one side −1 ≤ 2x− 4 2x− 4 ≤ 1

3 ≤ 2x 2x ≤ 5

Divide or multiply to find the variable. 32 ≤ x x ≤ 5

2

When dividing or multiplying by x ≥ 32


Write the answer for each in interval notation.[32 ,∞) (−∞, 5

2

]


32

52

[ ]

Write the answer in interval notation, and-Solution includes the part(s)

taking in consideration the conjunction. where both inequalities are true.

[32 ,

52

]


b. |x− 5| − 3 > 1

Process Inequality

Isolate the absolute value |x− 5| − 3 > 1

|x− 5| > 4

What numbers have absolute value > 4?−4 4

) (

Describe the above interval using compound inequality u < −4 or u > 4

Write the inequality using the expression x− 5 < −4 or x− 5 > 4


Terms containing the variable on one side x < 1 x > 9

Divide or multiply to find the variable.



Write the answer for each in interval notation. (−∞, 1) (9,∞)

Graph the answer on the number line.1 9

) (

Write the answer in interval notation, or-Solution will include the numbers

taking in consideration the conjunction. where either inequality is true.

(−∞, 1) ∪ (9,∞)


c. |2x− 4|+ 4 ≥ 2

Process Inequality

Isolate the absolute value |2x− 4|+ 4 ≥ 2

|2x− 4| ≥ −2

What numbers have absolute value ≥ −2?

Describe the above interval using compound inequality All the real numbers

Write the inequality using the expression


Terms containing the variable on one side




Write the answer for each in interval notation.


Write the answer in interval notation,

taking in consideration the conjunction. (−∞,∞)


d. |5− 3x|+ 9 < 4

Process Inequality

Isolate the absolute value |5− 3x|+ 9 < 4

|5− 3x| < −5

What numbers have absolute value < −5?

Describe the above interval using compound inequality No real numbers

Write the inequality using the expression


Terms containing the variable on one side




Write the answer for each in interval notation.


Write the answer in interval notation,

taking in consideration the conjunction. {}


Homework Exercises



1. (16 points) Solve the inequalities.a. (8 pts) |4x− 3|+ 7 ≤ 4

b. (8 pts) |2x− 5| − 6 ≥ −3


2. (4 points)(a) I feel comfortable solving absolute value inequalities.


b. (4 pts) Write down and solve an absolute value inequality.



2.4. NONLINEAR INEQUALITIES 79

2.4 Nonlinear Inequalities


Suppose we wanted to solve the inequality

1

x> 4x.

The first step we may think of is multiplying both sides by x to cancel thedenominator, but after that step we would not know whether to change theinequality sign or keep it the same since we do not know whether x is positiveor negative. This section goes over solving inequalities of this type by usingtest-points.

Main Concept 1. To solve an inequality that is not linear:

1. Change it to an equation and solve it.

2. Find the values where the inequality is not defined.

3. On the number line, graph the numbers where the inequality is notdefined from step 2 and the solutions to the equation from step 1.

4. The number line is now separated into different parts. Choose a numberin each part and check whether the inequality is true.

5. Write the answer using interval notation to include all the parts of thenumber line where the inequality is true.

Example 1. Solve the inequalities:

a. 1x > 4x

b. 3x2 + 7x + 5 > 1

c. x2+2xx−1 ≤ 2x + 3

http://youtu.be/1stgZevM-jE


Solution 1.

a. 1x > 4x

Process Inequality

Change it to an equation and solve it 1x = 4x

1 = 4x2

0 = 4x2 − 1

0 = (2x− 1)(2x + 1)

x = 12 , x = − 1

2

Find the values where the inequality is not defined. x = 0

On the number line, graph the numbers

where the inequality is not defined0− 1

212

from step 2 and the solutions to

the equation from step 1.

Choose a number in each part Choosing x = −1 to the left of − 12 :

and check whether 1−1 > 4(−1)

the inequality is true −1 > −4 (True)

Choosing −0.1 between − 12 and 0:

1−0.1 > 4(−0.1)

−10 > −0.4 (Not True)


Choosing 0.1 between 0 and 12 :

10.1 > 4(0.1)

10 > 0.4 (True)

Choosing 2 to the right of 12 :

12 > 4(2) (Not Ture)

Write the answer in interval notation including0− 1

212

all the parts of the number line where Since the inequality is strict,

the inequality is true we cannot include − 12 or 1

2 .

And since the inequality is not defined

at 0, we will not include it either.

Write solution in interval notation (−∞,− 12 ) ∪ (0, 1

2 )


b. 3x2 + 7x + 5 > 1

Process Inequality

Change it to an equation and solve it 3x2 + 7x + 5 = 1

3x2 + 7x + 4 = 0

(3x + 4)(x + 1) = 0

x = − 43 , x = −1

Find the values where the inequality is not defined. It is defined everywhere


where the inequality is not defined− 4

3 −1



Choose a number in each part Choosing x = −2 to the left of − 43 :

and check whether 3(−2)2 + 7(−2) + 5 > 1

the inequality is true 12− 14 + 5 > 1 (True)

Choosing −1.1 between − 43 and −1:

3(−1.1)2 + 7(−1.1) + 5 > 1

Check in calculator (Not True)

Choosing 0 to the right of −1:

3(0)2 + 7(0) + 1 > 5 (True)


Write the answer in interval notation including− 4

3 −1

all the parts of the number line where Since the inequality is strict,

the inequality is true. we cannot include − 43 or −1.

Write solution in interval notation. (−∞,− 43 ) ∪ (−1,∞)


c. x2+2xx−1 ≤ 2x + 3

Process Inequality

Change it to an equation and solve it x2+2xx−1 = 2x + 3

x2 + 2x = (2x + 3)(x− 1)

x2 + 2x = 2x2 − 2x + 3x− 3

0 = x2 − x− 3 which does not factor.

So use the Quadratic Formula:

x =1±√

(−1)2−4(1)(−3)2(1)

The approximate values:

x ≈ 2.3, x ≈ −1.3

Find the values where the x− 1 = 0,

inequality is not defined. x = 1


where the inequality is not defined

1−√13

21 1+

√13

2



Choose a number in each part Choosing x = −3 to the left of 1−√13

2 :

and check whether (−3)2+2(−3)−3−1 ≤ 2(−3) + 3

the inequality is true 3−4 ≤ −3 ( Not true)


Choosing x = 0 between 1−√13

2 and 1:

0−1 ≤ 3 (True)

Choosing x = 2 between 1 and 1+√13

2 :

22+2(2)2−1 ≤ 2 · 2 + 3 (Not True)

Choosing x = 3 to the right of 1+√13

2

9+62 ≤ 9 (True)

Write the answer in interval notation

1−√13

21 1+

√13

2

including all the parts of the number Since the inequality includes =,

line where the inequality is true. we have to include 1±√13

2 .

And since the inequality is not defined

at 1, we will not include it.

Write solution in interval notation. [ 1−√13

2 , 1) ∪ [ 1+√13

2 ,∞)


Homework Exercises



1. (32 points) Solve the inequalities:a. (6 pts) 3x2 + 5x− 2 ≥ x− 3

b. (8 pts) 2x+4x−3 ≤ 4 + 1

x+1


c. (8 pts) x2+5x−3x + 2x > x + 2

d. (10 pts) |3x2 + 5x− 1|+ 4 ≤ 7


2. (4 points)(a) I feel comfortable solving nonlinear inequalities.


b. (4 pts) Write down and solve an nonlinear inequality.



Chapter 3

Functions

3.1 The Cartesian Coordinate System

Textbook Information. For more information on this topic, moreexamples and more exercises see Section 2.1

For a video presentation of The Carteisian Coordinate System and the Dis-tence Formula click on this link.

3.1.1 Introduction

Main Concept 1. Points on the plane can be represented by anordered pair of real numbers. This pair describes how to get from theorigin to a point on the plane. This first number of the pair is called thex-coordinate and it indicates how far to go to the right (positive) or left(negative) of the origin. The second is called the y-coordinate and itindicates how far to go up (positive) or down (negative).

Ax

y

III

III IV

As it can be seen from the picture, the axis separate the plane into fourparts. We call these parts quadrants. They are labeled as shown above.

http://youtu.be/yNFiGMMJdWE

http://youtu.be/yNFiGMMJdWE

90 CHAPTER 3. FUNCTIONS

Example 1. Find the coordinates of the points shown on the graph below:

−6 −5 −4 −3 −2 −1 1 2 3 4 5

−6

−5

−4

−3

−2

−1

1

2

3

4

5

0

A

B

C

D

E

Solution 1. To answer these problems remember that the coordinates of apoint give a description of how to go from (0, 0) to the point on the plane.

a. To go from (0, 0) to A we have to go 1 to the right (the first coordinate is1) and 1 up (the second coordinate is 1). Therefore the coordinates of A are(1, 1).

b. B = (3, 2)

c. To go from the origin to C we have to go 4 to the left (the first coordinate is−4) and 1 down (the second coordinate is −1). So C = (−4,−1).

d. D = (−1, 5)

e. E = (4,−2)

Example 2. Graph the following points. Which quadrant or coordinateaxis does each lie?

A. (−1, 3)

B. (3,−1)

C. (0,−4)

D. (4, 2).

3.1. THE CARTESIAN COORDINATE SYSTEM 91

Solution 2. The points are graphed below:

−5 −4 −3 −2 −1 1 2 3 4

−5

−4

−3

−2

−1

1

2

3

4

0

A

B

C

D

a. A lies in the second quadrant.

b. B lies in the fourth quadrant.

c. C lies on the y-axis.

d. D lies in the first quadrant.

3.1.2 Distance Formula

If both the x and y axis use the same unit, then we can find the distance betweentwo points using the same unit.

Main Concept 2. Let A = (x0, y0) and B = (x1, y1), then the distancefrom A to B, d, is given by the formula: d =

√(x1 − x0)2 + (y1 − y0)2.

Three points lie on the same line if

Largest Distance = The sum of the other two.

Three point form a right triangle if the following holds:


Largest Distance 2 = Another Distance 2 + Third Distance 2.

Area of triangle =Base · Height

2

Example 3. Find the distance between the points: A = (1, 2) and B =(−1, 3)

Solution 3. This follows directly from the formula given above by substi-tuting x0 = 1, y0 = 2, x1 = −1 y1 = 3 :

d =√

(−1− 1)2 + (3− 2)2

=√

(−2)2 + 12

=√

5

Example 4. Determine whether the three points, A = (0,−1), B = (2, 3)and C = (3, 4), lie on a straight line.

Solution 4. Three points lie on a straight line if the largest distance distancebetween them is a sum of the other two distances. So we first need to find allthree distances between these points then check whether this last statement istrue.

|AB| =√

(2− 0)2 + (3− (−1))2

=√

4 + 16

=√

20

= 2√

5

|BC| =√

(3− 2)2 + (4− 3)2

=√

2

|AC| =√

(3− 0)2 + (4− (−1))2

=√

9 + 25

=√

34

To check whether three points lie on the same line, we check whether

Largest Distance = The sum of the other two.


From the three distances above, the largest one is√

34. Since√

34 6= 2√

5 +√

2the points A, B and C do not lie of a straight line.

Example 5. Determine whether the three points, A = (−2, 5), B = (12, 3)and C = (10,−11), form a right triangle. If so, find the area of the triangle.

Solution 5. Three point form a right triangle if the following holds:

largest distance 2 = other distance 2 + third distance 2.

As before we have to calculate the three distances between the points andcheck whether the above equation holds.

|AB| =√

(12− (−2))2 + (3− 5)2

=√

142 + (−2)2

=√

196 + 4

=√

200

= 10√

2

|BC| =√

(10− 12)2 + (−11− 3)2

=√

(−2)2 + 142

=√

200

= 10√

2

|AC| =√

(10− (−2))2 + (−11− 5)2

=√

122 + 162

=√

400

= 20

The largest distance is 20 so we need to check:

202 = (10√

2)2 + (10√

2)2

400 = 100 · 2 + 100 · 2

400 = 200 + 200

Which is true. So the three points form a right triangle.To find the area of the triangle remember that

Area of a triangle =base · height

2


For any right triangle the largest leg is the hypotenuse, and the other legs canbe taken to be one the base, the other the height. So in our case

Area of triangle =10√

2 · 10√

2

2

=200

2

= 100.

3.1.3 Midpoint Formula

For a video presentation of The Midpoint Formula click on this link.

Main Concept 3. Given A = (x0, y0) and B = (x1, y1),the pointmidway between A and B is called the midpoint. The coordinates of themidpoint are given by: M = (x0+x1

2 , y0+y1

2 ).

Example 6. Find the midpoint of the line segment joining A = (−3, 4) andB = (−4,−4)

Solution 6. The solution follows directly from the formula above by sub-stituting x0 = −3, y0 = 4, x1 = −4 and y1 = −4:

M =

(−3 +−4

2,

4 + (−4)

2

)

=

(−7

2, 0

).

http://youtu.be/Vm0eRBZ6Mi4


Homework Exercises



3. (13 points) Let A = (−1, 1), B = (2, 1), and C = (1,−2).a. (3 pts) Find the distance from A to B.

b. (3 pts) Find the distance from A to C.

c. (3 pts) Find the distance from B to C.

d. (4 pts) Determine algebraically whether the three points lie onthe same line.


4. (13 points) Let A = (0, 2), B = (−2, 4), and C = (1, 3).a. (3 pts) Find the distance from A and B.

b. (3 pts) Find the distance from B to C.

c. (3 pts) Find the distance from A to C.

d. (4 pts) Determine algebraically whether the three points form aright triangle.


5. (2 points) Find the midpoint between A = (−3, 4) and B = (1,−8).

6. (5 points)(a) I feel comfortable determining whether three points lie on a

straight line.


b. (5 pts) Randomly choose three points and determine whetherthey lie on a straight line.

7. (5 points)(a) I feel comfortable determining whether three points form a

right triangle.


b. (5 pts) Randomly choose three points and determine whetherthey form a right triangle.


8. (4 points)(a) I feel comfortable finding the midpoint between two points.


b. (4 pts) Randomly choose two points and find the midpointbetween the two.



3.2. INTRODUCTION TO FUNCTIONS 99

3.2 Introduction to Functions

Textbook Information. For more information on this topic, moreexamples and more exercises see section 3.1.

For a video presentation of this section click on this link.

3.2.1 Relations

Definition 1. A relation is a correspondence between two sets.

Examples of relations:

• The set of students and the model of the car they drive.

• The list of all the makes of cars and the name of student(s) who drive thatmake.

• Numbered bags of m&m’s (as is first bag, second bag, third bag ...) andthe number of candy in each bag.

Definition 2. The first set (whose elements are assigned to some

element(s) on the second set) is called the domain. The second set (whichhas the elements we assign) is called the range.

Example 1. For each relation given as an example above, find the domainand the range.

Solution 1.

• The set of all students is the domain, and the set of the models of the carsthey drive is the range.

• The set of all the makes of cars is the domain and the set of students isthe range.

• The numbered bags is the domain and the number of candy in each bagis the range.

3.2.2 Functions

http://youtu.be/TC2x9c0L4Hk


Definition 3. A function is a relation from a set X to a set Y that

associates to each element of X exactly one element from Y . The values inX are called inputs of the function, the values in Y corresponding to anelement from X are called outputs.

Example 2. For each of the relations given above, determine whether it is afunction.

Solution 2.

• Each student will drive one car, that is to each student we can associateone model. Therefore the relation is a function.

The only way this would not be a function is if one student drove two cars.

• Each make might correspond to more than one student. That is theremight be two or more students driving the same make. Therefore therelation is not a function.

• Since each bag can only correspond to one number (the number of candyit contains), this relation is a function.

Functions can be represented in different ways.

• Verbally.

• Geometrically as a table or Venn Diagram.

• Algebraically by a formula that describes how to get the output knowingthe input.

• Graphically by a pair of coordinates (x, y) where x is the input and y isthe output.

In this class we will focus on the last two ways of representing a function.

3.3. REPRESENTING A FUNCTION ALGEBRAICALLY 101

3.3 Representing a Function Algebraically


Textbook Information. For more information on this topic, moreexamples and more exercises see sections 3.1.

Functions represented algebraically are denoted by letter f, g, F,G, h, · · · , the in-puts are denoted by x and the outputs are denoted by f(x), g(x), F (x), G(x), h(x), · · ·depending on the letter used to represent the function.

Sometimes you might see the function written as a formula involving twovariables one for the input (usually x) and one for the output (usually y).

3.3.1 Evaluating a Function

Main Concept 1. When asked to find f(expression) substitute theexpression for x and simplify.

Example 1. Given f(x) = 3x2 − 2 find the following outputs of f .

a. f(−1)

b. −f(1)

c. f(1) + x

d. f(1 + x)

e. f(−x)

f. f(x+h)−f(x)h

Solution 1.

a. Substitute −1 for x:f(−1) = 3(−1)2 − 2

= 3− 2

1

b.−f(1) = −

(3(1)2 − 2

)= −(3− 2)

= −1

http://youtu.be/pDijaBtCWMw


c.f(1) + x = 3(1)2 − 2 + x

= 1 + x

d.f(1 + x) = 3(1 + x)2 − 2

= 3(1 + 2x + x2)− 2

= 3 + 6x + 3x2 − 2

= 3x2 + 6x + 1

e.f(−x) = 3(−x)2 − 2

= 3x2 − 2

f. To find f(x+h)−f(x)h first find f(x + h):

f(x + h) = 3(x + h)2 − 2

= 3(x2 + 2xh + h2)− 2

= 3x2 + 6xh + 3h2 − 2

It is given that f(x) = 3x2 − 2. Substitute these expressions in what you aretrying to find:

f(x + h)− f(x)

h=

(3x2 + 6xh + 3h2 − 2

)−(3x2 − 2

)h

=3x2 + 6xh + 3h2 − 2− 3x2 + 2

h

=6xh + 3h2

h

= 6x + 3h

3.3.2 Finding the Domain of a Function

Definition 1. The domain of a function that is represented alge-

braically includes all the numbers for which the function can be evaluatedand the output will be a real number.

