INTERMEDIATE A UNITS 4-5 · 2016-11-08 · Intermediate Algebra iii Intermediate Algebra Unit 4:...

Intermediate Algebra i

INTERMEDIATE ALGEBRA

~ UNITS 4-5 ~

Authors

Jennifer Boehlke, Belinda Eerdmans, Jean Hanson, Mark Hedin, Mary Jo Hughes, Jennifer Lehman,

Anne Roehrich, Stephanie Pogalz, Loren Tenold, Renee Voltin

Contributors

Sarah DeBoer, John FitzSimons, Jonathan Kell, Phyllis Kisch, Patrick Pangborn, Matt Rowe, Julie Rydberg

Editors (Electronic and Print)

Bruce DeWitt and Renee Voltin

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4th Edition, Printed: July, 2016

ii PREFACE

Preface

Mathematics has the extraordinary power to reduce complicated problems to simple rules and

procedures. Therein lies the danger in teaching mathematics: it is possible to teach the subject as

nothing but the rules and procedures – thereby losing sight of both the mathematics and of its practical

value. The revision process into the fourth edition of Intermediate Algebra, produced by teachers from

Anoka-Hennepin and math consultants from the local area, strives to refocus the teaching of

mathematics on concepts as well as procedures.

Balancing Skills and Conceptual Understanding

These materials stress conceptual understanding and multiple ways of representing mathematical ideas.

Our goal is to provide students with a clear understanding of the ideas of functions as a solid foundation

for subsequent courses in mathematics and other disciplines. When we designed this material, we

started with a clean slate; we focused on the key concepts, emphasizing depth of understanding.

Skills are developed in the context of problems and reinforced in a variety of settings, thereby

encouraging retention. This balance of skills and understanding enables students to realize the power of

mathematics in modeling.

Guiding Principles: Varied Problems and the Rule of Four

Since students usually learn most when they are active, the exercises in a text are of central importance.

In addition, research has shown that multiple representations encourage students to reflect on the

meaning of the material. Consequently, we have been guided by the following principles.

• Our problems are varied and some are challenging. Many cannot be done by following a

template in the text.

• The Rule of Four promotes multiple representations. Each concept and function is represented

symbolically, numerically, graphically and verbally.

• The components of these Intermediate Algebra materials make every effort to tie together

conceptual knowledge and procedural fluency. Functions as models of change is a central

theme, and algebra is integrated where appropriate.

• Problems involving real data are included to prepare students to use mathematics in other fields.

• To use mathematics effectively, students need skill in both symbolic manipulation and the use of

technology. The exact proportions of each may vary widely, depending on the preparation of the

student and the wishes of the instructor.

Intermediate Algebra iii

Intermediate Algebra

Unit 4: Situations that can be Modeled with Quadratic Functions

Students use various forms of quadratic equations to graph parabolas with a focus on using the line of

symmetry and the vertex. Though students will be working on converting between different forms of

equations for parabolas (multiplying binomials, completing the square, and factoring), the focus is on

extending their understanding from Unit 2 and recognizing the advantages of determining the location

of the vertex when graphing a parabola. Students analyze graphs of quadratic functions, use graphs and

tables to solve real-world situations, and translate between representations.

Identified Learning Targets:

4.1 I can graph quadratic functions and demonstrate understanding of significant

features of different forms of quadratic equations and their real-world situations.

4.2 I can translate quadratic equations from factored and vertex forms into standard

form.

4.3 I can translate quadratic equations from standard form into factored and vertex

forms.

Unit 5: Solving Quadratic Equations

Students understand that the quadratic formula can be found by completing the square and use the

symmetry in the graph of a parabola to identify how the x-coordinate for the vertex can be seen in

the quadratic formula. Students solve quadratic equations by factoring, finding square roots,

completing the square, and by using the quadratic formula. Students determine the number of

rational, real, and non-real solutions by factoring or solving the equation and also by using the graph.


5.1 I can use tables and graphs to solve quadratic equations including real-world

situations and translate between representations.

5.2 I can represent real-world situations with quadratic equations and solve using

appropriate methods, find real and non-real complex roots when they exist, and

recognize that a particular solution may not be applicable in the original context.

5.3 I can determine the number of real and non-real solutions for a quadratic equation.

5.4 I can represent relationships involving quadratic inequalities in multiple ways, find

solutions and interpret these solutions to solve real-world situations.

Embedded throughout the unit:

• I understand how to verify that an answer is a solution and can interpret the

solution in the context of the situation.

• I can demonstrate understanding of the real and non-real number systems and

mathematical operations using expressions from these number systems.

iv PREFACE

Table of Contents

� This book is organized in a workbook style manner. Each unit has the following structure:

o The resources and reference material for the entire unit (i.e. Unit 4: pages 3-22)

o Materials that may be used during the classroom experience (i.e. Unit 4: pages 23-78)

o A collection of practice problems to demonstrate understanding of a concept (i.e. Unit 4: pages 79-140)

� You can access the online resources, available through the Anoka-Hennepin District Moodle website at

anoka.k12.mn.us/intermediatealgebra and click on “Log in as a guest”

o Electronic access to the entire textbook (pdf format)

o Electronic access to the worked out solutions to the practice problems

o Web-based video reference to many of the topics that are addressed throughout this course.

� A graphing calculator is available through online access at Desmos.com or by downloading the free

Desmos app available for most electronic devices.

Unit 4: Situations that can be Modeled with Quadratic Functions

Text Practice Problems

2 Unit 4 Reference and Resource Material

23 79 4.1A Graphing Quadratic Equations in Standard Form

28 85 4.1B Graphing Quadratic Equations in

Intercept (Factored) Form

33 91 4.1C Graphing Quadratic Equations in Vertex Form

37 97 4.2A Multiplying Polynomials

41 101 4.2B Factored Form and Standard Form of Quadratic Equations

43 105 4.2C Vertex Form and Standard Form of Quadratic Equations

45 109 4.3A Factored Form of Quadratic Equations: Greatest Common Factor

48 113 4.3B Factored Form of Quadratic Equations: Polynomial with a = 1

53 119 4.3C Factored Form of Quadratic Equations: Polynomial with a ≠ 1

59 123 4.3D Factored form of Quadratic Equations: Special Patterns

63 127 4.3E Vertex Form of Quadratic Equations: a = 1

68 131 4.3ext Vertex Form of Quadratic Equations: a ≠ 1

73 135 Unit 4 Review Material

Intermediate Algebra v

Unit 5: Solving Quadratic Equations



175 237 5.1A Solving Quadratic Equations by Graphing

178 241 5.1B Answering Real-World Questions by Graphing Quadratic Functions

181 245 5.2A Factoring Review

182 247 5.2B Solving Quadratic Equations by Factoring: Part I

185 249 5.2C Solving Quadratic Equations by Factoring: Part II

188 251 5.2D Operations with Radical Expressions

195 255 5.2E Solving Quadratic Equations Using Square Roots to Find Rational Solutions

198 257 5.2F Solving Quadratic Equations Using Square Roots to Find Real Solutions

200 261 5.2G Operations with Complex Expressions

203 265 5.2H Solving Quadratic Equations Using Square Roots to Find Real or Complex

Solutions

204 269 5.2I (ext) Solving Quadratic Equations by Completing the Square to Find Rational

Solutions

206 271 5.2J (ext) Solving Quadratic Equations by Completing the Square to Find Real

Solutions

208 273 5.2K (ext) Solving Quadratic Equations by Completing the Square to Find Real or

Complex Solutions

209 277 5.2L Solving Quadratic Equations Using the Quadratic Formula to Find Real

Solutions

211 279 5.2M Solving Quadratic Equations Using the Quadratic Formula to Find Real or

Complex Solutions

215 281 5.2N Solving Quadratic Equations – Choosing the Best Method: Part I

218 289 5.2O Solving Quadratic Equations – Choosing the Best Method: Part II

220 293 5.3A Number and Type of Solutions: Part I

223 297 5.3B Number and Type of Solutions: Part II

225 299 5.4A Graphing Quadratic Inequalities

227 303 5.4B Solving Quadratic Inequalities


Appendices

Appendix A - Glossary

page 315 Glossary – Units 1 – 3

Appendix B - Selected Answers to Practice Problems

page 318 Unit 4 – Situations that can be Modeled with Quadratic Functions

332 Unit 5 – Solving Quadratic Equations

vi PREFACE

To Students: How to Learn from this Flexbook

� This book may be different from other math textbooks that you have used, so it may be helpful to

know about some of the differences in advance. At every stage, this book emphasizes the

meaning (in practical, graphical or numerical terms) of the symbols you are using. There is

much less emphasis on “plug-and-chug” and using formulas, and much more emphasis on the

interpretation of these formulas than you may expect. You will often be asked to explain your

ideas in words or to explain an answer using graphs.

� The book contains the main ideas of intermediate algebra concepts in plain English. Success in

using this book will depend on reading, questioning, and thinking hard about the ideas presented.

It will be helpful to read the text in detail, not just the worked out examples.

� There are few examples in the text that are exactly like the practice problems, so practice

problems can’t be done by searching for similar-looking “worked out” examples. Success with

the practice problems will come by grappling with the ideas of advanced algebra.

� Many of the problems in the book are open-ended. This means that there is more than one

correct approach and more than one correct solution. Sometimes, solving problem relies on

common sense ideas that are not stated in the problem explicitly but which you know from

everyday life.

� This book assumes that you have access to a calculator or computer that can graph functions and

find (approximate) roots of equations. There are many situations where you may not be able to

find an exact solution to a problem, but can use a calculator or computer to get a reasonable

approximation. An answer obtained this way can be as useful as an exact one. However, the

problem does not always state that a calculator is required, so use your own judgment.

� This book attempts to give equal weight to four methods for describing functions: graphical (a

picture), numerical (a table of values), algebraic (a formula) and verbal (words). Sometimes it’s

easier to translate a problem given in one form into another. For example, you might replace the

graph of a parabola with its equation, or plot a table of values to see its behavior. It is important

to be flexible about your approach; if one way of looking at a problem doesn’t work, try another.

� Students using this book will find discussing these problems in small groups helpful. There are

great many problems which are not cut-and-dried; it can help to attack them with the other

perspectives your colleagues can provide. If group work is not feasible, see if your instructor can

organize a discussion session in which additional problems can be worked on.

� You are probably wondering what you’ll get from the book. The answer is, if you put in a solid

effort, you will get a real understanding of functions as well as a real sense of how mathematics

is used in the age of technology.

1

SITUATIONS THAT CAN BE MODELED WITH QUADRATIC FUNCTIONS

Students use various forms of quadratic equations to graph parabolas with a focus on

using the line of symmetry and the vertex. Though students will be working on

converting between different forms of equations for parabolas (multiplying

binomials, completing the square, and factoring), the focus is on extending their

understanding from Unit 2 and recognizing the advantages of determining the

location of the vertex when graphing a parabola. Students analyze graphs of

quadratic functions, use graphs and tables to solve real-world situations, and translate

between representations.



4.1 I can graph quadratic

functions and demonstrate

understanding of

significant features of

different forms of

quadratic equations and

their real-world

situations.

23 79 4.1A Graphing Quadratic Equations in Standard

Form

28 85 4.1B Graphing Quadratic Equations in

Intercept (Factored) Form

33 91 4.1C Graphing Quadratic Equations in Vertex Form

4.2 I can translate quadratic

equations from factored

and vertex forms into

standard form.

37 97 4.2A Multiplying Polynomials

41 101 4.2B Comparing Factored Form to Standard Form

of Quadratic Equations

43 105 4.2C Comparing Vertex Form to Standard Form of

Quadratic Equations

4.3 I can translate quadratic

equations from standard

form into factored and

vertex forms.

45 109 4.3A Factored Form of Quadratic Equations:

Greatest Common Factor

48 113 4.3B Factored Form of Quadratic Equations:

Polynomial with a = 1

53 119 4.3C Factored Form of Quadratic Equations:

Polynomial with a ≠ 1

59 123 4.3D Factored form of Quadratic Equations:

Special Patterns

63 127 4.3E Vertex Form of Quadratic Equations: a = 1

68 131 4.3ext Vertex Form of Quadratic Equations: a ≠ 1


UNIT

4


4.1 I can graph quadratic functions and demonstrate understanding of significant features of different

forms of quadratic equations and their real-world situations.

4.2 I can translate quadratic equations from factored and vertex forms into standard form.

4.3 I can translate quadratic equations from standard form into factored and vertex forms.

Unit 4 Reference and Resource Material

2 UNIT 4 – SITUATIONS THAT CAN BE MODELED WITH QUADRATIC FUNCTIONS (CLASSWORK)

4.1 I can graph quadratic functions and demonstrate understanding of significant

features of different forms of quadratic equations and their real-world situations.

A quadratic function is a function that can be described by an equation of the form

2y ax bx c= + + , where 0a ≠ .

This is called the standard form of a quadratic function. No term in the polynomial function is of a degree higher

than 2. Quadratic functions are useful when working with area, and they often appear in motion problems that

involve gravity or acceleration.

Graphing Quadratics in Any Form

The graph of a quadratic function is a U-shaped graph, called a parabola. All

parabolas have similar key features. A parabola can be divided in half by a

vertical line. Because of this, parabolas have symmetry. The vertical line

dividing the parabola into two equal portions is called the line of symmetry.

All parabolas have a vertex, the ordered pair that represents the bottom (or

the top) of the curve. The vertex of a parabola has an ordered pair (h, k).

Because the line of symmetry is a vertical line, its equation has the form

x = h, where h is the x-coordinate of the vertex. A quadratic function will

have a term that has a degree of 2. The coefficient of the x2 term is called

“a”. This value tells us some things about the parabola.

• If a is positive, the parabola will open upward. The vertex will be a

minimum.

• If a is negative, the parabola will open downward. The vertex will

be a maximum.

• If a > 1 or a < –1, the parabola will be a vertical stretch about the line of symmetry.

• If –1 < a < 1, the parabola will be a vertical compression about the line of symmetry.

Remember that the domain is the set of all inputs (x-coordinates) and the range is the set of all outputs

(y-coordinates). The domain of every quadratic equation is all real numbers. The range of a parabola depends upon

whether the parabola opens up or down. If k is the y-coordinate of the vertex:

• If a is positive, the range will be y ≥ k.

• If a is negative, the range will be y ≤ k.

4.1 Example 1:

Determine the direction, shape, and y-intercept of the parabola formed by 234

2y x= − .

Solution:

The value of “a” is positive, therefore… the parabola opens upward.

The value of “a” is greater than 1, so… the parabola is vertically stretched about its line of symmetry.

The value of “c” is –4 so… the y-intercept is (0, –4).

Unit 4 Resources


Reference and Resource Material 3

.

Graph by Plotting Points and Determine the Characteristics of a Parabola

Finding the Axis of Symmetry from a Function Written in Standard Form

For a quadratic equation 2y ax bx c= + + , the x-coordinate of the vertex is always 2

b

a− . Since the axis of

symmetry passes through the vertex, this means the axis of symmetry is the vertical line 2

bx

a= − . The axis

of symmetry creates an imaginary vertical line that runs through the middle of the parabola and divides it into

two sides that are mirror images of one another.

Graphing Quadratic Functions in Standard Form

4.1 Example 3:

Graph 2 2 12y x x= + − .

Solution:

Step 1: Look at the coefficient of x2 to determine whether the parabola opens up or down. The coefficient of x2 is positive 1, so the parabola opens up.

Step 2: Find the x-coordinate of the vertex by using 2

bx

a= − . a = 1 and b = 2 so

21

(2)(1)x

−= = −

1x = −

4.1 Example 2:

Graph the parent graph for quadratic equations and describe at least 5 of its features.

Solution:

Create a table of values and plot the points

x (x, y)

–3

–2

–1

0

1

2

3

Possible features of this graph include:

� The y-values do not change by a constant amount.

� The shape of the graph is a parabola.

� The axis of symmetry is x = 0.

� The vertex is a minimum point.

� The domain is all real numbers.

� The range is all y-values greater than or equal to 0.

� The parabola will open up.

� The y-intercept will be (0, 0).



Step 3: Find the y-coordinate of the vertex by substituting the

x-value (from Step 2) into the original equation.

( ) ( )

( )

21 2 1 12

1 2 12

13

y

y

y

= − + − −

= + − −

= −

Step 4: Plot the vertex.

The vertex is ( 1, 13)− − . Recognize that the line of

symmetry is x = –1.

Step 5: Find and plot 2 points on one side of the axis of symmetry.

For x = 0, then y = (0)2 + 2 (0) – 12

y = –12

(0, –12)

For x = 1, then y = (1)2 + 2 (1) – 12

y = –9

(1, –9)

Step 6: Plot 2 matching points on the other side of the axis of symmetry.

x y

–3 –9

–2 –12

–1 –13 (vertex)

0 –12

–1 –9

Step 7: Sketch the parabola by drawing a smooth curve connecting the points.

-5 -4 -3 -2 -1 1 2 3

-16

-14

-12

-10

-8

-6

-4

-2

2

4

x

y

-5 -4 -3 -2 -1 1 2 3

-16

-14

-12

-10

-8

-6

-4

-2

2

4

x

y



.

4.1 Example 4:

Graph 24 8 7y x x= − + + .

Solution:

Step 1: Find the vertex.

� Use 2

bx

a= − to find the x-coordinate of the vertex a = –4, b = 8, c = 7

( ) ( )

8

2 4x = −

−

( )

8

8x = −

− 1x =

� Substitute the x-coordinate into the equation to determine the y-value.

( ) ( )2

4 1 8 1 7

4 8 7

11

y

y

y

= − + +

= − + +

=

plot the vertex at the point (1, 11)

Step 2: Then, find two points on one side of the axis of symmetry (x = 1). (For example, use x = 2 and x = 3).

�

( ) ( )2

4 2 8 2 7

4 4 16 7

16 16 7

7 plot point (2, 7)

y

y

y

y

= − + +

= − ⋅ + +

= − + +

=

�

( ) ( )2

4 3 8 3 7

4 9 24 7

36 24 7

5 plot point (3, 5)

y

y

y

y

= − + +

= − ⋅ + +

= − + +

= − −

Step 3: Because of symmetry,

you can plot the matching

points on the opposite

side of the axis of

symmetry.

Step 4: There should be a minimum of 5 points plotted on the graph of a quadratic equation.

Draw the curve.

Depending on the nature of the graph (the direction of the U-shape and the location of the vertex), a quadratic

function can have zero, one, or two x-intercepts. Here are some examples of parabolas with one, two, and zero

x-intercepts.

x y

–1 –5

0 7

1 11 (vertex)

2 7

3 –5



-4 -2 2 4

-4

-2

2

4

x

y

-4 -2 2 4

-4

-2

2

4

x

y

-4 -2 2 4

-4

-2

2

4

x

y

Graphing Quadratic Functions in Intercept (Factored) Form

4.1 Example 5:

Graph y = (x – 2)(x + 1).

Solution:

Step 1: Substitute 0 in for y: 0 = (x – 2)(x + 1)

Step 2: From section 4.1B, then either x – 2 = 0 or x + 1 = 0 Solve for

both possibilities.

x – 2 = 0 x + 1 = 0

x = 2 x = –1

So the x-intercepts are (2, 0) and (–1, 0)

Step 3: Find the x-coordinate of the vertex.

Because of the symmetrical nature of a parabola, knowing

the x-intercepts also helps us find the x-coordinate of the

vertex. If there are two x-intercepts, the vertex is exactly

halfway between them. In this case, the vertex is an equal

distance from x = 2 and x = –1, or at x = 0.5.

This is the same result as if one were to find the average of

the x-coordinates of the x-intercepts.

x = 2 and x = –1, ( )2 1 1

0.52 2

+ −= = , x = 0.5

Step 4: Find the y-coordinate of the vertex by substituting the x-value into

the original equation.

y = (0.5 – 2)(0.5 + 1) = (–1.5)(1.5) = –2.25

Step 5: Plot the vertex.

The vertex is (0.5, –2.25). Recognize that the line of

symmetry is x = 0.5.

Step 6: Find and plot a point on one side of the axis of symmetry.

For x = –2, then y = (–2 – 2)( –2 + 1) = (–4)( –1) = 4

(–2, 4)

Step 7: Plot a matching (reflecting) point on other side of the axis of symmetry.




.

4.1 Example 6:

Use a table to graph y = (x + 1)(x – 3). Solution:

x y = (x + 1)(x – 3) y

–2 (–2 + 1) (–2 – 3) = (–1)( –5) 5

–1 (–1 + 1) (–1 – 3) = (0)( –4) 0

0 (0 + 1) (0 – 3) = (1)( –3) –3

1 (1 + 1) (1 – 3) = (2)( –2) –4

2 (2 + 1) (2 – 3) = (3)( –1) –3

3 (3 + 1) (3 – 3) = (4)(0) 0

4 (4 + 1) (4 – 3) = (5)(1) 5

As you can see, y = (x + 1)(x – 3) gives a parabola. It is another form of a quadratic function.

Intercept (or factored) form is a quadratic function in the form: y = (x – m)(x – n).

In the example above y = (x + 1)(x – 3). If x + 1 = 0, then x = –1. If x – 3 = 0, then x = 3. You can see

from the graph that x = –1 and x = 3 are the x-intercepts.

4.1 Example 7:

Graph y = –2(x + 6)(x – 2).

Solution:

x-intercepts: 0 = –2(x + 6)(x – 2)

0 6

6

x

x

= +

− =

0 2

2

x

x

= −

=

( ) ( )6, 0 and 2, 0−

Vertex: Halfway between –6 and 2 is –2

( ) ( )

( ) ( )

( )

2 2 6 2 2

2 4 4

32

2, 32

y

y

= − − + − −

= − −

=

−

Additional Points: Substitute x = 0

( ) ( )

( ) ( )

( ) ( )

( )

2 6 2

2 0 6 0 2

2 6 2

24

0, 24

y x x

y

y

= − + −

= − + −

= − −

=

Reflect over axis of symmetry: (–4, 24)

Sketch the parabola.

-8 -6 -4 -2 2 4

-8

-4

4

8

12

16

20

24

28

32

36

40

x

y



Graphing Quadratic Functions in Vertex Form

The vertex form of a quadratic function is y = a (x – h)2

+ k. This form is very useful for graphing because it gives the vertex of the parabola explicitly.

The vertex is at the point (h, k). Remember the a-value tells us if the parabola opens up or down and the

general width of the parabola. The graph will shift left/right due to the quadratic term. For example, if the

quadratic term is (5x – 2)2, then the x-coordinate of the vertex is 2

5+ (set 5x – 2 = 0 and solve for x), and if the

quadratic term is (5x + 2)2, then the x-coordinate of the vertex is 2

5− (set 5x + 2 = 0 and solve for x).

4.1 Example 8:

Graph y = 2(x + 3)2 – 4.

Solution:

Step 1: Identify the vertex: (–3, –4)

Step 2: Plot the vertex and the line of symmetry ( 3x = − ).

Step 3: Find and plot 2 points on one side of the axis of symmetry.

For x = –2 then y = 2(–2 + 3)2 – 4 = –2 (–2, –2)

For x = –1 then y = 2(–1 + 3)2 – 4 = 0 (–1, 4)

Step 4: Plot 2 matching points on other side of the axis of

symmetry. (–1, 0) and ( –2, –3).




.

4.1 Example 9:

Andrew has 100 feet of fence to enclose a rectangular tomato patch. What should the dimensions of the

rectangle be in order for the rectangle to have the greatest possible area?

Solution:

Drawing a picture will help us find an equation to describe this situation:

The length and the width add up to 50, not 100, because two lengths and

two widths together add up to 100.

If the length of the rectangle is x, then the width is 50 – x.

Let y be the area of the rectangle. We know that the area is

(length) times (width), so y = x(50 – x).

Graph the parabola. The x-intercepts would be 0 and 50.

The vertex would be halfway between them at x = 25.

The y coordinate of the vertex would be y = 25(50 – 25) = 625.

Here’s the graph of that function, so we can see how the area of

the rectangle depends on the length of the rectangle.

We can see from the graph that the highest value of the area occurs

when the length of the rectangle is 25. The area of the rectangle for

this side length equals 625. (Notice that the width is also 25, which makes the shape a

square with side length 25.)

4.1 Example 10:

An arrow is shot straight up from a height of 2 meters with a velocity of 50 m/s. The function

( )2

4.9 5.1 129.45y t= − − + represents the height of the arrow in terms of time, t, in seconds.

What is the maximum height that the arrow will reach and at what time will that happen?

Solution:

Since this equation is in vertex form, we know some things about the graph already.

Because a is negative, the parabola opens downward, so the maximum height occurs at the vertex of the

parabola. Since the y-value of the vertex is 129.45, then the maximum height is 129.45 meters and it

occurs at 5.1 seconds (the x-coordinate of the vertex).

x

50 – x



Vocabulary

• A quadratic function results in a U-shaped graph called a parabola.

• The standard form of a quadratic equation is 2y ax bx c= + + .

• All parabolic functions have a vertical axis of symmetry (line of symmetry), an imaginary line that runs

through the vertex of the parabola and divides it into two sides that are mirror images of one another.

Any two points with the same y-value will be the same distance from this axis.

• When a quadratic function is given in standard form, 2y ax bx c= + + , the x-coordinate of the vertex is

always 2

b

a− . The axis of symmetry will be the vertical line

2

bx

a= −

• A coefficient is the number which is multiplied by one or more variables or powers of variables in the

term. There are two coefficients for the standard quadratic form: 2y ax bx c= + + . The coefficient of

the quadratic term (x2-term) is a and the coefficient of the linear term (x-term) is b. The value c is called

a constant term.

• The coefficient of the x2-term is called a.

• If a is positive, the parabola will open upward. The vertex will be a minimum.

• If a is negative, the parabola will open downward. The vertex will be a maximum.

• If a > 1 or a < –1, the parabola will be stretched vertically about the line of symmetry.

• If –1 < a < 1 and 0a ≠ , the parabola will be compressed vertically about the axis of

symmetry.

• The value of c is the y-intercept, or where the parabola will intersect the y-axis.

• The vertex is the highest or lowest point on a parabola. For a parabola that opens upward, the vertex is

the lowest point (minimum); for a parabola that opens downward, the vertex is the highest point

(maximum).

• The domain is the set of all inputs (x-coordinates). The domain of every quadratic equation is all real

numbers.

• The range is the set of all outputs (y-coordinates). The range of a parabola depends upon whether the

parabola opens up or down. If k is the y-coordinate of the vertex:

• If a is positive, the range will be y > k.

• If a is negative, the range will be y < k.

• Intercept (or factored) form is a quadratic function in the form: ( ) ( )y a x m x n= − − .