Main Concept 2. A function is not defined, or the output will notbe a real number if:


• We divide by zero.

• We take the even root of negative numbers.

• We take the logarithm of zero or a negative number. (This will becovered later in class.)

• In application problems consider what the variable x represents andits possible values. For example, length or time cannot be negative.

So to find the domain we find the set of numbers we need to put in asinputs, so that the above situations are avoided.

Example 2. Find the domain for each of the following functions:

a. f(x) = x−4x2−2x−3

b. g(x) = x2 + 3

c. h(x) =√

3x− 1

d. f(x) =√

3x− 1 + 1

e. g(x) = 3√

3x− 1

Solution 2.

a. This function has a denominator (x2 − 2x− 3) so :

x2 − 2x− 3 6= 0

(x− 3)(x + 1) 6= 0

x 6= 3 and x 6= −1

Therefore the domain will include all numbers besides 3 and −1:

−5 −4 −3 −2 −1 1 2 3 40

The domain is (−∞,−1) ∪ (−1, 3) ∪ (3,∞).

b. This function has no denominators, even roots or logarithms therefore thedomain is all the real numbers.

c. This function has an even root, so the inside of the root (3x − 1) has to bepositive.

3x− 1 ≥ 0


3x ≥ 1

x ≥ 1

3

Therefore the domain is [ 13 ,∞).

d. As in the previous example, this function has an even root so the inside ofthe root (3x− 1) has to be positive.

3x− 1 ≥ 0

3x ≥ 1

x ≥ 1

3

Therefore the domain is [ 13 ,∞).

Notice that the +1 outside of the even root does not affect the domain.

e. Since this function has no even roots or denominators, its domain is all thereal numbers.


Homework Exercises



1. (18 points) Let g(x) = 2x +√x2 − 4. Find each function value.

a. (2 pts) g(0)

b. (2 pts) g(1)

c. (2 pts) g(2)

d. (2 pts) g(−3)

e. (5 pts) g(−x)

f. (5 pts) g(2 + k)


2. (28 points) Find the domain of each function:a. (7 pts) f(x) = 2x +

√x + 3

b. (7 pts) f(x) = 1x2+1

c. (7 pts) f(x) = 1−xx2+5x+6

d. (7 pts) f(x) = x2 + 3 +√

2x + 8


3. (10 points) Given f(x) = x2 − x find:a. (5 pts) f(x + h)

b. (5 pts) f(x+h)−f(x)h for h 6= 0.


4. (4 points)(a) I feel comfortable evaluating functions.


b. (4 pts) Write down a function f(x) and evaluate it at x + h.

5. (4 points)(a) I feel comfortable finding the domain of a function.


b. (4 pts) Write down a function f(x) and find its domain.



3.4. REPRESENTING A FUNCTION GRAPHICALLY 109

3.4 Representing a Function Graphically


For a video presentation of the introductory material up to example 2 clickon the links to the two videos.

Video 1Video 2For a video presentation of the material after the end of example 2 click on

the link to the thrid video on the section.Video 3

Definition 1. The graph of a relation consists of all the points in

the plane with coordinates satisfying the relation. The first coordinate isan input and the second coordinate is the corresponding output.

Suppose a relation represents a function y = f(x), then a point (a, b) isin the graph of f(x) if f(a) = b.

Main Concept 1. Recall that a relation is a function if for each inputwe have one output. In the graphical representation this implies that eachvertical line intersects the graph in at most one point.

Example 1. Determine whether the following graphs represent functions:

a.

−4 −2 2 4

−2

2

4

6

0

http://youtu.be/B3XIYCkN_0c

http://youtu.be/7CbNrtvuq8Y

http://youtu.be/38JEkrBEOdY


b.

−5 −4 −3 −2 −1 1 2 3 4

−5

−4

−3

−2

−1

1

2

3

4

0

c.

−5 −4 −3 −2 −1 1 2 3 4

−5

−4

−3

−2

−1

1

2

3

4

0

d.

−3 −2 −1 1 2

−3

−2

−1

1

2

0


Solution 1.

a. If we imagine a vertical line passing through the x-axis we will see that it willtouch the graph in at most one point. So this graph represents a function.

b. Any vertical line will touch the graph in at most one point, so this graphrepresents a function.

c. Vertical lines on the positive side on the x-axis will touch the graph at twopoints. Therefore the graph is not a function.

d. A vertical line through x = 0 or any other number between −1 and 1 willtouch the graph at 2 points. Therefore this graph is not a function.

Reading the Graph of a Function

Main Concept 2. The graph of a function consists of points (x, y)for which y = f(x).

• f(expression) is the y-coordinate of the point on the graph with x-coordinate equal to expression.

• The domain of a function f is the set of inputs x where f is defined.In a graph this means that the domain of f is the part of the x axiscovered by the graph of f .

• The range of a function f is the set of its outputs y. In a graph thismeans that the range is the part of the y axis covered by the graph.

• The x-intercept(s) are/is the point(s) where the graph intersects thex-axis.

• The y-intercept is the point where the graph intersects the y-axis.

• The function is increasing for inputs in (a, b) if as we move from x = ato x = b the graph goes up.

• The function is decreasing for inputs in (a, b) if as we move from x = ato x = b the graph goes down.

• The function is constant for inputs in (a, b) if for all the values of x is(a, b) the value of f(x) is the same.

• A point (a, b) on the graph of f(x) is called a local maximum if f(x) ≤b for all x close to a.

• A point (a, b) on the graph of f(x) is called a local minimum if f(x) ≥b for all x close to a.


• A point (a, b) on the graph of f(x) is called an absolute maximum iff(x) ≤ b for all x in the domain of f .

• A point (a, b) on the graph of f(x) is called an absolute minimum iff(x) ≥ b for all x in the domain of f .

Example 2. Use the following graph to answer the questions:

−1 −0.5 0.5 1 1.5 2 2.5

−5

−4

−3

−2

−1

1

2

3

4

0

a. What are f(0), f (0.5) and f(1.25)?

b. What is the domain of f?

c. What is the range of f?

d. List the intercepts.

e. For what values of x does f(x) = −4?

f. For what values of x is f(x) > 0?

g. Where is the function increasing?

h. Where is the function decreasing?

i. Find the local maximums of f(x).

j. Find the local minimums of f(x).

k. Find the absolute maximums of f(x).

l. Find the absolute minimums of f(x).

Solution 2.


a. The values inside parenthesis give the values of the x-coordinates. From therefind the y-coordinate of the corresponding point on the graph.

−1 −0.5 0.5 1 1.5 2 2.5

−5

−4

−3

−2

−1

1

2

3

4

0

x=0

(0, 4)

Sof(0) = 4.

−1 −0.5 0.5 1 1.5 2 2.5

−5

−4

−3

−2

−1

1

2

3

4

0x=0.5

(0.5,−4)

Sof(0.5) = −4.

−1 −0.5 0.5 1 1.5 2 2.5

−5

−4

−3

−2

−1

1

2

3

4

0

x=1.25(1.25, 0)


So

f(1.25) = 0.

b. The domain of f will be the part of the x-axis where f is defined:

−1 −0.5 0.5 1 1.5 2 2.5

−5

−4

−3

−2

−1

1

2

3

4

0

Therefore the domain is

[0, 2].

c. The range of f includes the part of the y-axis covered by the graph of f :

−1 −0.5 0.5 1 1.5 2 2.5

−5

−4

−3

−2

−1

1

2

3

4

0

Therefore the range of f is

[−4, 4].

d. The x-intercepts are the points where the graph intersects the x-axis. So thex-intercepts of f are (0.25, 0), (0.75, 0), (1.25, 0), (1.75, 0).

The y-intercepts is the point where the graph intersects the y-axis. So they-intercepts of f is the point (0, 4)

e. Here we are finding the x-coordinate(s) of the point(s) that have y-coordinate−4:


−1 −0.5 0.5 1 1.5 2 2.5

−5

−4

−3

−2

−1

1

2

3

4

0

From the above graph we conclude that for x = 0.5 and x = 1.5, f(x) = −4.

f. To find the values of x where f(x) > 0 first find which part of the graph hasy > 0. Be careful not to include the points where y = 0:

−1 −0.5 0.5 1 1.5 2 2.5

−5

−4

−3

−2

−1

1

2

3

4

0

Then find the x-coordinates that correspond to this part of the graph:

−1 −0.5 0.5 1 1.5 2 2.5

−5

−4

−3

−2

−1

1

2

3

4

0


Therefore f(x) > 0 for x in

[0, 0.25) ∪ (0.75, 1.25) ∪ (1.75, 2].

g. The function is increasing when it goes up from left to right.

−1 −0.5 0.5 1 1.5 2 2.5

−5

−4

−3

−2

−1

1

2

3

4

0

Therefore the function is increasing on (0.5, 1) ∪ (1.5, 2).

h. The function is decreasing when it goes down as it moves from the left to theright.

−1 −0.5 0.5 1 1.5 2 2.5

−5

−4

−3

−2

−1

1

2

3

4

0

Therefore it is decreasing on (0, 0.5) ∪ (1, 1.5).

i. The local maximums are (0, 4), (1, 4) and (2, 4).

j. The local minimums are (0.5,−4), (1.5,−4).

k. The local maximums are also absolute maximums since their y-coordinatesare the largest in the graph.

l. The local minimums are also absolute minimums since their y-coordinatesare the smallest in the graph.


Example 3. Use the following graph to answer the questions:

−10−9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 8 9

−10

−9

−8

−7

−6

−5

−4

−3

−2

−1

1

2

3

4

5

6

7

8

9

0

a. What is the domain of f?

b. What is the range of f?

c. What is f(0)?

d. List the intercepts.

e. For what values of x is f(x) = −5?

f. For what values of x is f(x) < 1?

g. Where is f(x) increasing?

h. Where is f(x) decreasing?

i. Find the local minimum(s) of f(x).

j. Find the absolute minimum(s) of the function.

k. Find the local maximum(s) of the function.

l. Find the absolute maximum(s) of the function.


Solution 3.

a. As we move from left to right we record the x-coordinates of the startingand ending points. Since both the starting and ending points are part of thegraph their x-coordinates will be included. So the domain is [−1, 6].

b. As we move form down up, we record the y-coordinates of the starting andending points. Since the starting and ending points are part of the graph,we will include their y-coordinates. So the range is [−8, 3].

c. We find f(0) by going on the x-axis where x = 0 and move up or down to thepoint of the graph. The value of f(0) will be the y-coordinate of the point.So f(0) = 1.

d. The x-intercepts are the points where the graph intersects the x-axis. There-fore the x-intercepts for this graph will be approximated to (−0.2, 0) and(2.2, 0).

The y-intercept is the point where the graph intersects the y-axis. Thereforethe y-intercept for this graph is (0, 1).

e. The equation f(x) = −5 indicates that y = −5, so go on the y-axis wherey is −5 and move to the left/right to where the graph is. The values of xwhere f(x) = −5 will be the x-coordinates of these points. Make sure youinclude all the parts of the graph. For this graph f(x) = −5 for x = −1, 3and 5.

f. The inequality f(x) < 1 is the same as y < 1. The graph of the equationy = 1 is the horizontal line going through the y-axis at 1, and y < 1 belowthat line. Draw the line where y = 1 and record the x-coordinates of thestarting and ending points of the parts of the graph below y = 1. For ourexample we get the x-coordinates from −1 to 0 and from 2 to 6. We will useparenthesis (, or , ) for the endpoints that have y-coordinate equal to 1, sincein that case y is not less than 1 and use brackets [, or , ] for the endpointswhere the y-coordinate is less than 1. Therefore the answer will be

[−1, 0) ∪ (2, 6]

g. The function is increasing when it goes up as we go to the right. In the answerrecord the x-coordinates of the starting and ending points of the parts wherethe function goes up. Always use open intervals (that is parenthesis). So fis increasing (−1, 1) ∪ (3, 4).

h. The function is decreasing when it goes down as we go to the right. Inthe answer record the x-coordinates of the starting and ending points ofthe parts where the function goes down. Always use open intervals (that isparenthesis). So f is decreasing (−1, 3) ∪ (4, 6).

i. The local minimums of f(x) will be the coordinates of the low points (−1,−5),(3,−5) and (6,−8).


j. The absolute minimum(s) of f(x) will be the lowest point(s) of the graph:(6,−8).

k. The local maximums of f(x) will be the coordinates of the high points (1, 3)and (4,−4).

l. The absolute maximum(s) of f(x) will be the highest point(s) of the graph:(1, 3).

Connecting Algebraic and Graphical Representa-tions of a Function

Example 4. Let f(x) = xx+1

a. What is f(2)? To which point on the plane does this correspond?

b. Is the point(1, 1

2

)on the graph of f?

c. If f(x) = 2, what is x? To which point(s) on the plane does this correspond?

Solution 4.

a. Recall that to find f(expression) substitute x = expression and the outputis the same as the y-coordinate. So

f(2) =2

2 + 1

=2

3,

and this corresponds to the point(2, 2

3

)on the graph of f .

b. To check whether the point(1, 1

2

)is on the graph of f we check whether

f(1) = 12 :

f(1) =1

1 + 1

=1

2

Hence the point(1, 1

2

)is on the graph of f .

c. Since f(x) = 2 and f(x) = xx+1 we have:

2 =x

x + 1

Solve this equation to find x:


2(x + 1) = x

2x + 2 = x

x = −2

This does not make the denominator zero, so it is a solution to the equationand it corresponds to the point (−2, 2) on the graph of f .

Note 1. One might have gotten no solution to the equation then there willbe no points on the graph of f which have f(x) = 2. Similarly, one mighthave gotten more than one solution to the equation, then there would bemore than one point on the graph of f(x) with y = 2.


Homework Exercises



1. (27 points) Use the graph of f given below to answer the questions:

−6 −5 −4 −3 −2 −1 1 2 3 4 5 6

−5

−4

−3

−2

−1

1

2

3

4

5

0

a. (4 pts) What is the domain of f?

b. (4 pts) What is the range of f?

c. (2 pts) What is f(0)?

d. (3 pts) For what values of x is f(x) = 0?

e. (4 pts) For what values of x is f(x) < 0?

f. (3 pts) Where is f(x) increasing?

g. (3 pts) Were is f(x) decreasing?

h. (2 pts) Find the local minimums and the (absolute) minimum ofthe function.

i. (2 pts) Find the local maximums and the (absolute) maximumof the function.


2. (27 points) Use the graph of f given below to answer the questions:

−7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7

−7

−6

−5

−4

−3

−2

−1

1

2

3

4

5

6

7

0

a. (4 pts) What is the domain of f?

b. (4 pts) What is the range of f?

c. (2 pts) What is f(1)?

d. (3 pts) For what values of x is f(x) = −3?

e. (4 pts) For what values of x is f(x) < 1?

f. (3 pts) Where is f(x) increasing?

g. (3 pts) Where is f(x) decreasing?

h. (2 pts) Find the local minimum(s) and the absolute minimum ofthe function.

i. (2 pts) Find the local maximum(s) and the absolute maximumof the function.


3. (4 points)(a) I feel comfortable finding the domain of a function represented

graphically.


b. (4 pts) Draw the graph of a function and find its domain.

−7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7

−7

−6

−5

−4

−3

−2

−1

1

2

3

4

5

6

7

0

4. (4 points)(a) I feel comfortable finding the range of a function represented

graphically.


b. (4 pts) Use the graph you drew in the previous exercise and tofind its range.

5. (4 points)(a) I feel comfortable evaluating a function represented graphically.


b. (4 pts) Use the graph you previously drew to find f(2).


6. (4 points)(a) I feel comfortable solving an equation of the form f(x) = a

when the graph of f(x) is given.


b. (4 pts) Use the graph you previously drew to find where

f(x) = −1.

7. (4 points)(a) I feel comfortable solving inequalities of the form f(x) > a

when the grapg of f(x) is given.


b. (4 pts) Use the graph you previously drew to find where

f(x) < 1.

8. (4 points)(a) I feel comfortable finding where a function represented graph-

ically is increasing.


b. (4 pts) Use the graph you previously drew to find where f(x) isincreasing.


9. (4 points)(a) I feel comfortable finding where a function represented graph-

ically is decreasing.


b. (4 pts) Use the graph you previously drew to find where f(x) isdecreasing.

10. (4 points)(a) I feel comfortable finding the absolute minimums of a function

represented graphically.


b. (4 pts) Use the graph you previously drew to find the absoluteminimums of f(x).

11. (4 points) (a) I feel comfortable finding the localminimums of a function represented graphically.


b. (4 pts) Use the graph you previously drew to find the localminimums of f(x).


12. (4 points)(a) I feel comfortable finding the absolute maximums of a function

represented graphically.


b. (4 pts) Use the graph you previously drew to find the absolutemaximums of f(x).

13. (4 points) (a) I feel comfortable finding the localmaximums of a function represented graphically.


b. (4 pts) Use the graph you previously drew to find the localmaximums of f(x).



3.5. FUNCTION OPERATIONS 127

3.5 Function Operations

For a video presentation of the part of section up to ”Function Decomposi-tion” click on this link.

Textbook Information. For more information on this topic, moreexamples and more exercises see sections 3.1, 6.1.

3.5.1 Basic Algebraic Operations

Main Concept 1. Given two functions f(x) and g(x) the sum func-tion (f+g)(x) is defined as f(x)+g(x), the difference function (f−g)(x)is defined as f(x) − g(x), the product function (f · g)(x) is defined as

f(x) · g(x), and when g(x) 6= 0 the quotient function(

fg

)(x) is defined

as f(x)g(x) .

Example 1. Given f(x) = x2 − 3x and g(x) = x− 1 find:

a. (f + g) (x)

b. (f − g) (3)

c. (fg) (x)

d.(

fg

)(x)

Solution 1.

a.(f + g)(x) = f(x) + g(x)

= (x2 − 3x) + (x− 1)

= x2 − 2x− 1

b.(f − g)(3) = f(3)− g(3)

= (32 − 3(3))− (3− 1)

= 9− 9− 2

= −2

http://youtu.be/cjaYGELLzGM

http://youtu.be/cjaYGELLzGM


c.