• The point(s) where the parabola crosses the x-axis is called the x-intercept(s) of the parabola. They are

at points (m, 0) and (n, 0) because the x-axis is where y = 0.

• The vertex is exactly halfway between the two x-intercepts. This will always be the case, and you can

find the vertex using that property.

• The vertex form of a quadratic function is ( )2

y a x h k= − + .

Video Resources:

• http://www.montereyinstitute.org/courses/Algebra1/U10L1T1_RESOURCE/index.html

• CK-12 Foundation: 1002S Graph Quadratic Functions in Intercept Form

• CK-12 Foundation: 1007S Graph Quadratic Functions in Vertex Form



.

4.2 I can translate quadratic equations from factored and vertex form into standard form.

We have studied three different forms of quadratic functions and how to graph them and find the significant features of the parabola from each of them.

• Standard form: y = ax2 + bx + c

• Vertex form: y = a(x – h)2 + k

• Intercept or factored form: y = a(x – m) (x – n)

While these forms all look different and give different information, they are just different ways to express the same quadratic function.

The Product of a Monomial and a Polynomial

The distributive property can be used to multiply a polynomial by a monomial (one-termed polynomial). Just remember that the monomial must be multiplied by each term in the polynomial. Consider the expression

( )22 2 5 10x x x+ + .

This expression can be modeled with a sketch like the one below. This model is called an area model because the rectangular pieces represent the area created by the multiplication of the monomial and the polynomial.

(Length) × (Width)

( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

2

2

3 2

2 2 5 10

2 2 2 5 2 10

4 10 20

x x x

x x x x x

x x x

+ +

+ +

+ +

( ) ( )2 3 22 2 5 10 4 10 20x x x x x x+ + = + +

Remember that the distributive property says that multiplying a sum by a number is the same as multiplying each addend by the number and then adding: a(b + c) = ab + ac. It does not matter how many terms there are: a(b + c + d) = ab + ac + ad.

The Product of Two Binomials

Now let's explore multiplying two binomials (polynomials with two terms). Once again, we can draw an area model to help us make sense of the process. We'll use each binomial as one of the dimensions of a rectangle, and their product as the area.

Remember: Area = (Length) × (Width)

2x2

2x

5x 10

4x3 10x2 20x

4x3 10x2 20x



We can draw an area model to help us make sense of the process. We'll use each binomial as one of the dimensions of a rectangle, and their product as the area.

The model below shows (x + 4)(2x + 2). The (2x + 2) is displayed across the top with the (x + 4) along the side:

Multiply the length and width to find the area for each rectangular region. After multiplying, combine like terms, for a simplified answer of 2x2 + 10x + 8.

In summary: (x + 4)(2x + 2) = 2x2 + 10x + 8

4.2 Example 1:

Multiply ( ) ( )3 3 4x x+ − .

Solution:

In this example, the box method will be used. Because it is a binomial times a binomial we will create a box that is two wide by two long.

Multiply each term together and fill in the corresponding rectangular area in the middle section.

Now, combine like terms (circled).

( ) ( ) 2

2

3 3 4 3 4 9 12

3 5 12

x x x x x

x x

+ − = − + −

= + −

3x

x + 3

–4

3x

x + 3

–4

9x

–12

3x2

–4x

3x

x + 3

–4

9x

–12

3x2

–4x



.

4.2 Example 2:

Multiply ( ) ( )2 1 3x x+ + .

Solution:

We can also use algebra to determine the product of two binomials. Multiply each term in one binomial

by all the terms in the other binomial as shown below: (distributive property).

First, multiply the first term in the first set of parentheses by all the terms in the second set of

parentheses.

Now we’re done with the first term. Next, we multiply the second term in the first parenthesis by all

terms in the second parenthesis and add them to the previous terms.

Now we’re done with the multiplication and we can simplify: (2x)(x) + (2x)(3) + (1)(x) + (1)(3)

= 22 6 3x x x+ + +

= 22 7 3x x+ +

The order in which we multiply binomials does not matter. What matters is that we multiply each

term in one binomial by each term in the other binomial. Use the distributive property to multiply the

terms of the binomials together. The last step in multiplying polynomials is to combine like terms.

Remember that a polynomial is simplified only when there are no like terms remaining.

Translate Factored (Intercept) Form into Standard Form

Now that you can multiply polynomials, let’s use this to translate a quadratic function that is in factored (intercept)

form into standard form. Remember factored form is y = a(x – m)(x – n).

To convert this to standard form, you need to multiply the binomials (x – m)(x – n), combine like terms and then

distribute the value of a to the trinomial.

4.2 Example 3:

Rewrite y = (x – 3)(x + 4) in standard form.

Solution:

We will use both strategies above to multiply (x – 3)(x + 4). (the area model or distributive property).

( ) ( ) ( ) ( )

( ) ( )

2

2

3 4 4 3 4

4 3 12

3 4 12

x x x x x

x x x

x x x x

− + = + − +

= + − −

− + = + −

So standard form would be: 2 12y x x= + − .

x

x – 3

4

–3x

–12

x2

4x



x

x 5

3

+5x

+15

x2

+3x

Now multiply by –4

4.2 Example 4:

Rewrite ( ) ( )4 5 3y x x= − + + in standard form.

Solution:

We will use both strategies above to multiply ( ) ( )5 3x x+ + then multiply

that result by –4

( ) ( ) ( ) ( )

( ) ( )

2

2

2

4 5 3 4 3 5 3

4 3 5 15

4 12 20 60

4 5 3 4 32 60

x x x x x

x x x

x x x

x x x x

− + + = − + + +

= − + + +

= − − − −

− + + = − − −

So standard form would be: 24 32 60y x x= − − − .

Translate Vertex Form into Standard Form

You use a similar process to translate a quadratic function that is in vertex form into standard form. Remember

vertex form is ( )2

y a x h k= − + . To convert this to standard form, you need to square the binomials (x – h),

distribute the value of “a” to the trinomial and then combine like terms including k.

4.2 Example 5:

Rewrite y = (x + 4)2

– 8 in standard form.

Solution:

( )

( ) ( )

( ) ( )

2

2

2

4 8

4 4 8

4 4 4 8

4 4 16 8

8 8

y x

x x

x x x

x x x

y x x

= + −

= + + −

= + + + −

= + + + −

= + +

So standard form would be: 2 8 8y x x= + + .

4.2 Example 6: Rewrite y = 3(x – 5)2 + 10 in standard form.

Solution:

( )

( ) ( )

( ) ( )

2

2

2

2

3 5 10

3 5 5 10

3 5 5 5 10

3 5 5 25 10

3 15 15 75 10

3 30 85

y x

x x

x x x

x x x

x x x

y x x

= − +

= − − +

= − − − +

= − − + +

= − − + +

= − +

So standard form would be: 23 30 85y x x= − + .



.

4.2 Example 7:

Show that the following three quadratic functions are equivalent. 2 2 8y x x= − − , ( ) ( )4 2y x x= − + and ( )

21 9y x= − −

Solution:

( ) ( )2

2

4 2

2 4 8

2 8

y x x

x x x

y x x

= − +

= + − −

= − −

( ) ( )2

2

1 1 9

1 1 1 9

2 8

y x x

x x x

y x x

= − − −

= − − + −

= − −

So all three functions are equivalent to 2 2 8y x x= − − .

Vocabulary

• A polynomial is an expression made with constants, variables, and positive integer exponents of the

variables.

• A monomial is a polynomial with one term.

• A binomial is a polynomial with two terms.

Video Resources:

• CK-12 Foundation: Multiplying a Polynomial by a Polynomial

• http://www.montereyinstitute.org/courses/Algebra1/U08L2T3_RESOURCE/index.html

4.3 I can translate quadratic equations from standard form into factored and vertex forms.

There are three different forms of quadratic functions:

• Standard form: 2

y ax bx c= + +

• Vertex form: ( )2

y a x h k= − +

• Intercept or factored form: ( ) ( )y a x m x n= − −

While these forms all look different and give different information, they are just different ways to express the same

quadratic function. Knowing how to multiply polynomials allows you to translate vertex and intercept form to

standard form.

In order to translate from standard form into intercept form, you need to learn how to factor polynomials. The first

kind of factoring to learn is to factor out the greatest common factor. This means you will use the distributive

property to factor a monomial out of a polynomial.

Factoring Monomials

4.3 Example 1:

Find the greatest common factor (GCF) of 15b2 and 20b.

Solution:

The GCF = 5 • b = 5b.

215 20

3 5 2 2 5

3 5 2 2 5

b b

b b b

b b b

⋅ ⋅ ⋅ ⋅ ⋅ ⋅

⋅ ⋅ ⋅ ⋅ ⋅ ⋅



4.3 Example 2:

Find the greatest common factor (GCF) of 481c d and 245c d .

Solution:

The GCF = 3 • 3 • c • c • d = 9c2d

Factoring Polynomials

Remember that factoring is the process of breaking a number down into its multiplicative components.

Let's go through the process of factoring a polynomial, step by step.

4.3 Example 3:

Factor 245 18c d d− .

Solution:

GCF = 3 • d, simplified, 3d

Rewrite using the factored monomials in place of the original terms ( ) ( )23 15 3 6d c d−

Use the distributive property to pull out the GCF: ( )23 15 6d c −

To check that this is correct, we can multiply 3d • (15c2 – 6), checking to see if we get the original form

of the polynomial, 245 18c d d− .

The factored form of a polynomial is one in which the polynomial is written as a product of factors, and each non-

monomial factor has no common factors in its terms. For example, the factors of 6d(4d – 3) are 6, d, and 4d – 3.

The two terms of 4d – 3 have no common factors.

4.3 Example 4:

Factor 4 281 45c d c d+

Solution:

The GCF would be 29c d

Factor (divide) out the GCF: ( )2 29 9 5c d c +

4 281 45

3 3 3 3 3 3 5

3 3 3 3 3 3 5

c d c d

c c c c d c c d

c c c c d c c d

⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

245 18

3 3 5 1 2 3 3

3 3 5 1 2 3 3

c d d

c c d d

c c d d

−

⋅ ⋅ ⋅ ⋅ ⋅ − ⋅ ⋅ ⋅ ⋅

⋅ ⋅ ⋅ ⋅ ⋅ − ⋅ ⋅ ⋅ ⋅

4 281 45

3 3 3 3 3 3 5

3 3 3 3 3 3 5

c d c d

c c c c d c c d

c c c c d c c d

⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅



.

4.3 Example 5: State the GCF of each polynomial expression and then rewrite it in factored form.

Polynomial 6x + 9 a2 – 2a 4c3 + 4c 36a2b – 18ab

GCF 3 a 4c 18ab

Factored Form 3(2x + 3) a(a – 2) 4 (c2 + 1) 18ab(2a – 1)

Solutions are recorded within the table above.

Look at how we can “pull out” the common factor in each polynomial. We know that both terms in a polynomial

are divisible by their GCF, so we can rewrite each polynomial as a product of the GCF and the combined "left

over" factors of each monomial. The work is the same for polynomials that have more terms.

4.3 Example 6:

a) Factor 5 38 12 4y y y− + .

Solution: Factor (divide) out the GCF: ( )4 24 2 3 1y y y− +

b) Factor 5 4 318 48 56 86x x x x− + − .

Solution: Factor (divide) out the GCF: ( )4 3 22 9 24 28 43x x x x− + −

Factoring Polynomials when a = 1

The factors of 2x bx c+ + are always two binomials: (x + m)(x + n) such that n + m = b and mn = c.

4.3 Example 7:

Factor 2 5 6x x+ + .

Solution:

We are looking for an answer that is a product of two binomials: ( ) ( ) x x

We want two numbers, m and n, that multiply to 6 and add to 5. A good strategy is to list the possible

ways we can multiply two numbers to get 6 and then see which of these numbers add up to 5.

6·1 and 6 + 1 = 7

2·3 and 2 + 3 = 5

So the answer is (x + 2)(x + 3).

We can check to see if this is correct by multiplying (x + 2)(x + 3):

Check: ( ) ( )

( ) ( )

2

2

2 3 3 2 6

2 3 5 6

x x x x x

x x x x

+ + = + + +

+ + = + + The answer checks out.



4.3 Example 8:

Factor 2 6 8x x− + .

Solution:

We are looking for an answer that is a product of two binomials: ( ) ( ) x x

We want two numbers, m and n, that multiply to 8 and add to -6. When negative coefficients are

involved, we have to remember that negative factors may be involved also. The number 8 can be written

as the product of the following numbers:

8·1 and 8 + 1 = 9 but also (–8)( –1) –8 + –1 = –9

4·2 and 4 + 2 = 6 but also (–2)( –4) –2 + –4 = –6

The answer is (x – 2)(x – 4).

Check: ( ) ( )

( ) ( )

2

2

2 4 4 2 8

2 4 6 8

x x x x x

x x x x

− − = − − +

− − = − + The answer checks out.

In general, whenever b is negative and a and c are positive, the two binomial factors will have minus signs instead

of plus signs.

4.3 Example 9:

Factor 2 2 15x x+ − .

Solution:

We are looking for an answer that is a product of two binomials: (x )(x )

We want two numbers, m and n, that multiply to –15 and add up to 2. When negative coefficients are

involved, we have to remember that negative factors may be involved also. The number –15 can be

written as the product of the following numbers:

15(–1) and 15 + –1 = 14 but also (–15)(1) –15 + 1 = –14

5(–3) and 5 + –3 = 2 but also (–5)(3) –5 + 3 = –2

The answer is (x + 5)(x – 3).

Check: ( ) ( )

( ) ( )

2

2

5 3 3 5 15

5 3 2 15

x x x x x

x x x x

+ − = − + −

+ − = + − The answer checks out.



.

Factoring Polynomials when a ≠ 1

4.3 Example 10:

Factor 26 2x x− − .

Solution: Test all of the factors of a and c. Check to see what combination gives you outer and inner product

that will add up to b. Remember the order will make a difference when you multiply outer and inner.

Factors of a and c Product Simplified

a = 6 c = –2

1·6 and (1)(–2) (1x + 1)(6x – 2) = 26x + 4x – 2

1·6 and (–1)(2) (1x – 1)(6x + 2) = 26x – 4x – 2

1·6 and (2)( –1) (1x + 2)(6x – 1) = 26x +11x – 2

1·6 and (–2)(1) (1x – 2)(6x + 1) = 26x – 11x – 2

2·3 and (1)( –2) (2x + 1)(3x – 2) = 26x – x – 2

2·3 and (–1)(2) (2x – 1)(3x + 2) = 26x + x – 2

2·3 and (2)( –1) (2x + 2)(3x – 1) = 26x + 4x – 2

2·3 and (–2)(1) (2x – 2)(3x + 1) = 26x – 2x – 2

Notice all of the simplified products give 26x and –2. That’s because we chose factors of a and c.

However, only one combination gives the correct b value of –1: (2x + 1)(3x – 2).

The factored form of the expression 26 2x x− − is (2x + 1)(3x – 2).

4.3 Example 11:

Factor 26 19 15x x− + .

Solution:

Again, we will test all of the factors of a and c and see what combination makes outer and inner product

that will add up to b. A positive c value means that both factors of c have the same sign: either both are

positive or both are negative. Since b < 0 then both factors of c must be negative.

Factors of a and c Product Simplified

a = 6 c = 15

1·6 and (–1)( –15) (1x – 1)(6x –15) = 26x –21x + 15

1·6 and (–15)( –1) (1x – 15)(6x – 1) = 26x –90x + 15

2·3 and (–1)( –15) (2x – 1)(3x – 15) = 26x –33x + 15

2·3 and (–15)( –1) (2x – 15)(3x – 1) = 26x –43x + 15

1·6 and (–3)( –5) (1x – 3)(6x – 5) = 26x –23x + 15

1·6 and (–5)( –3) (1x – 15)(6x – 3) = 26x –93x + 15

2·3 and (–3)( –5) (2x – 3)(3x – 5) = 26x –19x + 15

2·3 and (–5)( –3) (2x – 15)(3x – 3) = 26x –51x + 15

The factored form of the expression 26 19 15x x− + is (2x – 3)(3x – 5).



Factoring Polynomials: Special Patterns

There are a couple of special quadratics that, when factored, have a pattern.

Pattern: Perfect Square Trinomial

4.3 Example 12:

Find ( )2

a b+ .

Solution:

( ) ( ) ( )

( )

2

2 2

2 2 22

a b a b a b

a ab ab b

a b a ab b

+ = + +

= + + +

+ = + +

4.3 Example 13:

Find ( )2

a b− .

Solution:

( ) ( ) ( )

( )

2

2 2

2 2 22

a b a b a b

a ab ab b

a b a ab b

− = − −

= − − +

− = − +

A perfect square trinomial is an expression 2 22a ab b+ + or 2 22a ab b− +

• The first term is a perfect square: ( )2

a .

• The second term is twice as much as the product of the first and third terms: 2ab.

• The third term is a perfect square: ( )2

b .

4.3 Example 14:

Factor 2 8 16x x+ + .

Solution:

• First term: ( )22

x x=

• Second term: ( )8 2 4x x=

• Third term: ( )2

16 4=

The factored form of 2 8 16x x+ + is ( )2

4 .x +

Check: ( ) ( ) 2

2

4 4 4 4 16

8 16

x x x x x

x x

+ + = + + +

= + +

This is a perfect square trinomial:

2 22a ab b+ +

( )22 22a ab b a b+ + = +


2 22a ab b− +

( )22 22a ab b a b− + = −


a2

+ 2ab + b2

( ) ( )

( ) ( )

( )

2

2 2

22

8 16

2 4 4

4 4

8 16 4

x x

x x

x x

x x x

+ +

+ +

+ +

+ + = +

i i



.

4.3 Example 15:

Factor 29 12 4x x− + .

Solution:

• First term: ( )229 3x x=

• Second term: ( ) ( )12 2 2 3x x− = −

• Third term: ( )2

4 2= −

The factored form of 29 12 4x x− + is ( )2

3 2 .x −

Check: ( ) ( ) 2 23 2 3 2 9 6 6 4 9 12 4x x x x x x x− − = − − + = − +

Pattern: Difference of Squares

Multiply (a + b) (a – b)

( ) ( ) 2 2 2 2a b a b a ab ab b a b+ − = + − − = −

This is a difference of squares: 2 2a b−

A binomial that identified as difference of squares is an expression 2 2a b−−−−

• The first term is a perfect square: 2a

• The third term is a perfect square 2b

4.3 Example 16:

a) Factor 2 81x − .

Solution:


• Second term: ( )2

81 9− = −

The factored form of 2 81x − is ( ) ( )9 9x x− +

Check: ( ) ( ) 2 29 9 9 9 81 81x x x x x x− + = + − − = −

b) Factor 225 16x − .

Solution:


• Second term: ( )2

16 4− = −

The factored form of 225 16x − is ( ) ( )5 4 5 4x x+ −

Check: ( ) ( ) 2 25 4 5 4 25 20 20 16 25 16x x x x x x+ − = + − − = −

It is important to note that if you forget these formulas or do not want to use them, you

can still factor these quadratics the same way you learned previously.

This is a binomial that is the

difference of squares:

a2

– b2

( ) ( )

( ) ( )

( ) ( )

2

2 2

2

81

0 9

9 9

81 9 9

x

x x

x x

x x x

−

+ −

− +

− = − +



4.3 Example 17: Complete the square to translate the equation to vertex form and then find the vertex:

y = x2 – 6x + 1

Solution:

• ( )2 6 1y x x= − + Find the “c” value to create a perfect square trinomial.

• ( )2 6 9 1 9y x x= − + + − Balance the equation.

• ( )2

3 8y x= − − Rewrite the perfect square trinomial as a binomial squared.

• Vertex: ( )3, 8−

Vocabulary

• Factors are expressions that multiply together to produce another expression and

factoring is the process of breaking an expression down into its multiplicative factors.

• The greatest common factor (GCF) of a polynomial can be a number, a variable, or a

combination of numbers and variables that appear in every term of the polynomial.

• The distributive property is used to rewrite the polynomial as the product of the GCF

and the other parts of the polynomial. It says that a(b + c) = a • b + a • c which is the

factored form of a polynomial.

• A perfect square trinomial is a quadratic expression in the form:

2 22a ab b+ ++ ++ ++ + or 2 22a ab b− +− +− +− + .

• A difference of squares is a quadratic expression in the form: 2 2a b−−−− .

Video Resources:

• CK-12 Foundation: 0906S Greatest Common Monomial Factors

• CK-12 Foundation: Greatest Common Monomial Factors

• Khan Academy: Factoring Quadratic Expressions

• James Sousa: Ex: Factor Trinomials When A is NOT Equal to 1 - Grouping Method

• Khan Academy: U09_L2_TI_we1 Factoring Special Products 1

• James Sousa: Factoring a Difference of Squares

4.1A Graphing Quadratic Equations in Standard Form

4.1 I can graph quadratic functions and demonstrate understanding of significant features of 23

. different forms of quadratic equations and their real-world situations.

What are the characteristics of the parent graph of a quadratic function?

1) Complete the table and plot the points to sketch the graph of � = ��

x –3 –2 –1 0 1 2 3

y

The shape of a quadratic function is called a ________________.

The highest or lowest point on the curve is the ______________.

The _______________________ is the vertical line passing

through the vertex.

The domain is ________________________________.

The range is __________________________________.

2) What does a tell you about the shape of the graph of � = �� + � + �?

Graph each of the equations using a graphing utility to complete the information in the table below.

Equation Positive or negative a?

Does the graph open up or down? | |

Is the graph a vertical stretch

or compression of � = ��?

27y x=

21

2y x=

21

6y x= −

23

2y x= −

a) When a is positive, the parabola ___________________________________.

b) When a is negative, the parabola ___________________________________.

c) When |�| < �, the graph, compared to the parent graph, ( stretches vertically / compresses vertically ).

d) When |�| > �, the graph, compared to the parent graph, ( stretches vertically / compresses vertically ).

Section

4.1A

-10 -8 -6 -4 -2 2 4 6 8 10

-10

-8

-6

-4

-2

2

4

6

8

10

x

y

circle the correct situation

circle the correct situation



3) Explore how the a- and b-values of the equation y = ax2 + bx + c affect the graph.


Use the MAX or MIN option in the CALC menu to find the coordinates of the vertex.

Equation

Coordinates of the vertex. (min or max?)

2

bx

a= −

(Show your work.)

Substitute 2

bx

a= − in for x to find y.

(Show your work.)

2 2 1y x x= + +

a = b = c =

2 2 1y x x= − − −

a = b = c =

24 8 2y x x= − + +

a = b = c =

22 8 6y x x= − +

a = b = c =

23 18 20y x x= − +

a = b = c =

a) The equation 2

bx

a= − represents the ______________________________________ of the graph.

b) To find the vertex, ___________________________________________________________________

___________________________________________________________________________________

___________________________________________________________________________________

4) Explore how the c-value of the equation y = ax2 + bx + c affects the graph.


Equation What is the y-intercept?

� = �� + 3

a = b = c =

� = −2�� + 4

a = b = c =

� = �� + 3� + 2

a = b = c =

� = −�� − 2� + 1

a = b = c =

a) To find the y-intercept, ______________________________________________________________

b) The graph of a quadratic equation will sometimes, always, or never (circle one) have an x-intercept.

c) The graph of a quadratic equation will sometimes, always, or never (circle one) have a y-intercept.




Ao

S:

Vertex

:

y-intercep

t:

y =

2x

2 +

8x

+ 4

u

p/d

ow

n

Ao

S:

Ver

tex

:

y-in

terc

ept:

Stan

dard

Fo

rm P

artner M

atch U

p

Draw

lines co

nn

ecting

the g

raph

s with

their eq

uatio

ns. A

oS

:

Vertex

:

y-intercep

t: A

oS

:

Ver

tex

:

y-in

terc

ept:

Ao

S:

Ver

tex

:

y-in

terc

ept:

y = x

2 + 4

x +

4

up

/do

wn

Ao

S:

Vertex

:

y-in

tercept:

y = -3

x2 +

6x

- 3

up

/do

wn

Ao

S:

Vertex

:

y-intercep

t:

y =

-x

2 +

2x

- 3

u

p/d

ow

n

Ao

S:

Ver

tex

:

y-i

nte

rcep

t:



5) Fill in the following table.

Equation

Does the graph open up or down? Vertex y-intercept

a) � = �� − 8� + 15 Up / Down

b) � = −�� + � − 6 Up / Down

c) 21

4 32

y x x= − + Up / Down

d) 22

4 33

y x x= − − + Up / Down

e) � = �� − 9 Up / Down

f) 24 2 1y x x= − + Up / Down

g) 2 1

42

y x x= + − Up / Down

6) Graph 22 8 5y x x= − − − 7) Graph 2 4 3y x x= − +

a) Min or Max: ________________ a) Min or Max: ________________________

b) Axis of symmetry: ____________ b) Axis of symmetry: ____________________

c) Vertex: _____________________ c) Vertex: _____________________________

d) y-intercept: __________________ d) y-intercept: __________________________

e) Domain: ____________________ e) Domain: ____________________________

f) Range: _____________________ f) Range: _____________________________

g) Opens (up or down): ___________ g) Opens (up or down): __________________

x y

x y




8) A ball is thrown up in the air from 3 meters above

the ground. The height of the ball over time, (t) in

seconds, can be modeled by

( ) 24.9 14 3h t t t= − + + .

a) Find the maximum height of the ball.

b) How long does it take the ball to reach the maximum height?

c) Find the length of time the ball is in the air.

9) Given the quadratic equation 2 11 10y x x= − + . Answer the following questions.

a) Does the relation have a maximum or minimum value? Explain your thinking.

b) What is the y-intercept? Explain your thinking.

10) The vertex of a quadratic relation is at (2, 5). Does the relation have a maximum or minimum value? Explain.

4.1B Graphing Quadratic Equations in Intercept (Factored) Form


Exploring Quadratics in Factored Form

The quadratics we have explored so far have all been in the form

� = �� + �� + � . Recall what an equation in this form tells us about its graph.

1) Fill in the table below with information you have collected on your calculator. Work through the first

example with your teacher and then repeat the same steps for the remainder of the examples.

Equation in the form

( ) ( )y a x m x n= − − x-intercepts Axis of Symmetry Vertex

a) ( ) ( )1 3 5y x x= − −

b) ( ) ( )1 3 2y x x= − − +

c) ( ) ( )2 1 5y x x= + −

d) ( )3y x x= −

e) ( ) ( )1 2 1 3 4y x x= − − +

f) ( ) ( )3 3y x x= + +

g) ( ) ( )3 2 1 2 7y x x= − +

� = �� + � + �

The a value determines:

The c value determines:

Section

4.1B




Draw a conclusion about the relationship between the factored form of a quadratic and the x-intercepts of its graph. Record it in the space below. Include an example.

y = a(x – m)(x – n)

2) a) From your chart in #1, what is the relationship between the x-intercepts and the x-value of the vertex of

the parabola?

b) How can the x-intercepts be used to determine the x-value of the vertex?

Use 2 8 15y x x= − + as an example. The x-intercepts are ( )3,0 and ( )5,0 .

c) How can we use this information to find the vertex of the parabola?

3) For the graph shown, choose the equation that the

graph represents.

[A] ( ) ( )1 4y x x= + +

[B] ( ) ( )2 3y x x= + −

[C] ( ) ( )2 3y x x= − +

[D] ( ) ( )2 3y x x= − −

4) Regarding the previous question, explain your

thinking and why you chose the equation you did.



5) Water is shot straight up out of a water soaker toy. The quadratic function 16 ( 2)y x x= − − models the

height in feet of a water droplet after x seconds. How long is the water droplet in the air?

a) Graph the function, label the axes. What should you use as a scale on the vertical and horizontal axes?

b) At what time does the water spray reach its

maximum height?

c) What is the maximum height?

d) What do the x-intercepts represent for this

situation?

e) What are the domain and range of the

function?

f) How does the situation restrict the domain and range?

g) Does this model make sense in this context? Why or why not?




6) Kevin wants to fence in his backyard using a rectangular region so that his dogs can have a safe play area.

He has purchased 140 feet of fencing and will use the back of his house as one side of the enclosed yard. The

equation that would give the area of the fenced in region is

( ) ( )140 2A x x x= − .

a) Sketch the graph that you create using a graphing utility. Be sure to label and include ordered pairs for the significant features of the graph.

b) What is the maximum area that can be enclosed?

c) Find the dimensions that produce the maximum area. Use at least two methods. Show/explain your methods.

backyard x x

140 – 2x

HOUSE



#7 – 10: Find the following for each equation.

7) ( ) ( )3 4y x x= − −

a) x-intercept(s): ______________________

b) Axis of Symmetry: __________________

c) Vertex (Min or Max value): ___________

d) Domain: __________________________

e) Range: ____________________________

f) Graph the equation.

x y

8) ( ) ( )1

3 72

y x x= + −

a) x-intercept(s): ______________________



d) Domain: __________________________

e) Range: ___________________________


x y

9) ( ) ( )2 1 5y x x= − +

a) x-intercept(s): ______________________



d) Domain: __________________________

e) Range: ____________________________


x y

10) ( ) ( )4 1 2 3y x x= − + −

a) x-intercept(s): ______________________

b) Axis of Symmetry: _________________

c) Vertex (Min or Max value): __________

d) Domain: __________________________

e) Range: ___________________________


x y

4.1C Graphing Quadratic Equations in Vertex Form



1) The following quadratic equations are in standard form. Using a graphing utility,

sketch a quick sketch of the graph. Find the vertex of each of the following

functions.

( ) 22 4 2r x x x= − − − ( ) 24 8 6s x x x= + −

( ) 2 10 22t x x x= − + ( ) 23 12 7w x x x= − − −

2) The following quadratic functions are written in vertex form. Use a graphing utility to graph r(x) listed below along with r(x) from 1) above. What do you notice? Do the same for s(x), t(x), and w(x).

( ) ( )2

2 1r x x= − + ( ) ( )2

4 1 10s x x= + −

( ) ( )2

5 3t x x= − − ( ) ( )2

3 2 5w x x= − + +

3) A quadratic equation, 2y ax bx c= + + , when written as ( )2

y a x k k= − + is in vertex form. Make a

conjecture about the relationship between the vertex of a quadratic function (as found in question 2)) and the values of a, h and k in the vertex form of a quadratic equation.

4) For each quadratic function, identify the values for a, h and k. Using these values and the conjecture you

made in question 3), record what you suspect are the coordinates of the vertex for the graph of each function. Compare these values to the information you have gathered in problems 1) and 2).

( ) ( )2

2 1r x x= − + ( ) ( )2

4 1 10s x x= + −

( ) ( )2

5 3t x x= − − ( ) ( )2

3 2 5w x x= − + +

Section

4.1C



5) Describe how to find the vertex of a quadratic function that is written in vertex form. 6) Find the vertex of the graph of each of the following quadratic functions. Verify your answers by graphing.

( ) ( )2

2 6 8f x x= − − + ( ) ( )2

2 3g x x= + −

( ) ( )2

2 3q x x= − ( ) ( )2

3 5 1p x x= − + −

Vertex Form of a Quadratic Equation:

A quadratic equation, 2y ax bx c= + + , when written as ( )2

y a x h k= − + is in vertex form, helping to

identify the coordinates of the vertex of the graph at ( ),h k and the vertical stretch of the graph when

compared to the parent function 2y x= .

7) The Big Buns Bakery sells more bagels when it reduces its prices, but then its profit changes. The function

( ) ( )2

1000 0.55 300f x x= − − + models the bakery’s daily profit in dollars from selling bagels, where x is

the price of a bagel in dollars. The bakery wants to maximize the profit.

a) What is the domain of the function? Can x be negative? Explain.

b) Find the daily profit for selling bagels for $0.40 each; for $0.85 each.

c) What price should the bakery charge to maximize its profits from bagels?

d) What is the maximum profit?

e) What happens if they charge more than $1.10 for a bagel? Explain.




# 8 – 11: Find the following for each equation.

8) ( )2

5 2y x= − +

a) Vertex: __________________________

b) Opens (up or down): _______________

c) Axis of Symmetry: ________________

d) Maximum or minimum? ____________

e) Domain: _________________________

f) Range: __________________________

g) Graph the equation.

x y

9) ( )2

2 4y x= − + +

a) Vertex: __________________________

b) Opens (up or down): ________________

c) Axis of Symmetry: _________________

d) Maximum or minimum? _____________

e) Domain: _________________________

f) Range: ___________________________

g) Graph the equation.

x y

10) ( )21

1 32

y x= − − −

a) Vertex: __________________________

b) Opens (up or down): _______________

c) Axis of Symmetry: ________________

d) Maximum or minimum? ____________

e) Domain: _________________________

f) Range: __________________________

11) ( )2

3 5y x= + −

a) Vertex: __________________________

b) Opens (up or down): ________________

c) Axis of Symmetry: _________________

d) Maximum or minimum? _____________

e) Domain: _________________________

f) Range: ___________________________



12) A ball is hit into the air. Its height H (in meters above the ground) after t seconds is modeled by the function

( ) ( )2

4.9 2.9 42.H t t= − − +

a) In which direction does the parabola open?

How do you know?

b) Determine the initial height of the ball. Show

the analysis that leads to your answer.

c) What are the coordinates of the vertex? What

does the vertex represent in this situation?

d) Determine one other coordinate on the curve

and interpret its meaning.

e) When does the ball hit the ground? Use a

graphing utility.

f) State the domain and range for this model.

Explain the reasoning for your answers.

g) Sketch the flight path of the ball.

4.2A Multiplying Polynomials

4.2 I can translate quadratic equations from factored and vertex forms into standard form. 37

.

1) Find the total area for the joined rectangles below in at least two different ways. a) b)

c) d)

e) f)

2) Draw the area model to represent the following multiplication problems.

a) ( )8 2x − b) ( )3 4x y+

c) ( )3x x y− + d) ( )22 3 5 1x x x− −

�� + 3x –8

3

�� –4x + 8

–2

−7�� + x –5

2x

Section

4.2A



3) One equation to represent the first picture in problem #1 on the previous page is 2(3 + 6) = 6 + 12. This is an example of the Distributive Property. Go back and write equations like this for each of the images on the previous page if you don’t have them already.

A monomial is the product of non-negative integer powers of variables. Consequently, a monomial has NO variable in its denominator. It has one term. (mono implies one)

13, 3x, –57, x², 4y², –2xy, or 520x²y² (notice: no negative exponents, no fractional exponents)

A binomial is the sum of two monomials. It has two unlike terms. (bi implies two)

3x + 1, x² – 4x, 2x + y, or y – y²

An equivalent equation or expression is when two expressions or equations simplify to represent the same thing 7(x – 8) is equivalent to 7x – 56

�� = −2�� − 12� is equivalent to �� = −2�� + 24�

4) Draw an area model and write an equivalent equation for each of the following functions using the distributive property.

a) ( ) ( )5 3f x x= + b) ( ) ( )24 2f x x x= − −

c) ( ) ( )15f x x x= + d) ( ) ( )2 12f x x x= − −

5) i) Sketch a graph of each of the functions below. ii) Translate the following quadratic functions (written in intercept form) into standard form by using the

distributive property. iii) Graph both equations on your calculator to check that they represent the same quadratic relationship.

a) ( ) ( )2 3f x x x= + b) ( ) ( )7h x x x= − +



.

6) Find the total area for the joined rectangles below in at least three different ways.

a) b)

c) d)

e) f)

3x –4 2y

6x

5y



7) One representation of the multiplication shown in the area model in 6a) on the previous page is

( ) ( )5 2 6 4 5(6 4) 2(6 4)

5 6 5 4 2 6 2 4.

+ + = + + +

= ⋅ + ⋅ + ⋅ + ⋅

This is an example of the distributive property. Go back and write equations like this for each of the images above if you haven’t already done so.

8) i) Draw rectangles like the ones from #6 to represent each equation (as if it were a “find the area”

problem). ii) Write an equivalent equation for each of the following functions using the distributive property.

a) ( ) ( ) ( )5 8f x x x= − − b) ( ) ( ) ( )8 5f x x x= − +

c) ( ) 2(4 3)( 2)f x x x= − − + d) ( ) 2( 1)f x x= +

9) A square garden is surrounded by a walkway of width x.

a) Find the area of the garden.

b) Find the area of the walkway and the garden combined.

c) Assume the contractor uses 1’ x 1’ square paving bricks. Determine the number of paving bricks needed to create the walkway.

10) Find an expression for the area of each shape in both factored and standard form.

a) b)

2x - 6

x + 3

2x – 3

2x +5

4.2B Comparing Factored Form to Standard Form of Quadratic Equations


.

There are several ways that a quadratic function can be expressed: factored form, vertex form, and standard form. When written in each form, different significant features can be identified.

#1 – 4: Find the x-intercepts, convert to standard form, find the y-intercept and

graph. Graph both the factored form and standard form of each function

on your graphing utility to check that they represent the same quadratic

relationship.

1) ( ) ( ) ( )4 3f x x x= + + 2) ( ) ( ) ( )3 4f x x x= − +

a) x-intercept(s): _______________________ a) x-intercept(s): _______________________

b) standard form: ______________________ b) standard form: ______________________

c) y-intercept: _________________________ c) y-intercept: _________________________

d) make a table and graph: d) make a table and graph:

x f(x)

x f(x)

3) ( ) ( ) ( )4 5f x x x= − − 4) ( ) ( ) ( )3 4 2 4f x x x= − − −





x f(x)

x f(x)

Section

4.2B



5) Use this process to change the following equations, written in factored (intercept) form, into standard form.

a) ( ) ( ) ( )4 4 3f x x x= − + b) ( ) ( ) ( )3 2 9f x x x= − + −

c) ( ) ( ) ( )5 2 1 6f x x x= + − d) ( ) ( ) ( )6 2 5 3 8f x x x= + −

6) Four hundred people came to last year’s winter play at Sunnybrook High School. The ticket price was $5.

This year, the Drama Club is hoping to earn enough money to take a trip to a Broadway play. They estimate that for each $0.50 increase in the price, 10 fewer people will attend their play. This situation can be modeled

by ( ) ( ) ( )400 10 5 0.50M x x x= − + where x is

the number of $0.50 increases and m(x) is the amount of money earned.

a) How much should the tickets cost in order to

maximize the income from this year’s play?

b) What is the maximum income the Drama Club can expect to make?

Example: Translate �� = 3�� + 3�� − 8� into standard form. Solution: Ignoring the 3 and expanding (x + 3)(x – 8), using the distributive property,

gives �� = 3�� − 5� − 24�.

Then using the distributive property gives �� = 3�� − 15� − 72.

4.2C Comparing Vertex Form to Standard Form of Quadratic Equations


.

Draw area models like the ones from previous sections to represent the following equations: [A] Use the distributive property to translate from vertex form to standard form.

[B] Graph both the vertex and standard forms of the function on your graphing utility to check that they represent the same quadratic relationship.

NOTE: (x + y)2 is the same as (x + y)(x + y).

1) ( ) ( )2

3 9f x x= − − 2) ( ) ( )2

5 2 3f x x= − +

a) vertex: ____________________________ a) vertex: ____________________________

b) standard form: ______________________ b) standard form: _______________________

c) y-intercept: _________________________ c) y-intercept: __________________________


x f(x)

x f(x)

3) ( ) ( )2

3 4f x x= − 4) ( ) ( )2

2 3 4f x x= − + −

a) vertex: ____________________________ a) vertex: ______________________________

b) standard form: ______________________ b) standard form: ________________________

c) y-intercept: _________________________ c) y-intercept: ___________________________


x f(x)

x f(x)

Section

4.2C



5) Determine whether the equations below are equivalent using both a graphical approach and an algebraic approach.

( )2

2 1 y x= − + and 2 4 5y x x= − +

Graphically

( )2

2 1 y x= − + 2 4 5y x x= − +

x y

x y

Algebraically

Does: ( )2 22 1 equal 4 5y x y x x= − + = − +

( )?

2 2 2 1 4 5x x x− + = − +

4.3A Factored Form of Quadratic Equations: Greatest Common Factor

4.3 I can translate quadratic equations from standard form into factored and vertex forms. 45

.

1) Make a list of factors for 36.

2) Make a prime factor tree for each expression below.

28 36 x2

3) The GCF is the greatest factor that two expressions have in common. What is the greatest common factor of

28 and 36? (Ex: The GCF of 16 and 24 is 8.)

4) Find the GCF of:

a) 30 and 45

b) 210 and 168

c) 30x4 and 12x2y d) 2x and 6x2

e) 15xy and 8 f) 9a2b3 and 18a4

Section

4.3A



Example: Factor out the GCF and rewrite as a

product of two factors.

5) Factor each polynomial.

a) 6 9x− + b) 220 4t t−

c) 27x2 – 12x + 15 d) –18x3 + 60x – 36x2

e) 72m5n – 45m3n4

6) Write the equation in factored form. Using a graphing utility, record the graph for both equations. Finish by

identifying the y-intercept and the vertex.

Factor completely. Graph both standard and factored

forms on your graphing utility (they

should match). Sketch the graph.

Use the graphing utility to find

the y-intercept and the vertex.

23 24 45y x x= − +

y-intercept:

vertex :

7) Jerry says that ( ) 218 12f x x x= + factors into ( ) ( )3 6 4f x x x= + . Is this completely factored? Explain.

To factor using the GCF…

Step 1: Identify the GCF of the terms.

Step 2: Divide the GCF from the rest of the terms.

Step 3: Rewrite as a product (use parentheses).

639 2++ xx



.

Tic – Tac – Toe

Pick a problem from the top board to mark your spot and factor.

Find your solution on the bottom board to mark your spot. If you do not find your solution, then your partner gets

the spot.

x2 – 6x

144x8y2 + 27x4y + 9x2 10m3n2 – 25m2n3 8a3 + 4a2 + 12a

15x3y – 35x2y2 + 40xy3

2x2 + 8x 7a4 – 35a3 – 14a2

49m5n3 + 28mn4

8m3n4 – 22m5n6

6a8 + 10a6 – 3a4 45x2 – 20x 12x5y2 + 9x4y2 – 6x3y2

36a3 – 24a4 + 60a5

72m7n + 24n 40x2 – 100xy – 80y2 8x5 – 15x3

9x2(16x6y2 + 3x2y + 1)

5x(9x – 4) 7a2(a2 – 5a – 2) 5m2n2(2m – 5n)

2x(x + 4)

12a3(3 – 2a + 5a2) 24n(3m7 + 1) 20(2x2 – 5xy – 4y2)

5xy(3x2 – 7xy + 8y2)

7mn3(7m4 + 4n)

3x3y2(4x2 + 3x – 2) x(x – 6)

x3(8x2 – 15)

a4(6a4 + 10a2 – 3) 2m3n4(4 – 11m2n2) 4a(2a2 + a + 3)

4.3B Factored Form of Quadratic Equations: Polynomials with a = 1


1) The standard form function rule ( ) 2 5 6f x x x= + + is equivalent (the same as) to the

factored form function rule ( ) ( ) ( )3 2f x x x= + + as seen in the area model below:

NOTE: factored form and intercept form are the same thing.

[1] Circle all the pieces ( ��, ��, �) of the area model to the left that represent the standard form function rule.

[2] Put a box around all the pieces

�� + "�and�� + &� of the area model to the left that represent the factored form function rule.

[3] Where does the 5x come from?

2) Complete each area model below and then write the equivalent equations in both standard form and factored

form. a) b) c)

d) e) f)

g) h)

Section

4.3B

4.3B Factored Form of Quadratic Equations: Polynomial with a=1


.

3) Complete each area model below and then write the equivalent equations in both standard form and intercept

form. State the x-intercepts.

a) ( ) ( ) ( )2 25 10 5 5 2 1 5 ___ ___x x x x x x+ + = + + = + +

b) ( ) ( ) ( )23 21 36 ___ ___ x x− + − = =

c�c�c�c� ( ) ( ) ( )26 18 60 ___ ___ x x− + + = =

x2 x

1

2x

x

x2 x

__x

x

x-intercepts =

x-intercepts =

x-intercepts =

x2 x

__x

x



4) Create an area model to factor each trinomial.

a) 2 7 12x x+ + b) 2 12 20x x+ +

c) 2 8 12x x− + d) 2 2 15x x− −

e) 2 4 45x x− − f) 2 6 8m m− +

g) 2 5 6x x+ + h) 2 5 6x x− +

i) 2 5 6a a− − j) 2 2 1x x+ +

4.3B Factored Form of Quadratic Equations: Polynomial with a=1


.

5) Given the quadratic equation: ( ) 2 6 5f x x x= + +

a) Use 2

b

a− to find the x-coordinate of the vertex.

b) Write in factored form.

c) Use factored form to find the x-intercepts.

d) Find the x-coordinate halfway between the x-intercepts. What does this value represent?

e) Do your answers from part a and d match? Explain.

f) Graph the function.

x f(x)



6) Jason jumped off of a cliff into the ocean in Acapulco while vacationing with some friends. His height as a

function of time could be modeled by the function ( ) 216 16 32h t t t= − + + , where t is the time in seconds

and h is the height in feet above the ocean.

a) How high is the cliff that Jason jumped off of?

b) Write the equation in factored form? (Hint: you may need to factor out a GCF first).

c) When does Jason land in the ocean? Explain your reasoning.

d) When does Jason reach his

maximum height? What is his

maximum height?

e) Create a graph that shows Jason's

height at time t.

f) What is the domain and the range?

(Remember to take the context of the

situation into consideration.)

4.3C Factored Form of Quadratic Equations: Polynomial with a≠1


.

1) Fill in the missing operations for the problems below:

a) ( ) ( )212 7 10 3 ___2 4 ___5x x x x− − =

b) ( ) ( )212 7 10 3 ___2 4 ___5x x x x+ − =

c) ( ) ( )212 23 10 3 ___2 4 ___5x x x x− + =

d) ( ) ( )212 23 10 3 ___2 4 ___5x x x x+ + =

2) The function rule ( ) 22 7 3f x x x= + + is the same as ( ) ( ) ( )2 1 3f x x x= + +

as seen in the area model below:

[1] Circle all the pieces ( ��, ��, �) of the area model to the right that represent the standard form function rule.

[2] Put a box around all the pieces,

( ) ( ) and ax m bx n+ + , of the area model to the left

that represent the factored form function rule.

[3] Where does the 7x come from?

3) Complete each area model below and then write the equivalent equations in both standard form and intercept form. a) b) c)

d) e)

Section

4.3C



4) Write the equation in factored form. Using a graphing utility, record the graph for both equations. Finish by

identifying the vertex and the x-intercept(s).

Factor completely

Graph both standard and factored forms

on a graphing utility (they should match).

Sketch the graph.

Use a graphing utility to find

the vertex and the

x-intercepts.

26 7 5y x x= − −

x-intercept(s):

vertex :

5) Using an area model, go from standard form to intercept form (and graph both forms on your graphing utility

to check that both represent the same quadratic relationship).

a) 22 14 20x x+ + b) 25 60 140x x− −

c) 23 24 48x x− + + d) 22 7 3x x+ +

e) 23 17 10x x+ + f) 28 18 9x x+ +

g) 23 2 8t t+ − h) 22 11 6x x− −

i) 26 15x x− − j) 218 9 14n n+ −

k) 235 20 15u u− − l) 220 38 12m m− +

m) 25 7 6t t− + + n) 23 7 2x x− − −



.

6) Determine if the equations represent the same parabola and explain.

a) 23 10 3y x x= + + and ( )( )3 1 3y x x= + +

b) 212 4 8y x x= − − and ( )( )2 3 4 2 1y x x= − +

7) You and a friend are hiking in the mountains. You want to climb to a ledge that is 30 ft. above you. The

height of the grappling hook you throw is given by the function ( ) 216 38 5h t t t= − + + .

a) What is the height from which you throw the grappling hook?

b) Write the equation in factored form. (Hint: it might be easier if you factor out a value of –1 first)

c) When does the grappling hook land on the ground?

d) What is the maximum height that the grappling hook reaches? Was your throw successful? Explain

your reasoning.

e) Create a graph that shows the grappling

hook at time t.

f) What is the domain and range?

(Remember to take the context of the

situation into consideration.)



This page intentionally left blank



.

Activity:

4.3D Factored Form of Quadratic Equations: Special Patterns


.

1) The polynomials below follow special patterns when in factored form.

Complete each multiplication.

( ) ( )4 4x x+ − ( ) ( )3 3x x+ −

( ) ( )2 5 2 5x x+ − ( ) ( )a b a b+ −

What do you notice about the resulting polynomial?

2) Let’s write the general rule.

Difference of Two Squares: ( ) ( )a b a b+ − =

3) Factor.

24 36u − 216 49m −

29 4p + 2 2 25d e −

4) Multiply. What do you notice?

a) ( ) ( ) ( )2

5 5 5 x x x+ = + + =

b) ( ) ( ) ( )2

5 5 5 x x x− = − − =

c) ( ) ( ) ( )2

2 3 2 3 2 3 x x x+ = + + =

5) Write the general rule.

Perfect Square Trinomials:

( ) ( ) ( )2

a b a b a b+ = + + =

� − �� = � − �� − �� =

Section

4.3D



6) Factor. Look for GCFs.

a) 236 60 25t t+ + b) 28 24 18z z− +

c) 24 28 49r r+ + d) 28 100m −

e) 227 36 12t t− + f) 312 27w w−

g) 4 2 2 49 6x x y y+ + h) 2 26 24a b−

7) You can also factor these problems using an area model like the one from previous sections.

• Find the intercept form of each of the following quadratic representations (given in standard form). Remember to factor out the GCF if needed.

• State the x-intercepts.

a) ( ) 2 10 25f x x x= + + b) ( ) 24 12 9f x x x= + +

c) ( ) 2 10 25f x x x= − + d) ( ) 24 12 9f x x x= − +

e) ( ) 2 25f x x= − f) ( ) 24 49f x x= −



.

8) Sometimes using an area model doesn’t seem to fit very well. But it can still be used, even when there are multiple variables.

a) ( ) 2 2 22f x a x abx b= + + b) ( ) 2 2 6 9f x t u tu= − +

c) ( ) 2 2f x x y= − d) ( ) 2 236 25f x n m= −

9) Compare the following by graphing both f(x) and g(x) on the same graph by first writing the function into intercept (factored) form, then finding the vertex of each graph.

a) ( ) 2 2 3f x x x= − − b) ( ) 22 4 6g x x x= − −

( ) ( ) ( ) f x = ( ) ( ) ( )___ g x =

x-intercepts = ___________ x-intercepts = ___________

Vertex = ___________ Vertex = ___________

10) Compare the following by graphing both f(x) and g(x) on the same graph by first writing the function into

intercept (factored) form. Then, find the vertex of each graph.

a) ( ) 2 4f x x= − b) ( ) 23 12g x x= −

( ) ( ) ( ) f x = ( ) ( ) ( )___ g x =

x-intercepts = ___________ x-intercepts = ___________

Vertex = ___________ Vertex = ___________



11) You can also factor these problems using an area model like the one from previous sections. Find the intercept form of each of the following quadratic representations (given in standard form).

a) ( ) 2 6 9f x x x= + + b) ( ) 23 18 27f x x x= + +

c) ( ) 2 22f x x bx b= + +

d) ( ) 2 6 9f x x x= − + e) ( ) 2 14 49f x x x= − +

f) ( ) 2 22f x x bx b= − + g) ( ) 2 2 22f x a x abx b= − +

h) ( ) 2 20 9 9f x x x x= + − = − i) ( ) 24 9f x x= −

j) ( ) 2 16f x x= − k) ( ) 29 25f x x= −

l) ( ) 2 81f x x= − m) ( ) 2100 1f x x= −

n) ( ) 2 2f x x b= − o) ( ) 216 36f x x= −

4.3E Vertex Form of Quadratic Equations: a = 1


.

What value for c will make the expression a perfect square trinomial? Complete the

square for each expression. Write the result as a binomial squared.

�� + 6� � � �� _____��

#1 – 6: Use the area model to complete the square in order to create the square of a binomial.

1) ( ) ( ) ( )22 8 _____ ____ ____ ____x x x x x+ + = + + = +

2) ( ) ( ) ( )22 10 _____ ____ ____ ____x x x x x− + = − − = −

3) ( ) ( ) ( )22 20 _____ ___ ____ ___ ____ ___ ____x x x x x+ + = =

Section

4.3E

x

x

-10



#1 – 6 (continued): Use the area model to complete the square in order to create the square of a binomial.

4) ( ) ( ) ( )22 12 _____ ___ ____ ___ ____ ___ ____x x x x x− + = =

5) ( ) ( ) ( )22 7 _____ ___ ____ ___ ____ ___ ____x x x x x+ + = =

6) ( ) ( ) ( )22 11 _____ ___ ____ ___ ____ ___ ____x x x x x− + = =

#7 – 9: Complete the square and write the square of the binomial that is created.

7) ( ) ( ) ( )22 14 _____ ___ ____ ___ ____ ___ ____x x x x x− + = =

8) ( ) ( ) ( )22 24 _____ ___ ____ ___ ____ ___ ____x x x x x+ + = =

9) ( ) ( ) ( )22 9 _____ ___ ____ ___ ____ ___ ____x x x x x− + = =



.

x

x

6

Recall the problem from the beginning of this section. Start with the expression x2 + 6x and write it in vertex (completed square) form.