(fg) (x) = f(x)g(x)

= (x2 − 3x)(x− 1)

= x3 − x2 − 3x2 + 3x

= x3 − 4x2 + 3x

d. (f

g

)(x) =

f(x)

g(x)

=x2 − 3x

x− 1

=x2 − 3x

x− 1

3.5.2 Composition of Functions

Definition 1. Given two functions f(x) and g(x), the composite of

f with g is denoted by f ◦ g and is defined by

f ◦ g(x) = f(g(x)).

Main Concept 2. Given f(x) and g(x) to find f ◦ g(x) first writeit as f ◦ g(x) = f(g(x)), then starting with x ”unravel” the expression outof the parenthesis. So, find g(x) and substitute that in f(g(x)). Then,in the formula for f(x), substitute for x the expression inside parenthesis.Simplify the expression.

Example 2. Suppose f(x) = 2x2 + 3 and g(x) = 4x3 + 1. Find:

a. f ◦ g(1)

b. g ◦ f(1)

c. f ◦ f(−2)

d. g ◦ g(−1)

Solution 2.


a.

f ◦ g(1) = f(g(1))

= f(4 · 13 + 1)

= f(5)

= 2 · 52 + 3

= 53.

b.

g ◦ f(1) = g(f(1))

= g(5)

= 4 · 125 + 1

= 501.

c.

f ◦ f(−2) = f(f(−2))

= f(11)

= 2 · 121 + 3

= 245.

d.

g ◦ g(−1) = g(g(−1))

= g(−3)

= 4 · (−27) + 1

= −107.

Main Concept 3. To find the domain of the composition of twofunctions first find the formula for the composite function (do not simplifyit), then find its domain.

Example 3. Suppose f(x) = 2x2 + 3 g(x) = 4x3 + 1. Find

a. f ◦ g

b. g ◦ f

Then find the domain of each composite.Solution 3.


a.f ◦ g(x) = f(g(x))

= f(4x3 + 1)

= 2(4x3 + 1)2 + 3

At this point it can be seen that the domain is all the real numbers. Simpli-fying this more:

= 2(16x6 + 8x3 + 1) + 3

= 32x6 + 16x3 + 5.

b.g ◦ f(x) = g(f(x))

= g(2x2 + 3)

= 4(2x2 + 3)3 + 1

So the domain of g ◦ f is all the real numbers. Simplifying:

= 4(4x4 + 12x2 + 9)(2x2 + 3) + 3

= 4(8x6 + 12x4 + 24x4 + 36x2 + 18x2 + 27) + 3

= 4(8x6 + 36x4 + 54x2 + 27) + 3

= 32x6 + 144x4 + 216x2 + 108 + 3

= 32x6 + 144x4 + 216x2 + 111.

Example 4. Find the domain of f ◦ g(x) for f(x) = 1x+2 , g(x) = 4

x−1Solution 4. As stated above, to find the domain of a composition, find its

formula (do not simplify).

f ◦ g(x) = f

(4

x− 1

)

=1

4x−1 + 2

.

From here one can see that since the domain cannot include the values wherethe denominators are zero, it cannot include the values where 4

x−1 + 2 = 0 orx− 1 = 0. Therefore:

4

x− 1+ 2 6= 0 multiply by (x− 1) for x 6= 1

4 + 2(x− 1) 6= 0

4 + 2x− 2 6= 0

2x 6= −2


x 6= −1.

Andx− 1 6= 0

x 6= 1.

Hence the domain includes all the real numbers besides −1 and 1. Writtenin interval notation Domain = (−∞,−1) ∪ (−1, 1) ∪ (1,∞).

Note 1. If we continued to simplify the formula for the composite in theabove example we would get:

=1

4x−1 + 2

=1

4x−1 + 2x−2

x−1

=1

2x+2x−1

=x− 1

2x + 2.

This function and the composite in the example do not have the same domain!Example 5. Let f(x) = 1

x and g(x) =√x− 1. Find:

a. f ◦ g

b. f ◦ fThen find the domain of each composite function.

Solution 5.

a.f ◦ g(x) = f(

√x− 1)

=1√x− 1

The domain of f ◦ g(x) cannot include√x− 1 = 0 and has to include only

values for which x − 1 ≥ 0. That is x 6= 1 and x ≥ 1. This is the same asx > 1. In interval notation: Domain = (1,∞).

b.

f ◦ f(x) = f

(1

x

)=

11x

.

So the domain cannot include 1x = 0 or x = 0. Since 1

x 6= 0 for any realnumber x, the first condition has no restrictions, and the only restriction onthe domain is x 6 0. In interval notation Domain = (−∞, 0) ∪ (0,∞).

Simplifying the composite:

=11x

= x.


3.5.3 Decomposing Functions

For a video presentation of the rest of this section click on this link.

Main Concept 4. In some instances, given a function we need towrite it as a composition. In this case consider the “inside” and “outside”functions. The “outside” function is the last operation performed whenevaluating the function, and the “inside” function is the rest of the functionasked to decompose.

Once the “outside” and “inside” functions are found compare them tothe “inside” and “outside” functions of the composition. So, if asked towrite the function as a composition f ◦ g(x) then f(x) will be equal to the“outside” function and g(x) will be equal to the ”inside” function.

Example 6. Find two functions f(x) and g(x) so that the function H(x) =(x− 1)2 as the composition f ◦ g(x).

Solution 6. When evaluating the function H(x) = (x−1)2 squaring will bethe last operation. So x2 is the “outside” function, and what is being squared,x− 1, is the “inside” function. Therefore f(x) = x2 and g(x) = x− 1. One mayalso check that indeed f ◦ g(x) = H(x):

f ◦ g(x) = f(g(x))

= f(x− 1)

= (x− 1)2

= H(x).

Example 7. Find two functions f and g such that H(x) = (x2 + 1)50 is thecomposite f ◦ g(x).

Solution 7. Consider the “inside” and “outside” functions for H(x) andfor f ◦ g(x) = f(g(x)) and match them. So the “inside” of H(x) (x2 + 1) is the“inside” for f(g(x)) that is g(x), and the “outside of H(x) (x50) is the outsidefor f(g(x)) that is f(x). Hence for f(x) = x50 and g(x) = x2 + 1 we haveH(x) = f ◦ g(x).

Example 8. Find functions f and g such that f ◦ g(x) = 1x+1 .

Solution 8. As in the previous example look at the “inside” and “outside”functions. Here we are using x + 1 as an input for 1

x . So x + 1 is the “inside”and 1

x is the “outside” function. That is f(x) = 1x and g(x) = x + 1.

http://youtu.be/sc9DRBih8LU


Homework Exercises



1. (12 points) Given f(x) = 1√x+2

and g(x) = 2x + 1 find the following

values:a. (3 pts) (f + g)(−1)

b. (3 pts) (f − g)(0)

c. (3 pts) (fg)(2)

d. (3 pts)(

fg

)(1)

2. (15 points) Let f(x) = 3x2 + 2 and g(x) = xx−2

a. (3 pts) Find f ◦ g(1)

b. (7 pts) Find g ◦ f(x).

c. (5 pts) Find the domain of g ◦ f(x).


3. (13 points) Given f(x) =√3x−4x and g(x) = x2 + 1 find the following

values:a. (3 pts) (f + g)(x)

b. (3 pts)(

fg

)(x)

c. (7 pts) Find g ◦ f(x).


4. (5 points) Given H(x) =√

3x− 4, find two functions f(x) and g(x)such that H(x) = f ◦ g(x)

5. (4 points)(a) I feel comfortable performing basic algebraic operations with

functions.


b. (4 pts) Write down two functions f(x) and g(x) and find (f ·g)(x).

6. (4 points)(a) I feel comfortable composing two functions.


b. (4 pts) Write down two functions f(x) and g(x) and find f ◦g(x)and g ◦ f(x).

7. (4 points)(a) I feel comfortable decomposing a function.


b. (4 pts) Write down a function and write it as a composition oftwo other functions.

3.6. ONE-TO-ONE FUNCTIONS 137

3.6 One-to-One Functions



Definition 1. A function is one-to-one if any two different inputs in

the domain correspond to different outputs in the range.

3.6.1 Determining Whether a Function is One-to-One

Main Concept 1. When a function is represented graphically, todetermine whether it is a one-to-one function we check that each verticalline intersects the graph at only one point.

Example 1. From the graph of each function determine whether it is one-to-one.

a.

−8 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7

−3

−2

−1

1

2

3

4

5

6

7

8

9

10

11

12

13

0

http://youtu.be/kEyCPuwvVJY


b.

−4 −3 −2 −1 1 2 3

−7

−6

−5

−4

−3

−2

−1

1

2

3

4

5

6

0

c.

−10−9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 8 9

−10

−9

−8

−7

−6

−5

−4

−3

−2

−1

1

2

3

4

5

6

7

8

9

0

Solution 1.


a. If a horizontal line goes through the y-axis, it will cross the graph at twopoints. Therefore the function is not one-to-one.

b. If a horizontal line goes through the y-axis, it will cross the graph at twopoints. Therefore the function is not one-to-one.

c. If a horizontal line goes through the y-axis, it will always cross the graph atonly one point. Therefore the function is one-to-one.

Main Concept 2. When a function is represented algebraically, todetermine whether it is a one-to-one function we check that we are able tosolve for x as a function of y and get a unique answer.

Example 2. Given f(x) = 2x+1x−1 determine algebraically (without graph-

ing) whether the function is one-to-one.

Solution 2. Instead of f(x) we can write y:

y =2x + 1

x− 1

Solve this equation for x:

y(x− 1) = 2x + 1

yx− y = 2x + 1

yx− 2x = 1 + y

x(y − 2) = 1 + y

x =1 + y

y − 2

Since we were able to solve for x and get a unique answer, the function isone-to-one.

Example 3. Given f(x) = x2+5 determine algebraically (without graph-ing) whether the function is one-to-one.

Solution 3. Replacing f(x) by y gives:

y = x2 + 5

Solving for x:

y − 5 = x2

±√

y − 5 = x

We got two possible answers for x, so the function is not one-to-one.


Homework Exercises



1. (10 points) For each of the following functions determine whether thefunction is one-to-one:

a. (5 pts) f(x) = 2x2 − 4

b. (5 pts) f(x) = 3x−14x

2. (4 points)


(a) I feel comfortable determining whether a function is one-to-one.


b. (4 pts) Write down a function and determine whether it isone-to-one.




3.7 Inverse of a Function



Definition 1. Given a one-to-one function f and an output of f y,

then there is a unique input x so that f(x) = y. This correspondence fromoutputs y to inputs x defines a function called the inverse of f and denotedby f−1.

Main Concept 1. To find the inverse of a function you first weed tocheck that the function is one-to-one as shown above. Then in the uniquesolution for x replace y by x. This will give you the formula of the inversefunction.

Example 1. Find the inverse of each function given below.

a. f(x) = 5x+23x−1 , x 6= 1

3

b. f(x) = x2 for x ≥ 0

Solution 1.

a. First check that the function is one-to-one. Solve y = 5x+23x−1 for x:

y(3x− 1) = 5x + 2

3xy − y = 5x + 2

3xy − 5x = 2 + y

x(3y − 5) = 2 + y

x =2 + y

3y − 5.

So the function is one-to-one and replacing y by x gives f−1(x) = 2+x3x−5 .

http://youtu.be/gXLAOkJDFWI

3.7. INVERSE OF A FUNCTION 143

b. Check the function is one-to-one:

y = x2

±√y = x

It may seem like this function is not one-to-one, but notice that x ≥ 0, so√y = x. Therefore f−1(x) =

√x.


Homework Exercises



1. (20 points) For each of the following functions find the inverse:a. (3 pts) f(x) = 2x− 4

b. (5 pts) f(x) = (x− 2)3 + 4

c. (7 pts) f(x) = 8+x5x−1

d. (5 pts) f(x) =3√x+14

3.7. INVERSE OF A FUNCTION 145

2. (4 points)(a) I feel comfortable finding the inverse of a function.


b. (4 pts) Write down a one-to-one function and find its inverse.




3.8 Even and Odd Functions



Main Concept 1. In some functions f(−x) and f(x) are the same,in others they are opposite and in the rest of the functions they are neitherthe same nor opposites.

A function f is even if f(−x) = f(x).A function f is odd if −f(−x) = f(x).If f(−x) is not equal to f(x) or −f(x), then the function is neither odd

nor even.

Example 1. Determine whether each of the following functions is even, odd,or neither.

a. f(x) = x2 − 5

b. g(x) = x3 − 1

c. h(x) = 5x3 − x

d. F (x) = |x|

e. f(x) =√2x2+4−x3+x .

Solution 1.

a. To check whether a function is even check whether f(−x) = f(x):

f(−x) = (−x)2 − 5

= x2 − 5

= f(x).

Therefore the function f(x) = x2 − 5 is even. Since you know this is even,you do not have to check whether it is odd.

http://youtu.be/hwY6ktIv5SU

3.8. EVEN AND ODD FUNCTIONS 147

b. Check whether g(x) = x3 − 1 is even, that is check g(−x) = g(x):

g(−x) = (−x)3 − 1

= −x3 − 1

6= g(x).

Therefore g(x) is not even.

Check whether g(x) is odd, that is check if −g(−x) = g(x):

−g(−x) = −((−x)3 − 1)

= −(−x3 − 1)

= x3 + 1

6= g(x).

So g(x) is neither odd nor even.

c. Check whether h(x) is even, h(−x) = h(x):

h(−x) = 5(−x)3 − (−x)

= −5x3 + x

6= h(x).

So, h(x) is not even.

Check if it is odd, −h(−x) = h(x):

−h(−x) = −(5(−x)3 − (−x)

= −(−5x3 + x)

= 5x3 − x

= h(x)

Therefore the function is odd.

d. Check if F is even, F (−x) = F (x):

F (−x) = | − x|

= |x|

= F (x)

Therefore the function F (x) = |x| is even.


e. Check whether f(x) is even:

f(−x) =

√2(−x)2 + 4

−(−x)3 + (−x)

=

√2x2 + 4

x3 − x.

So, f(x) is not even.

Check if it is odd:

−f(−x) = −√

2x2 + 4

x3 − x

=1

−1

√2x2 + 4

x3 − x

=

√2x2 + 4

−x3 + x.

So, f(x) is odd.

3.8. EVEN AND ODD FUNCTIONS 149

Homework Exercises



1. (30 points) Determine (algebraically) whether each function is even,odd or neither.

a. (3 pts) f(x) = 2x2 + 4

b. (3 pts) f(x) = 2x3 + 4x

c. (8 pts) f(x) = 24x2+3x

d. (8 pts) f(x) = 5x2x2+3


e. (8 pts) f(x) =3√x2+4x2x2+1 .

2. (4 points) (a) I feel comfortable determining whether afunction is even, odd or neither.


b. (4 pts) Write down a function and determine whether it is even,odd or neither.



Library of Functions 151

Library of Functions

Function Properties Graph

Constant Function If b 6= 0, f has no x-intercept, y-intercept at (0, b).f(x) = b, b is a number No local minima or maxima

Domain:(−∞,∞), Range: {b}

Identity Function f has x and y-intercept at (0, 0),f(x) = x It is always increasing

f has no local minima or maximaDomain=Range: (−∞,∞)

Square Function f has x- and y-intercept at (0, 0),f(x) = x2 It is decreasing (−∞, 0), increasing (0,∞)

It has a minimum at (0, 0)Domain: (−∞,∞), Range: [0,∞)

Cube Function f has x- and y-intercept at (0, 0)f(x) = x3 It is always increasing

It has no local maxima or minimaDomain=Range:(−∞,∞)

Square Root Function f has x- and y-intercepts at (0, 0)f(x) =

√x It is increasing (0,∞)

It has a minimum at (0, 0)Domain= Range: [0,∞)

Cube Root Function f has x- and y-intercepts at (0, 0)f(x) = 3

√x It is always increasing

It has no local minima or maximaDomain=Range: (−∞,∞)

152 Library of Functions

Function Properties GraphReciprocal Function f has no x- or y-intercepts

f(x) = 1x It is decreasing (−∞, 0) ∪ (0,∞)

It has no local minima or maximaDomain=Range: (−∞, 0) ∪ (0,∞)

Horizontal Asymptote y = 0Vertical Asymptote x = 0

Absolute Value Function f has x- and y-intercepts at (0, 0)f(x) = |x| It is decreasing (−∞, 0), increasing (0,∞)

It has a minimum at (0, 0), no local maximaDomain: (−∞,∞), Range: [0,∞)

3.9. USING TRANSFORMATIONS TO GRAPH FUNCTIONS 153

3.9 Using Transformations to Graph Functions



Suppose f(x) is one of the main functions in the Library of Functions givenabove. A transformation of f(x) is a function which is obtained by adding,subtracting or multiplying constant(s) and x or f(x). For a, b, c and d constantsa transformation of f(x) will be of the form:

a · f(bx + c) + d.

Main Concept 1. A transformation is called vertical if it is obtainedby adding, subtracting, multiplying or dividing f(x) and constant(s). Inthe above formula

a · f(bx + c) + d

multiplying by a and adding d are vertical transformations. Vertical trans-formations are done to y.

A transformation is called horizontal if it is obtained by adding, sub-tracting, multiplying or dividing x and constant(s). In

a · f(bx + c) + d.

b and c are horizontal transformations. We apply the horizontal transfor-mations to x in the same way as if we are solving for x.

When graphing a function of the form g(x) = af(bx+ c) +d where f(x)is one of the main functions in the library of function the following ordershould be followed:

1. Graph the main function involved and find the two or three importantpoints on this graph by setting x = 0, x = 1 and x = −1. Follow theseas you move the graph.