Vertex form x2 + 6x = (x2 + 6x +_______ ) – ________ = ��________�� _________

#10 – 11: Use the area model to complete the square in order to write an equation in vertex form.

10) 2 8y x x= +

Vertex form y = x2 + 8x

y = (x2 + ____x + _____ ) – ________

y = ��________��– _____

y = _________________

What is the vertex of the parabola? __________________

11) 2 10y x x= −

Vertex form y = x2 – 10x

y = (x2 – ____x + _____ ) – ________

y = ��________��– _____

y = _________________

What is the vertex of the parabola? __________________

?? What number do we

need to add to make

this a square?



Complete the square in order to write the standard form equation into vertex form.

� � �� 4� 1

� � )�� 4�_______* 1 − ________

� � ��________��_____________

#12 – 14: Complete the square to translate the standard form equation into a vertex form equation.

12) 2 8 3y x x= + −

13) 2 5 4y x x= − −

14) 2 10 20y x x= − +



.

15) Write the equation in vertex form. Working with the new function, create a table of values, graph the function, and label the vertex.

( ) 2 4 6f x x x= − −

x f(x)

16) Given the quadratic equation: ( ) 2 2 3f x x x= + − .

a) Use 2

b


b) Write in vertex form.

c) Use vertex form to find the vertex.

d) Do your answers from part a) and c) match?

e) Make a table and graph the function.

x f(x)

4.3ext Vertex Form of Quadratic Equations: a ≠≠≠≠ 1


In this section, we will again, simplify a quadratic expression by completing the square.

In these problems, a is not equal to 1. Complete the square for each expression. Write

the result as a binomial squared plus (or minus) a constant.

Completing the square when a +1.

Start with an equation in Standard Form:

2 0ax bx c+ + =

…and turn it into this:

2( ) 0a x d e+ + =

Example 1:

Write the expression in vertex form by completing the square. 24 24x x+

Step 1: Factor out the 4. ( )24 6x x+

Step 2: Complete the square.

Looking at the original expression: Looking at the simplified expression:

( )

2224

92 2 4

b

a

= =

( )

226

92 2 1

b

a

= =

Add 9 inside the parentheses. Add 9 inside the parentheses.

( )

( )

( )

2

2

2

4 6 ___

4 6 9 4(9)

4 6 9 36

x x

x x

x x

+ +

+ + −

+ + −

** REMEMBER to multiply the 9 by 4 (a = 4) ** This is the actual value that has been added to the expression. Subtract the resulting value to keep the expression

equivalent.

Step 3: Write the expression in vertex form.

( )

( ) ( )

( )

2

2

4 6 9 36

4 3 3 36

4 3 36

x x

x x

x

+ + −

+ + −

+ −

Example 2: Let’s look at where d and e above come from.

Start with a generic quadratic expression: 2

ax bx c+ +

Move c out of the way and factor a out of the first two terms.

( )2

2

ax bx c

ba x x c

a

+ +

+ +

Add

2

2

b

a

inside the parentheses.

**Remember to multiply it by a when you subtract it

from c on the outside of the parentheses.

2

2 2

2

____

2 2

ba x x c

a

b b ba x x c a

a a a

+ + +

+ + + −

"Complete the Square"

2 2

2 4

b ba x c

a a

+ + −

You will notice that we have: 2( )a x d e+ + where:

2

and2

4

bd

a

be c

a

=

= −

Section

4.3ext



.

#1 – 6: Complete the square of a quadratic expression. Confirm the equivalence of the original and

equivalent expressions by using a graphing utility. Sketch the graph of both the original equation

and the vertex form of the equation using y1 = ‘original equation’ and y2 = ‘equivalent equation’.

1) 25 20y x x= + 2) 23 30y x x= +

( )25 _____y x= + _____________________

( ) ( )25 ____ _____ 5 ___y x= + + − ________________________

( )( )5 ____ ____ _____y x x= + + − _____________________________

( )2

5 ____ ______y x= + − _____________________________

Equivalent equation: _____________________ Equivalent equation: _____________________

3) 23 4y x x= + 4) 24 10y x x= −

2 43

3y x x

= +

_____________________

( )2 43 ____ 3 ___

3y x x

= + + −

________________________

( )( )3 ____ ____ _____y x x= + + − _____________________________

( )2

3 ____ ______y x= + − _____________________________


5) 22 7y x x= − + 6) 23 5y x x= − −

2 72

2y x x

= − −

_____________________

( )( )2 72 ____ 2 ___

2y x x

= − − + − −

________________________

( )( )2 ____ ____ _____y x x= − − − − _____________________________

( )2

2 ____ ______y x= − − _____________________________




Example 3:

Start with a quadratic equation: 23 24 10y x x= + +

Slide the 10 out of the way. 23 24 10y x x= + +

Factor 3 (the lead coefficient) out of the first 2 terms. 23( 8 ) 10y x x= + +

Complete the square using 2 8x x+ .

To complete the square, add 16 inside the parentheses. Multiply 16 by the 3 to offset the

actual value before subtracting it outside the

parentheses.

2

2

2

3( 8 16) 10 3(16)

3( 8 16) 10 48

3( 8 16) 38

y x x

x x

x x

= + + + −

= + + + −

= + + −

Write the expression as a perfect square (vertex form). 23( 4) 38y x= + −

The vertex for the graph of this expression is… Vertex: ( )4, 38− −

#7 – 12: Rewrite each equation into vertex form (((( ))))2

y a x h k= − += − += − += − + by completing the square. Note that

you must subtract off anything you add when completing the square. Sketch the graph of the two

equations (the original equation and the equivalent equation written in vertex form).

7) 25 30 20y x x= + +

2 305 20

5y x x

= + +

( ) ( )25 6 _____ 20 5 _____y x x= + + + −

What does the 5 represent? _______________ Why do two 5’s appear? ____________________________

( )( )5 _____ _____ ___ ______y x x= + +

( )2

5 _____ ___ ______y x= +

Equivalent equation: __________________________What is the vertex? __________________________

8) 22 24 5y x x= + −

____________________________________y =

____________________________________y =

____________________________________y =

____________________________________y =




.

#7 – 12 (continued): Rewrite each equation into vertex form (((( ))))2

y a x h k= − += − += − += − + by completing the square.

Note that you must subtract off anything you add when completing the square. Sketch the graph

of the two equations (the original equation and the equivalent equation written in vertex form).

9) 25 8 20y x x= + +

Equivalent equation: __________________________What is the vertex? _________________________

10) 26 10 7y x x= − − +

Equivalent equation: __________________________What is the vertex? _________________________



#7 – 12 (continued): Rewrite each equation into vertex form (((( ))))2

y a x h k= − += − += − += − + by completing the square.

Note that you must subtract away any value you add when completing the square. Sketch the

graph of the two equations (the original equation and the equivalent equation written in vertex

form).

11) 22 5 4y x x= − −


12) 25 7 20y x x= − +


Unit 4 Review Material

Review Material 73

.

1) Use the factors of the equation ( ) ( )3 2 1y x x= + − to find the following

information:

a) x-intercept _________________

b) y-intercept _________________

c) graph the equation

#2 – 4: Use the graph of equation to find the requested information in the table.

2)

3)

( ) 23 12f x x x= − +

4)

2 9 8y x x= + +

Vertex

Axis of

Symmetry

Domain

Range

x-intercept

y-intercept

Unit 4

REVIEW



5) The height (in feet) of a soccer ball that is kicked can be modeled by the equation 28 24y x x= − + , where x

is the time (in seconds) after the ball is kicked. For each of the following, explain your thinking of how you found your answer.

a) After how many seconds did the ball reach its maximum height?

b) What is the maximum height of the soccer ball?

c) How long was the soccer ball in the air?

6) Identify the following information from the equation ( )2

4 7 3y x= − + .

a) Vertex _________________________

b) Axis of Symmetry _______________

c) Domain ________________________

d) Range __________________________

e) Sketch a graph

7) Convert the following equations from factored form to standard form.

a) ( ) ( )6 4y x x= − + ______________________ b) ( ) ( )2 1 5 3y x x= − + ____________________

c) ( )2

3y x= − ____________________________ d) ( ) ( )3 2 1y x x= − + − ____________________

x y


Review Material 75

.

8) Use the factors of the equation ( ) ( )2 1 2y x x= − + +

to find the following information:

a) x-intercept(s) ____________________

b) y-intercept _______________________

c) graph the equation

9) Convert the equations into standard form.

a) ( )2

6 7y x= + b) ( )2

0.4 8 5y x= − − + c) ( )2

2 3 2y x= + −

__________________ ______________________ ______________________

10) Factor the following polynomials completely.

a) 2 42x x+ − b)

2 36x −

c) 2 12 20x x+ + d) 25 14 8x x+ +

e) 212 36 21x x− − f) 218 32x x−

g) 23 9x x−



11) Complete the square to rewrite the equation of each function into vertex form.

a) ( ) 2 8 5f x x x= + + b) ( ) 23 4 8f x x x= + −

12) Which choice would represent the graph below (circle the correct choice)?

[A] 2 4 3y x x= − − −

[B] 2 4 3y x x= − −

[C] 2 4 3y x x= − + +

[D] 2 4 3y x x= − +

a) How do you know?

b) Could you narrow your answers down without doing any extra work? Explain the strategy you used.

13) You are on a team that is building a roller coaster. According to the design, the path of the first hill can be

modeled by 20.065 5.3 1.85y x x= − + + where x is time (in seconds) and y is height (in feet). Answer the

following questions. For each of the following, explain your thinking of how you found your answer.

a) What is the maximum height of the hill?

b) How long did it take the rollercoaster to reach the top of the hill?

c) How long was the rollercoaster in the air before it reached a height of 0 feet?

d) What is the initial height of the rollercoaster?

e) Explain the domain and range in the context of this situation.

-1 1 2 3 4 5

-2

-1

1

2

3

4

5

6

x

y


Review Material 77

.

14) Two rock climbers try an experiment while scaling a steep rock face. They each carry rocks of similar size and shape up a rock face. One climbs to a point 400 ft. above the ground and the other to a place below her at 300 ft. above the ground. The higher climber drops her rock and 1 second later the lower climber drops his. Note that the climbers are not vertically positioned. No climber is

injured in this experiment. Indicate units of measure on all answers.

a) Use the model 2( ) 16( )h t t h k= − − + to write the two

functions that can be used to model the relationship between the heights, ,� and ,�, of the rocks, in feet, after t seconds.

b) When are the rocks at the same height? Explain how to use the graph to find this information.

c) Assuming the rocks fall to the ground without hitting anything on the way, which of the two rocks will reach the ground first? When does it hit the ground? Explain how you found your answer.

d) What is the vertical distance between the rocks when the first rock lands?

e) How much time passes between the two rocks landing? Be specific and use algebra to justify your answer.

NO

GR

AP

HIN

G C

AL

CU

LA

TO

R N

O G

RA

PH

ING

CA

LC

UL

AT

OR

WORK SPACE

h1

h2

Name ______________________________ Period __________


4.1 I can graph quadratic functions and demonstrate understanding of significant features of different 79


1) The shape of the graph of a quadratic equation is called a _____________________.

a) What is another name for

x-intercepts? _____________

b) What is the vertical line that goes

through the vertex called?

___________________________

c) How do you find the y-intercept?

___________________________

___________________________

2) Fill in each box with the appropriate

label.

3) When the vertex is the highest point on

the graph, the point is called the

________________________.

4) When the vertex is the lowest point on

the graph, the point is called the

________________________.

#5 – 10: Create a table of values for the quadratic function, and then graph each quadratic function. On

the graph mark the locations of the following significant features of the graph: axis of symmetry,

vertex and the y-intercept. Find the information related to the features of each graph.

5) ( ) 2 2 3f x x x= − − 6) ( ) 22 4 5f x x x= − + +

axis of symmetry: ____________________________ axis of symmetry: __________________________

y-intercept:_______vertex: _____________________ y-intercept:_______vertex: ___________________

domain:__________range: _____________________ domain:__________range: ___________________

opens (up or down): __________________________ opens (up or down): _________________________

Vertical stretch, compression, or same width _______ Vertical stretch, compression or same width ______

Name ______________________________ Period __________


80 UNIT 4 – SITUATIONS THAT CAN BE MODELED WITH QUADRATIC FUNCTIONS (PRACTICE)

#5 – 10 (continued): Create a table of values for the quadratic function, and then graph each quadratic

function. On the graph mark the locations of the following significant features of the graph: axis of

symmetry, vertex and the y-intercept. Find the information related to the features of each graph.

7) ( ) 211 4

2f x x x= − − 8) ( ) 24 8 1f x x x= + +

axis of symmetry:____________________________ axis of symmetry: __________________________

y-intercept:_______vertex: ____________________ y-intercept:_______vertex: ___________________

domain:__________range: ____________________ domain:__________range: ___________________

opens (up or down): __________________________ opens (up or down): ________________________

Vertical stretch, compression, or same width ______ Vertical stretch, compression or same width _____

9) ( ) 2 2f x x x= − + 10) ( ) 22 4 4f x x x= − − +

axis of symmetry:____________________________ axis of symmetry: __________________________

y-intercept:_______vertex: ____________________ y-intercept:_______vertex: ___________________

domain:__________range: ____________________ domain:__________range: ___________________

opens (up or down): __________________________ opens (up or down): ________________________

Vertical stretch, compression, or same width ______ Vertical stretch, compression or same width _____

Name ______________________________ Period __________




#11 – 12: Without graphing, looking only at the function model, identify the key features of the graph of

the given function.

11) a) ( ) 24 8f x x x= − + − b) ( ) 22 3 1f x x x= + −

y-intercept: ____________________________ y-intercept: ____________________________

axis of symmetry: _______________________ axis of symmetry: _______________________

opens (up or down): _____________________ opens (up or down): _____________________

12) a) ( ) 22 12 7f x x x= + + b) ( ) 22 6 1f x x x= − + −

y-intercept: ____________________________ y-intercept: ____________________________

axis of symmetry: _______________________ axis of symmetry: _______________________

opens (up or down): _____________________ opens (up or down): _____________________

13) A juggler tosses a ring into the air. The height, in feet, of the ring above the

ground can be modeled by the function ( ) 216 16 6f x x x= − + + , where x is the

time, in seconds, after the ring is tossed.

a) Use a graphing utility to graph and determine a table of values for the

function. Graph the function and label the axes.

b) Record the greatest height the ring reaches.

c) Record the time it takes to reach that height.

At which significant feature does this occur?

d) Record the length of time the ring is in the air.

At which significant feature does this occur?

e) Record the y-intercept and explain the

meaning of this point in the context of the

problem.

1 2 3

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

0 x

f(x)

Name ______________________________ Period __________



14) The height, in feet, of a fireworks shell can be modeled by ( ) 216 224h x x x= − + ,

where x is the time, in seconds, after it is fired.

a) Use a graphing utility to graph and determine a table of values for the function.

Graph the function and label the axes.

b) How high does the shell get before it begins

to fall?

c) Record the time it takes to reach the highest

point. At which significant feature does this

occur?

d) Record the length of time the shell is in the

air. Which significant feature of the graph

helps you to determine this time?

15) A frogsay is launched into the air with a cannon. Its

trajectory can be simulated with the equation:

216 64 80y x x= − + + where y is the height in feet and

x is the time, measured in seconds, after it is launched.

a) Graph the function, label the axes.

b) Record the highest point the frogsay reaches. At

which significant feature does this occur?

c) Record the time it takes to

reach the maximum height.

d) Record the length of time the

frogsay is in the air.

Name ______________________________ Period __________




16) Standard form of a quadratic equation is 2

= + +y ax bx c .

a) What two pieces of information about the graph does the value of a provide?

b) Explain how the value of b helps you to graph the function.

c) What information about the graph does the value of c provide?

d) How can you find the x-coordinate of the vertex of the graph?

e) How can you find the y-coordinate of the vertex of the graph?

17) Identify the key features of the graph of each function.

a)

b) 2 4 3y x x= + − c)

y-intercept: ________________ y-intercept: ________________ y-intercept: ________________

axis of symmetry: ___________ axis of symmetry: __________ axis of symmetry: ___________

vertex: ____________________ vertex: ___________________ vertex: ____________________

opens (up or down): ___________ opens (up or down): ___________ opens (up or down): __________

vertical (compress, stretch, or same)

____________________________


____________________________


___________________________

domain: ___________________ domain: __________________ domain: ___________________

range: ____________________ range: ____________________ range: _____________________

Section 4.1A

Name ______________________________ Period __________




Name ______________________________ Period __________




1) Fill in each box with the appropriate

label.

2) The quadratic equation 2 2 15y x x= + − and the quadratic equation ( ) ( )3 5y x x= − + are equivalent,

which means their graphs are the same.

a) Describe how you would graph 2 2 15y x x= + − .

b) Describe how you would graph ( ) ( )3 5y x x= − + .

c) How is graphing a quadratic equation in standard form and graphing a quadratic equation in intercept

(factored) form similar and how is it different?

Name ______________________________ Period __________



3) A small designer purse company sells purses for $50 each and currently sells 450 purses per month on

average. The company did some research and realized that for each $2 decrease in price, they can sell 50

more purses per month. How much should the company charge for the purse so they can maximize

monthly income?

Income is given by the equation: Income = (price) •••• (number sold)

Let x = the number of $2 price decreases (note a negative value of x would represent a price increase).

So, then the new price would be given by price = 50 – 2x and the corresponding number that would sell

would be given by number sold = 450 + 50x.

Hence, the equation that models the projected income would be given by: Income = (50 – 2x)(450 + 50x)

a) Find the x-intercepts. Explain what

they mean in the context of the

problem?

b) How many $2 price decreases should

the company make in order to

maximize the income? What price

will the purses then sell for? How

many purses will they sell at this

price? What will the income be at

this price?

c) Make a graph of the income using the

information that you have collected so

far. Label the axes.

d) What is the domain and range?

(Make sure that you take into account the context of the situation)

-10 -5 5 10 15 20 25 300

5000

10000

15000

20000

25000

30000

0x

y

Name ______________________________ Period __________




4) The height of a ball (measured in meters) thrown straight up in the air can be modeled by the equation

( ) ( )4.9 3h t t t= − − where t is the time in seconds after it is thrown.

a) Graph the function, label the axes.

b) How long will the ball be in the air?

c) Find the time it takes to reach the maximum

height.

d) Find the maximum height of the ball.

e) What are the domain and range of the function?

f) How does the situation restrict the domain and range?

#5 – 12: Graph the quadratic function. Label the x-intercepts, vertex, axis of symmetry and state the

domain and range. State the maximum/minimum value.

5) ( ) ( )2 6y x x= − − 6) ( ) ( )4 1 1y x x= + −

x-intercept(s): ___________________________ x-intercept(s): ___________________________

Axis of Symmetry_______Max/Min: ________ Axis of Symmetry________Max/Min: _______

domain: _____________ range: _____________ domain: ___________ range: _______________

Name ______________________________ Period __________



#5 – 12 (continued): Graph the quadratic function. Label the x-intercepts, vertex, axis of symmetry and

state the domain and range. State the maximum/minimum value.

7) ( ) ( )0.5 3 7y x x= − + − 8) ( )3 5y x x= −

x-intercept(s): _________________________ x-intercept(s): ____________________________

Max/Min: ____________________________ Max/Min: _______________________________

Axis of Symmetry: ____________________ Axis of Symmetry: ________________________

domain: ____________ range: ____________ domain: ____________ range: ________________

9) ( ) ( )5 1y x x= + + 10) ( ) ( )2 3 1y x x= − − +

x-intercept(s): _________________________ x-intercept(s): __________________________

Max/Min: ____________________________ Max/Min: _____________________________

Axis of Symmetry: _____________________ Axis of Symmetry: _____________________

domain: ____________ range: ____________ domain: ____________ range: ______________

-4 -2 2 4 6 8 10

-5

5

10

15

x

y

-2 2 4 6

-20

-15

-10

-5

5

x

y

Name ______________________________ Period __________




#5 – 12 (continued): Graph the quadratic function. Label the x-intercepts, vertex, axis of symmetry and

state the domain and range. State the maximum/minimum value.

11) ( ) ( )1 5y x x= − − + 12) ( ) ( )1

3 12

y x x= − +

x-intercept(s): _________________________ x-intercept(s): __________________________

Max/Min: ____________________________ Max/Min: _____________________________

Axis of Symmetry: _____________________ Axis of Symmetry: ______________________

domain: ____________ range: ____________ domain: ____________ range: _____________

#13 – 16: Find the x-intercepts, axis of symmetry, vertex, and the direction the parabola opens.

13) ( ) ( )6 2y x x= + + 14) ( ) ( )3 5y x x= − + +

x-intercept(s): _______________ x-intercept(s): _________________

axis of symmetry: ____________ axis of symmetry: ______________

vertex ______________________ vertex: _______________________

opens (up or down): __________ opens (up or down): ____________

15) ( ) ( )2 1 7y x x= − + 16) ( ) ( )1

3 52

y x x= − − +

x-intercept(s): _______________ x-intercept(s): _________________

axis of symmetry: ____________ axis of symmetry: ______________

vertex: _____________________ vertex: _______________________

opens (up or down): __________ opens (up or down): ____________

Name ______________________________ Period __________



17) Intercept (factored) form of a quadratic equation is (((( )))) (((( ))))y a x m x n= − −= − −= − −= − − .

a) What two items about the graph does the value a give?

b) How can you find the x-intercepts of the graph when the equation is written in intercept (factored) form?

c) How can you find the x-coordinate of the vertex of the graph?

d) How can you find the y-coordinate of the vertex of the graph?

e) Explain how you would decide what values to put in the x column when making a table to graph a

quadratic function in intercept (factored) form.

#18 – 19: Write the equation of the quadratic function. Label the vertex, axis of symmetry and x-

intercepts. State the domain and range. State the maximum/minimum.

18) y = __________________________________ 19) y = __________________________________


maximum or minimum value: ____________ maximum or minimum value: ______________

-6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

8

x

y

-8 -7 -6 -5 -4 -3 -2 -1 1 2 3

-14

-13

-12

-11

-10

-9

-8

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

x

y

Section 4.1B

Name ______________________________ Period __________




1) Fill in each box with the appropriate label.

2) Vertex form of a quadratic equation is

( )= − +2

y a x h k .

a) What two items about the graph does

the value of a give?

b) How can you find the coordinates of

the vertex?

c) Explain how you decide what values

to put in the x column when making a

table of values to help graph a

quadratic function in vertex form.

3) Identify a, h, and k in the following functions:

a) ( ) ( )2

3 1 2f x x= − − b) ( ) ( )2

0.5 3 1g x x= − + +

#4 – 9: Graph the quadratic function, identify the vertex and identify the domain and range.

4) ( )2

3 2 1y x= − + 5) ( )2

1 2y x= + +

vertex: _______________________________ vertex: _________________________________

opens (up or down): ____________________ opens (up or down): ______________________

maximum or minimum value: _____________ maximum or minimum value: ______________


Name ______________________________ Period __________



#4 – 9 (continued): Graph the quadratic function, identify the vertex and identify the domain and range.

6) ( )2

2 3 1y x= − + − 7) ( )2

2 1y x= − + +

vertex: _________________________ _____ vertex: ________________________________




8) ( )2

2 5y x= − − 9) ( )21

5 32

y x= − −

vertex: _________________________ _____ vertex: ________________________________




Name ______________________________ Period __________




10) Give the indicated form of a quadratic equation and explain how to find the vertex using that form. H

ow

do

yo

u f

ind

th

e v

erte

x

usi

ng

th

is f

orm

?

Lis

t th

e si

gn

ific

an

t fe

atu

res

yo

u

can

id

enti

fy d

irec

tly

from

th

e

form

th

e eq

uati

on

is

in.

Eq

ua

tio

n

Fo

rm

Ver

tex

Inte

rcep

t (F

act

ore

d)

Sta

nd

ard

Name ______________________________ Period __________



11) Which form of a quadratic equation do you like the most? Explain why you prefer it.

12) The path that a diver follows is given by ( )2

0.4 4 14y x= − − + where x is the horizontal distance (in feet)

from the edge of the diving board and y is the height (in feet) from the water surface.

a) Graph the function, label the

axes.

b) What is the vertex?

c) Explain what the x and y-values of the vertex mean

for this situation.

d) What does the x-intercept mean for this situation?

13) While playing basketball this weekend Frank shoots an air-ball. The height h in feet of the ball is given by

( ) ( )2

16 1 24h t t= − − + where t is time in seconds. Graph the function and label the axes.

a) How long will it take the ball to hit

the ground?

b) What is the maximum height of the

ball?

c) What are the domain and range of

the function?

d) How does the situation restrict the domain and

range?

Name ______________________________ Period __________




#14 – 19: Determine whether the graph opens up or down, identify the vertex, and identify the axis of

symmetry.

14) ( )2

2 4y x= − + 15) ( )2

5 1y x= − + −

Opens: _____________________________ Opens: _____________________________

Axis of Symmetry: ____________________ Axis of Symmetry: ____________________

Vertex: _____________________________ Vertex: _____________________________

16) ( )2

5 10y x= + 17) ( )2

2 1 12y x= + −

Opens: _____________________________ Opens: _____________________________


Vertex: _____________________________ Vertex: _____________________________

18) ( )2

2 3 1y x= − − + 19) ( )2

3 14 74y x= − − −

Opens: _____________________________ Opens: _____________________________


Vertex: _____________________________ Vertex: _____________________________

Section 4.1C

Name ______________________________ Period __________




Name ______________________________ Period __________



.

#1 – 2: Find the product of the expression by finding the area of the figures below.

1) ( )3 4x + 2) ( )2 2 3x +

#3 – 4: Draw a figure that represents the following product as an area of a rectangle. Find the area.

3) ( )4 2x + 4) ( )3 2 1x +

#5 – 16: Multiply.

5) ( )3 4x − 6) ( )5 2a− +

7) ( )2 3b b + 8) ( )2k k− −

9) ( )23 4 3x x+ − 10) ( )22 5 1x x− − +

11) ( )25 2 3 1x x+ + 12) ( )22 3 4n n n− −

13) ( )24 8 7 3x x− + + 14) ( )25 2 8 4x x− − +

15) ( )28 4 4 1x x− − + − 16) ( )23 2 7 2x x x− − +

x 4

3

2x 3

2

Name ______________________________ Period __________



#17 – 18: Find the product of the expression by finding the area of the figures below.

17) ( ) ( )4 15x x+ + 18) ( ) ( )3 2 8x x+ +

18)

#19 – 20: Draw a figure that represents the following product as an area of a rectangle. Find the area.

19) ( ) ( )5 2x x+ + 20) ( ) ( )4 2 1x x+ +

#21 – 30: Multiply the following binomials.

21) ( ) ( )3 5x x+ + 22) ( ) ( )3 5x x− −

23) ( ) ( )3 3x x+ − 24) ( ) ( )2 8x x+ −

25) ( )2

5x + 26) ( )2

4x −

27) ( ) ( )2 4 1x x+ − 28) ( ) ( )3 1 7x x+ −

29) ( ) ( )1 6x x− − − 30) ( ) ( )2 5 4x x+ −

Name ______________________________ Period __________



.

#31 – 33: Find the area of each shape.

31) 32)

33) The dimensions of a rectangular window are 3x + 10 and 2x + 6. What is the area of the window?

34) Revenue for a video game store can be calculated by multiplying the price of the video game by the number

of video games sold. The price of a video game the store sells can be modeled by the equation

( ) 3 2C x x= + . The number of video games sold can be modeled by the equation ( ) 8 7N x x= − . Write a

model for the revenue earned from the sale of the video games.

Section 4.2A

x – 2

2x + 7

4x – 2

x + 2

Name ______________________________ Period __________




Name ______________________________ Period __________



.

#1 – 8: Write each equation in standard form.

1) ( )23 2 6y x x= + − 2) ( )22 4 5y x x= − −

3) ( )2 5 8y x x= − + − 4) ( )25 5 1y x x= − − +

5) y = (x – 6)(x + 2) 6) y = (x – 3)(x – 1)

7) y = 3(x + 4)(x + 2) 8) ( )( )2 5 4y x x= − + −

#9 – 12: Find the x-intercepts, convert to standard form, find the y-intercept and graph the equation.

9) ( ) ( ) ( )2 1f x x x= + + 10) ( ) ( ) ( )2 3 4f x x x= − +





x f(x)

x f(x)

Name ______________________________ Period __________



#9 – 12 (continued): Find the x-intercepts, convert to standard form, find the y-intercept and graph.

11) ( ) ( ) ( )2 4 1f x x x= − − − 12) ( ) ( ) ( )3 7f x x x= − − +

a) x-intercept(s): ______________________ a) x-intercept(s): ______________________


c) y-intercept: _________________________ c) y-intercept: ________________________


x f(x)

x f(x)

13) The population of a city is modeled by ( ) ( ) ( )0.5 50 10P t t t= + + where P(t) is

the population in thousands and t is the time in years from 2000.

a) What was the population in the year 2000?

b) What will the population be in the year 2020?

c) Make a table of values and graph. Label your axes.

t P(t)

Name ______________________________ Period __________



.

14) A ball, thrown from the top of a building, follows the path ( ) ( ) ( )4.9 2 7h t t t= − + − , where h(t) is the

height of the ball in meters and t is time in seconds.

a) When will the ball hit the ground?

b) How tall is the building?

c) Make a table and graph. Label your axes.

t h(t)

Name ______________________________ Period __________



#15 – 16: Write each equation in standard form. Use a graphing calculator to check your answer. Graph

each equation. Enter the original equation as Y1 and the standard form equation as Y2. If you

expanded the equation correctly, the second parabola should be graphed on top of the first

equation. If you see two parabolas, go back and check your work.

15) ( ) ( )2 3 4y x x= − + 16) ( ) ( )3 1 2y x x= − + +

a) Standard Form: a) Standard Form:

b) Sketch the graph. b) Sketch the graph.

Section 4.2B

Name ______________________________ Period __________



.


1) ( )2

3 1y x= − + 2) ( )2

2 3 6y x= − − −

3) ( )2

4 8y x= − + − 4) ( )2

8 3y x= − +

5) ( )2

2 7 10y x= − + − 6) ( )2

2.4 5.1 3y x= − +

#7 – 8: Determine whether the equations in each pair are equivalent.

7) ( )2

2 3 7y x= − − 8) ( )2

2 4 8y x= + +

22 12 11y x x= − + 22 16 32y x x= − +

Name ______________________________ Period __________



#9 – 10: Find the vertex, convert to standard form, find the y-intercept and graph the equation.

9) ( )2

1 16y x= − − 10) ( )2

2 3 7y x= + +

a) vertex: ____________________________ a) vertex: ______________________________

b) standard form: _____________________ b) standard form: ________________________

c) y-intercept: ________________________ c) y-intercept: ___________________________


x f(x)

x f(x)

#11 – 14: Match each equation with the correct statement.

11) 2 5 3y x x= + + [A] The vertex is at ( )2, 3 .

12) ( )2

2 3y x= − + [B] The y-intercept is ( )0, 5 .

13) ( )2

2 5 3y x= − + [C] The y-intercept is ( )0, 3 .

14) 2 3 5y x x= + + [D] The vertex is at ( )5, 3 .

Name ______________________________ Period __________



.

15) The Galleria, in BCE Place in Toronto, has many beautiful parabolic arches. One of the arches can be

modeled by the function ( ) ( )2

0.5 6.8 26f x x= − − + . The x-axis represents the floor in the Galleria and

the y-axis represents the height above the floor. Distances are in meters.

a) Write the function in standard form.

b) What is the height of the arch at its center?

c) The y-intercept represents the lowest point at one side of the base of an arch. What is this height?

#16 – 17: Expand each equation to standard form. Use a graphing utility to check your answer. Graph

each equation. Enter the original equation as Y1 and the standard form equation as Y2. If you

expanded the equation correctly, the second parabola should be graphed on top of the first

equation. If you see two parabolas…go back and check your work.

16) ( )2

6 2y x= − − + 17) ( )2

2 1 4y x= + −

a) Standard Form: a) Standard Form:

b) Sketch the graph. b) Sketch the graph.

Name ______________________________ Period __________



18) The height, h, of a baseball thrown off a bridge can be modeled by the equation ( ) ( )2

4.9 4 130h t t= − − +

where the height is measured in meters and t is the time in seconds since the ball was thrown.

a) At what height was the ball released?

b) How high was the ball thrown?

c) How long did the ball take to reach its highest point?

d) How high is the bridge?

e) Make a table and graph. Label your axes.

t h(t)

19) Which of the following represent the same parabola as ( )2

2 3 2y x= − − ?

[A] ( ) ( )2 4y x x= − − −

[B] ( ) ( )3 4 2y x x= − +

[C] ( ) ( )2 4 2y x x= − −

[D] ( ) ( )2 3 1y x x= − +

Section 4.2C

Name ______________________________ Period __________



.

#1 – 10: Factor the greatest common factor from each of the following binomials and write the expression

in factored form.

1) 3 6x + 2) 5 20x −

3) 16 8b + 4) 24 6n −

5) 24 28w − 6) 30 25p q+

7) 23 9a + 8) 2 2x x−

9) 214 7y y+ 10) 25 7t t+

#11 – 20: Factor the greatest common factor for each of the following trinomials and write the expression

in factored form.

11) 227 9 3p p− + 12) 245 5 40b b− +

13) 26 9 36x x− + 14) 212 16 4w w+ −

15) 225 10 30w w− + + 16) 235 14 70m m− + −

17) 3 216 16 28v v v− − 18) 2 34 5 3x x x− +

19) 2 3 650 45 20m m m− + − 20) 7 4 250 10 50n n n− + +

Name ______________________________ Period __________



#21 – 22: Determine whether the equations in each pair are equivalent.

21) 26 12 24y x x= − − 22) 212 16 8y x x= − + −

( )26 2 8y x x= − − ( )24 3 4 2y x x= − − +

#23 – 26: Follow the directions in the box for each problem.

Factor completely. Graph both standard and factored forms

on your graphing utility (they should

match). Sketch the graph.

Use the graphing utility to

find the y-intercept and the

vertex.

23) 22 16 8y x x= − −

y-intercept:

vertex :

24) 23 21 12y x x= − +

y-intercept:

vertex :

25) 22 12 4y x x= − + −

y-intercept:

vertex :

26) 26 24 6y x x= − + +

y-intercept:

vertex :

Name ______________________________ Period __________



.

#27 – 30: Find the greatest common factor of the following.

27) 224x yz and 236xy 28) 230 pqr , 2 218 p qr , 26 pr

29) 15pq3, 10p2qr, 25q2r3 30) 3 2 318x y z , 4 2 327x y z , 281xy z

#31 – 40: Factor the greatest common factor for each of the following polynomials.

31) 5 490 60 70b b+ + 32) 321 21 12b b− + −

33) 628 70 63r r+ + 34) 5 241 1 0 1 00n n n− + +

35) 4 3 25 7 8x x x+ + 36) 7 6 542 48 1 8 n n n+ +

37) 2 335 42 49 21a a a+ + − 38) 3 4 5 656 32 80 4x x x x− − + +

39) 6 5 3 239 15 6 9d d d d− + − 40) 7 5 4 228 35 70 1 4b b b b+ − +

Section 4.3A

Name ______________________________ Period __________




Name ______________________________ Period __________

4.3B Factored Form of Quadratic Equations: Polynomial with a = 1


.

#1 – 4: Fill in the blanks with the correct sign.

1) x2 + 6x + 8 = (x ___ 4)(x ___ 2) 2) x2 – 6x + 8 = (x ___ 4)(x ___ 2)

3) x2 + 2x – 8 = (x ___ 4)(x ___ 2) 4) x2 – 2x – 8 = (x ___ 4)(x ___ 2)

#5 – 10: Factor completely.

5) 2 5 24b b− − 6) 2 1 7 70m m+ +

7) 2 1 2 35n n− + 8) 2 3 54a a− −

9) 2 6n n+ − 10) 2 8 12v v− +

#11 – 14: Write in factored form. Identify the x-intercepts.

11) 2 16 60y x x= + + 12) 2 2y x x= + −

x-intercepts: ________________ x-intercepts: ________________

13) 2 13 40y x x= + + 14) 2 14 45y x x= − +


Name ______________________________ Period __________




15) 2 10 16k k+ +

16) 2 5 4n n+ + 17) 2 6 5x x− +

18) 2 2 48a a+ −

19) 2 5 6v v− + 20) 2 5 14x x− −

21) 2 8 20x x+ −

22) 2 15 56a a+ + 23) 2 4 3x x+ +

24) 2 6 16x x+ −

25) 2 9 18x x+ − 26) 2 15 26x x+ +

Name ______________________________ Period __________



.

27) Given the quadratic function: ( ) 2 10 9f x x x= + +

a) Use 2

b




d) Find the x-coordinate halfway between the x-intercepts.

e) Do your answers from part a and d match?


x f(x)

# 28 – 29: Factor completely (including the GCF).

28) 23 42 147b b− + 29) 25 20 225x x+ −

Name ______________________________ Period __________



#30 – 31: Write each function in factored form. Working with the new function, create a table of values

and graph the function, identify and label the x-intercepts.

Write in Factored Form Make a table Graph and label the x-intercepts

30) ( ) 2 11 24f x x x= − +

x f(x)

31) ( ) 2 6 16f x x x= − −

x f(x)

#32 – 33: Write in factored form. Identify the x-intercepts.

32) 24 24 160y x x= − − 33) 23 9 30y x x= + −


Name ______________________________ Period __________



.

34) Given the quadratic function: ( ) 22 28 80f x x x= − + .

a) Use 2

b




d) Find the x-coordinate halfway between the x-intercepts.

e) Do your answers from part a and d match?


x f(x)

Name ______________________________ Period __________



35) A cannonball is launched upwards from the surface of the moon. Its height is described by the function

( ) 25 40 45h t t t= − + + , where h is measured in meters and t is measured in seconds.

a) How high is the cannonball at 0, 1 and 2 seconds?

b) From what height was the cannonball launched?

c) Write in factored form and find when the cannonball hits the ground. Explain your reasoning.

Why can’t you use the both x-intercepts when answering this question?

d) Find the maximum height of the cannonball and when it occurs.

e) Graph the height of the cannonball.

t h(t)

Section 4.3B

Name ______________________________ Period __________

4.3C Factored Form of Quadratic Equations: Polynomial with a ≠ 1


.

#1 – 4: Fill in the blanks with the correct sign.

1) ( ) ( )26 13 5 2 ___ 1 3 ___ 5x x x x+ + = 2) ( ) ( )26 13 5 2 ___ 1 3 ___ 5x x x x− + =

3) ( ) ( )26 7 5 2 ___ 1 3 ___ 5x x x x+ − = 4) ( ) ( )26 7 5 2 ___ 1 3 ___ 5x x x x− − =


5) 23 28 32n n− + 6) 25 32 35k n− +

7) 27 51 54x x+ + 8) 22 5 12b b− −

9) 25 31 28x x− − 10) 2

3 17 56p p+ −

11) 26 25 25a a− + 12) 28 14 5x x+ +

Name ______________________________ Period __________



#13 – 14: Determine whether the equations in each pair are equivalent and explain.

13) 22 28y x x= − − 14)

26 11 10y x x= + −

( ) ( )2 7 4y x x= − + ( ) ( )3 2 2 5y x x= − +

#15 – 18: Write each equation in factored form. Using a graphing utility, record the graph for both

equations. Finish by identifying the vertex and the x-intercepts.

Write in Factored Form

Graph both standard and factored forms

on a graphing utility (they should match).

Sketch the graph.


the vertex and the

x-intercepts.

15) 22 5 3y x x= − −

vertex :

x-intercept(s):

16) 23 4 4y x x= − −

vertex :

x-intercept(s):

17) 28 5 3y x x= + −

vertex :

x-intercept(s):

18) 26 25 24y x x= + +

vertex :

x-intercept(s):

Name ______________________________ Period __________



.

#19 – 22: Factor completely (including the GCF).

19) 210 105 50x x− + 20) 214 52 30v v+ +

21) 210 4 14r r+ − 22) 212 28 160n n+ −

23) Which of the following represent the same parabola as 212 4 5 ?y x x= − −

[A] ( ) ( )3 1 4 5y x x= − +

[B] ( ) ( )3 5 4 1y x x= − +

[C] ( ) ( )6 1 2 5y x x= − +

[D] ( ) ( )6 5 2 1y x x= − +

24) The height of a ball above the ground after it is thrown upwards from the top of a building at 40 feet per

second can be modeled by the function, ( ) 216 40 24h t t t= − + + where the height, h, is given in feet and the

time t is in seconds.

a) How high is the building from where the ball

was thrown?

b) Write in factored form and find when the ball

hits the ground.

c) Find the maximum height of the ball and

when it occurs.

d) Graph the height with respect to time of the

ball.

Name ______________________________ Period __________




25) 23 5 2x x+ + 26) 22 7 3x x+ + 27) 27 9 2x x− +

28) 23 16 5x x− + 29) 26 11 3x x+ + 30) 28 6 1b b+ +

31) 28 9 1b b− + 32) 26 13 2b b+ + 33) 212 13 3b b− +

34) 23 14 8b b+ + 35) 25 4 1k k+ − 36) 22 5 3k k− −

37) 23 2 5k k+ − 38) 27 13 2k k− − 39) 210 3 1k k− −

40) 26 7 3v v− − 41) 25 18 8v v+ − 42) 24 15 9v v− +

43) 23 16 12v v+ − 44) 28 14 15v v− + + 45) 29 24 16x x+ +

Section 4.3C

Name ______________________________ Period __________



.

#1 – 4: Fill in the blanks.

1) ( ) ( )2 4 __ __ __ - __ x − = + 2) ( ) ( )24 9 2 ___ 3 2 ___ 3x x x− =

3) ( ) ( )2 22 121 ___ 11 ___ 11x x x x− + = 4) ( ) ( )24 12 9 2 ___ 3 2 ___ 3x x x x+ + =


5) 2 20 100x x− + 6) 2 16x −

7) 29 25x − 8) 2 14 49x x+ +

9) 264 169x − 10) 29 24 16x x+ +

11) 2 25x − 12) 216 81x −

#13 – 14: Determine whether the expressions in each pair are equivalent and explain.

13) 249 28 4x x− + 14) 29 16x −

( ) ( )7 2 7 2x x− − ( ) ( )3 8 3 8x x+ −

Name ______________________________ Period __________



#15 – 18: Write each equation in factored form. Enter the equations into a graphing utility, sketch the

graphs. Finish by identifying the vertex and the x-intercept(s).

Write in Factored Form

Graph both standard and factored

forms on your graphing utility

(they should match). Sketch the graph.


the vertex and the

x-intercepts.

15) 2 144y x= −

vertex :

x-intercept(s):

16) 2 8 16y x x= + +

vertex :

x-intercept(s):

17) 2 12 36y x x= + +

vertex :

x-intercept(s):

18) 225 4y x= −

vertex :

x-intercept(s):

Name ______________________________ Period __________



.

#19 – 26: Write in factored form.

19) 298 288x − 20) 22 32 128x x+ +

21) 2 216 9x y− 22) 2 2 100x y −

23) 2 24 4a ab b+ + 24) 2 24 20 25x xy y+ +

25) 4 2 2 46 9x x y y+ + 26) 4 2 2 44 20 25k k w w+ +

Section 4.3D

Name ______________________________ Period __________




Name ______________________________ Period __________



.

#1 – 2: Use the diagrams to help complete the square for each of the following.

1) ( )22 8 ________ ___ x x x+ + = + 2) ( )

22 10 ________ ___ x x x+ + = +

#3 – 4: Draw a diagram to help complete the square for each of the following.

3) ( )22 4 ________ ___ x x x+ + = + 4) ( )

22 20 ________ ___ x x x+ + = +

#5 – 10: Fill in the blanks.

5) ( ) ( ) ( )22 2 __ __ __ __x x x x x+ + = + + = + 6) ( ) ( ) ( )

22 6 __ __ __ __x x x x x− + = + + = +

7) ( ) ( ) ( )22 18 __ __ __ __x x x x x+ + = + + = + 8) ( ) ( ) ( )

22 16 __ __ __ __x x x x x+ + = + + = +

9) ( ) ( ) ( )22 12 __ __ __ __x x x x x− + = − − = − 10) ( ) ( ) ( )

22 24 __ __ __ __x x x x x+ + = + + = +

x

x x2 4x

4x 4

4

? 5x

x

x x2 5x

5

5

?

Name ______________________________ Period __________



#11 – 14: Find the value of c that will make the expression a perfect square trinomial.

11) 2 14x x c+ + 12) 2

12x x c− +

13) 29x x c− + 14)

25x x c+ +

15) Given the quadratic function: ( ) 2 8 5f x x x= + −

a) Use 2

b


b) Write in vertex form by completing the square.

c) Identify the vertex using your function from part b.

d) Do your answers from part a and c match?

e) Graph the function.

x f(x)

Name ______________________________ Period __________



.

#16 – 19: Write each function in vertex form. Working with the new function, create a table of values,

graph the function, and label the vertex.

Write in Vertex Form Make a table Graph and label the vertex

16) ( ) 2 6 2f x x x= + −

x f(x)

17) ( ) 2 14 13f x x x= − +

x f(x)

18) ( ) 2 8 15f x x x= + −

x f(x)

19) ( ) 2 2 6f x x x= − −

x f(x)

Name ______________________________ Period __________




a) Use 2

b






x f(x)

Section 4.3E

Name ______________________________ Period __________

4.3ext Vertex Form of Quadratic Equations: a ≠ 1


.


a) Use 2

b






x f(x)

f) What is the axis of symmetry?

g) Give the domain and range of the function.

Name ______________________________ Period __________



#2 – 5: Write each function in vertex form. Working with the new function, create a table of values,

graph the function, and label the vertex.

Write in vertex form Make a table Graph and label the vertex

2) ( ) 2 2 4f x x x= − − +

x f(x)

3) ( ) 25 30 5f x x x= + +

x f(x)

4) ( ) 23 6 4f x x x= − − +

x f(x)

5) ( ) 22 8 3f x x x= − − −

x f(x)

Name ______________________________ Period __________



.

#6 – 9: Complete the square to convert the standard form quadratic function into vertex form.

6) ( ) 22 5 2f x x x= + − 7) ( ) 23 4 2f x x x= + +

8) ( ) 25 4 14f x x x= + − 9) ( ) 22 9f x x x= +

10) Given ( ) 22 4 30f x x x= + − :

a) Transform f (x) to intercept form by factoring.

b) What are the x-intercepts of the function?

c) Transform f (x) to vertex form by completing the square.

d) What is the vertex of the function?


x f(x)

Name ______________________________ Period __________



11) Given ( ) ( ) ( )3 4 2f x x x= − +

a) What are the x-intercepts of the function?

b) Transform f (x) to standard form. c) Transform f (x) to vertex form.

d) What is the vertex of the function?


x f(x)

12) The maximum area for a fence constructed out of 60 feet of fencing for a rectangular shaped section of land

is given by the function ( ) 2 30A x x x= − + .

a) Write the function in vertex form.

b) What would be the dimensions of the rectangle that would give the

maximum area?

c) What would be the maximum area?

Section 4.3

(extension)

Name ______________________________ Period __________


Review Material 135

3)

y

= –

2(x

– 1

)2 –

3

#1

– 3

: F

ind

th

e re

qu

este

d i

nfo

rmati

on

for

each

eq

ua

tio

n.

2)

y

= 0

.5(x

– 3

) (x

+ 1

)

1)

y

= 2

x2 –

4x

+ 2

Op

ens

up

or

do

wn?

Ver

tica

l S

tret

ch o

r

Ver

tica

l C

om

pre

ssio

n o

f

the

par

ent

gra

ph?

Ver

tex

Ax

is o

f S

ym

met

ry

y-in

terc

ept

GR

AP

H

Wh

at a

re t

he

x-in

terc

epts

?

Wh

at i

s th

e d

om

ain

?

Wh

at i

s th

e ra

ng

e?

Name ______________________________ Period __________




4) ( ) ( )2 1 5y x x= − + 5) ( ) ( )2 6 4y x x= − − −

6) ( )2

9 4 12y x= + − 7) ( )2

2 1 3y x= − − +

#8 – 14: Write each equation in factored form.

8) 2 11 24y x x= − + 9) 25 13 6y x x= + −

10) 210 25 15y x x= − + 11) 24 20 25y x x= + +

12) 29 12 4y x x= − + 13) 216 100y x= −

14) 210 3 4y x x= − −

Name ______________________________ Period __________


Review Material 137

#15 – 18: Write each equation in vertex form by completing the square.

15) 2 14 8y x x= − + 16) 2 3 2y x x= − −

17) 24 24 40y x x= + + 18) 23 3 1y x x= − +

#19 – 23: Tim kicks a ball off the ground. After t seconds, its height (in feet) is given by the formulas:

( ) 216 64h t t t= − + � equivalent � ( ) ( ) ( )16 0 4h t t t= − + −

19) What is the maximum height reached by the ball? Record your thinking.

20) How many seconds does it take to reach the maximum height? Record your thinking.

21) How long is the ball in the air? Record your thinking.

22) What is the domain? Explain in the context of the problem.

23) What is the range? Explain in the context of the problem.

Name ______________________________ Period __________



#24 – 28: The height (in feet) of an object shot from a cannon after t seconds, can be modeled by the

function:

( ) 216 20 6h t t t= − + +

24) What is the maximum height the object reaches? Record your thinking.

25) How many seconds does it take to reach the maximum height? Record your thinking.

26) How high will the object be after 1 second? Record your thinking.

27) How long is the object in the air? Record your thinking.

28) What is the starting height of the object? Explain.

Name ______________________________ Period __________


Review Material 139

5 10 15 200

200

400

600

800

1000

1200

1400

1600

0

29) An amateur rocketry club is holding a competition. Vertical motion can be modeled using the equation

( ) 20 016h t t v t h= − + ⋅ + where 0v represents the initial velocity and 0h represents the initial height. The

rocket is launched from a platform that is 4 feet above the ground with an initial velocity of 315 ft/sec.

Indicate units of measure on all answers.

a) Write an equation that models the height of the

rocket at time t.

b) Sketch the graph that models the height of the rocket

at time t.

c) Find the x-intercept. Explain the meaning of this

point in the context of the situation.

d) Find the y-intercept. Explain the meaning of this

point in the context of the situation.

e) Find the vertex. Explain the meaning in the context

of the problem.

f) Find the height of the rocket at time t = 3 seconds.

g) There is cloud cover at 1000 ft. During what time

interval will the rocket not be visible to the

spectators?

h) What is the domain and range? Explain the meaning

in the context of the problem.

Domain:

Range:

Time (in seconds)

Ro

cket

Hei

gh

t (i

n f

eet)

Unit 4 Review

WORK SPACE

Name ______________________________ Period __________




141

SOLVING QUADRATIC EQUATIONS

Students solve quadratic equations by factoring, finding square roots, completing the

square, and by using the quadratic formula. Students determine the number of

rational, real, and non-real solutions by factoring or solving the equation and also by

using the graph. Students understand that the quadratic formula can be found by

completing the square and use the symmetry in the graph of a parabola to identify

how the x-coordinate for the vertex can be seen in the quadratic formula.



5.1 I can use tables and

graphs to solve quadratic

equations including real-

world situations and

translate between

representations.

175 237 5.1A Solving Quadratic Equations by Graphing

178 241 5.1B Answering Real-World Questions by

Graphing Quadratic Functions

5.2 I can represent real-world

situations with quadratic

equations and solve using

appropriate methods, find

real and non-real

complex roots when they

exist, and recognize that a

particular solution may

not be applicable in the

original context.

181 245 5.2A Factoring Review

182 247 5.2B Solving Quadratic Equations by Factoring:

Part I

185 249 5.2C Solving Quadratic Equations by Factoring:

Part II

188 251 5.2D Operations with Radical Expressions


5.1 I can use tables and graphs to solve quadratic equations including real-world situations and translate between representations.

5.2 I can represent real-world situations with quadratic equations and solve using appropriate methods, find real and non-real complex roots when they exist, and recognize that a particular solution may not be applicable in the original context.


5.4 I can represent relationships involving quadratic inequalities in multiple ways, find solutions and interpret these solutions to solve real-world situations.

Embedded throughout the unit:

• I understand how to verify that an answer is a solution and can interpret the solution in the context of the situation.

• I can demonstrate understanding of the real and non-real number systems and mathematical operations using expressions from these number systems.

UNIT

5

142


195 255 5.2E Solving Quadratic Equations Using Square

Roots to Find Rational Solutions

5.2 I can represent real-world

situations with quadratic

equations and solve using

appropriate methods, find

real and non-real

complex roots when they

exist, and recognize that a

particular solution may

not be applicable in the

original context.