2. Multiply the y-coordinates by a.

3. Subtract c from x-coordinates of the points in the previous step.

4. Divide the x-coordinates of the points in the previous step by b.

5. Add d to the y-coordinates of the points in the previous step.

http://youtu.be/lSm7-JOTQxo


Example 1. Sketch the graph of the following functions:

a. f(x) =√

1− x + 2

b. f(x) = 2|x− 1|+ 3

c. f(x) = 3x−2 + 1

Solution 1.

a. The main function being used is g(x) =√x. This has (0, 0) and (1, 1) as the

main points. To these points we will apply the order of transformations::

1. Subtract 1 from the x-coordinates. So the points change to (−1, 0) and(0, 1).

2. Divide the x-coordinates by −1. So the points change to (1, 0) and (0, 1).

3. Add 2 to the y-coordinates. So the points change to (1, 2) and (0, 3).

−5 −4 −3 −2 −1 1 2 3 4

−5

−4

−3

−2

−1

1

2

3

4

0


b. The main function being used is y = |x|. This has (−1, 1), (0, 0) and (1, 1)as the main points. Make sure you distinguish its graph from that of y = x2.To these points we will apply the transformations:

1. Multiply the y-coordinates by 2. So the points now become (−1, 2), (0, 0)and (1, 2).

2. Add 1 to the x-coordinates. So the points become (0, 2), (1, 0), (2, 2).

3. Add 3 to the y-coordinates. So the points become (0, 5), (1, 3), (2, 5).

−10−9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 8 9

−10

−9

−8

−7

−6

−5

−4

−3

−2

−1

1

2

3

4

5

6

7

8

9

0


c. The main function being used is g(x) = 1x . This has x = 0 and y = 0 as

asymptotes and the main points are (1, 1) and (−1,−1). To the asymptotesand points the following transformations are applied in the given order:

1. Mutiply the y-coordinates by 3. The asymptotes will not change and thepoints will change to (1, 3), (−1,−3).

2. Add 2 to the x-coordinates. The asymptote y = 0 does not change, theother asymptote will change to x = 2 and the points will be (3, 3) and(1,−3).

3. Add 1 to the y-coordinates. The asymptote x = 2 will not change theother asymptote will be at y = 1, and the points (3, 4) and (1,−2).

−7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6

−7

−6

−5

−4

−3

−2

−1

1

2

3

4

5

6

0

g(x) = 1x

h(x) = 3x

i(x) = 3x−2

f(x) = 3x−2 + 1

A

B

B′

A′

B′′

A′′

B′′′

A′′′


Homework Exercises

Name:This homework has 24 points.

1. (20 points) Sketch the graph of each function. Show all the transfor-mations used and their order.

a. (5 pts) f(x) =√x + 3− 1

b. (5 pts) g(x) = 4|x + 1|+ 2


c. (5 pts) h(x) = (−2x + 3)2 − 1

d. (5 pts) F (x) = −(x + 2)3 − 1


2. (4 points)(a) I feel comfortable applying transformations to graph functions.


b. (4 pts) Write down a function and apply transformations tograph it.



Chapter 4

Main Algebraic Functions

4.1 Linear Functions

Textbook Information. For more information on this topic, moreexamples and more exercises see sections 2.3, 4.1, 4.2.

For a video presentation of Graphing Linear Functions and Finding theEquation of a Linear Function click on this link.

For a video presentation of Average Rates of Change click on this link.For a video presentation of Applications of Linear Functions click on this

link.

Definition 1. A linear function is one that can be written in the

form y = mx+ b, where m and b are real numbers. The number m is calledthe slope and the number b is called the y-intercept.

4.1.1 Graphing Linear Functions

Main Concept 1. The graph of a linear function is a line and oneneeds two points to graph a line. If the formula for the function is given,the easiest way to graph the function is:

• Plot the y-intercept,

• Write the slope as a fraction riserun

http://youtu.be/98NIl3734c8

http://youtu.be/98NIl3734c8

http://youtu.be/eAb81m6rDwg

http://youtu.be/PK9PAw0d2jg

http://youtu.be/PK9PAw0d2jg

162 CHAPTER 4. MAIN ALGEBRAIC FUNCTIONS

• Plot a second point by staring at the y-intercept and moving rise inthe y-direction, and run in the x-direction.

• Connect the two points to get the graph of the function.

Example 1. Graph f(x) = −2x + 7.Solution 1. First find the y-intercept at by making x = 0 to get the point

(0, 7). Plot the y-intercept on the graph (point A). The slope is −2, written as afraction −21 , therefore we get the second point on the line (point B) by startingat A and moving 2 units down and 1 to the right. Draw the graph as the linethat connects the two points.

−10−9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 8 9

−10

−9

−8

−7

−6

−5

−4

−3

−2

−1

1

2

3

4

5

6

7

8

9

0

A

B

4.1.2 Finding the Equation of a Linear Function

Main Concept 2. Since a linear equation is of the form y =slope x+y-intercept, to find a linear equation the slope and the y-interceptare needed.

The slope:

4.1. LINEAR FUNCTIONS 163

1. may be given in the problem,

2. may be found by using the formula m = y2−y1

x2−x1,

3. may be found by using the facts:

i. Parallel lines have the same slope.

ii. Perpendicular lines have opposite reciprocal slopes.

One can find the value of b by solving for it after substituting the valueof the slope and a point on the line.

Finding the Equation of a Line Given a Point and the Slope

Example 2. Find the equation of the line with slope 4 and through the point(−1, 2).

Solution 2. The general equation for a line is y = mx + b. Here we aregiven that m = 4, so the equation is of the form y = 4x + b. We know that thepoint (−1, 2) satisfies the equation, so we can substitute it to get:

2 = 4(−1) + b

we can solve this for b:2 = −4 + b

b = 6.

Hence the equation of the line will be y = 4x + 6.

Finding the Equation of a Line Passing Through Two Given Points

Example 3. Find the equation of the line through the points (−3, 0) and(2, 1).

Solution 3. As in the previous example, the general equation of a line is

y = mx + b.

We need to find the slope and the y-intercept. How can we find the slope giventwo points?

m =y2 − y1x2 − x1

=1− 0

2− (−3)

=1

5.

Now that we know the slope the equation becomes

y =1

5x + b


and we need to find b. Substitute any of the points given:

0 =1

5(−3) + b

3

5= b.

Therefore the equation of the line is

y =1

5x +

3

5

Finding the Equation of a Line Parallel to a Given Line

Example 4. Find the equation of the line parallel to y = −3x+ 3 and passingthrough the point (2, 0)

Solution 4. Notice that the problem involves two lines:

• L1: The line we are trying to find which passes through (2, 0) and isparallel to

• L2: The line y = −3x + 3

As stated above, parallel lines have the same slope, so the line L1 has thesame slope as the line L2. What’s the slope?

m = −3

Therefore the line L1 has equation of the form

y = −3x + b.

Using the point (2, 0) we find the value of b:

0 = −3(2) + b

6 = b

Therefore the equation of the line L1 is

y = −3x + 6.

Example 5. Find the equation of the line going through (−5, 9) and parallelto the one through (0, 2) and (2, 1).

Solution 5. We have two lines involved:

• L1: The line we are trying to find the equation of and that goes through(−5, 9) and is parallel to:


• L2: The line through (0, 2) and (2, 1).

To find the equation of L1 we need its slope. Since L1 and L2 are parallel,they have equal slopes. We can find the slope for L2 by the formula m = y2−y1

x2−x1:

m =1− 2

2− 0

= −1

2.

This gives also the slope for L1. The equation for L1 has the form

y = −1

2x + b.

Now substitute the point L1 goes through to find b:

9 = −1

2(−5) + b

9 =5

2+ b

9− 5

2= b

18

2− 5

2= b

13

2= b.

The equation for L1:

y = −1

2x +

13

2.

Finding the Equation of a Line Perpendicular to a Given Line

Example 6. Find the equation of the line through (−1, 0) and perpendicularto y = 2x− 3.

Solution 6. Again we have two lines involved:

• L1: The line we are trying to find which goes through (−1, 0) and isperpendicular to

• L2: The line y = 2x− 3.

Since the two lines are perpendicular and the slope of L2 is 2, the slope ofthe line L1 is its opposite reciprocal:

m = −1

2.


The equation of L1 has form

y = −1

2x + b

Substitute (−1, 0) to find b:

0 = −1

2(−1) + b

0 =1

2+ b

−1

2= b

Hence L1 has equation: y = − 12x−

12 .

4.1.3 Using Average Rates of Change to Identify LinearFunctions

Sometimes function are given in tables (most of the real application modelingproblems), how can we recognize whether they represent a linear function inthis case?

Finding the Average Rate of Change

Example 7. Suppose that in a controlled experiment studying fruit fly popu-lation change over 20 days it was found that the equation P (t) = (t−20)2 +400gives the population P after t days.

a. What is the change of the population from t = 0 to t = 10?

b. What is the change of the population from t = 0 to t = 20?

Solution 7.

a. This is the same as finding P (10)− P (0):

P (10)− P (0) = ((10− 20)2 + 400)− ((0− 20)2 + 400)

= ((−10)2 + 400)− ((−20)2 + 400)

= (100 + 400)− (400 + 400)

= −300

So, from t = 0 to t = 10 the population decreased by 300 flies.


b. This is the same as finding P (20)− P (0):

P (20)− P (0) = ((20− 20)2 + 400)− ((0− 20)2 + 400)

= 400− (400 + 400)

= −400

So, from t = 0 to t = 20 the population decreased by 400 flies.

It is understandable that the population change from t = 0 to t = 20 islarger than the population change from t = 0 to t = 10, since more time haspassed, but we need to measure the population change in a more objective way,not depending on how much time has passed.

Definition 2. Given a function f(x) and two numbers a and b such

that f(x) is defined on [a, b], define the average rate of change of f(x)from a to b as the quotient:

f(b)− f(a)

b− a.

Example 8. Using the information on the previous example, answer thefollowing questions:

a. What is the average rate of change of P from t = 0 to t = 10?

b. What is the average rate of change of P from t = 0 to t = 20?

Solution 8.

a. Find P (10)−P (0)10−0 . The difference P (10) − P (0) was found in the previous

example so:

P (10)− P (0)

10− 0=−300

10

= −30

Therefore in the time interval from t = 0 to t = 10 the number of files wasdecreasing by an average of 30 flies per day.

b. Similarly from the previous example we have P (20)− P (0) = −400 so:

P (20)− P (0)

20− 0=−400

20

= −20

Therefore in the time interval from t = 0 to t = 20 the number of fruit flieswas decreasing by an average of 20 flies per day.


Example 9. Find the average rate of change of f(x) = 3x2 from:

a. 1 to 3.

b. 1 to 5.

c. 1 to 7.

Solution 9.

a. The average rate of change of f from 1 to 3 is f(1)−f(3)1−3 :

=3(1)2 − 3(3)2

−2

=3− 27

−2

=−24

−2

= 12

b.

f(1)− f(5)

1− 5=

3(1)2 − 3(5)2

1− 5

=3− 75

−4

=−72

−4

= 18

c.

f(1)− f(7)

1− 7=

3(1)2 − 3(7)2

1− 7

3− 147

−6

=−144

−6

= 24


The Average Rate of Change and Linear Functions

Main Concept 3. Linear functions have a constant average rate ofchange between any two points, and the value of this average rate of changeis the same as the slope.

Example 10. A company that produces air-fresheners wants to measure itscost to produce a certain number of air fresheners. The company collected to fol-lowing data on the number of products sold (in millions) and the correspondingcost (in thousand dollars).

Number of Products in millions Cost in thousand dollars20 501030 751035 8760

a. Will the model describing the cost as a function of the products sold be alinear function?

b. Interpret each average rate of change calculated above in the problem. Whatdoes it say about the products sold and the cost?

c. Can we predict the cost if 15 million products were sold?

The company decided to collect more data to further investigate the resultsand the following table was obtained:

Number of Products in millions Cost in thousand dollars20 501030 751035 876040 921560 13815

d. Will the new model describing the cost as a function of the products sold bea linear function?

Solution 10.

a. To determine whether the model is a linear function we need to find theaverage rate of change for each consecutive pair and check whether it is thesame:

The average rate of change from (20, 5010) to (30, 7510) is

7510− 5010

30− 20

= 250thousand dollars

million products.



8760− 7510

35− 30

=1250

5


million products.

Since the average rates of change are the same, the model is a linear function.

b. The average rates of change calculated above measure how the cost (in thou-sand dollars) changes for one million of products sold. It means that for onemillion of products sold, the cost goes up by 250 thousand dollars. This isthe same as the cost going up $0.25 with each product sold.

c. Since from the previous part we concluded that the model describing the cost(C) in thousands of dollars is a linear function of the products sold (p) withan average rate of change of 250, then the function C = mp + b will haveslope 250. To find the y-intercept b we can substitute any of the three pointsgiven on the table. If we use (20, 5010):

C = 250p + b

5010 = 250 · 20 + b

5010 = 5000 + b

10 = b.

So C = 250p + 10, and if 15 million products were sold the cost would be250 · 15 + 10 = 3760 thousand dollars or $3, 760, 000.

d. We compute the other average rates of change to get:


9215− 8760

40− 35

=455

5


million products.

Since this is not the same as the average rate of change between the firstthree points, the new model is not a linear function.


4.1.4 More Applications of Linear Functions

Example 11. Book value is the value of an asset that a company uses tocreate its balance sheet. Some companies depreciate their assets using straightline depreciation so that the value of the asset declines by a fixed amount eachyear. The amount of the decline depends on the useful life that the companyplaces on the asset. Suppose a company just purchased a fleet of new cars forits sales force at a cost of $28,000 per car. The company chooses to depreciateeach vehicle using the straight-line method over 7 years. This means that eachcar will depreciate by $28,000

7 = $4000 per year.

a. Write a linear function that expresses the book value V of each car as afunction of its age, x.

b. Graph the linear function.

c. What is the book value of each car after 3 years?

d. Interpret the slope.

e. When will the book value of each car be $8000?

Solution 11.

a. In this situation x is the age of the car in years and y is its value. From theproblem we know that the value of the car when it is bought is 28000 and itsvalue in seven years is zero. This gives us two points the line passes through:(0, 28000) and (7, 0) and from these two points we can find the equation ofthe line. The slope

m =y2 − y1x2 − x1

=0− 28000

7− 0

= −4000

The y-intercept is given to be 28000, so the equation of the line would bey = −4000x + 28000. Therefore V (x) = −4000x + 28000.

b. This is the same as finding V (3).

V (3) = −4000 · 3 + 28000

= −12000 + 28000

= 16000

c. The slope of −4000 indicates that the value of the car decreases each yearby $4000.


d. Here we have V (x) = 8000 and we are trying the find x.

8000 = −4000x + 28000

−20000 = −4000x

5 = x

Therefore the value of the car will be $8000 after 5 years.

Example 12. The quantity supplied of a good is the amount of a productthat a company is willing to make available for sale. The quantity demandedof a good is the amount of product that consumers are willing to purchase.Suppose that the quantity supplied, S, and the quantity demanded, D, of waterfilters each month are given by the following functions:

S(p) = 60p− 900

D(p) = −15p + 2850

a. The equilibrium price of a product is defined as the price at which quantitysupplied equals quantity demanded. That is, the equilibrium price is the priceat which S(p) = D(p). Find the equilibrium price. What is the equilibriumquantity, the amount demanded (or supplied), at the equilibrium price?

b. Determine the prices for which quantity supplied is greater than quantitydemanded. That is, solve the inequality S(p) > D(p)

Solution 12.

a. The problem gives

S(p) = 60p− 900, D(p) = −15p + 2850

and asks to solve the equation

S(p) = D(p).

This solution will give the equilibrium price, from there we can find theequilibrium quantity by substituting the price.

S(p) = D(p)

60p− 900 = −15p + 2850

75p = 3750

p = 50

So, equilibrium occurs when the price is $50 and the equilibrium quantitywill be S(5) = D(5) = 60(50)− 900 = 2100.


b. This is the same as solving the inequality:

S(p) > D(p)

60p− 900 > −15p + 2850

75p > 3750

p > 50

Therefore for prices higher than $50, in (50,∞), the quantity supplied isgreater than the quantity demanded.


Homework Exercises



1. (6 points) Find the equation of the line through the points A = (1, 1)and B = (3,−1)

2. (6 points) Find the equation of the line parallel to x − 2y = −5 andthrough the point (0, 0).


3. (6 points) Find the equation of the line perpendicular to 2x+ 3y = −1and through the point (−1, 3).

4. (3 points) Find the equation of the line graphed below:

−5 −4 −3 −2 −1 1 2 3 4 5 6

−5

−4

−3

−2

−1

1

2

3

0

5. (4 points) Find the average rate of change for f(x) = x2 +3 from x = 1to x = 4.


6. (6 points) Find the equation of the line through the points A = (3, 5)and B = (−1, 2)

7. (6 points) Find the equation of the line parallel to 2x − y = 5 andthrough the point (0, 3).

8. (6 points) Find the equation of the line perpendicular to x − 4y = 5and through the point (−2, 2).

9. (4 points) Find the average rate of change for f(x) = 4x2 + 3x − 2from x = 0 to x = 4.


10. (6 points) The following data represents the price p and quantitydemanded q per day for 24” LCD monitors.

Price, p (in dollars) Quantity demanded q150 100200 80250 60300 40

a. (2 pts) Show by using average rates of change that the quantityq is a linear function of p.

b. (4 pts) Find the linear function that describes q as a function ofp.


11. (7 points) Suppose that a company has just purchased a new com-puter for $3000. The company chooses to depreciate the computer using thestraight line method over 4 years.

a. (4 pts) Write a linear function that expresses the book value Vof the computer as a function of its age x.

b. (1 pt) What is the book value after 2 years?

c. (2 pts) When will the computer have a book value of $2000?


12. (4 points) (a) I feel comfortable finding the equation ofa line through two points.


b. (4 pts) Randomly choose two points and find the equation of theline through the two.

13. (4 points) (a) I feel comfortable finding the equation ofa line parallel to another line.


b. (4 pts) Randomly choose one point that is different from the twopoints you chose in exercise 12 and find the equation of the line parallel to theline you found in exercise 12 and through this point.