198 257 5.2F Solving Quadratic Equations Using Square

Roots to Find Real Solutions

200 261 5.2G Operations with Complex Expressions

203 265 5.2H Solving Quadratic Equations Using Square

Roots to Find Real or Complex Solutions

204 269 5.2I (ext) Solving Quadratic Equations by Completing

the Square to Find Rational Solutions

206 271 5.2J (ext) Solving Quadratic Equations by Completing

the Square to Find Real Solutions

208 273 5.2K(ext) Solving Quadratic Equations by Completing

the Square to Find Real or Complex

Solutions

209 277 5.2L Solving Quadratic Equations Using the

Quadratic Formula to Find Real

Solutions

211 279 5.2M Solving Quadratic Equations Using the

Quadratic Formula to Find Real or

Complex Solutions

215 281 5.2N Solving Quadratic Equations –

Choosing the Best Method: Part I

218 289 5.2O Solving Quadratic Equations –

Choosing the Best Method: Part II

5.3 I can determine the

number of real and non-

real solutions for a

quadratic equation.

220 293 5.3A Number and Type of Solutions: Part I

223 297 5.3B Number and Type of Solutions: Part II

5.4 I can represent

relationships involving

quadratic inequalities in

multiple ways, find

solutions and interpret

these solutions to solve

real-world situations.

225 299 5.4A Graphing Quadratic Inequalities

227 303 5.4B Solving Quadratic Inequalities




5.1 I can use tables and graphs to solve quadratic equations including

real-world situations and translate between representations.

Solving a quadratic equation means finding the x-values that will make the quadratic

function equal a specified value. This value is often zero so that would mean finding the

x-intercepts, the points where the graph of the function crosses the x-axis. The solutions

to a quadratic equation are also called the roots. The x-intercepts of the function are

often called the zeros of the function. In this section we will learn how to find solutions

by graphing.

Identify the Number of Solutions of a Quadratic Equation

Three different situations can occur when graphing a quadratic function:

Case 1: The parabola crosses the x-axis at two points.

An example of this is ( ) 2 6f x x x= + − :

Looking at the graph, we see that the parabola crosses the x-axis at x = –3

and x = 2. We can also find the solutions by converting it to intercept or

factored form. We can factor: ( ) ( )2 6 3 2x x x x+ − = + − which means that

the x-intercepts are –3 and 2 as shown.

Given the equation ( ) 0f x = , when the graph of a quadratic function

crosses the x-axis at two points, we get two distinct real solutions to the

quadratic equation.

Case 2: The parabola touches the x-axis at one point.

An example of this is ( ) 2 2 1f x x x= − + :

We can see that the graph touches the x-axis at x = 1. Again, we can also

find the solution by converting it to factored form. We can factor:

( )22 2 1 1x x x− + = −

which means that there is only one x-intercept and it

is 1.

Given the equation ( ) 0f x = , when the graph of a quadratic function

crosses the x-axis in one point, we get one real solution to the quadratic

equation.

Case 3: The parabola does not cross or touch the x-axis.

An example of this is ( ) 2 4f x x= + :

We can see that this parabola does not have an x-intercept; 2 4x + does not

factor, so we cannot convert it to factored form.

Given the equation ( ) 0f x = , when the graph of a quadratic function does

not cross or touch the x-axis, the quadratic equation has no real solutions.

Unit 5 Resources

-6 -4 -2 2 4 6

-2

2

4

6

8

x

y

-6 -4 -2 2 4 6

-2

2

4

6

8

x

y

-6 -4 -2 2 4 6

-8

-6

-4

-2

2

x

y


144 UNIT 5 – SOLVING QUADRATIC EQUATIONS

Solve Quadratic Equations by Graphing

So far we have found the solutions to quadratic equations using factoring. However, in real life very few functions

factor easily. As you just saw, graphing a function gives a lot of information about the solutions. Sometimes it is

helpful to use a graphing utility to find approximate solutions to a quadratic equation.

5.1 Example 1:

Given the function ( ) 2 3f x x x= − + − , consider the equation ( ) 0f x = . Find the solutions to the

equation. Let’s use a table to sketch a graph.

Solution:

Based on the symmetry in the table and

the values of the function, it indicates that

( )f x would not have a value of 0 at any

point.

The graph shows that the parabola does

not cross the x-axis. This quadratic

equation has no real solutions.

5.1 Example 2:

Given the function ( ) 2 4 4f x x x= − + − , consider the equation ( ) 0f x = . Find the solutions to the

equation. Let’s use a table to sketch a graph.

Solution:

The table shows that when x = 2 that

( ) 0f x = , so the x-intercept is ( )2, 0 .

The graph shows that the parabola touches

the x-axis at x = 2. The solution is x = 2.

5.1 Example 3:

Given the function ( ) 2 3f x x= − , consider the equation ( ) 0f x = . Find the solutions to the equation.

Let’s use a table to sketch a graph.

Solution:

The table indicates that ( )f x would be 0

between –2 and –1 and again between

1 and 2. The graph shows that the

parabola crosses the x-axis at about

–1.7 and 1.7.

A graphing utility will give a better

approximation of the x-intercepts, or zeros:

( )1.732, 0− and (1.732, 0).

x ( )f x

–3 –15

–2 –9

–1 –5

0 –3

1 –3

2 –5

3 –9

x ( )f x

–2 –16

–1 –9

0 –4

1 –1

2 0

3 –1

4 –4

x ( )f x

–3 6

–2 1

–1 –2

0 –3

1 –2

2 1

3 6

-4 -3 -2 -1 1 2 3 4

-8

-6

-4

-2

2

x

y



Vocabulary

• The solutions of the quadratic equation (((( )))) 0f x ==== are the x-intercepts of the parabola and

are often called the roots and zeros.

Video Resources

• CK-12 Foundation: 1003 Solving Quadratic Equations by Graphing

• CK-12 Foundation: 1002S Lesson Solving Quadratic Equations by Graphing

5.2 I can represent real-world situations with quadratic equations and solve by using appropriate

methods, find real and non-real complex roots when they exist, and recognize that a particular

solution may not be applicable in the original context.

Solve by Factoring

5.2 Example 1:

The height of a ball that is thrown straight up in the air from the top of a building 14.7 meters

above the ground with a velocity of 9.8 meters per second is given by the function

( ) 24.9 9.8 14.7h t t t= − + + , where t is the time in seconds. How long does it take the ball to

hit the ground?

Solution:

In the previous unit, we learned to solve this problem by looking at the graph. The x-coordinate of the

vertex (t in this situation) would be 9.8

12 9.8

b

a− = − = .

Substitute 1 for t into the equation

( ) ( ) ( )2

4.9 1 9.8 1 14.7 19.6h t = − + + = . So the vertex is at (1,

19.6). Since we are looking for when the ball hits the ground, we

would substitute h = 0 and look for the t-intercepts. The

intercepts on the t-axis can be found by factoring.

Equation: ( ) 24.9 9.8 14.7h t t t= − + +

Factor out the greatest common factor: ( ) ( )24.9 2 3h t t t= − − −

Factor into binomials: ( ) ( ) ( )4.9 1 3h t t t= − + −

The t-intercepts would be at 1t = − and 3t = when ( ) 0h t =

Using the graph, the ball hits the ground at 3t = seconds.

The value 1t = − would not make sense for this situation as time cannot be a negative value.

Seconds

Hei

gh

t

Ball Toss Graph



To find x-intercepts, we convert a quadratic function in the form y = ax2 + bx + c to factored form,

y = (x – m)(x – n). x-intercepts have a y-value of 0 so we substitute 0 in for y. The Zero Product Property states

that if ab = 0, then either a = 0 or b = 0 (or both) so we find the x-intercepts by solving x – m = 0 and

x – n = 0. Converting to factored form allowed us to find the values of x that made ax2 + bx + c = 0 a true

statement. Values that make an equation true are called solutions. Solutions to the equation f(x) = 0 are also

called zeros or roots.

5.2 Example 2:

Solve by factoring. x2 – 9x + 18 = 0

Solution:

x2 – 9x + 18 = 0

(x – 6)(x – 3) = 0

When two numbers are multiplied together and one of them is zero, the product is always zero.

So: x – 6 = 0 or x – 3 = 0

x = 6 x = 3

Check the answer:

x2 – 9x + 18 = 0 x2 – 9x + 18 = 0

check x = 6 (6)2 – 9(6) + 18 = 0 and check x = 3 (3)2 – 9(3) + 18 = 0

36 – 54 + 18 = 0 9 – 27 + 18 = 0

0 = 0 0 = 0

5. 2 Example 3:

Solve by factoring. 6�� + � − 4 = 11

Solution:

6�� + � − 4 = 11

6x2 + x – 15 = 0 Rewrite in standard form

(2x – 3)(3x + 5) = 0 Factor

2x – 3 = 0 or 3x + 5 = 0 Set the factors equal to 0

2x = 3 3x = −5 Solve

= � = � �

� = − = −� �

Check the answer:

6�� + � − 4 = 11 6�� + � − 4 = 11

check � = 1 �� 6 �1 �

�� + �1 �

�� − 4 = 11 and check � = −1 �� 6 �−1 �

�� + �−1 �

�� − 4 = 11

6(2.25) + (1.5) – 4 = 11 16 ��− 1 �

�− 4 = 11

11 = 11 11 = 11



5.2 Example 4:

Solve by factoring. 10x2 – 25x = 0

Solution:

Here is an example of a quadratic equation without a constant term. The proper way to factor this is to

take out the GCF.

10x2 – 25x = 0

5x(2x – 5) = 0

5x = 0 or 2x – 5 = 0

x = 0 = � = � �

�

Check the answer:

Check 0x = ( ) ( )

2

2

10 25 0

10 0 25 0 0

0 0 0

0 0

x x− =

− =

− =

=

and check 1

22

x =

( ) ( )

2

2

10 25 0

1 110 2 25 2 0

2 2

10 6.25 25 2.5 0

0 0

x x− =

− =

− =

=

5.2 Example 5:

Solve by factoring. 4x2 – 12x + 9 = 0

Solution:

( )( )

( )

2

2

4 12 9 0

2 3 2 3 0

2 3 0

x x

x x

x

− + =

− − =

− =

The factors are the same. When factoring a perfect square trinomial, the factors will always be the same.

In this instance, the solutions for x will also be the same. Solve for x.

2 3 0

2 3

x

x

− =

=

3 1

12 2

x = == == == =

Check the answer:

Check 1

12

x =

( )

2

2

4 12 9 0

1 14 1 12 1 9 0

2 24 2.25 18 9 0

x x− + =

− + =

− + =

When the two factors are the same, we call the solution for x a double root because it is the same

solution twice.



5.2 Example 6:

One leg of a right triangle is 3 feet longer than the other leg. The hypotenuse is 15 feet. Find the

dimensions of the triangle.

Solution:

Let x = the length of the short leg of the triangle; then the other leg will measure x + 3.

Use the Pythagorean Theorem: a2 + b2 = c2, where a and b are the lengths of the legs and c is the length

of the hypotenuse. When we substitute the values from the diagram, we get (x)2 + (x + 3)2 = 152.

In order to solve this equation, we need to get the polynomial in standard form. We must first distribute,

combine like terms, then factor and solve.

(x)2 + (x + 3)2 = (15)2

x2 + x2 + 6x + 9 = 225

2x2 + 6x + 9 = 225

2x2 + 6x – 216 = 0

2(x2 + 3x – 108) = 0

2(x – 9)(x + 12) = 0

Set each factor equal to zero and solve:

x – 9 = 0 or x + 12 = 0

x = 9 x = –12

It makes no sense to have a negative value (–12) for the length of a side of the triangle, so the answer

must be x = 9. That means the short leg is 9 feet and the long leg is 12 feet.

Check: 92 + 122 = 152 81 + 144 = 225, so the answer checks.

The dimensions of the triangle are: (x), (x + 3), (15) and x = 9, so… 9 feet, 12 feet and 15 feet

Vocabulary

• A solution is a value that makes an equation true. Solutions can also be called zeros or roots.

• Double root is a solution that is repeated twice.

Video Resources

• Khan Academy: Solving Quadratic Equations by Factoring.avi

• CK-12 Foundation: Solving Real-World Problems by Factoring



Solve by Finding Square Roots: Perfect Squares

A bird was flying through the air at a height of 1600 feet carrying a worm in its beak. If the worm slips from the

bird’s beak, how long would it take the worm to hit the ground, assuming there was no air resistance? Remember

that projectile motion can be modeled by a quadratic function of the form:

h(t) = –16t2 + v0t + h0.

Since the worm is dropped, v0 would be 0 and h0 would be 1600. The

function would be:

h(t) = –16t2 + 1600

Looking at the graph of this function, the worm would hit the ground

where the parabola crosses the x-axis, also called the

x-intercept. The height of the worm would be 0 so h(t) = 0 .

–16t2 + 1600 = 0

We could solve by factoring but there is another method that will

work well for this particular problem. Let’s solve for t

algebraically using inverse operations. Remember, the inverse of a

square is a square root.

2

2

2

2

16 1600 0

1600 1600

16 1600

16 16

t 100

t 100

10

10

t

t

t

t

− + =

− −

− = −

− −

=

=

=

= ±

In this problem t represents time, which cannot be a negative number. The worm would hit the ground 10

seconds after the bird dropped it.

We were able to use the square root method to solve this quadratic equation because the value of b is 0. Without

a b term the equation becomes ax2 – c = 0. This allows us to rearrange the equation so that the x2 term is by itself

and then take the square root of both sides.

In general, the solution to a quadratic equation of the form ax2 – c = 0 is:

� = �� or� = −��

Another way to write this is � = ±��

-12-10 -8 -6 -4 -2 2 4 6 8 10 12-100

100

200

300

400

500

600

700

800

900

1000

1100

1200

1300

1400

1500

1600

1700

t

h(t)

Height of Worm



5.2 Example 7:

Solve for x using the square root property. �� − 81 = 0

Solution: x2 – 81 = 0

+ 81 + 81 Add 81 to both sides

x2 = 81

2 81x = Take the square root of each side

9x =

x = 9 or x = –9

Check the answer: (9)2 – 81 = 0 and (–9)2 – 81 = 0

81 – 81 = 0 81 – 81 = 0

5.2 Example 8: Solve for x using the square root property. 2�� − 18 = 0

Solution: 2x2 – 18 = 0


2x2 = 18

2 2 Divide both sides by 2

2 9x =

2 9x = Take the square root of each side

3x =

x = ±3 Check the answer: 2(3)2 – 18 = 0 and 2(–3)2 – 18 = 0

2(9) – 18 = 0 2(9) – 18 = 0

0 = 0 0 = 0

This method also works for equations in vertex form, a(x – h)2 + k = 0. In this case we rearrange the equation so

that the (x – h)2 term is by itself, take the square root of both sides and then solve for x.

5.2 Example 9:

Solve for x using the square root property. �� − 4�� − 9 = 0

Solution: (x – 4)2 – 9 = 0

(x – 4)2 = 9 Add 9 to both sides

( )2

4 9x − = Take the square root of both sides

|x – 4| = 3

(x – 4) = ±3

x – 4 = 3 or x – 4 = –3 Solve for x

x = 7 or x = 1

Check the answer: �� − 4�� − 9 = 0 �� − 4�� − 9 = 0

check x = 7 �7 − 4�� − 9 = 0 and check x = 1 �1 − 4�� − 9 = 0

�3�� − 9 = 0 �−3�� − 9 = 0

0 = 0 0 = 0



5.2 Example 10:

Solve for x using the square root property. ( )2

2 5 8 0x + − =

Solution: 2(x + 5)2 – 8 = 0

2(x + 5)2 = 8 Add 8 to both sides

(x + 5)2 = 4 Divide both sides by 2

( )2

5 4x + =

5 2x + =

x + 5 = ±2 Take the square root of both sides

x + 5 = 2 or x + 5 = –2 Solve for x

x = –3 or x = –7

Check the answer:

Check 3x = −

( )

( )

( )

2

2

2

2 5 8 0

2 3 5 8 0

2 2 8 00 0

x + − =

− + − =

− =

=

and check 7x = −

( )

( )

( )

2

2

2

2 5 8 0

2 7 5 8 0

2 2 8 00 0

x + − =

− + − =

− − =

=

Video Resources:

• CK-12 Foundation: 1004 Solving Quadratic Equations Using Square Roots

Solve by Finding Square Roots: Non-Perfect Squares

5.2 Example 11:

Solve !"�#+ 3 = 27

Solution:

!"�#+ 3 = 27

!"�# = 24 Subtract 3 from both sides

�� = 384 Multiply both sides by 16

2 384x = Take the square root of both sides

384x =

� = ±√384 = ±√64 · 6 = ±8√6 Simplify

= &√'or = −&√'

Check the answer: !"�#+ 3 = 27

!"�#+ 3 = 27

check � = 8√6 ()√#*"�# + 3 = 27 and check � = −8√6

(+)√#*"�# + 3 = 27

#,∙#�# + 3 = 27

#,∙#�# + 3 = 27

�),�# + 3 = 27

�),�# + 3 = 27

24 + 3 = 27 24 + 3 = 27

27 = 27 27 = 27



To simplify radical expressions that are not perfect squares, we use the radical rule:

√�. = √� · . = √� · √.

Example:

To simplify √48, find a perfect square factor to break it down and simplify

√48 = √16 · 3 = √16 · √3 = 4√3

5.2 Example 12:

Solve for x. x2 – 50 = 0

Solution:

x2 – 50 = 0


x2 = 50

2 50x =

50x =

� = ±√50 Take the square root of each side

To simplify √50:

• We can use a calculator to get an approximate answer. On the calculator your answer should be

√50 ≈ 7.071067811865… We will have to round the answer. If we round to the hundredths place

√50 ≈ 7.07.

• We can simplify to get a more accurate exact answer. To simplify the square root, the square numbers

must be “pulled out.” Look for factors of 50 that are square numbers: 4, 9, 16, 25... 25 is a factor of 50,

so break the factors apart.

√50

√25 · 2

√25 · √2

5√2 x = √� or x = − √�.

Check the answer: �� − 50 = 0 �� − 50 = 0

check x = 5√2 (5√2*� − 50 = 0 and check x = −5√2 (−5√2*� − 50 = 0

(5√2*(5√2* − 50 = 0 (−5√2*(−5√2* − 50 = 0

5 ∙ 5 ∙ √2 ∙ √2 − 50 = 0 �−5� ∙ �−5� ∙ √2 ∙ √2 − 50 = 0

25 ∙ 2 − 50 = 0 25 ∙ 2 − 50 = 0

50 − 50 = 0 50 − 50 = 0

0 = 0 0 = 0



5.2 Example 13:

Solve �� − 5 = 27

Solution:

�� − 5 = 27

�� = 32 Add 5 to both sides

2 32x = Square root of both sides -- Simplify

32x =

� = ±√32 = ±√16 · 2 = ±4√2

= 0√� or = −0√�

Check the answer: �� − 5 = 27 and �� − 5 = 27

(4√2*� − 5 = 27 (−4√2*� − 5 = 27

4 ∙ 4 ∙ √2 ∙ √2 − 5 = 27 �−4� ∙ �−4� ∙ √2 ∙ √2 − 5 = 27

32 − 5 = 27 32 − 5 = 27

27 = 27 27 = 27

5.2 Example 14:

Solve for x. 3�� − 5�� + 7 = 43

Solution:

3�� − 5�� + 7 = 43

3�� − 5�� = 36 Subtract 7 from both sides

�� − 5�� = 12 Divide both sides by 3

( )2

5 12x − = Take the square root of both sides

5 12x − =

� − 5 = ±√12

� − 5 = ±√4 · 3 Simplify

� − 5 = ±2√3

� = 5 ± 2√3 Add 5 to both sides

= + �√ or = − �√

Check the answer: 3�� − 5�� + 7 = 43 3�� − 5�� + 7 = 43

3(5 + 2√3 − 5*� + 7 = 43 and 3(5 − 2√3 − 5*� + 7 = 43

3(2√3*� + 7 = 43 3(2√3*� + 7 = 43

3(2 ∙ 2 ∙ √3 ∙ √3* + 7 = 43 3(2 ∙ 2 ∙ √3 ∙ √3* + 7 = 43

3�4 ∙ 3� + 7 = 43 3�4 ∙ 3� + 7 = 43

36 + 7 = 43 36 + 7 = 43

43 = 43 43 = 43



5.2 Example 15:

Consider the following right triangle. What is the length of the triangle's hypotenuse?

Solution:

Because we know the length of two sides and need to know the length of the third side, to find the

hypotenuse, we must use the Pythagorean Theorem.

Pythagorean Theorem: a2 + b2 = c2

(2√5*� + (3√4*� = ��

�4 · 5� + �9 · 4� = ��

20 + 36 = ��

56 = ��

256 c=

56 c=

±√56 = c ±√4 ∙ √14 = c ±2√14 = c

Because c represents a length, a negative value is not a realistic value. The hypotenuse is �√�0 units.

Vocabulary

• Simplifying the square root of a non-perfect square number can be done using the radical rule:

√�. = √� · . = √� · √.. A simplified answer is in exact form.

• A calculator will give an approximate value of the square root of a non-perfect square number.

Video Resources:

• CK-12 Foundation: 1004S Solving Quadratic Equations Using Square Roots

• Khan Academy: Solving Quadratics by Square Roots

2√5

3√4



Solve by Finding Square Roots: Complex Numbers

The graph of the quadratic equation 2 = �� + 16 is:

Notice the parabola does not cross the x-axis. It does not have

x-intercepts.

If you substitute 0 in for y:

�� + 16 = 0

�� = −16

2 16x = −

16x − = −

� = ±√−16

We have previously learned we cannot find √−16 because you

cannot take the square root of a negative number. There is no

real number that, when multiplied by itself, equals –16. Let’s

simplify 16± − .

16 16 1 16 1 4 1± − = ± ⋅ − = ± ⋅ − = ± ⋅ −

In order to take the square root of a negative number we are going to assign √−1 as the imaginary unit, 3. The

unit 4 represents an imaginary number. Now, we can use 4 to take the square root of a negative number.

16 16 1 16 1 4 4i i± − = ± ⋅ − = ± ⋅ − = ± ⋅ = ±

Hence, the solutions to the original equation are 4i± .

All complex numbers have the form � + .4, where a and b are real numbers. a is the real part of the complex

number and b is the coefficient of the imaginary part. If b = 0, then only the a is left and the number is a real

number. If a = 0, then only the b part remains and the number is called a pure imaginary number.



The imaginary number,4, is defined as 4 = √−1

Note4� = −1 because 4� = 4 · 4 = √−1 · √−1 = (√−1*� = −1

In order to solve some of the equations you will come upon throughout the rest of your math career, the number

system needs to be expanded from the REAL number system to the COMPLEX number system which

includes imaginary numbers.

Substitute the solutions in the original equation:

5.2 Example 16:

Solve for x using the square root property. 3�� + 27 = 0

Solution:

3�� + 27 = 0

3�� = −27 Subtract 27 from both sides

�� = −9 Divide both sides by 3

� = ±√−9 Take the square root of both sides

� = ±√9 · −1 Simplify

� = ±34 = 3 or = −3. Check the answer: 3�� + 27 = 0 3�� + 27 = 0

check � = 34 3�34�� + 27 = 0 and check � = −34 3�−34�� + 27 = 0

3�34��34� + 27 = 0 3�−34��−34� + 27 = 0

3�94�� + 27 = 0 3�94�� + 27 = 0

3�−9� + 27 = 0 3�−9� + 27 = 0

−27 + 27 = 0 −27 + 27 = 0

An imaginary number, when squared, gives a negative result.

1i = −= −= −= −

It helps us answer the question: 2 1x = −= −= −= −

2

2

1

1

1

1 or 1

or

x

x

x

x x

x i x i

= −

= −

= −

= − = − −

= = −

( ) ( )

( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( )

2 2

2 2

2

2

check the solution check the solution

1 1

1 1

1 1

1 1 1

1 1

1

x i x i

x x

i i

i i i

i i

i i

i

= = −

= − = −

= − − = −

= − − − = −

− − = −

= −

= −



5.2 Example 17:

Solve for x using the square root property. �� + 162 = 0

Solution:

�� + 162 = 0

�� = −162 Subtract 36 from both sides

� = ±√−162 Take the square root of both sides

� = ±√−1 · 162

� = ±4√162

� = ±4√81 ∙ 2 Simplify

� = ±94√2 = 53√� or = −53√�.

Check the answer: �� + 162 = 0 �� + 162 = 0

check � = 94√2 (94√2*� + 162 = 0 and check � = −94√2 (−94√2*� + 162 = 0

(94√2*(94√2* + 162 = 0 (−94√2*(−94√2* + 162 = 0

81 ∙ 4� ∙ 2 + 162 = 0 81 ∙ 4� ∙ 2 + 162 = 0

162�−1� + 162 = 0 162�−1� + 162 = 0

−162 + 162 = 0 −162 + 162 = 0

5.2 Example 18:

Solve for x using the square root property. �� − 8�� = −25

Solution: �� − 8�� = −25

�� − 8� = ±√−25 Take the square root of both sides

� − 8 = ±√−1 ∙ √25

� − 8 = ±54 Simplify

� = ±54 + 8 Add 8 to both sides

� = 8 ± 54 = & + 3 or = & − 3. Check the answer: �� − 8�� = −25 �� − 8�� = −25

check � = 8 + 54 �8 + 54 − 8�� = −25 and check � = 8 − 54 �8 − 54 − 8�� = −25

�54�� = −25 �−54�� = −25

�54��54� = −25 �−54��−54� = −25

5 ∙ 5 ∙ 4 ∙ 4 = −25 �−5� ∙ �−5� ∙ 4 ∙ 4 = −25

254� = −25 254� = −25

−25 = −25 −25 = −25



5.2 Example 19:

Solve for x using the square root property. 2�3� − 5�� + 10 = −30

Solution:

2�3� − 5�� + 10 = −30

2�3� − 5�� = −40 Subtract 10 from both sides

�3� − 5�� = −20 Divide by 2 on both sides

3� − 5 = ±√−20 Take the square root of both sides

3� − 5 = ±24√5 Simplify √−20 = √−1 ∙ 4 ∙ 5 = 24√5

3� = 5 ± 24√5 Add 5 to both sides

� = 6±�7√6� Divide both sides by 3

= 8�3√ or = +�3√

.

Check the answer: 2�3� − 5�� + 10 = −30

check � = �68�7√6� � 2 �3 ∙ 68�7√6� − 5�� + 10 = −30

2(5 + 24√5 − 5*� + 10 = −30

2(24√5*� + 10 = −30

2�4 ∙ 4� · 5� + 10 = −30

2�−20� + 10 = −30

−40 + 10 = −30

−30 = −30

check � = �6+�7√6� � 2 �3 ∙ 6+�7√6� − 5�� + 10 = −30

2(5 − 24√5 − 5*� + 10 = −30

2(−24√5*� + 10 = −30

2�4 ∙ 4� · 5� + 10 = −30

2�−20� + 10 = −30

−40 + 10 = −30

−30 = −30



Vocabulary

• An imaginary number is a number that has the imaginary unit, 4 to represent √−1 .