14. (4 points) (a) I feel comfortable finding the equation ofa line perpendicular to another line.


b. (4 pts) Randomly choose one point and find the equation of theline perpendicular to the line you found in exercise 12 and through this point.


15. (4 points) (a) I feel comfortable finding the average rateof change of a function from x = a to x = b.


b. (4 pts) Randomly choose one function f(x), a value for a and avalue for b. Find the average rate of change of f(x) from a to b.



4.2. QUADRATIC FUNCTIONS 181

4.2 Quadratic Functions


Definition 1. A quadratic function is one of the form

y = ax2 + bx + c,

where a, b and c are real numbers and a 6= 0.

For a video presentation of Completing the Square and Graphing QuadraticFunctions click this link.

For a video presentation on Finding Minimum and Maximum Values ofQuadratic Functions and Quadratic Models click on this link.

Before we go into the main properties of quadratic functions it is important toreview the completion of the square.

4.2.1 Completing the Square

The process of “Completing the Square” is the process that changes an expres-sion of the form ax2 + bx + c to a(x + d)2 + e. In this section we will use thecompletion of square to change a quadratic function of the form y = ax2+bx+cto the form y = a(x+ d)2 + e. This will make it easy to recognize the necessarytransformations to graph the function.

Main Concept 1. To complete the square in a function of the fromy = ax2 + bx + c:

1. Place terms involving x on one set of parenthesis: y = (ax2 + bx) + c

2. Factor a from the terms inside parenthesis: y = a(x2 + bax) + c

3. Take half of the coefficient multiplying x, and square it:(

b2a

)24. Add the quantity above inside the parenthesis and add/subtract as

necessary to keep the equation the same as the original: y =

a(x2 + b

ax +(

b2a

)2)− a(

b2a

)2+ c

http://youtu.be/noCnk_b8sGg

http://youtu.be/noCnk_b8sGg

http://youtu.be/gw8Xwj3nPHw

http://youtu.be/gw8Xwj3nPHw


5. Factor the terms inside parenthesis on the right side:

y = a

(x +

b

2a

)2

− b2

4a+ c.

Example 1. Complete the square for y = 2x2 + 4x + 6Solution 1. Following the steps listed above we get:

1. Place terms involving x on one set of parenthesis: y = (2x2 + 4x) + 6

2. Factor a from the terms inside parenthesis: y = 2(x2 + 2x) + 6

3. Take half of the coefficient multiplying x, and square it:(22

)2= 1

4. Add the quantity above inside the parenthesis and add/subtract as necessaryto keep the equation the same as the original: y = 2(x2 + 2x + 1)− 2 + 6


y = 2(x + 1)2 + 4

4.2.2 Graph a Quadratic Function Using its Vertex andIntercepts

Once a quadratic function is written in the form f(x) = a(x+ d)2 + e then is iteasy to see that the graph of the function will be a transformation of f(x) = x2.

The vertex of a quadratic function is its minimum or maximum point.The axis of symmetry of a quadratic function is the vertical line goingthrough the vertex.

Example 2. Graph the function f(x) = 2x2 + 8x + 5. Indicate its vertexand its axis of symmetry.

Solution 2. As before, to graph the function we need to complete thesquare. To do this one can follow the steps we listed above:

1. Place terms involving x on one set of parenthesis: y = (2x2 + 8x) + 5

2. Factor a from the terms inside parenthesis: y = 2(x2 + 4x) + 5

3. Take half of the coefficient multiplying x, and square it:(42

)2= 4

4. Add the quantity above inside the parenthesis and add/subtract as necessaryto keep the equation the same as the original: y = 2(x2 + 4x + 4)− 8 + 5


y = 2 (x + 2)2 − 3


Now use transformations to graph the function:

−5 −4 −3 −2 −1 1 2 3 4

−5

−4

−3

−2

−1

1

2

3

4

0

In the previous example notice the coordinates of the vertex.

Main Concept 2. A quadratic function with completed square formy = a(x+h)2 +k has vertex with coordinates (−h, k) and axis of symmetryx = −h.

Sometimes a more “speedy” way of graphing a quadratic function, is byfinding the vertex, the axis of symmetry and intercepts. If the vertex and theintercept are the same point, pick two points one on each side of the vertex.

Example 3. Graph f(x) = −3x2 + 6x + 1 by finding its vertex and theintercepts.

Solution 3. To find the vertex we need to complete the square:

y = −3(x2 − 2x) + 1

y = −3(x2 − 2x + 1) + 3 + 1

y = −3(x− 1)2 + 4

Therefore the vertex has coordinates:

(1, 4)

andAxis of Symmetry: x = 1.

Recall that the x-intercept(s) is (are) found by setting y = 0:

0 = −3x2 + 6x + 1

solve this using the quadratic formula:

x =−b±

√b2 − 4ac

2a


x =−6±

√36 + 12

−6

x =−6± 4

√3

−6

x =3± 2

√3

3.

These round to 2.1547 and −0.1547Recall that the y-intercept is found by making x = 0:

y = 1

−5 −4 −3 −2 −1 1 2 3 4

−5

−4

−3

−2

−1

1

2

3

4

0

4.2.3 Finding Minimum/Maximum value of a QuadraticFunction

It can be easily seen from most of the examples we have worked that the vertexis the minimum or maximum value of the function. Also notice that whenthe coefficient multiplying x2 determines whether the function will look likef(x) = x2

−5 −4 −3 −2 −1 1 2 3 4

−5

−4

−3

−2

−1

1

2

3

4

0


so its vertex is its minimum point. On the other hand if a is negative thevertex is the maximum.

−5 −4 −3 −2 −1 1 2 3 4

−5

−4

−3

−2

−1

1

2

3

4

0

Example 4. Find the minimum or maximum value of the function f(x) =−3x2 + 4x + 5

Solution 4. Since here x2 is multiplied by a negative number, the functionwill have a maximum point equal to its vertex. Its maximum value will be they coordinate of its vertex. To find the vertex complete the square:

y = (−3x4 + 4x) + 5

y = −3

(x2 − 4

3x

)+ 5

y = −3

(x2 − 4

3x +

4

9

)+

4

3+ 5

y = −3

(x− 2

3

)2

+19

3

Therefore the vertex is(23 ,

193

), and the vertex

(23 ,

193

)is a maximum.

4.2.4 Quadratic Models

Example 5. The marketing department at Texas Instruments has found that,when certain calculators are sold at a price of p dollars per unit, the quantitysold as a function of the price p is

Q(p) = −150p + 21000

a. Find a function that describes the revenue R in terms of the price p.

b. What unit price should be established to maximize revenue? If this price ischarged, what is the maximum revenue?

Solution 5.


a. Since the revenue is the quantity times the price we will have:

R(p) = Q(p)p

R(p) = −150p2 + 21000p

b. Since the revenue function is a quadratic function, its maximum point willbe its vertex.Complete the square:

R(p) = −150(p2 − 140p)

R(p) = −150(p2 − 140p + 4900) + 735000

R(p) = −150(p− 70)2 + 735000

Therefore the vertex is (70, 735000), and the maximum revenue of $735000is obtained when the unit price is $70.

Example 6. A farmer has 2000 yards of fence to enclose a rectangular field.What are the dimensions of the rectangle that encloses the most area?

Solution 6.Given Find

Circumference = 2000yards length = l of the rectangle with maximum areawidth = w of the rectangle with maximum area

Area is the quantity to be maximized, so it needs to be written as a function.The area of a rectangle with dimensions l and w is given by A = lw. To be ableto find the maximum value of A we need to write it as a function of one variable.We let l = x and write w as a function of x using the fact that the length of thefence (2000 yards) will cover the circumstance of the field, so 2000 = 2x + 2w,that is w = 1000− x. The area of the rectangle is given by

A = wx = 1000x− x2 = −x2 + 1000x.

Again, since this is a quadratic function its maximum point is its vertex.Complete the square:

A = −(x2 − 1000x)

A = −(x2 − 1000x + 250000) + 250000

A = −(x− 500)2 + 250000

The problem asks to find the dimensions that will enclose the largest areaand from the completion of the square we see that x, which is the length, shouldbe 500 yards, and the width, which is w = 1000− x, should also be 500 yards.


Homework Exercises



1. (20 points) Graph each function. Indicate the coordinates of its vertex,intercepts and the axis of symmetry.

a. (5 pts) f(x) = 3x2 + 6x− 4

b. (5 pts) g(x) = −2x2 + 5x + 6


c. (5 pts) f(x) = 2x2 + 4x− 4

d. (5 pts) g(x) = −3x2 + 3x + 4

2. (8 points) Determine, without graphing, whether each quadratic func-tion given has a minimum or maximum value. Find the minimum or maximumvalue.

a. (4 pts) f(x) = 3x2 − 9x

b. (4 pts) g(x) = −x2 + x + 3


3. (7 points) The price (in dollars) and the quantity sold for a certainproduct x are related by the equation:

p = −1

4x + 200

a. (2 pts) Find the function that describes the revenue R as afunction of the quantity sold x.

b. (1 pt) What is the revenue if 250 products were sold?

c. (3 pts) How many products should be sold to maximize revenue?

d. (1 pt) What is the maximum revenue?


4. (6 points) The price (in dollars) and the quantity sold for a certainproduct x are related by the equation:

p = −1

5x + 600.

Find the number of products that should be sold to maximize revenue.

5. (6 points) A farmer with 400 meters of fencing wants to enclose arectangular garden that is adjacent to a river. If he decides not to use fencingon the part of the garden along the river, what is the maximum area he canenclose?


6. (4 points) (a) I feel comfortable finding the vertex of aquadratic function.


b. (4 pts) Randomly choose a quadratic function and find its vertex.

7. (4 points) (a) I feel comfortable determining the maximumor minimum of a quadratic function.


b. (4 pts) For the quadratic function chosen above, find its maximumor minimum. State its coordinates and whether it is a maximum or a minimum.

8. (4 points) (a) I feel comfortable finding the intercepts ofa quadratic function.


b. (4 pts) For the quadratic function chosen above, find the x- andy-intercepts.


9. (4 points) (a) I feel comfortable graphing a quadraticfunction.


b. (4 pts) Graph the quadratic function chosen above.



4.3. OTHER POLYNOMIAL FUNCTIONS 193

4.3 Other Polynomial Functions

Textbook Information. For more information on this topic, moreexamples and more exercises see sections 5.1, 5.5

For a video presentation of Long Division click this link.For a video presentation on the Zeros of Polynomials click on this link.For a video presentation on Graphing Polynomials click on this link.

4.3.1 Power Functions

Definition 1. A power function is a function of the form

f(x) = anxn,

where an is a real number and n is a nonnegative integer.

Power functions are easy to graph and analyze.

Main Concept 1. The graph of f(x) = xn when n is even is similarto the graph of f(x) = x2 and the graph of f(x) = xn when n is odd issimilar to the graph of f(x) = x3.

4.3.2 Introduction to General Polynomials

Definition 2. A polynomial function is a sum of power functions:

f(x) = anxn + an−1x

n−1 + · · ·+ a1x + a0,

where an, an−1, . . . , a1, a0 are real numbers, an 6= 0 and n in a nonnegativeinteger called the degree of f .

Example 1. Determine which of the following functions are polynomials. Forthose that are state the degree; for those that are not, state the reason they arenot.

a. f(x) = 2− 3x2

b. g(x) =√x

http://youtu.be/dgB68bWf_go

http://youtu.be/c42HO8aECPE

http://youtu.be/9mQXdQXM2Bw


c. h(x) = x2−2x3−1

d. F (x) = 0

e. G(x) = 8

f. H(x) = −2x3(x− 1)2

Solution 1.

a. This is a polynomial of degree 2.

b. It is not a polynomial since the power involved√x = x

12 is not an integer.

c. It is not a polynomial since it involves a denominator.

d. It is not a polynomial since in a polynomial the leading term cannot be zero.

e. It is a polynomial and since 8 = 8x0 it has degree zero.

f. It is a polynomial. When we multiply it out the see the power functionsinvolved we see that it has degree 5.

4.3.3 Long Division of Polynomials

Given two polynomials one can add, subtract, multiply and divide them. Thefirst three operations are straight forward, in this part of the section we willlook closer into dividing two polynomials.

Definition 3. Let p(x) and q(x) be two polynomials. The quotient

p(x)q(x) has proper form p(x)

q(x) = f(x) + r(x)q(x) with deg(r(x)) < deg(q(x)).

The numerator p(x) is called the dividend, q(x) is called the divisorand the polynomial f(x) is called the quotient.

In order to write a quotient of polynomials in its proper form, we divide thepolynomials using long division (similar to dividing integers). The process oflong division is best described when working with an example:

Example 2. Write the quotients in their proper form:

a.8x2 − x + 2

4x2 − 1

b.3x2 + 4x− 1

x + 3

c.2x5 − x3 + 2

x3 − 1


Solution 2.

a. Write the polynomials with their degrees descending and in the form:

Divisor Dividend

4x2 − 1 8x2 − x + 2

Divide the terms of highest degree and place the quotient in the quotientline:

Quotient

Divisor Dividend

2

4x2 − 1 8x2 − x + 2

Multiply the divisor by the expression placed in the quotient line in theprevious step, that is multiply 4x2 − 1 by 2. Place this expression under thedividend.

2

4x2 − 1 8x2 − x + 2

8x2 − 2

Subtract the above quantity from the dividend:

2

4x2 − 1 8x2 − x + 2

−(8x2 − 2)

Add like terms and write the resulting expression as shown:


2

4x2 − 1 8x2 − x + 2

−(8x2 − 2)

−x + 4

Continue this process until the resulting expression at the end of the columnon the right has degree less than the degree of the divisor. For this example,at this point 4x2 does not divide −x, so the long division process is finishedhere. Therefore the quotient

8x2 − x + 2

4x2 − 1

can be written in its proper form:

2 +−x + 4

4x2 − 1.

b. Write the polynomials with their degrees descending and in the form:

x + 3 3x2 + 4x− 1

Divide the terms of highest degree:

3x

x + 3 3x2 + 4x− 1

Multiply the divider by the quantity in the previous step.

3x

x + 3 3x2 + 4x− 1

3x2 + 9x


3x

x + 3 3x2 + 4x− 1

−(3x2 + 9x)



3x

x + 3 3x2 + 4x− 1

−(3x2 + 9x)

−5x− 1

Continue this process until the resulting expression at the end of the columnon the right has degree less than the degree of the divisor. Now we assumethe dividend to be −5x− 1, and the divisor is still x + 3.

Divide the terms of highest degree, that is divide −5x by x, and place thisnumber in the quotient line:

3x− 5

x + 3 3x2 + 4x− 1

−(3x2 + 9x)

−5x− 1

Multiply the divider by the quantity placed in the quotient line in the previ-ous step:

3x− 5

x + 3 3x2 + 4x− 1

−(3x2 + 9x)

−5x− 1

−5x− 15


3x− 5

x + 3 3x2 + 4x− 1

−(3x2 + 9x)

−5x− 1

−(−5x− 15)



3x− 5

x + 3 3x2 + 4x− 1

−(3x2 + 9x)

−5x− 1

−(−5x− 15)

14

Therefore the proper form of: 3x2+4x−1x+3 is

3x− 5 +14

x + 3.

c. Following the steps listed in the previous two examples yields:

2x2 − 1

x3 − 1 2x5 − x3 + 2

−(2x5− 2x2)

−x3+2x2+2

−(−x3 + 1)

2x2 + 1

Therefore the proper form of this quotient is

2x5 − x3 + 2

x3 − 1= 2x2 − 1 +

2x2 + 1

x3 − 1.

4.3.4 Zeros of Polynomials and Their Multiplicities

Main Concept 2. If f is a polynomial function and r is a real numberfor which f(r) = 0, then r is called a zero of f , or a root of f . If r is azero of f , then:

a. r is an x-intercept of the graph of f .

b. (x− r) is a factor of f .


Example 3. Find a polynomial of degree 3 with −4, −2, and 3 as zeros.Solution 3. From the previous theorem, since −4 is a root x− (−4) = x+4

is a factor. Similarly x+2 and x−3 are factors. So f has to (x+4)(x+2)(x−3)as a factor. This already gives us a polynomial of degree 3, so f is a constantmultiple of it. That is f(x) = a(x + 4)(x + 2)(x− 3).

Sometimes a linear factor divides a polynomial more than once. For example(x− 1) divides f(x) = (x− 1)2(x + 4) twice.

Definition 4. If (x− r)m is a factor of a polynomial f and (x− r)m+1

is not a factor of f , then r is called a zero of multiplicity m of f .

Example 4. Identify the zeros and their multiplicities for f(x) = −2(x −2)(x + 1)3(x− 3)4.

Solution 4. The function has been factored and its zeros are x = 2 withmultiplicity one, x = −1 with multiplicity 3 and x = 3 with multiplicity 4.

4.3.5 Rational Zeros Theorem

When the polynomial is in the factored form, it is easy to see the zoroes andtheir multiplicities, but when it is given in the genral form we need to factor it.Some polynomials we know how to factor, but for others we will need to to usethe Rational Zeros Theorem to factor them.

Main Concept 3. Rational Zero Theorem Suppose f(x) = anxn+

an−1xn−1 + · · ·+ a1x + a0 is a polynomial, then the rational zeros of f are

of the form pq where p is a factor of a0 and q is a factor of an.

We use the theorem when we are trying to find all the rational zeroesof a polynomial:

1. We first make a list of:

a. the factors of the constant term,

b. the factors of leading term and

c. all the fractions of the form Factor of the Constant TermFactor of the Leading Term

.

2. From the fractions listed, we check which ones make the polynomialequal to zero.

Example 5. Find the rational zeros of f(x) = x3 + 2x2 − x− 2.Solution 5. This is equivalent to finding the rational solutions to x3+2x2−

x − 2 = 0. To solve this, one has to factor. This equation can be factored bygrouping

(x3 + 2x2) + (−x− 2) = 0


x2(x + 2) + (−1)(x + 2) = 0

(x2 − 1)(x + 2) = 0

and using the difference of squares formula:

(x− 1)(x + 1)(x + 2) = 0.

Now from here we can find the zeroes to be

x− 1 = 0 so x = 1

x + 1 = 0 so x = −1

x + 2 = 0 so x = −2.

But, one can also solve this by using the Rational Zero Theoremgiven above. As stated there we

1. list

a. factors of the constant term, that is factors of −2: −2,−1, 1, 2

b. factors of the leading term, that is factors of 1: −1, 1

c. fractions of the form Factors of constant termFactors of leading term

: −2,−1, 1, 2.

2. Check which numbers listed in step c make f(x) equal to zero.

f(−2) = (−2)3 + 2(−2)2 − (−2)− 2= 0

f(−1) = (−1)3 + 2(−1)2 − (−1)− 2= 0

f(1) = 13 + 2(1)2 − 1− 2= 0

f(2) = 23 + 2(2)2 − 2− 2 6= 0.

Therefore the rational zeroes are −1, 1, 2.The theorem will not give us multiplicites for the zeros, to do that we have

to use the theorem to find all the rational zeros and then use long division.Example 6. Find the rational zeros and their multiplicities for f(x) =

x3 + x2 − 5x + 3.Solution 6. Using the Rational Zeros Theorem we

1. list

a. factors of the constant term: −3,−1, 1, 3

b. factors of the leading term: −1, 1

c. possible zeros: −3,−1, 1, 3.


2. Check which numbers in the last step are zeros:

f(−3) = (−3)3 + (−3)2 − 5(−3) + 3= 0

f(−1) = (−1)3 + (−1)2 − 5(−1) + 3 6= 0

f(1) = (1)3 + (1)2 − 5(1) + 3= 0

f(3) = (3)3 + (3)2 − 5(3) + 3 6= 0

So we see that x = 1 and x = −3 are the rational zeros, but these two zerosare the zeros of a quadratic (x − 1)(x + 3). The function we have either hasother zeros, or one of these zeros listed has multiplicity greater than 1. To findwhat is the case we find the general form of the quadratic function that has thesane zeros we found: (x − 1)(x + 3) = x2 + 2x − 3. Since these are also zerosof f(x) = x3 + x2 − 5x + 3, the quadratic we have divides f(x). So we do longdivision to find the other factors:

x− 1

x2 + 2x− 3 x3 + x2 − 5x + 3

−(x3+2x2−3x)

−x2 − 2x + 3

−(−x2− 2x+ 3)

0

Therefore

x3 + x2 − 5x + 3 = (x− 1)(x2 + 2x− 3)

= (x− 1)(x− 1)(x + 3)

= (x− 1)2(x + 3),

and f(x) has x = 1 as a zero of multiplicity 2 and x = −3 as a zero ofmultiplicity 1.

4.3.6 Zeros, Their Multiplicity and the Graph of a Poly-nomial

Going back to graphing, it is easy to see that the zeros of a polynomial will bex-intercepts. At any x-intercept, the graph may go through the x-axis or justtouch it. For polynomial functions, the multiplicity of a zero recognizes whetherthe function goes through the x-axis or touches it.


−10 −8 −6 −4 −2 2 4 6 8

−20

−18

−16

−14

−12

−10

−8

−6

−4

−2

2

4

6

8

10

12

14

16

18

0

Crosses the x-axis Touches the x-axis

Main Concept 4. Suppose r is a zero of f(x) of multiplicity m. Thenthe graph of f(x) goes through the x-axis is m is odd, and it bounces offthe x-axis if m is even.

4.3.7 Graphing a Polynomial Function Using its Sign Chart

The x-intercepts, and their multiplicities can be used to graph polynomials.

Main Concept 5. To graph a polynomial using its sign chart:

1. Find the x- and y-intercepts.

2. Use the x-intercepts to separate the x-axis into intervals. Check whetherthe function will be positive or negative in each of the intervals by plug-ging in values for x.

3. For each interval in the previous step determine whether the functionwill be above the x-axis (positive) or below the x-axis (negative).


4. Graph the function.

5. Check your graph is consistent with the multiplicities of its roots.

Note 1. This graph is only an approximation of the graph of f since we donot know and, at this point, cannot find the local minimums/maximums.

Example 7. Graph f(x) = x2(x− 2) using its sign chart.Solution 7. Go through the steps listed in the Main Concept above:


The x-intercepts are found by making y = 0:

0 = x2(x− 2)

This is already factored so x2 = 0 or x− 2 = 0. That is x = 0 or x = 2. Thegraph has x-intercepts at (0, 0) and (2, 0).

The y-intercept is found by making x = 0: f(0) = 02(0− 2) = 0. Thereforethe y-intercept is at (0, 0)

2. Use the x-intercepts to separate the x-axis into intervals. Check whether thefunction will be positive or negative in each of the intervals by plugging invalues for x.

The x-axis is separated in three parts: to the left of 0, between 0 and 2 andto the right of 2. If we choose x = −2 to the left of zero we get f(−2) =(−2)2(−2− 2) = 4(−4) which is negative. So the function is negative to theleft of 0. If we choose 1 between 0 and 2 we get f(1) = 12(1 − 2) = 1(−1)which is negative. So the function is negative between 0 and 2. If we choosex = 4 to the right of 2 we get f(4) = 42(4− 2) = 16(2) which is positive. Sothe function is positive to the right of 2.

−4 −2 2 4

−4

−2

2

4

0

Negative Negative Positive

3. For each interval in the previous step determine whether the function will beabove the x-axis (positive) or below the x-axis (negative).


−4 −2 2 4

−4

−2

2

4

0

Below the x-axis Below Above


−4 −2 2 4

−4

−2

2

4

0


The root x = 0 has multiplicity 2 so the graph should bounce off the x-axisthere. The root x = 2 has multiplicity 1 so the graph should go through thex-axis there. This agrees with the graph given above.

Example 8. Graph f(x) = −6x3 + 15x2 + 12x− 9 using its sign chart.Solution 8. Go through the steps listed in the Main Concept above:


The x-intercepts are found by making y = 0:

0 = −6x3 + 15x2 + 12x− 9.

To solve this equation we can use the Rational Zero Theorem:

i. List possible zeros:

a. Factors of -9: -9, -3, -1, 1, 3, 9.

b. Factors of -6: -6, -3, -2, -1, 1, 2, 3, 6.

c. Possible zeros: -9, -3, - 32 , -1, − 12 , − 1

3 , 13 , 1

2 , 1, 32 , 3, 9.


ii. Find which of the numbers listed is a zero.

Checking each we see that f(12

)= 0, f(−1) = 0 and f(3) = 0. Since

f(x) has degree 3 these are all its zeros.

So the graph has x-intercepts at(12 , 0), (−1, 0) and (3, 0).

The y-intercept is found by making x = 0: f(0) = (1−0)(0+1)(0−9) = −9.Therefore the y-intercept is at (0,−9)

2. Use the x-intercepts to separate the x-axis into intervals. Check whether thefunction will be positive or negative in each of the intervals by plugging invalues for x.

The x-axis is separated in three parts: to the left of −1, between −1 and 12 ,

between 12 and 3, and to the right of 3. If we choose x = −2 to the left of −1

we get f(−2) = −6(−2)3 + 15(−2)2 + 12(−2) − 9 which is positive. So thefunction is above the x-axis to the left of −1. If we choose 0 between −1 and12 we get f(0) = −9 which is negative. So the function is below the x-axisbetween −1 and 1

2 . If we choose 1 between 12 and 3 we get f(1) = −6 + 15 +

12−9 which is positive. So the function is above the x-axis between 12 and 3.

If we choose x = 4 to the right of 3 we get f(4) = −16(4)3+15(4)2+12(4)−9which is negative. So the function is below the x-axis to the right of 3.

3. For each interval in the previous step determine whether the function will beabove the x-axis (positive) or below the x-axis (negative).

−4 −2 2 4

−4

−2

2

4

0

Above the x-axis Below Above Below



−4 −2 2 4

−10

10

20

0


All the roots have multiplicity 1 so the graph should go through the x-axisin each of the x-intercepts. This agrees with the graph given above.

4.3.8 Graphing a Polynomial Function Using its End Be-havior

Another way of graphing a polynomial function is by using its end behavior.

Definition 5. Let f(x) be a function. The direction that the graph

of f approaches as x becomes large and gets close to ∞, or small and getsclose to −∞ is called the end behavior of f .

Main Concept 6. Let f(x) = anxn +an−1x

n−1 + · · ·+a1x+a0. Theend behavior of f(x) is the same as the end behavior of y = anx

n.Using the end behavior, one can graph a polynomial f(x) by following

the steps:

1. Find the x- and y-intercepts of f .

2. Determine whether the graph crosses or touches the x-axis at each x-intercept.

3. Determine and draw the end behavior.

4. Put all the information together to obtain the graph of f .

Example 9. Graph each function below using their end behavior:

a. f(x) = x3 + x2 − 12x

b. f(x) = −2x4 + 2x3 + 4x2


Solution 9. Follow the steps given above:

a. f(x) = x3 + x2 − 12x


To find the x-intercepts:

0 = x3 + x2 − 12x

0 = x(x2 + x− 12)

0 = x(x + 4)(x− 3)

x = 0, x = −4, x = 3.

Find the y-intercept: f(0) = 03 + 02 − 12 · 0 f(0) = 0.


The x-intercept at 0 has multiplicity 1, so the graph crosses the x-axis at0.

The x-intercept at −4 has multiplicity 1, so the graph crosses the x-axisat −4.

The x-intercept at 3 has multiplicity 1, so the graph crosses the x-axis at3.


The leading term is x3 so the end behavior is the same as y = x3.


−5 −4 −3 −2 −1 1 2 3 4

−10

−9

−8

−7

−6

−5

−4

−3

−2

−1

1

2

3

4

5

6

7

8

9

0



−10−9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 8 9

−20

−18

−16

−14

−12

−10

−8

−6

−4

−2

2

4

6

8

10

12

14

16

18

0

b. f(x) = −x2(2x− 4)(x + 1)


To find the x-intercepts solve:

0 = −2x4 + 2x3 + 4x2

One can solve this using the Rational Zero Theorem, or by factoring. Sincewe worked with the Rational Zero Theorem in Example 6 and becausefactoring is easier in this case, we will solve this by factoring. First wecan factor −2x2 as the greatest common factor:

0 = −2x2(x2 − x− 2).

Then we can factor x2 − x− 2 as (x− 2)(x + 1). Therefore the equationto be solve is

0 = −2x2(x− 2)(x + 1),

which gives

x = 0, x = 2, x = −1.

(0, 0), (2, 0), (−1, 0).


The y-intercept:

f(0) = 02(0− 4)(0 + 1)

= 0

(0, 0).


The multiplicity of x = −1 is odd (1) so the graph will go through thex-axis.

Since the multiplicity of x = 0 is even (2) the graph will bounce off (touch)the x-axis there.

The multiplicity of x = 2 is odd (1) so the graph will go through thex-axis.


The end behavior is the leading term: y = −2x4:

−4 −3 −2 −1 1 2 3

−5

−4

−3

−2

−1

1

2

3

4

5

0



−4 −3 −2 −1 1 2 3

−5

−4

−3

−2

−1

1

2

3

4

5

0


Homework Exercises



1. (7 points) Let f(x) = 2x3 + 7x2 − 4xa. (2 pts) Find the x and y intercepts of f(x).

b. (3 pts) Draw the sign chart.

c. (2 pts) Graph the function.


2. (7 points) Let g(x) = −3x2(x + 1)2(x + 3)5

a. (2 pts) Find the x and y intercepts of g(x).




3. (7 points) Let h(x) = (2x− 3)3(x + 5)2(x− 2)a. (2 pts) Find the x and y intercepts of h(x).




4. (7 points) Let f(x) = −3(x− 2)(3x + 5)3(x− 4)a. (2 pts) Find the x and y intercepts.

b. (1 pt) For each x intercept determine whether the functioncrosses or touches the x axis.

c. (2 pts) Determine the end behavior of the function.

d. (2 pts) Graph the function.


5. (7 points) Let F (x) = 4x3 − 2x2 − 10x− 4a. (2 pts) Find the x and y intercepts.





6. (7 points) Let g(x) = 3(2x + 5)(x− 3)(x2 + 5x + 4)a. (2 pts) Find the x and y intercepts.





7. (4 points) (a) I feel comfortable finding the possiblerational zeroes of a polynomial function.


b. (4 pts) Randomly choose a polynomial function of degree 4 orgreater and find its possible rational zeros.

8. (4 points) (a) I feel comfortable finding the end behaviorof a polynomial function.


b. (4 pts) Determine the end behavior of the polynomial you chosein exercise 7.

9. (4 points) (a) I feel comfortable dividing two polynomialfunctions using long division.


b. (4 pts) Randomly choose two polynomial functions and dividethem using long division.


4.4 Rational Functions


There are video presentations on the main concept on graphing a rationalfunction and the last example shown in this section. For a video on the mainconcept on graphing rational functions and parts a and b of the last exampleclick on this link.

For a video presentation on parts c, d, and e of the last example click onthis link.

4.4.1 Introduction to Rational Functions

Definition 1. A rational function is a function of the form

F (x) = p(x)q(x) , where p(x) and q(x) are polynomials and q(x) is not the

zero polynomial.The domain of F (x) is the set of all real numbers except those for which

the denominator q(x) is zero.

Example 1. Find the domain of the rational functions:

a.

R(x) =2x2 − 4

x + 5

b.

R(x) =1

x2 − 4

c.

F (x) =x3

x2 + 1

d.

f(x) =x2 − 1

x− 1

Solution 1. For each of the functions above, and any function involvingfractions, the domain cannot include the numbers when the denominator is zero.

a. x + 5 6= 0, so x 6= −5. Therefore the domain of R(x) is {x|x 6= −5}, or(−∞,−5) ∪ (−5,∞).

http://youtu.be/7F9Id2624kE



http://youtu.be/Axli-9vxQaA

http://youtu.be/Axli-9vxQaA

4.4. RATIONAL FUNCTIONS 221

b. x2 − 4 6= 0, x2 6= 4, so x 6= ±2. Therefore the domain is {x|x 6= −2, x 6= 2},or (−∞,−2) ∪ (−2, 2) ∪ (2,∞).

c. x2 + 1 6= 0, x2 6= −1. Since x2 is always positive it will always be differentfrom −1. Therefore the domain is the set of all the real numbers.

d. x− 1 6= 0, so x 6= 1 and the domain is {x|x 6= 1} or (−∞, 1) ∪ (1,∞).

Note 1. In the last example notice that the functions x2−1x−1 and x+1 are not

equal, since their domains are different. So we can simplify x2−1x−1 = (x−1)(x+1)

x−1to x + 1 when x− 1 (the expression we are canceling) is nonzero.

4.4.2 Graphing Rational Functions

Definition 2. A vertical line x = c is a vertical asymptote of a

function R(x) if the value of |f(x)| gets large as x gets close to c.

Main Concept 1. A rational function of proper form f(x) + r(x)q(x) has

x = c as a vertical asymptote if q(c) = 0 and r(c) 6= 0. To find the verticalasymptotes, factor, cancel same factors and make the denominator equal tozero.

−10−8 −6 −4 −2 2 4 6 8

−10

−8

−6

−4

−2

2

4

6

8

0

Definition 3. A non-vertical line is an asymptote of a function R(x)

if R(x) approaches the line as |x| gets large.

Non-vertical asymptotes of the form y = L are called a horizontal asymp-totes.


−10−8 −6 −4 −2 2 4 6 8

−10

−8

−6

−4

−2

2

4

6

8

0

Non-vertical asymptotes of the form y = mx + b, for m 6= 0, are calledoblique or slant asymptotes.

−10−8 −6 −4 −2 2 4 6 8

−10

−8

−6

−4

−2

2

4

6

8

0

Main Concept 2. Suppose R(x) is a rational function with proper

form f(x) + r(x)q(x) . Then y = f(x) is a non-vertical (horizontal or oblique)

asymptote of R(x) if f(x) is a linear function. If f(x) is not a linear functionthen R(x) has no non-vertical asymptotes. To find the non-vertical asymp-totes find the proper form of the function by long division, y = Quotient isthe non-vertical asymptote.

Example 2. Find the (vertical, horizontal, and oblique) asymptotes foreach of the rational functions:

a. R(x) = 5x2

3+x

b. H(x) = x−3x2−4

c. F (x) = x−1x2+5x+4

d. G(x) = x2+3x+2x2−4 .


Solution 2. In order to find the asymptotes of a rational function, firstwrite it in its proper form.

a. Long division gives:

5x− 15

x + 3)

5x2

− 5x2 − 15x

− 15x15x + 45

45

So 5x2

x+3 = 5x− 15 + 45x+3 , its vertical asymptote: x + 3 = 0, x = −3. its non-

vertical asymptote y = 5x − 15. The non-vertical asymptote is an obliqueasymptote.

b. Since x2 does not divide x, the proper form of this function is 0 + x−3x2−4 .

Therefore its vertical asymptotes are the solutions to x2 − 4 = 0, x = ±2.

Its non-vertical asymptotes are y = 0 (horizontal asymptote).

c. As in the previous example, the proper form of F (x) is 0+ x−1x2+5x+4 . Therefore

its vertical asymptotes are the solutions to x2+5x+4 = 0, (x+4)(x+1) = 0,x = −4 and x = −1.

Its non-vertical asymptotes are y = 0 (horizontal asymptote).

d. 1

x2 − 4)

x2 + 3x + 2− x2 + 4

3x + 6

So x2+3x+2x2−4 = 1 + 3x+6

x2−4 , it has x = 2 as a vertical asymptote, and y = 1 as ahorizontal asymptote.

Main Concept 3. When graphing a rational function it is helpful tofind:

1. the domain, by setting the denominator not equal to zero.

2. the proper form, by long division.

3. the vertical asymptotes and horizontal or oblique asymptote, by usingthe previous steps.

4. the points where the function may cross its non-vertical asymptote(s),by making the function equal to its non-vertical asymptote and solvingthe equation.


5. the intercepts. Find the x-intercepts, by making the function equal tozero and solving for x and find the y-intercepts, by finding y when x = 0.