• All complex numbers have the form � + .4, where a and b are real numbers. a is the real part of the

complex number and b is the coefficient of the imaginary part.

Completing the Square

If you have a quadratic equation of the form �� − 2�� = 5, we can solve it by taking the square root of each side.

�� − 2�� = 5

� − 2 = ±√5

� = 2 ± √5

The solutions are � = 2 + √5or � = 2 − √5.

Unfortunately, quadratic equations are not usually written in this nice form. In this section, you will learn the

method of completing the squares in which you take any quadratic equation and rewrite it in a form so that you

can take the square root of both sides.

Complete the Square of a Quadratic Expression

The purpose of the method of completing the squares is to rewrite a quadratic expression so that it contains a

perfect square trinomial that can be factored as the square of a binomial using the square root property (previous

section). We previously completed the square to convert a quadratic equation from standard form to vertex form.

Recall that to complete the square on an expression that is in the form �� + .�, add �9��. Then the expression

becomes �� + .� + �9�� which is a perfect square trinomial that can be factored as �� + 9

��.

Complete the Square to Solve a Quadratic Equation

We can also use completing the square to solve a quadratic equation. When we convert one side of the equation

to �� + 9��

�, we are able to use the square root method to solve it. A key to this method is remembering that when

working with equations, if you add a value such as �9��to one side of an equation, you must add the same value to

the other side of the equation.



5.2 Example 20:

Solve the following quadratic equation by completing the square. �� + 12� = 3

Solution:

�� + 12� = 3

�� = 6� = 36 Find�9��

�

�� + 12� + ' = 3 + ' Add this constant to both sides of the equation

�� + 6�� = 39 Factor the perfect square trinomial

� + 6 = ±√39 Take the square root of both sides

� = −6 ± √39 Solve for �

The exact solutions are � = −6 + √39or � = −6 − √39.

The approximate solutions are x = 0.245 or x = –12.245

Use a calculator to check the approximate solutions.

�� + 12� = 3

�0.245�� + 12�0.245� ≈ 3

�−12.245�� + 12�−12.245� ≈ 3

5.2 Example 21:

Solve the following quadratic equation by completing the square. �� + 15� + 12 = 0

Solution:

�� + 15� + 12 = 0

�� + 15� = −12 Move the constant to the other side.

��6� �� = ��6

, Find�9��

�� + 15� + ��6, = −12 + ��6

, Add to both sides

�� + �6� �

� = �<<, Factor the perfect square trinomial

� + �6� = ±��<<

, Take the square root of both sides

� = − �6� ± √�<<

� Simplify and solve for x

The exact solutions are � = − �6� + √�<<

� or � = − �6� − √�<<

�

The approximate solutions are x = –0.8479 or x = –14.1521

Check the answer:

�� + 15� + 12 = 0

�−0.8479�� + 15�−0.8479� + 12 ≈ 0

�−14.1521�� + 15�−14.1521� + 12 ≈ 0



If the coefficient of the� term is not one, we must divide that number from the whole equation before

completing the square.

5.2 Example 22:

Solve the following quadratic equation by completing the square. 3�� − 12� = −1

Solution:

3�� − 12� = −1

�� − 4� = − �� Divide all terms by the coefficient of the �� term which is 3

�� − 4� + 4 = − ��+ 4 Add�9��

�to both sides

�� − 2�� = �� Factor the perfect square trinomial

� − 2 = ±�� Take the square root of both sides

� = 2 ±�� Solve for x

The approximate solutions are x = 0.0851 or x = 3.9149.

Check the answer:

3�� − 12� = −1

3(0.0851)2 – 12(0.0851) ≈ –1

3(3.9149)2 – 12(3.9149) ≈ –1

Complete the Square to Solve a Quadratic Equation with Complex Solutions

Again, some quadratic equations do not have real solutions. We can use the completing the square method to

find complex roots as well.

5.2 Example 23:

Solve the following quadratic equation by completing the square. �� − 4� = −6

Solution:

�� − 4� = −6

�� − 4� + 4 = −6 + 4 Add�9�� to both sides of the equation

�� − 2�� = −2 Factor the perfect square trinomial

� − 2 = ±√−2 Take the square root of both sides

� = 2 ± 4√2 Solve for �

The exact solutions are � = 2 + 4√2 or � = 2 − 4√2.

Check the answer: �� − 4� = −6 �� − 4� = −6

(2 + 4√2*� − 4(2 + 4√2* = −6 (2 − 4√2*� − 4(2 − 4√2* = −6

4 + 44√2 + 4��2� − 8 − 44√2 = −6 4 − 44√2 + 4��2� − 8 + 44√2 = −6

4 − 2 − 8 = −6 4 − 2 − 8 = −6

−6 = −6 −6 = −6



Solve Real-World Problems Using Quadratic Functions by Completing the Square

5.2 Example 24:

An arrow is shot straight up from a height of 8 feet with an initial velocity of 160 ft/s. How long will it

take for the arrow to hit the ground?

Solution:

ℎ�>� = −16>� + ?@> + ℎ@

ℎ�>� = −16>� + 160> + 8 initial velocity is 160 ft/s and initial height is 8 ft

−16>� + 160> + 8 = 0 “hits the ground”... so the height at some time is 0

−16>� + 160> = −8 subtract 8 from both sides

>� − 10> = 0.5 divide by –16 must get a lead coefficient of 1

>� − 10> + 25 = 0.5 + 25 complete the square (–10 ÷ 2)2 (–5)2 25

�> − 5�� = 25.5 rewrite the trinomial into factored form

> − 5 = ±5.0498 take the square root of both sides

> = 5 ± 5.0498 solve for x by adding 5 to both sides

The possible solutions are –0.0498 or 10.0498. Since time cannot be negative the solution is 10.0498.

The arrow will hit the ground after approximately 10 seconds.

Vocabulary

• To complete the square on an expression that is in the form �� + .�, add �9��. Then the expression

becomes �� + .� + �9�� which is a perfect square trinomial that can be factored as �� + 9

��.

Video Resources

• CK-12 Basic Algebra: Solving Quadratic Equations by Completing the Square

Solve by using the Quadratic Formula

So far we have four methods to solve a quadratic equation:

• Graphing

• Factoring

• Square root property

• Completing the square

There is one more method for solving a quadratic equation and that is by using the quadratic formula.



Deriving the Quadratic Formula

Let’s complete the square for a quadratic equation given in standard form: 2

2

2

22 22

2 2

2 2

2

0

0 Move the constant to the right side of the equation

Divide each term by

Add to both sides to complete the square24 4

4Fac

2 4

ax bx c

ax bx c

b cx x a

a a

b b c b bx x

a a aa a

b b acx

a a

+ + =

+ = −

−+ =

− + + = +

− + =

2

2

2

2

2

tor and simplify

4Take the square root of both sides

2 4

4Solve for

2 4

4Simplify the expression

2

b b acx

a a

b b acx x

a a

b b acx

a

−+ = ±

−= − ±

− ± −=

The solutions to any quadratic equation in standard form�� + .� + � = 0 can be found by

using the Quadratic Formula:

� = −. ± √.� − 4��2�

5.2 Example 25:

Solve the following equation using the quadratic formula. �� + 10� + 9 = 0

Solution:

Identify the coefficients of the equation: � = 1,. = 10 and � = 9

Substitute these values into the quadratic formula: � = +9±A�9�"+,��

� = +�@±A��@�"+,��B��

Simplify: � = +�@±√�@@+�#� = +�@±√#,

� = +�@±)�

� = +�@8)� or � = +�@+)

�

� = +�� = −1 or � = +�)

� = −9

The solutions are x = –1 or x = –9.

Check the answer: �� + 10� + 9 = 0 �� + 10� + 9 = 0

�−1�� + 10�−1� + 9 = 0 �−9�� + 10�−9� + 9 = 0

1 − 10 + 9 = 0 81 − 90 + 9 = 0



5.2 Example 26:

Solve the following equation using the quadratic formula. �� − 4� + 7 = 5�� − 5� + 6

Solution:

• Rearrange the equation so that all the terms are on one side: −4�� + � + 1 = 0

• Identify the coefficients of the equation: � = −4,. = 1 and � = 1

• Substitute these values into the quadratic formula: � = +�±A��"+,�+,��+,�

• Simplify: � = +�±√�8�#+) = +�±√�<

+)

The exact solutions are = +�8√�C+& or = +�+√�C

+& .

The approximate solutions are x = –0.3904 or x = 0.6404.

Check the answer: �� − 4� + 7 = 5�� − 5� + 6

�−0.3904�� − 4�−0.3904� + 7 = 5�−0.3904�� − 5�−0.3904� + 6

0.1524 + 1.5616 + 7 = 0.7621 + 1.9520 + 6 .

8.7140 ≈ 8.7141 .

�� − 4� + 7 = 5�� − 5� + 6

�0.6404�� − 4�0.6404� + 7 = 5�0.6404�� − 5�0.6404� + 6

0.4101 − 2.5616 + 7 = 2.0506 + 3.2020 + 6 .

4.8485 ≈ 4.8486 .

5.2 Example 27:

Solve the following equation using the quadratic formula. 2�� + 50 = 0

Solution:

• Identify the coefficients of the equation: � = 2,. = 0 and � = 50

• Substitute these values into the quadratic formula: � = @±A�@�"+,��6@��

• Simplify: � = @±√@+,@@, = ±√+,@@

, = ±�@7, = ±54

The exact solutions are = 3 or = − 3 Check the answer: 2�� + 50 = 0 2�� + 50 = 0

2�54�� + 50 = 0 2�−54�� + 50 = 0

2 ∙ 25 ∙ 4� + 50 = 0 2 ∙ 25 ∙ 4� + 50 = 0

−50 + 50 = 0 −50 + 50 = 0

0 = 0 0 = 0



5.2 Example 28:

Solve the following equation using the quadratic formula. 9�� − 30� + 26 = 0

Solution:

• Identify the coefficients of the equation: � = 9,. = −30 and � = 26

• Substitute these values into the quadratic formula: � = �@±A�+�@�"+,�B��#��B�

• Simplify: � = �@±√B@@+B�#�) = �@±√+�#

�) = �@±#7�) = 6±7

�

The exact solutions are = 83 or = +3

Check the answer: 9�� − 30� + 26 = 0 9�� − 30� + 26 = 0

9 �687� �� − 30 �687� � + 26 = 0 9 �6+7� �� − 30 �6+7� � + 26 = 0

9 �687� � �687� � − 30 �687� � + 26 = 0 9 �6+7� � �6+7� � − 30 �6+7� � + 26 = 0

�25 + 104 − 1� − �50 + 104� + 26 = 0 �25 − 104 − 1� − �50 − 104� + 26 = 0

25 + 104 − 1 − 50 − 104 + 26 = 0 25 − 104 − 1 − 50 + 104 + 26 = 0

25 − 1 − 50 + 26 = 0 25 − 1 − 50 + 26 = 0

0 = 0 0 = 0

One type of application problem that can be modeled using quadratic equations involves two objects that are

moving away in perpendicular directions from each other. Here is an example of this type of problem.

5.2 Example 29:

Two cars leave an intersection. One car travels north; the other travels east. When the car traveling

north had gone 30 miles, the distance between the cars was 10 miles more than twice the distance

traveled by the car heading east. Find the distance between the cars at that time.

Solution:

Because all distances are related to the distance of the car headed east, let � represent the distance

traveled by the car heading east. The distance between the two cars would be

2� + 10. Let’s make a sketch.

We can use the Pythagorean Theorem to find an equation:

�� + .� = ��

�� + �30�� = �2� + 10�� Substitute side lengths into the Pythagorean Theorem

�� + 900 = 4�� + 40� + 100 Multiply the binomials

0 = 3�� + 40� − 800 Simplify

• Identify the coefficients of the equation: � = 3,. = 40 and � = −800

• Substitute these values into the quadratic formula: � = +,@±A�,@�"+,��+)@@��

The approximate solutions are x = –24.3 or x = 10.97. Since �represents a distance, x = –24.3 would

not be a reasonable solution.

The distance between the two cars is about �� + �D = � miles.



Vocabulary

• The solutions to any quadratic equation in standard form�� + .� + � = 0 can be found by using the

Quadratic Formula � = +9±√9"+,�� .

Video Resources

• CK-12 Basic Algebra: Using the Quadratic Formula

• CK-12 Foundation: 1008S The Quadratic Formula

• CK-12 Foundation: 1008 The Quadratic Formula

Best Method for Solving a Quadratic Equation

Suppose you need to solve the quadratic equation −16>� + 22> + 3 = 0 in order to determine after how

many seconds a pebble shot up into the air by a slingshot will hit the ground. How could you solve such an

equation? Could you use graphing, factoring, taking the square root, completing the square, and/or the

quadratic formula? How would you choose which method to use?

Which Method to Use?

Usually you will not be told which method to use. You will have to make that decision yourself. However,

here are some guidelines to which methods are better in different situations.

• Graphing – a good method to visualize the parabola and easily see the intersections. Not always

precise.

• Factoring – best if the quadratic expression is easily factorable. It is always worthwhile to check if

you can factor because this is the fastest method.

• Taking the square root – is best used with the form �� − � = 0 (there is no � term).

• Completing the square – can be used to solve any quadratic equation. It is a very important method

for rewriting a quadratic function in vertex form. This method is best when � = 1

and . is even.

• Quadratic formula – is the method that always works for solving a quadratic equations.



5.2 Example 30: Solve by Completing the Square

The length of a rectangular pool is 10 meters more than its width. The area of the pool is 875

square meters. Find the dimensions of the pool.

Solution:

Begin by drawing a sketch. The formula for the area of a rectangle is E = FG.

E = �� + 10� 875 = �� + 10� 875 = �� + 10�

In this problem � = 1 and . = 10 so completing the square is a reasonable method to use.

�� + 10� = 875

�� + 10� + � = 875 + �

�� + 5�� = 900

� + 5 = ±30

� = −5 ± 30

The solutions to the quadratic equation are –35 and 25. Since � represents length, the only reasonable length

would be 25. The dimensions of the pool are 25 meters by 35 meters.

5.2 Example 31: Solve by Taking the Square Root

In order to find the stopping distance H of a car with a constant deceleration� applied when at a

speedI, the following formula can be used. H = J"��

Find the maximum speed of a car if it must stop before 30 feet and if the constant deceleration is 20

feet per second squared.

Solution:

Substitute the appropriate values into the formula:

30 = J"��@�

1200 = I�

In this problem . = 0 so taking the square root of both sides is the easiest method to use.

±√1200 = I

±34.64 ≈ I

Since negative speed does not make sense, the maximum speed is 34.64 feet per second.



5.2 Example 32: Solve by Factoring

The product of two consecutive even integers is 80. Find the integers.

Solution:

Let the first even integer be K. Then the next even integer is 2 larger:K + 2. Their product is:

K�K + 2� = 80

K� + 2K − 80 = 0 .

This can be solved by factoring. �K + 10��K − 8� = 0 .

so K = −10 or K = 8

There are two possible pairs of consecutive even integers that have a product of 80:

–10 and –8, or 8 and 10.

5.2 Example 33: Solve by using the Quadratic Formula

Remember that the equation that represents the height of an object that is thrown straight up in the air is

2 = −4.9>� + ?@> + ℎ@ when the height is given in meters.

An arrow is shot straight up from a height of 2 meters with an initial velocity of 50 m/s. At what time

will the arrow hit the ground again?

Solution:

Substitute the appropriate values into the formula:

2 = −4.9>� + 50> + 2

The height of the arrow on the ground is when 2 = 0.

−4.9>� + 50> + 2 = 0

Let’s use the quadratic formula to solve this problem.

� = −4.9,. = 50 and � = 2

� = −50 ± A�50�� − 4�−4.9��2�2�−4.9�

� = −50 ± √2539.2−9.8

� = −0.04 or � = 10.24

The arrow will hit the ground about 10.2 seconds after it is shot.

Video Resources

• CK-12 Foundation: 1009 Solving Real-World Problems Using Quadratic Equations

• - James Sousa: Solving Quadratic Equations by Factoring

• - James Sousa: Completing the Square to Solve Quadratic Equations

• CK-12 Foundation: 1009S Solving Real-World Problems Using Quadratic Equations




How many solutions are there for a quadratic equation? Remember, when you

graph a quadratic function it is a parabola and if the function is set equal to zero,

the number of x-intercepts will determine the number of real solutions the equation

( ) 0f x = will have.

Case 1: The parabola has two -intercepts. This situation occurs when the

parabola crosses the �-axis twice. Using the quadratic formula, the value inside the

square root is positive. So the two solutions are

� = +98√9"+,�� and � = +9+√9"+,��

�� .

Case 2: The parabola has one -intercept. This situation occurs when the

vertex of the parabola just touches the �-axis. This is called a repeated root,

or double root. In this case, the x-intercept is also the vertex. Using the

quadratic formula, the value inside the square root is zero.

The one solution is � = +9±√@�� = − 9

�� .

Case 3: The parabola has no -interept. This situation occurs when the

parabola does not cross the �-axis. The value inside the square root is

negative, so there are no real roots. The solutions to this type of situation are

imaginary.

The quadratic formula is� = +9±√9"+,�� . The value inside the square root, .� −

4�� is called the discriminant. It helps to determine the number of real solutions

the quadratic equation has.

The Discriminant Theorem

Given the discriminant, L = .� − 4��

• If D > 0, the parabola will have two �-intercepts. The quadratic equation will have two real solutions.

• If D = 0, the parabola will have one �-intercept. The quadratic equation will have one real solution.

• If D < 0, the parabola will have no �-intercepts. The quadratic equation will have no real solutions

but it will have two complex solutions.

If D = perfect square, the equation is factorable.

Case 1

Case 2

Case 3



5.3 Example 1:

Determine the number of real solutions to−3�� + 4� + 1 = 0.

Solution:

By finding the value of its discriminant, you can determine the number of�-intercepts the parabola has

and thus the number of real solutions.

Find the discriminant

L = .� − 4��

L = �4�� − 4�−3��1� L = 16 + 12 = 28

Because the discriminant is positive, the parabola has two real -intercepts and thus two real

solutions.

5.3 Example 2:

Determine the number of solutions to −2�� + � = 4.

Solution:

Before we can find its discriminant, we must write the equation in standard form:

−2�� + � − 4 = 0

Find the discriminant:

L = �1�� − 4�−2��−4� L = 1 − 32 = −31

The value of the discriminant is negative; there are no real solutions to this quadratic equation.

There are two complex/imaginary solutions. The parabola does not cross the -axis.

5.3 Example 3:

Determine the number of solutions for�� − 2� + 1 = 0.

Solution:


L = �−2�� − 4�1��1� L = 4 − 4 = 0

Because the discriminant is zero, the parabola has one real x-intercept and thus one real solution.



5.3 Example 4:

Marcus kicks a football in order to score a field goal. The height of the ball is given by the equation 2 =− ��

#,@@�� + � where y is the height of the ball above the ground (in feet) and x is the horizontal distance

from the ball to the goalpost. If the goalpost is 10 feet high, can Marcus kick the ball high enough to go

over the goalpost? What is the farthest distance that Marcus can kick the ball from and still make it over

the goalpost?

Solution:

We want to know if it is possible for the height of the ball to equal 10 feet at some real horizontal distance

from the goalpost.

10 = − 326400�� + �

0 = − 326400�� + � − 10


L = �1�� − 4M− 326400N �−10�

L = 1 − 0.2 = 0.8

Since the discriminant is positive, we know that it is possible for the ball to go over the goal post, if

Marcus kicks it from an acceptable distance x from the goalpost.

To find the value of x that will work, we need to use the quadratic formula:

1 0.810.56

322

6400

x− ±

= ≈

−

feet or 189.44 feet

What does this answer mean? It means that if Marcus is approximately.44 feet or approximately 10.56

feet from the goalpost, the ball will just barely go over it. Are these the only distances that will work?

No; those are just the distances at which the ball will be exactly 10 feet high, but between those two

distances, the ball will go even higher than that. (It travels in a downward-opening parabola from the

place where it is kicked to the spot where it hits the ground.) This means that Marcus will make the goal

if he is anywhere between 10.56 and 189.44 feet from the goalpost.

Vocabulary

• The discriminant is the value under the radical in the Quadratic Formula, .� − 4�� . The discriminant

gives us information to determine the number and type of solution(s) a quadratic equation has.

• If .� − 4�� O 0, the equation has two separate real solutions.

• If .� − 4�� = 0, the equation has one real solution, a double root.

• If .� − 4�� P 0, the equation has only non-real solutions (2 imaginary solutions).

Video Resources

• CK-12 Foundation: 1010 The Discriminant

• http://sciencestage.com/v/20592/a-level-maths-:-roots-of-a-quadratic-equation-:-discriminant-:-

examsolutions.html

• CK-12 Foundation: 1010S The Discriminant

• Khan Academy: Discriminant for Types of Solutions for a Quadratic

• CK-12 Basic Algebra: Discriminant of Quadratic Equations



Possible Solutions of a Quadratic Equation

2 Real Solutions 1 Real Solution

(Double Root)

No Real Solutions

2 Imaginary Solutions

�� − 4 = 0

�� = 4

� = ±2

�� + 4� + 4 = 0

�� + 2�� + 2� = 0

� = −2

�� + 4 = 0

�� = −4

� = ±24

5.4 I can represent relationships involving quadratic inequalities in multiple ways, find solutions

and interpret these solutions to solve real-world situations.

Earlier you learned to graph linear inequalities in two variables. You learned to determine whether it was a solid or

dashed line and what side should be shaded. The process for graphing a quadratic inequality is similar.

5.4 Example 1:

Graph 2 1y x> − .

Solution:

Start by graphing 2 1y x= − . Factored form of 2 1x − is

( ) ( )1 1x x− + . The x-intercepts are 1 and –1. The vertex

is halfway between these two points at

x = 0.

The parabola would be dashed as it is a > sign so the

solutions are close to the boundary, but will not be on the boundary

itself.

Pick a point not on the parabola such as (0, 0) and test it in the inequality:

( )

2

2

1

0 0 1

0 1

y x> −

> −

> −

The point (0, 0) makes the inequality true, therefore all

points within that region will make the inequality true. Shade the region

that contains the point (0, 0).

x y

–2 3

–1 0

0 –1

1 0

2 3

-3 -2 -1 1 2 3

-3

-2

-1

1

2

3

4

5

6

7

8

x

y



5.4 Example 2:

Graph 22 3 4y x x≥ − + + .

Solution:

Graph 22 3 4y x x= − + + . Use 2

b

a− to find

the x-coordinate of the vertex.

( )

3 3

2 2 4

− = −

The parabola would be solid as it is a ≥ sign.

Pick a point not on the parabola such as (0, 0) and test it in the

inequality:

( ) ( )

22 3 4

0 2 0 3 0 4

0 4

y x x≥ − + +

≥ − + +

≥

False – shade the region of the graph that does not include (0, 0).

Graphing Quadratic Inequalities in Two Variables

The general procedure for graphing quadratic inequalities in two variables is as follows:

1. Graph the parabola.

a. Standard form: y = ax2

+ bx + c where the x-coordinate of the vertex can be

found using 2

b

a−

b. Factored form: y = (x – m)( x – n), m and n are x-intercepts and the x-coordinate

of the vertex is halfway between m and n

c. Vertex form: y = (x – h)2

+ k where (h, k) is the vertex

2. Determine if the parabola should be dashed or solid.

a. Dashed: the equals sign is not included (P or O)

b. Solid: the equals sign is included (≤ or ≥)

3. Use a test point to determine which region to shade.

a. If the test point makes a true inequality, shade the region that includes the test

point.

b. If the test point makes a false inequality, shade the region that does NOT include

the test point.

Quadratic inequalities are inequalities that have one of the following forms 2 0ax bx c+ + > and 2 0ax bx c+ + < .

We can solve these inequalities by using the techniques that we have learned about solving quadratic equations.

x y

–1 –1

0 4

¾ 5.125

1 5

2 2 -3 -2 -1 1 2 3

-4

-2

2

4

6

8

x

y



5.4 Example 3:

a) Solve 2 6 0x x+ − >

b) Solve 2 6 0x x+ − <

Solution:

Consider the graph of the equation 2 6y x x= + −

Notice that the parabola intersects the x-axis at –3 and 2.

From the graph, we notice the following

• If x < –3 then y > 0

• If –3 < x < 2 then y < 0

• If x > 2 then y > 0

Therefore, 2 6 0x x+ − > whenever x < –3 or x > 2 and 2 6 0x x+ − < when –3 < x < 2.

5.4 Example 4:

Find the solution set of the quadratic inequality 2 2 8 0x x+ − > .

Solution:

To find the solution, first factor:

( ) ( )2 2 8 4 2x x x x+ − = + −

The x-intercepts are found by solving x + 4 = 0 and x – 2 = 0 which

is x = –4 and x = 2. The vertex would be halfway between x-

intercepts which is –2. The parabola opens up so you can sketch the

parabola.

So 2 2 8 0x x+ − > when x < –4 or x > 2.

Vocabulary

• Quadratic Inequality: A term describing a squared function that is specified to

be smaller or larger than a given value.

Video Resources

• James Sousa: Solving Quadratic Inequalities

• http://www.virtualnerd.com/algebra-2/quadratics/inequalities/graphing-solving-inequalities/inequality-

definition

• http://www.virtualnerd.com/algebra-2/quadratics/inequalities/graphing-solving-inequalities/graph-

inequality

Notice that:

If x < –3 then y > 0

If –3 < x < 2 then y < 0

If x > 2 then y > 0

5.1A Solving Quadratic Equations by Graphing

5.1 I can use tables and graphs to solve quadratic equations including real-world situations 175 and translate between representations.

-10 -8 -6 -4 -2 2 4 6 8 10

-10

-8

-6

-4

-2

2

4

6

8

10

x

y

-10 -8 -6 -4 -2 2 4 6 8 10

-10

-8

-6

-4

-2

2

4

6

8

10

x

y

-10 -8 -6 -4 -2 2 4 6 8 10

-12

-10

-8

-6

-4

-2

2

4

6

8

x

y

To solve the quadratic equation f (x) = 0, means to find the values of x that make the

equation true. When y = 0 in an ordered pair, the point is on the x-axis. A solution to the

quadratic equation f (x) = 0, is also called an x-intercept or zero or root.