6. the sign chart of the function, by placing the domain and the x-interceptson the number-line and choosing a number in each part to determinewhether the function is positive or negative.

At the end put the above information together and graph the function.

Example 3. Graph the rational functions:

a. R(x) = x−1x2−4

b. F (x) = x2−1x

c. g(x) = x4+1x2

d. G(x) = 3x2−3xx2+x−12

e. h(x) = 2x2−5x+2x2−4

Solution 3.

a. First we will find:

1. The domain: x2 − 4 6= 0, x2 6= 0, x 6= ±2, so the domain is the set{x|x 6= 2, x 6= −2}

2. The proper form: Since x2 does not divide x, the proper form of thefunction is 0 + x−1

x2−4 = 0 + x−1(x−2)(x+2) .

3. • The vertical asymptotes: The lines x = 2 and x = −2 are the verticalasymptotes.

• The non-vertical asymptote: The line y = 0 is the horizontal asymp-tote.

4. The points where the function may cross the non-vertical asymptotes.Since y = 0 is the non-vertical asymptote, the function will cross it whenit is equal to zero:

x− 1

x2 − 4= 0

x− 1 = 0

x = 1.

5. Intercepts: If x = 0, y = 14 . So the point

(0, 1

4

)is the y-intercept. If y = 0,

we get x− 1 = 0, x = 1. Therefore the point (1, 0) is its x-intercept.


6. The x-intercepts and the domain separate the x-axis into different inter-vals.

Choosing a number to the left of −2 gives that the function is negative inthat interval, on the interval (−2, 1) the function is positive, on the interval(1, 2) the function is negative, and on the interval (2,∞) the function ispositive.

Now we put the information together to graph R(x) = x−1x2−4 :

b. For F (x) = x2−1x we get:

1. The domain: x 6= 0, {x|x 6= 0}, (−∞, 0) ∪ (0∞).

2. Long division gives:

x

x x2 − 1

−x2

−1

so the proper form of the function is x + −1x .


3. • The line x = 0 is the vertical asymptote of F (x).

• The line y = x is the oblique asymptote of F (x).

4. The graph will cross the non-vertical asymptotes if x2−1x = x:

x2 − 1 = x2

−1 = 0

which has no solution. So the graph does not cross the non-vertical asymp-tote.

5. If x = 0, the function is not defined, so it has no y-intercepts. Settingy = 0 we get x2 − 1 = 0, x = ±1. So the points (1, 0) and (−1, 0) are itsx-intercepts.

6. The x-intercepts and the vertical asymptotes separate the x-axis into dif-ferent intervals. Checking each part we get:

The function is negative on the interval (−∞, 1), it is positive on (−1, 0),negative on (0, 1) and positive on (1,∞).

Combing the above information gives the graph:

c. For g(x) = x4+1x2 :


1. The domain: x2 6= 0, x 6= 0, {x|x 6= 0}.

2. Long division gives:

x2

x2 x4 + 1

−x4

1

3. • The line x = 0 is the vertical asymptote.

• Since y = x2 is not a line, g does not have any horizontal or obliqueasymptotes. Its end behavior will be the same as y = x2.

4. The function crosses its non-vertical asymptote if

x4 + 1

x2= x2

x4 + 1 = x4

1 = 0

which has no solution. So the graph does not cross the non-vertical asymp-tote.

5. If x = 0, the function is undefined so it has no y-intercepts. Setting y = 0gives x4 + 1 = 0, so x4 = −1 which has no solution and the function hasno x-intercepts.

6. since we have no x-intercepts and x = 0 is the vertical asymptote, thex-axis is separated into two parts and it’s positive on both of them.

Combing the above information gives the graph:


d. For G(x) = 3x2−3xx2+x−12 :

1. The domain: x2 + x− 12 6= 0, (x + 4)(x− 3) 6= 0, {x|x 6= 3, x 6= −4}.2. Long division gives:

3

x2 + x− 12 3x2 − 3x

−(3x2 +3x− 36)

−6x + 36

This can be simplified to 3 + −6(x−6)(x+4)(x−3) .

3. • The lines x = 3 and x = −4 are the vertical asymptotes.

• The line y = 3 is its horizontal asymptote.

4. Since the equation3x2 − 3x

x2 + x− 12= 3


has solution

3x2 − 3x = 3x2 + 3x− 36

−6x = −36

x = 6

the function G(x) = 3x2−3xx2+x−12 crosses its horizontal asymptote y = 3 when

x = 6.

5. If x = 0, y = 0, so the point (0, 0) is its y-intercept. Setting y = 0 gives3x2 − 3x = 0, so 3x(x − 1) = 3x(x − 1) = 0 therefore the x-intercepts ofG(x) are (0, 0), (1, 0).

6. The sign chart gives that the function is positive on the intervals (−∞,−4),negative on (−4, 0), positive on (0, 1), negative on (1, 3) and positive on(3,∞).

From this information we get the graph:

e. For h(x) = 2x2−5x+2x2−4 :

1. The domain: x2 − 4 6= 0, (x + 2)(x− 2) 6= 0, {x|x 6= 2, x 6= −2}.


2. Long division gives: 2

x2 − 4)

2x2 − 5x + 2− 2x2 + 8

− 5x + 10

This can be simplified to 2 + −5(x−2)(x+2)(x−2) = 2 + −5

x+2 if x 6= 2.

3. • The line x = −2 is the vertical asymptote.

• The line y = 2 is its horizontal asymptote.

4. The graph will cross the non-vertical asymptote if

2x2 − 5x + 2

x2 − 4= 2

2x2 − 5x + 2 = 2x2 − 8

−5x = −10

x = 2

Since x = 2 makes the denominator zero, the equation has no solutionand the graph does not cross its non-vertical asymptote.

5. If x = 0, y = − 12 , so the point

(0,− 1

2

)is its y-intercept. Setting y = 0

gives 2x2 − 5x + 2 = 0, so (2x− 1)(x− 2) = 0, x = 12 , and x = 2. But f

is not defined for x = 2. Therefore the x-intercept of h(x) is(12 , 0).

6. From the sign chart we have that the function is positive on the intervals(−∞,−2), negative on

(−2, 1

2

), positive on

(12 , 2), and positive on (2,∞).

The graph is:


Homework Exercises



1. (10 points) Given f(x) = x2+2x−3x+5 :

a. (1 pt) Find the domain.

b. (2 pts) Find the proper form.

c. (1 pt) List the vertical and non-vertical asymptotes.

d. (2 pts) Find the intercepts.

e. (2 pts) Give the sign chart of the function.

f. (2 pts) Graph the function.

−10−9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 8 9

−10

−9

−8

−7

−6

−5

−4

−3

−2

−1

1

2

3

4

5

6

7

8

9

0


2. (10 points) Given f(x) = x2−3x+2x2−4x :







−10−9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 8 9

−10

−9

−8

−7

−6

−5

−4

−3

−2

−1

1

2

3

4

5

6

7

8

9

0


3. (10 points) Given f(x) = x−4x2+5 :







−10−9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 8 9

−10

−9

−8

−7

−6

−5

−4

−3

−2

−1

1

2

3

4

5

6

7

8

9

0


4. (10 points) Given f(x) = x2−9x2+4 :







−10−9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 8 9

−10

−9

−8

−7

−6

−5

−4

−3

−2

−1

1

2

3

4

5

6

7

8

9

0


5. (4 points)(a) I feel comfortable finding the vertical asymptotes of a rational

function.


b. (4 pts) Randomly choose a rational function and find its verticalasymptotes.

6. (4 points) (a) I feel comfortable finding the non-verticalasymptote of a rational function.


b. (4 pts) Find the non-vertical asymptote of the rational functionchosen in exercise 5.

7. (4 points) (a) I feel comfortable determining whether thenon-vertical asymptote of a rational function intersects the function.


b. (4 pts) Determine whether the non-vertical asymptote foundabove intersects the graph of the rational function chosen in exercise 5.


8. (4 points) (a) I feel comfortable finding the intercepts ofa rational function.


b. (4 pts) Find the intercepts of the rational function chosen inexercise 5.

9. (4 points) (a) I feel comfortable drawing the sign chart ofa rational function.


b. (4 pts) Determine the sign chart of the rational function chosenin exercise 5.


10. (4 points) (a) I feel comfortable graphing a rationalfunction.


b. (4 pts) Graph the rational function chosen in exercise 5.



4.5. PIECEWISE-DEFINED FUNCTIONS 239

4.5 Piecewise-defined Functions

Sometimes a function is defined differently on different parts of its domain.Functions defined this way are called piecewise-defined functions. The equa-tions used and the part of the domain they are used on are written on the sameline:

f(x) =

A function if x · · · ,

Another Function if x · · · ,More functions may follow if x · · · .

For example:

f(x) =

1− x if x ≤ −1,

2 if −1 < x < 1.

1 + 2x if x ≥ 1

Main Concept 1. To evaluate piecewise-defined functions first findwhich interval on the right does the value of x lies, then use the functionon the same line and substitute the value of x.

To graph a piecewise-defined function graph each part for the corre-sponding values of x.

Example 1.

a. Use the formula of f given to find the values below.

f(x) =

1− x if x ≤ −1,

3 if −1 < x < 1.

x2 if x ≥ 1

i. f(−3)

ii. f(−1)

iii. f(0)

iv. f(1).

b. Graph the function.

c. Use the graph to find the range of f .

Solution 1.

a. i x = −3 lies on x ≤ −1, so we will use the function f(x) = 1 − x andf(−3) = 1− (−3) = 4.

ii x = −1 lies on x ≤ −1. Use the function f(x) = 1 − x to find f(−1) =1− (−1) = 2.


iii x = 0 lies on −1 < x < 1. Use f(x) = 3 and find f(0) = 3.

iv x = 1 lies on x ≥ 1. Use f(x) = x2 to find f(1) = 12 = 1.

b. To graph f(x) first graph the function 1−x for x ≤ −1. This will be a line, sowe only need two points. Choose one of the points to be the point at x = −1where the graph ends, the other one can be any other point with x < −1 letschoose x = −2. So we will have the points A = (−1, 1− (−1)) = (−1, 2) andB = (−2, 1− (−2)) = (−2, 3) and the line through them ending at (−1, 2) aspart of the graph:

−10 −8 −6 −4 −2 2 4 6 8

−10

−8

−6

−4

−2

2

4

6

8

0

A

B

The graph of f(x) = 3 is a horizontal line so we will take this line for valuesof x: −1 < x < 1. Notice here that we are not including x = −1 or x = 1.To indicate this fact on the graph use open circles in these two points.


−10 −8 −6 −4 −2 2 4 6 8

−10

−8

−6

−4

−2

2

4

6

8

0

A

B C D

To graph f(x) = x2 recall the shape of the square function and graph it forvalues that are greater than or equal to 1 (including 1). So f(1) = 1 will givethe point E = (1, 1) where the function starts.


−10 −8 −6 −4 −2 2 4 6 8

−10

−8

−6

−4

−2

2

4

6

8

This is the graph of f(x).


Homework Exercises



1. (12 points) Use the function given below to answer the questions:

f(x) =

x+12 if x ≤ −1,

x2 + 3 if −1 < x < 3,

x− 5 if x ≥ 3.

a. (3 pts) What is f(2)?

b. (3 pts) What is f(−1)?

c. (6 pts) Graph f(x).


2. (4 points) (a) I feel comfortable evaluating piecewisedefined functions.


b. (4 pts) Randomly choose a piecewise-defined function f . Findf(−5), f(2) and f(7).

3. (4 points) (a) I feel comfortable graphing piecewise definedfunctions.


b. (4 pts) Graph the function f chosen in the previous exercise.



Chapter 5

Exponents and Logarithms

5.1 Exponential Functions

For a video presentation on this section from its beginning up to Example 2click on this link.

For video presetation reviewing Examples 2 and 3 click on this link


Given a number a and a positive integer n we write an for the product or a withitself n times: a · a · · · a︸︷︷︸

n times

.

Note 1. By definition a0 = 1 for any nonzero number a.Given a number a and a positive integer n we write a−n for 1

an . Thereforethe expression an is now defined for any number a and any integer n.

Moreover given a real number a and a rational number written as a fractionnd then we write a

nd for d

√an (whenever this is well defined). The expression ak

can be defined for any real numbers a and k.

Definition 1. An expression of the form ak, where a and k are real

numbers, is called an exponential expression. The number a is calledthe base and the number k is called the exponent.

5.1.1 Evaluating Exponential Expressions

Example 1. Evaluate the following expressions:

http://youtu.be/rjWlPnC0X0w

http://youtu.be/rjWlPnC0X0w

http://youtu.be/utuN2_HzBUI

246 CHAPTER 5. EXPONENTS AND LOGARITHMS

a. 41

b. 2512

c. 4−32

d. (−2)52

e. −252 .

Solution 1. The solutions follow from the remarks before the example.

a. 41 is the same as 4 multiplied by itself once. That is 41 = 4.

b. Recall that and = d

√an, so

2512 =

2√

251 =√

25 = 5.

c. Recall that a−n = 1an , so

4−32 =

1

432

=1√43

=1

23

1

8.

d.(−2)

52 =

√−25

which is not defined. So (−2)52 is not defined.

e.−2

52 = −

√25

= −√

32

= −4√

2.

One may use the calculator to evaluate exponential expressions of any form.

Main Concept 1. Laws of Exponents If s, t, a, b are real numberswith a > 0 and b > 0 then

as · at = as+t Notice that all the terms have the same base,

(as)t

= ast

5.1. EXPONENTIAL FUNCTIONS 247

(ab)s

= asbs

as

at= as−t

1s = 1

a−s =1

as

a0 = 1

Definition 2. An exponential function is a function of the form

f(x) = ax

where a is positive real number (a > 0) and a 6= 1. The domain of f is theset of all real numbers, and its range is the set of positive numbers (0,∞).

5.1.2 Graphing Exponential Functions

Main Concept 2. The graph of f(x) = ax for a > 1 is given below:

And the graph of f(x) = ax for 0 < a < 1 is:


Note 2. Both graphs have y-intercept at (0, 1), they have no x-intercept, andthey have a horizontal asymptote of equation y = 0. The difference between thetwo is that for a > 1 the graph is increasing, and for 0 < a < 1 it is decreasing.


Example 2. Graph h(x) = 2−x − 3.Solution 2. Start with the graph of f(x) = 2x and apply to transformations

in order:

1. Horizontal multiplication: Divide x by −1.

2. Vertical addition: Subtract 3 from y.

One of the numbers most used in exponential functions is e. This is definedas the value that the expression

(1 + 1

n

)napproaches as n gets large. We will

approximate e by 2.71.Example 3. Graph the function g(x) = −ex−3.Solution 3. Start with the graph of f(x) = ex and apply to transformations

in order:

1. Vertical multiplication: Multiply y by −1.

2. Horizontal addition: Add 3 to x.


Homework Exercises



1. (15 points) Graph each exponential function using transformations.a. (5 pts) f(x) = 3−x+2

b. (5 pts) g(x) = −2x−1 + 3

c. (5 pts) F (x) =(12

)2x−3+ 3


2. (4 points) (a) I feel comfortable graphing an exponentialfunction using transformations.


b. (4 pts) Randomly choose an exponential function and graph itusing transformations.




5.2 Logarithmic Functions

For a video presentation on the Definition of a Logarithm, Evaluating Loga-rithms and finding the Domain of Functions Involving Logarithms click on thislink,

and on this link.For a video presentation on Graphing Logarithmic Functions click on this

link.For a video presentation on Properties of Logarithms and writing them in

Expanded form click on this link.For a video presentation on Properties of Logarithms and writing them in

Condensed form and the Change of Base Formula click on this link.


Recall that one-to-one functions have inverses. The exponential functions ax

for a > 0 and a 6= 1 are one-to-one, so they have inverses. Some times we needto solve equations like 2x = 3, to do this the inverse function will be crucial.

Also recall that if y = ax is our exponential function, its inverse will bedefined by x = ay. So if a > 0, a 6= 1 the values for x (the domain) will bepositive (0,∞).

5.2.1 Definition of Logarithm

Definition 1. Let a > 0 and a 6= 1, and x > 0, then the logarithm

base a of x denoted by loga(x) is the power we need to raise a to get x.That is loga(x) = y if x = ay (the inverse of y = ax). The number a iscalled the base and x is called the argument.

Note 1. The logarithms with bases e and 10 are the most common, so theyhave special notation. The notation for loge(x) is ln(x) and the notation forlog10(x) is log(x).

Example 1. Change the following exponential expressions to equivalentexpressions involving a logarithm.

a. 1.23 = m

b. eb = 9

c. a4 = 24

http://youtu.be/PtGy8WYvMpU



http://youtu.be/CNqKK8noTn8

http://youtu.be/mTzPClypf0k

http://youtu.be/mTzPClypf0k

http://youtu.be/-hbWXV4EB1c

http://youtu.be/-hbWXV4EB1c

http://youtu.be/ZrG86vIdPcA

http://youtu.be/ZrG86vIdPcA

5.2. LOGARITHMIC FUNCTIONS 253

Solution 1. Using the fact that loga(x) = y is equivalent to ay = x, we canchange the exponential expressions to logarithmic expressions and vice versa.

a. 1.23 = m is equivalent to log1.2(m) = 3.

b. eb = 9 is equivalent to loge(9) = b or ln(9) = b.

c. a4 = 24 is equivalent to loga(24) = 4.

Example 2. Change the following logarithmic expressions to equivalentexponential expressions.

a. loga(4) = 5

b. loge(b) = −3

c. log3(5) = c

Solution 2.

a. loga(4) = 5 is equivalent to a5 = 4.

b. loge(b) = −3 is equivalent to e−3 = b.

c. log3(5) = c is equivalent to 3c = 5.

Evaluating Logarithmic Expressions

Example 3. Find the value of each logarithmic expression:

a. log3(81)

b. log2

(18

)Solution 3.

a. The value of log3(81) is the same as the power of 3 that will give 81. Since34 = 81, log3(81) = 4.

b. The value of log2

(18

)is the same as the power of 2 that will give 1

8 . Since18 = 1

23 = 2−3, log2

(18

)= −3.

Determining the Domain of a Logarithmic Function

As stated above, for a function y = loga x the domain is x > 0, (0,∞). So whenfinding the domain of a function you should consider:

1. The denominators (they cannot be zero).

2. Even roots (radicands have to be greater than or equal to zero).

3. Logarithms (arguments have to be greater than zero).


Example 4. Find the domain for each function below:

a. f(x) = log3(x− 2) + 3√

6x2 + 4x

b. F (x) = log2

(x+3x−1

)c. h(x) = log2(5x− 20)− 4x +

√2x− 3

Solution 4. Use the rules stated above to come with the conditions thatwill determine the domain. To solve more complicated inequalities (not linear),draw a sign-chart.

a. This function has no denominators, no even roots but is has a logarithm.So x− 2 (the argument) has to be greater than zero:

x− 2 > 0

x > 2

Therefore the domain is (2,∞).

b. The function has denominators, so x − 1 6= 0; and it has a logarithm, sox+3x−1 > 0. The first condition gives x 6= 1. To solve the inequality x+3

x−1 > 0

draw the sign chart for x+3x−1 . In the sign chart include where x+3

x−1 is equal tozero and where its denominator is zero:

x + 3

x− 1= 0

x + 3 = 0

x = −3

And:

x− 1 = 0

x = 1.

Place each number on the sign chart and check where is x+3x−1 > 0:

−5 −4 −3 −2 −1 1 2 3 40

If we choose x = −5 to the left of −3 we get that x+3x−1 = −2

−3 which is positive:


−5 −4 −3 −2 −1 1 2 3 40

⊕

If we choose x = 0 between −3 and 1 we have: x+3x−1 = 3

−1 which is negative:

−5 −4 −3 −2 −1 1 2 3 40

⊕

Choosing x = 3 to the right of 1 gives:

x + 3

x− 1=

6

2

which is positive:

−5 −4 −3 −2 −1 1 2 3 40

⊕ ⊕

Therefore x+3x−1 is greater than zero in (−∞,−3) ∪ (1,∞).

The part of the number line that satisfies the condition x 6 1 and the conditionx+3x−1 > 0 is:

−5 −4 −3 −2 −1 1 2 3 40

(−∞,−3) ∪ (1,∞).

And the domain is (−∞,−3) ∪ (1,∞).

c. This function has no denominators. It has√

2x− 3 so 2x− 3 ≥ 0; and it haslog2(5x− 20) so 5x− 20 > 0.

Solving these inequalities we get:

2x− 3 ≥ 0

2x ≥ 3

x ≥ 3

2.


And:5x− 20 > 0

5x > 20

x > 4.

Graph them on the number line and determine where both are true:

−4 −3 −2 −1 1 2 3 4 50

[(

Since the part of the number line where both inequalities are true is (4,∞),the domain is (4,∞).

Graphing Logarithmic Functions

Main Concept 1. The graph of a logarithmic function y = loga x fora > 1:

1. has domain (0,∞)

2. has range all the real numbers

3. goes through the point (1, 0)

4. has x = 0 as a vertical asymptote

5. is increasing.

-5 -4 -3 -2 -1 0 1 2 3 4 5

-3

-2

-1

1

2

3

(a,1)

(1/a,1)

And the graph of a logarithmic function y = loga x for 0 < a < 1:

1. has domain (0,∞)

2. has range all the real numbers

3. goes through the point (1, 0)

4. has x = 0 as a vertical asymptote


5. is decreasing.

-5 -4 -3 -2 -1 0 1 2 3 4 5

-3

-2

-1

1

2

3

(a,1)

(1/a,1)

Example 5. Graph f(x) = − ln(x− 2).Solution 5. Start with the graph of lnx and go through the transforma-

tions in order:

y = ln(x)

-5 -4 -3 -2 -1 0 1 2 3 4 5

-3

-2

-1

1

2

3

(1/e,1)

(e,1)

Vertical Multiplication: Multiply y by −1:

-5 -4 -3 -2 -1 0 1 2 3 4 5

-3

-2

-1

1

2

3

(1/e,1)

(e,-1)

Horizontal Addition: Add 2 to x.

-1 0 1 2 3 4 5 6 7 8

-3

-2

-1

1

2

3

(1/e+2,1)

(e+2,-1)


5.2.2 Properties of Logarithms

List of Properties

Main Concept 2. Let a, M and N be positive real numbers a 6= 1,and r be any real number. Then:

1. aloga M = M

2. loga ar = r

3. loga (MN) = loga M + loga N

4. loga

(MN

)= loga M − loga N

5. loga (Mr) = r loga M

Using these properties we may simplify logarithmic expressions by writing onelogarithm as a sum or difference of others, or (more useful) by writing a sum ordifference of logarithms as one.

Writing a Logarithmic Expression as a Sum or Difference of Loga-rithms

Example 6. Write the logarithms below as a sum or difference of logarithms(in expanded form). Express all powers as factors.

a. log2

(x2 3√x− 1

)for x > 1.

b. log6

(x4

(x2+3)2

)for x > 0

c. ln(

x2√x−2(x+1)2

)for x > 2.

Solution 6. Use the rule above to expand the expression:

a. Since x2 and 3√x− 1 are multiplying inside the logarithm from rule 3 above

it follows that

log2(x2 3√x− 1) = log2(x2) + log2( 3

√x− 1).

We can write 3√x− 1 as (x− 1)

13 and from rule 5 above we have

log2(x2) + log2((x− 1)13 ) = 2 log2(x) +

1

3log2(x− 1).

This is as simplified as possible since there are no multiplication/division orposers inside the logarithms. Moreover it will have domain x > 1 so it willbe defined for x > 1.


b. From rule 4 we have:

log6

(x4

(x2 + 3)2

)= log6(x4) + log6((x2 + 3)2)

From rule 5 it follows that this will be the same as

4 log6(x) + 2 log6(x2 + 3).

Since the expressions inside the logarithms will not factor, this is the mostsimplified expanded form.

Notice also that the expression will be defined for x > 0.

c. From rule 4 we have

ln

(x2√x− 2

(x + 1)2

)= ln(x2

√x− 2)− ln((x + 1)2).

Using rules 3 and 5 we simplify to:

ln(x2) + ln(√x− 2)− 2 ln(x + 1).

Since√x + 1 = (x + 1)

12 , using rule 5 we simplify to:

2 ln(x) +1

2ln(x− 2)− 2 ln(x + 1).

Writing a Logarithmic Expression as a Single Logarithm

Using the properties above, we can also write a logarithmic expression containinga sum or difference of logarithms, as one containing one logarithm.

Example 7. Write each expression as a single logarithm (in condensedform).

a 3 ln 2 + ln(x2)

b 12 loga 4− 2 loga 5

c −2 loga 3 + 3 loga 2− loga(x2 + 1)

Solution 7.

a. Using rule 5 in Main Concept 2 gives:

3 ln(2) + ln(x2) = ln(23) + ln(x2)

= ln(8) + ln(x2).

Using rule 3 simplifies this to:

ln(8x2).


b. Using rule 5:

−2 loga 3 + 3 loga 2− loga(x2 + 1) = loga(3−2) + loga(23)− loga(x2 + 1).

5.2.3 Changing the Base of Logarithms

Calculators only directly evaluate logarithms with base 10 or e. To evaluatelogarithms of any base using a calculator the following theorem is useful:

Main Concept 3. Change of Base Theorem Let a, b and M bepositive real numbers with a 6= 1 and b 6= 1. Then

loga M =logb M

logb a

This is especially useful when b is 10 or e. To evaluate log2 3 is a calculatoruse the above theorem to write it as log2 3 = log 3

log 2 .Example 8. Approximate to four decimal places each of the following

logarithms.

a log5 89

b log√2

√5

Solution 8. Using the change of base formula, write the expressions interms of log or ln:

a.

log5 89 =log 89

log 5

Inputting this in the calculator:


Approximating to four decimals gives:

log5 89 ≈ 2.7889.

b. Using the change of base formula:

log√2

√5 =

log√

5

log√

2

Inputting this in the calculator:

Approximating to four decimals gives:

log√2

√5 ≈ 2.3219.


Homework Exercises



1. (10 points) Find the domain for:

a. (5 pts) f(x) = log3(x+2)x−1

b. (5 pts) f(x) = log(x + 2) +√

2x− 5


2. (15 points) Graph each function:a. (5 pts) f(x) = log3(2x + 2)− 1

b. (5 pts) g(x) = 2 log 12(x− 1) + 2

c. (5 pts) F (x) = − log2(x + 2)− 1


3. (9 points) Write each expression in expanded form:

a. (3 pts) log2

(x−1x+2

)

b. (3 pts) log5

(x2+2x+1

2x−1

)

c. (3 pts) logb

(x2

x−1

)


4. (9 points) Write each expression in condensed form:a. (3 pts) log(x) + log(x + 2)− log(2x + 4)

b. (3 pts) 4 log2(3x− 2) + log2(3x)

c. (3 pts) 12 ln(x2 + 2x)− 1

3 ln(2x)

5. (6 points) Use the change of base formula and a calculator to evaluateeach expression to 3 decimal places:

a. (2 pts) log2(6)

b. (2 pts) log√2(5)

c. (2 pts) log 13(4)


6. (4 points) (a) I feel comfortable finding the domain offunctions involving demoninator(s), even root(s) and logarithms.


b. (4 pts) Randomly choose function involving at least one de-nominator, one even root and one logarithm. Find the domain of this function.

7. (4 points) (a) I feel comfortable graphing logarithmicfunctions using transformations.


b. (4 pts) Randomly choose a logarithmic function and graph itusing transformations.


8. (4 points) (a) I feel comfortable expanding logarithmicexpressions.


b. (4 pts) Randomly write down a logarithmic expression containingone logarithm and products, quotients and powers in the argument. Expandthis logarithmic expression.

9. (4 points) (a) I feel comfortable condensing logarithmicexpressions.


b. (4 pts) Take your answer from the previous exercise and write itin condensed form.




5.3 Exponential and Logarithmic Equations

For a video presentation on Exponential and Logarithmic Equations click onthis link.


5.3.1 Solving Exponential Equations

When solving exponential equations use the fact that loga aM = M


a. 23x−1 = 32

b. 22x−1 − 15 = 7

Solution 1. To solve exponential equations first isolate the exponentialexpression. Then take the appropriate logarithm to cancel the base. Simplifyas possible.

a. In the equation 23x−1 = 32 the left side is an exponential expression of base2. So to cancel the base we take log2 of each side.

log2

(23x−1

)= log2 32

3x− 1 = log2 32

Since 32 = 25 this can simplify to:

3x− 1 = log2(25)

3x− 1 = 5

3x = 6

x = 2.

Therefore the solution is {2}.

b. The exponential expression should be first isolated on one side:

22x−1 = 22

Take log2 of each side:

http://youtu.be/GqC4VoD6NRk

http://youtu.be/GqC4VoD6NRk

5.3. EXPONENTIAL AND LOGARITHMIC EQUATIONS 269

log2

(22x−1

)= log2 22

2x− 1 = log2 22

Since 22 cannot be written as a power of 2, the logarithm cannot be simplified.Solving for x:

2x = log2 22 + 1

x =log2 22 + 1

2

The solution is

{ log2 22 + 1

2}

5.3.2 Solving Logarithmic Equations

When solving logarithmic equations first condense all logarithms into one andisolate it on one side. Then use the fact that aloga M = M . Since logarithmshave restrictions on domains, make sure you check that the solutions make sensein the original equation.

Example 2. Solve:

a. log2(2x + 1) = 3

b. log3 4 = 2 log3 x

c. log2(x + 2) + log2(1− x) = 1

Solution 2.

a. In this equation we only have one logarithm that has been isolated. So weneed to raise both sides as powers of 2 to cancel log2:

2log2(2x+1) = 23

2x + 1 = 8

Now solve this as a linear equation:

2x = 7

x =7

2.

Make sure that when we substitute 72 in the original equation what’s inside

the logarithm (2x+ 1) is strictly grater than zero. Since this is the case, thesolution is

{7

2}.


b. For this equation first place all logarithms on one side:

log3 4− 2 log3 x = 0

Condense the left side to one logarithm:

log3 4− log3(x2) = 0

log3

(4

x2

)= 0

Raise both sides as powers of 3 to cancel log3:

3log3( 4x2 ) = 30

4

x2= 1

Solve this equation:

4 = x2

±2 = x.

Make sure you check that x = 2 and x = −2 do not give zero or negativeinside the logarithm(s) in the original equation. Because −2 will give anegative inside the logarithm, it is not a solution and the solution is

{2}.

c. Condense the left side into one logarithm:

log2 ((x + 2)(1− x)) = 1

Raise both sides as powers of 2 to cancel log2:

2log2((x+2)(1−x)) = 21

Simplify:

(x + 2)(1− x) = 2

Multiply and bring all terms to one side:

x− x2 + 2− 2x = 2

0 = x2 + x


Factor:

0 = x(x + 1)

x = 0 and x = −1

Check that what’s inside the logarithm(s) in the original equation is strictlygreater than zero. Since both answers satisfy this, the solution is:

{−1, 0}.


Homework Exercises



1. (35 points) Solve each equation:a. (5 pts) 2x−4 + 2 = 6

b. (5 pts) 35x−2 − 1 = 2

c. (5 pts) 2 · 53x−4 + 5 = 55


d. (5 pts) 4x − 1 = 15

e. (5 pts) 3x+2 − 1 = 5

f. (5 pts) log2(√

2x + 3− 1) = 2

g. (5 pts) log3(x + 2)− log3(2x + 1) = 2


2. (4 points) (a) I feel comfortable solving exponentialequations.


b. (4 pts) Write down an exponential equation and solve it.

3. (4 points) (a) I feel comfortable solving logarithmicequations.


b. (4 pts) Write down a logarithmic equation containing at least 2logarithms and solve it.



5.4. EXPONENTIAL GROWTH AND DECAY 275

5.4 Exponential Growth and Decay


In this section we will look at more situations that use exponential functions asmodels.

A population that grows or decays uninhibited uses the function P (t) =P0e

kt to determine its population at time t. In the equation P (t) is the popula-tion at time t, P0 is the initial population and k is the constant rate of growthor decay.

A positive value of k indicates growth, and a negative value of k indicatesdecay.

5.4.1 Solving Exponential Growth and Decay ProblemsWhen the Population Function is Given

Example 1. A colony of bacteria grows according to the law of uninhibitedgrowth described by the function P (t) = 90e0.05t, where N is measured in gramsand t is measured in days.

a. Determine the initial amount of bacteria.

b. What is the growth rate of the bacteria?

c. What is the population after 5 days?

d. How long will it take for the population to reach 140 grams?

e. What is the doubling time for the population?

Solution 1.

a. The initial amount of bacteria is the population when t = 0. So this is thesame as finding P (0).

P (0) = 90e0.05·0

P (0) = 90e0

P (0) = 90.

b. The growth rate is the constant k that multiplies t in the exponent. Bylooking as the formula

P (t) = 90e0.05·t

we see that k = 0.05.


c. The population after 5 days is P (5):

P (5) = 90e0.05·5.

Since the initial population was given as a whole number, we will approximateall population values to whole numbers. Inputting the above expression inthe calculator yields:

Rounding this to the whole number:

= P (5) = 116.

Therefore there will be 116 grams of bacteria after 5 days.

d. To find when will the population reach 140 grams find the value of t for whichP (t) = 140.

140 = 90e0.05t

14

9= e0.05t

ln

(14

9

)= ln(e0.05t)

ln

(14

9

)= 0.05t

ln(149

)0.05

= t


Since the times given in the problem are in (whole number) days, the answershould be rounded to the whole number. Therefore after 9 days there will be140 grams in the culture.

e. The doubling time for the population is the time it takes for it to double.Since the initial population started with 90 grams, the doubling time will bethe time it takes for a population to reach 180.

P (t) = 180

180 = 90e0.05t

2 = e0.05t

ln 2 = ln e0.05t

ln 2 = 0.05tln 2

0.05= t

Rounding this to the whole number gives that the doubling time for thepopulation is 14 days.


5.4.2 Finding the Population Function in Exponential Growth/Decay

Example 2. A colony of bacteria increases according to the law of uninhibitedgrowth.

a. If the number of bacteria doubles in 4 hours, find the function that gives thenumber of cells in the culture.

b. How long will it take for the size of the colony to triple?

c. How long will it take for the population to double again, that is to increasefour times?

Solution 2.

a. Since the population is increasing according to the law of uninhibited growth,P (t) = P0e

kt. Also since the population is doubling in 4 hours, P (4) = 2P0:

P0ek·4 = 2P0

e4k = 2

ln e4k = ln 2

4k = ln 2

k =ln 2

4

Therefore the population is given by the function P (t) = P0eln 24 t.

b. To find when the colony will triple, find the time when P (t) = 3P0:

P0eln 24 t = 3P0

eln 24 t = 3

ln 2

4t = ln 3

t =4 ln 3

ln 2


Rounding to the nearest hour yields that it will take 6 hours for the popula-tion to triple.

c. Similarly to the previous part, here find the time t for which P (t) = 4P0:

P0eln 24 t = 4P0

eln 24 t = 4

ln 2

4t = ln 4

t =4 ln 4

ln 2

It will take 8 hours for the population to increase four times.

Example 3. Traces of burned wood along with ancient stone tools in anarchaeological dig in Chile were found in 2012 contained approximately 1.67%of the original amount of carbon 14. If the half-life of carbon 14 is 5600 years,approximately when was the wood burned?


Homework Exercises



1. (9 points) A colony of bacteria increases according to the law ofuninhibited growth.

a. (3 pts) If there are 1000 mosquitoes initially and there are 1800after 1 day, find the population function.

b. (3 pts) What is the size of the colony after 3 days?

c. (3 pts) How long will it take for the size of the colony to triple?


2. (9 points) The half-life of radium is 1690 (that is after 1690 years itssize reduces to half the original). If 10 grams is present now, how much will bepresent in 100 years?

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