To solve a quadratic equation ( )g x k= , where 0k ≠ , we need to find the values of x

where the function is equal to k. These values will not be on the x-axis but will be on the

horizontal line y = k.

This section provides the opportunity to explore these concepts.

1) Find the solutions to the quadratic equation f (x) = 0. The

graph of f (x) is provided to the right.

When f (x) = 0, x = _____ and x = ____.

These values of x are called the solutions to the equation

f(x) = 0.

2) Find the solutions to the quadratic equation f (x) = 7. The

graph of f (x) is provided to the right.

When f (x) = 7, x = _____ and x = ____.

These values of x are called the solutions to the equation

f(x) = 7.

3) Graph the function f (x) below. Use the graph to identify the

solutions to f (x) = 0. Verify your results algebraically.

a) ( ) 2 2 3f x x x= + − b) ( ) 22 4 2f x x x= − − −

Solution(s): ____________________ Solution(s): ____________________

� Verify: � Verify:

Section

5.1A


176 UNIT 5 – SOLVING QUADRATIC EQUATIONS (CLASSWORK)

-10 -8 -6 -4 -2 2 4 6 8 10

-10

-8

-6

-4

-2

2

4

6

8

10

x

y

-10 -8 -6 -4 -2 2 4 6 8 10

-10

-8

-6

-4

-2

2

4

6

8

10

x

y

4) Graph the function f (x) below. Use the graph to identify the solutions to f (x) = k. Verify your results

algebraically.

a) ( ) 2 9f x x x= − − , where ( ) 3f x = b) ( ) 2 2 5f x x x= − + , where ( ) 5f x =

Solution(s): ____________________ Solution(s): ____________________


#5 – 7: Use your graphing utility to solve each equation by graphing (sketch the graph). If needed, round

your answer to the nearest hundredth.

5) 2 5 12x x− = 6) 23 20 12x x= − − 7) 2 4 9x x− + = −

Solution(s): Solution(s): Solution(s):



8) Given each graph of ( )y f x= , determine how many real solutions the equation ( ) 0f x = would have and

state those solutions.

Number of real solutions: _____ Number of real solutions: _____ Number of real solutions: _____

Solution(s): _________________ Solution(s): _________________ Solution(s): _________________

9) Given each graph of ( )y f x= , identify the number of real solutions for the equation ( ) 4f x = . Name the

real solutions.

Number of real solutions: _____ Number of real solutions: _____ Number of real solutions: _____

Solution(s): _________________ Solution(s): _________________ Solution(s): _________________

5.1B Answering Real-World Questions by Graphing Quadratic Functions


1) While playing basketball this weekend Frank shot an air-ball. The ball left Frank’s hands at a height of 8

feet with an initial velocity of 24 ft/sec. Use a graphing utility, as needed. Round solutions to the nearest

hundredth.

a) Draw a diagram to model this situation.

b) Write the function that models the height of the ball in feet, after t seconds.

c) What is the maximum height of the ball?

d) How long does it take for it to reach that height?

e) What is the height of the ball, ℎ�>�, when it hits the ground?

f) How long will it take to hit the ground?

2) Abigail wants to make a wish while throwing a coin off a bridge into a stream. When she throws it, the coin

is 112 feet above the water with an initial velocity of 50 ft/sec. Use your graphing utility, as needed. Round

solutions to the nearest hundredth.

a) Draw a diagram to model this situation. Label the diagram.

b) Write the equation that represents the distance the coin is above the water, after t seconds.

c) What is the greatest height of the coin?

d) How much time will it take for the coin to hit the water?

Section

5.1B

Projectile Motion Model: ( ) 20 0

1

2h t gt v t h= − + +

( )

0

0

height of the object

time in seconds

height of the object after seconds

initial (starting) velocity that the object is thrown;

if the object is dropped, 0

32 feet/seconforce of gravity

h

t

h t t

v

v

g

=

=

=

=

=

=

2

2

0

d

9.8 meters/second

initial (starting) heighth

=



3) The profits of Mr. Unlucky’s company can be represented by the equation 23 18 4y x x= − + − , where y is the

amount of profit in hundreds of thousands of dollars and x is the number of years of operation. He realizes

his company is on the downturn and wishes to sell before he ends up in debt. Use your graphing utility, as

needed. Round solutions to the nearest hundredth.

a) When will Unlucky’s business show the maximum profit?

b) What is the maximum profit?

c) At what time will it be too late to sell his business? (When will he start losing money?)

4) Mr. Jackson had a rectangular shaped garden where the length was 2 feet less than the width. The area of the

garden was 420 square feet. Use your graphing utility, as needed. Round solutions to the nearest hundredth.

a) Draw a picture that represents the situation described above. Label the diagram.

b) What is the formula for the area of a rectangle?

c) Write an equation to represent the area of the garden.

d) Write the equation in standard form.

e) Sketch the graph, showing the important points.

f) What are the dimensions of the garden?



5) Jocelyn and Kelly are team captains for a Destination Imagination competition. The teams are supplied with

rocket assembly kits and have time to design and build their rocket. After 1 hour, all teams will launch their

rockets to see which rocket will reach the highest height. Height will be measured in feet, and time will be

measured in seconds since launch. The height of the rocket from Jocelyn’s team can be modeled by the

function ( ) 216 180J t t t= − + . The height of the rocket from Kelly’s team can be modeled by the function

( ) 216 240K t t t= − + . Use your graphing utility, as needed. Round solutions to the nearest hundredth.

a) Whose rocket won the Destination Imagination competition? Explain how you determined your answer.

b) After how many seconds does each rocket land?

c) How long after the first rocket landed, did the second rocket land?

5.2A Factoring Review

5.2 I can represent real-world situations with quadratic equations and solve using appropriate 181 methods, find real and non-real complex roots when they exist, and recognize that a

particular solution may not be applicable in the original context.

Work through the problems below and identify the key characteristics of the expression

that give you clues on to how to factor it. Record the key characteristics, the factoring

method, and then factor completely.

1) 212 14x x− + 2) 2 81x −

3) 2 14 49x x+ + 4) 2 11 24x x+ +

5) 22 13 15x x− + 6) 26 15 36x x− −

7) How do you find the x-intercepts once an equation is in factored form. Use 2 11 24x x+ + , from problem 4)

above to help you explain your method.

Section

5.2A

5.2B Solving Quadratic Equations by Factoring: Part I


The Zero Product Property states (in algebraic terms, of course!) something that we all have known for a long time: if the product of two numbers is 0, then at least one of the factors is 0.

Zero Product Property:

If 0ab = , then either 0a = or 0b = , or both a and b are 0.

#1 – 2: Complete each table of information.

1) Table of values:

x

y

Factored (intercept) form of the

equation:

( ) ( )5 3y x x= − +

Solutions for the equation:

( ) ( )0 5 3x x= − +

Graph: x-intercept(s):

Verify the solution(s):

2) Table of values:

x

y

Factored (intercept) form of the

equation:

( ) ( )1 5y x x= − −

Solutions for the equation:

( ) ( )0 1 5x x= − −

Graph:

x-intercept(s):

Verify the solution(s):

3) Write two or more conjectures about the relationship of the information in the charts above.

Section

5.2B




#4 – 7: Write each equation in factored form and use it to find the x-intercepts for each quadratic

equation, check your answers, and then match each equation with the graphs pictured.

4) 2 9y x= − 5) 2 4y x= − +

x-intercepts: ______________ x-intercepts: _______________ � Verify: � Verify:

Matches with graph _______ Matches with graph _________

6) 2 4 12y x x= + − 7) 22 4 6y x x= − + +

x-intercepts: ______________ x-intercepts: _______________ � Verify: � Verify:

Matches with graph _______ Matches with graph _________

D C

B A



#8 – 13: Solve by Factoring. Verify your solution(s).

8) 2 2 15 0x x− − = 9) 22 5 3 0x x+ − = 10) 227 3 0t − =

11) 225 20 0n n+ = 12) 22 7 15x x+ = 13) 212 7 10x x+ =

14) Find the x-intercepts by factoring. 23 17 10y x x= − +

5.2C Solving Quadratic Equations by Factoring: Part II



1) A relief package is released from a helicopter at 1650 feet. The height of the

package can be modeled by the equation 216 1650h t= − + , where h is the

height of the package in feet and t is the time in seconds. The pilot wants to

know how long it will take for the package to hit the roof of a building 50 feet off

the ground.

a) Write the equation that you are trying to solve. ______________________

b) Solve the equation by factoring.

2) A certain number added to its square is 30. Find the number.

3) The height of a flare fired from the deck of a ship in distress can be modeled by ( ) 216 104 56h t t t= − + + ,

where h is the height of the flare above the water in feet and t is the time in seconds. Find the time it takes

the flare to hit the water.

a) Write the equation that you are trying to solve. _______________________________________


c) What is the domain and range? (Take in account the context of the situation.)

4) The height of a triangle is 5 less than its base. The area of the triangle is 42 square inches. Find its base and

height.

a) Draw a picture to represent the situation.

b) Write the equation that you are trying to solve. ______________________

c) Solve the equation by factoring.

d) What is the domain and range? (Take in account the context of the situation.)

Section

5.2C



5) The product of two consecutive odd integers is 99. Find the integers.



c) Do both of these answers make sense in this problem? Why or why not?

6) The height of a rocket launched upward from a 188 foot cliff is modeled by ( ) 216 48 188,h t t t= − + + where

h is the height in feet and t is the time in seconds. The rocket landed in a tree, 28 feet off the ground.




7) The length of a rectangle exceeds its width by 4 inches. Find the dimensions of the rectangle this describes.

Its area is 96 square inches.

a) Draw a picture.

b) Write the equation that you are trying to solve. _______________________________________


d) Do both of these answers make sense in this problem? Why or why not?

8) Robert threw a rock off a bridge into the river. The distance from the rock to the river is modeled by the

equation ( ) 216 16 60h t t t= − − + , where h is the height in feet and t is the time in seconds. Find how long it

took the rock to hit the surface of the water.







9) Joe’s rectangular garden is 6 meters long and 4 meters wide. He wishes to double the area of his garden by

increasing its length and width by the same amount. Find the number of meters by which each dimension

must be increased.

a) Draw a picture.

b) Write the equation that you are trying to solve. _______________________________________


d) Do both of these answers make sense in this problem? Why or why not?

10) In a trapezoid, the smaller base is 3 more than the height, the larger base is 5 less than 3 times the height, and

the area of the trapezoid is 45 square centimeters. Find, in centimeters, the height of the trapezoid.

a) Draw a picture.

b) What is a formula you could use to help with this problem? _____________________________

c) Write the equation that you are trying to solve. _______________________________________

d) Solve the equation by factoring.

e) Do both of these answers make sense in this problem? Why or why not?

5.2D Operations with Radical Expressions


Choose positive values for a and b and use decimal approximations in a calculator to

explore whether the following statements are true or false. Repeat with at least one other

set of values for a and b. Justify your answer. Caution: finding some values for which the

statement is true does not make the statement true. However, finding a single example

where the statement is not true proves the entire statement false.

When completed, compare your results with other groups in class. Make a conjecture for each statement, based on

this data.

Equation

Conjecture

First set of values:

a = b =

Second set of values:

a = b =

Equation

Conjecture

1) a b a b⋅ = ⋅

True False (circle one)


a b a b⋅ = ⋅

True False

2) a b a b+ = +



a b a b+ = +

True False

3) a b a b− = −



a b a b− = −

True False

4) a a

b b=



a a

b b=

True False

5) 2 2a b a b+ = +



2 2a b a b+ = +

True False

6) 3 4a a a+ =



3 4a a a+ =

True False

Section

5.2D




Simplifying radicals

A very common radical expression is a square root. One way to think of a square root is the number that will

multiply by itself to create a desired value. For example: √2 is the number that will multiply by itself to equal 2.

And in like manner √16 is the number that will multiply by itself to equal 16. In this case the value is 4 because

4 x 4 = 16. Because 4 x 4 = 16, 16 is a perfect square. When the square root of a perfect square number is taken

you get a whole number value. Otherwise, an irrational number is produced.

Example 1:

Simplify 20

Answer:

20 4 5 2 5= ⋅ = Answer: 2 5

20 is equivalent to 2 5 but 2 5 is considered to be in simplest radical form.

#1 – 6: Simplify the following. Show your work.

1) 20 2) 50 3) 18

4) 75 5) 48 6) 700

Simplifying within a radical

When there is an expression inside of the square root, the radical acts like a parenthesis and all computation must

occur prior to taking the square root.

Example 2:

Simplify 25 13−

Answer:

25 13 12 4 3 2 3− = = ⋅ = Answer: 2 3

#7 – 9: Simplify the following. Show your work.

7) 100 36− 8) 50 5− 9) 46 10−



Adding and subtracting involving radicals

Just as variables can only be added to like variables, square roots can only be added to like square roots.

Example 3: Example 4:

Simplify 3 2 5 2+ Simplify 3 5 5 2+

Solution: Solution:

3 2 5 2 8 2+ = This expression is already simplified. Why?

#10 – 17: Simplify (some of these may already be simplified).

10) 2 7 4 7+ 11) 12 18+ 12) 20 45−

13) 3 5− 14) 7 7− 15) 12 16−

16) (4 2 3) (5 4 3)− + + 17) (9 5 6) (2 6)− − −

Products and quotients involving radicals:

Things inside and outside of radicals cannot simply be multiplied or divided.

Example 5: Example 6:

Simplify 3 12 Simplify 32

2

Solution: Solution:

3 12 3 4 3 3 2 3 6 3= ⋅ = ⋅ = 32 16 2 4 2

2 22 2 2

⋅= = =

#18 – 20: Simplify (some of these may already be simplified).

18) 4 20 19) 72

9 20)

4 28

2

+




#21 – 28: Simplify each expression.

21) ( )2

3 5 22) ( )2

10 7− 23) ( )2

2 12− 24) ( )3 2 7 5− 25) ( )( )3 2 3 4 3 3+ +

26) ( )2

2 3 6+ 27) ( )2

4 2 7− 28) ( ) ( )2

2 5 2 3 8 5 2 3− + + +

#29 – 30: Verify that the given answer (value of x) is a solution to the equation.

29) 22 6 48; 3 3x x− = = − 30) 22 12 18 16; 3 2 2x x x− + = = +




A B

F G H

L M N

I J K

C D E

O P

5.2E Solving Quadratic Equations Using Square Roots to Find Rational Solutions



Absolute value can be defined as the from zero on a number line.

The solutions to 3x = are values that are ____ units away from _____.

1) Solve the following: 8x =

2) Solve the following: 3 7x − =

3) Solve the following: 34 2 2y− =

4) Is ( )f x x= a function?

Create a table of values and graph the points.

x x

0

1

–1

4

–4

9

–9

Section

5.2E

-10 -8 -6 -4 -2 2 4 6 8 10

-10

-8

-6

-4

-2

2

4

6

8

10

x

y



5) The square root or ______________________ sign √stands for the _________________square root of a

real number.

6) The √4 stands for the positive value or ____________.

7) To find the negative square root you need to write −√4, which equals____________.

8) Complete the following table of values and graph the points.

9) What conclusions can you draw about the solutions to ( ) 2f x x= when compared to the solutions of

( )f x x= ?

x 2x

0

1

–1

4

–4

9

–9

Absolute Value Square Root Theorem

For all real numbers , √�� = |�|




#10 – 15: Solve the following equations. Check your solutions.

10) 2 72x = 11) 23 18 0x − =

12) ( )2

2 3 16x − = 13) 24 78 29x − = −

14) ( )2

5 4 32x + − = 15) ( )2

2 1 5 3x− − + =

16) A square and a triangle have the same area. The triangle has base of 7 cm and a height of 6 cm. What is the

length of a side of the square?

5.2F Solving Quadratic Equations Using Square Roots to Find Real Solutions


#1 – 10: Solve by using square roots. Leave your answer in simplest radical form.

Check your solutions.

1) 23 108x = 2) 23 90x =

3) 22 5 41x + = 4) 2 12 87x− − = −

5) 2

6 48

x− + = − 6) ( )

22 3 8x − =

7) ( )2

3 2 18x− + = − 8) 26 240 0x − =

9) ( )21

6 74

x − = 10) ( )2

2 5 81x − =

Section

5.2F




#11 – 12: As you work to answer the following questions, use the falling object model, 2

016h t h= − += − += − += − + ,

where h is the height from the ground, t is the time in seconds, and 0h is the initial height from

the ground in feet.

11) An apple drops from the top of a tree that is 32 feet tall.

a) Identify the value of 0h : ___________________________

b) How long will it take for the apple to reach the ground? Round your answer to the nearest tenth of a

second.

12) A stunt woman working on the set of an action movie is to fall out of a window 150 feet above the ground.

For the stunt woman’s safety, an air cushion 28 feet wide by 36 feet long by 9 feet high is positioned on the

ground below the window.

a) Identify the value of 0h : ___________________________

b) For how many seconds will the stunt woman fall before she reaches the cushion? Round your answer to

the nearest tenth of a second.

c) A movie camera operating at a speed of 27 frames per second records the stunt woman’s fall. How many

frames of film show the stunt woman falling?

5.2G Operations with Complex Expressions


1) What does 9 mean? Write a complete sentence.

2) Solve using the square root property: 2 36 0x − =

3) Solve using the square root property: 2 1 0x + =

In order to solve some of the equations you will come upon throughout the rest of your math career, the number

system needs to be expanded from the REAL number system to the COMPLEX number system which

includes imaginary numbers.

Substitute the solutions in the original equation:

4) Simplify the following.

a) i b) 2i c) d) e) 5i f) 6i

5) How can you find the value of ni ?


a) 7i b) 22i c) 56i d) 79i

Section

5.2G

An imaginary number, when squared, gives a negative result.

1i = −= −= −= −

It helps us answer the question: 2 1x = −= −= −= −

i 2 = –1

3

2

___

____

i i i i

i i

i

= ⋅ ⋅

= ⋅

= ⋅

=

4

2 2

___ ___

____

i i i i i

i i

= ⋅ ⋅ ⋅

= ⋅

= ⋅

=

2

2

1

1

1

1 or 1

or

x

x

x

x x

x i x i

= −

= −

= −

= − = − −

= = −

( ) ( )

( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( )

2 2

2 2

2

2

check the solution check the solution

1 1

1 1

1 1

1 1 1

1 1

1

x i x i

x x

i i

i i i

i i

i i

i

= = −

= − = −

= − − = −

= − − − = −

− − = −

= −

= −




7) Joe and Sophie were given the following problem: Simplify 9− and 9− . Joe says that both answers

are –3. Sophie thinks Joe is wrong. Who is correct? Explain your thinking.

8) a) Complete the statement: 2i = _______ .

Consider the following simplification: ( ) ( ) ( )2

2 1 1 1 1 1 1 1i = − = − − = − ⋅ − = =

Notice, this does not equal the value of i2.

b) Identify the error from the simplification shown above?

c) Important point to remember:

____________________________________________________________

Remember that a b a b⋅ = ⋅ . We can use this property with i to simplify square roots of negative numbers.

9) Simplify 49− .


a) 4− b) 90− c) 5 80− d) 3 225− −

e) 4 9− ⋅ − f) 2 6− ⋅ − g) 3 8− ⋅ − h) 10 5− ⋅ −

A complex number is any number that can be written in the standard form a + bi, where a and b are real

numbers and i is the imaginary unit.

We can perform operations with complex numbers.


a) ( ) ( )7 3 9 12i i− + − + b) ( ) ( )6 5 8i i+ − −

c) ( ) ( )3 1 2 81+ − − − − d) ( ) ( )11 12 2 27+ − + − −

12) Explain how to add or subtract complex numbers in your own words.




a) 7 4i i⋅ b) 4 7 5 3− − ⋅ c) 2 3 3 12− − ⋅ −

d) ( )3 1 5i+ e) ( )4 6 16− − − f) ( )10 2 5− + −


a) ( ) ( )7 4 3 5i i+ − − b) ( )2

1 8+ − c) ( )2

5 5 4 3i− −

d) 8

2

i e)

9 36

3

− − f)

14 21

7

i+

15) Belle is getting ready to take a test on this learning target. What advice would you give her about

simplifying and performing operations on complex numbers?

5.2H Solving Quadratic Equations Using Square Roots to Find Real or Complex

Solutions



Not all quadratic equations have real number solutions. Solve the following equations

using the square root method to find their real or complex solutions. Verify your

solutions using a calculator.

1) ( )2

2 3 8x − = 2) ( )2

3 2 18x− + = −

3) ( )2

2 35 81 0x − + = 4) ( )21

8 74

x − =

5) ( )2

5 7 135x − = − 6) ( )2

1 24 75x + − =

7) ( )2

4 5 9x + = − 8) ( )2

2 6 45 53x− − − =

Section

5.2H

5.2I (ext.) Solving Quadratic Equations by Completing the Square to Find Rational

Solutions


Juan and Stephen solved the following equations by completing the square. When they

checked their answers they realized they had made a mistake. Describe and correct the

error(s) in their work.

2

2

2

2

2

Juan's Work

Line 1 5 24 0

Line 2 5 24

Line 3 5 25 24 25

Line 4 ( 5) 49

Line 5 ( 5) 49

Line 6 5 7

Line 7 5 7 or 5 7

Line 8 5 7 or 5 7

Line 9 2

x x

x x

x x

x

x

x

x x

x x

x

+ − =

+ =

+ + = +

+ =

+ =

+ =

+ = + = −

= − + = − −

= or 12x = −

( )

( )

2

2

2

2

Stephen's Work

Line 1 6 16

Line 2 6 9 16

Line 3 3 16

Line 4 3 16

Line 5 3 4

Line 6 3 4 or 3 4

Line 7 7 or 1

x x

x x

x

x

x

x x

x x

− =

− + =

− =

− =

− =

− = − = −

= = −

#1 – 4: Solve each equation by completing the square.

1) 27 6x x= + 2) 2 3

4x x− =

Section

5.2I ext

5.2I (ext.) Solving Quadratic Equations by Completing the Square to Find Rational

Solutions



#1 – 4 (continued): Solve each equation by completing the square.

3) 2 3 4x x− = 4) 2 4 12 0x x+ − =

5) The product of two consecutive positive odd integers is 783. Write an equation that models this

situation and then find the two integers by completing the square.

5.2J(ext.) Solving Quadratic Equations by Completing the Square to Find Real

Solutions


Juan and Stephen are at it again! They solved the following equations by completing the

square. When they checked their answers they realized they had made a mistake.

Describe and correct the error(s) in their work.

2

2

2

2

2

Juan's Work

Line 1 2 4 0

Line 2 2 4

Line 3 2 1 4 1

Line 4 ( 1) 5

Line 5 ( 1) 5

Line 6 1 5

Line 7 1 5 or 1 5

Line 8 1 5 or 1 5

x x

x x

x x

x

x

x

x x

x x

− − =

− =

− + = +

+ =

+ =

+ =

+ = + = −

= − + = − −

( )

( )

2

2

2

2

Stephen's Work

Line 1 14 11

Line 2 14 49 11

Line 3 7 11

Line 4 7 11

Line 5 7 11

Line 6 7 11 or 7 11

Line 7 7 11 or 7 11

x x

x x

x

x

x

x x

x x

− =

− + =

− =

− =

− =

− = − = −

= + = −


1) 2 2 9x x+ = 2) 2 12 38x x− =

Section

5.2J ext

5.2J(ext.) Solving Quadratic Equations by Completing the Square to Find Real

Solutions



#1 – 6 (continued): Solve each equation by completing the square.

3) 2 3 1 0x x+ − = 4) 2 4 2 35x x x− = +

5) 22 10 7x x− = − 6) 23 7 4x x+ = −

7) A foul ball leaves the end of a baseball bat and travels according to the

function 2( ) 16 64 3h t t t= − + + where h is the height of the ball in feet and t

is the time in seconds. How long will it take for the ball to reach a height of

58 feet in the air? Solve by completing the square.

5.2K(ext.) Solving Quadratic Equations by Completing the Square to Find Real or

Complex Solutions


Juan and Stephen thought they would give it one more try. They solved the following

equations by completing the square. When they checked their answers they realized

once again that they had made a mistake. Describe and correct the error(s) in their work.

( )

( )

2

2

2

2

Stephen's Work

Line 1 8 15

Line 2 8 16 15

Line 3 4 15

Line 4 4 15

Line 5 4 15

Line 6 4 15 or 4 5

Line 7 4 15 or 4 i 15

x x

x x

x

x

x

x x

x i x

− = −

− + = −

− = −

− = −

− = −

− = − − = −

= + = −

( )

( )

2

2

2

2

Juan's Work

Line 1 6 14

Line 2 6 9 14 9

Line 3 6 5

Line 4 6 5

Line 5 6 5

Line 6 6 5 or 6 i 5

Line 7 6 5 or 6 i 5

x x

x x

x

x

x i

x i x

x i x

+ = −

+ + = − +

− = −

− = −

− =

− = − = −

= + = −


1) 2 20 104 0x x+ + = 2) 2 4 29x x− = −

3) 2 16 76 0x x− + = 4) 2 17 200 13 43x x x− + = −

5) 23 24 1x x− = −

Section

5.2K

ext

5.2L Solving Quadratic Equations Using the Quadratic Formula to Find Real

Solutions



In sections 5.2I, J and K you solved quadratic equations by completing the square for

each equation separately. By completing the square once for the general equation2 0ax bx c+ + = , you can develop a formula that gives the solutions of any quadratic

equation. This formula is called the quadratic formula. The derivation of this formula

can be found in the resource section at the beginning of this unit with a link to a video of

this process.

The Quadratic Formula

Let a, b, and c be real numbers such that a ≠ 0. The solutions

of the quadratic equation 2 0ax bx c+ + = are:

2 4

2

b b acx

a

− ± −=

Use the quadratic formula to solve the following equation.

1) Solve 29 6 1x x+ =

Step 1: Write the equation in standard form.

Step 2: Identify the values of a, b and c.

a =________ b = __________ c = __________

Step 3: Substitute the values from Step 2 into the quadratic formula and simplify.

2 4

2

b b acx

a

− ± −= =

Juan and Stephen are still working hard. Again, each made at least one mistake. Find them and correctly solve

the problem.

Juan a = ______ b = ______ c = ______ Stephen a = ______ b = ______ c = ______

2�2 − 5� − 12 = 0

� =5 ± A−5� − 4�2��−12�

2�2�

=5 ± √−25 + 96

4

=5 ± √71

4

2�2 − 5� − 12 = 0

� =−5 ± A�−5�� − 4�2��−12�

2�2�

=−5 ± √25 − 96

4

=−5 ± √−71

4

=−5 ± 4√71

4

Section

5.2L


Solutions


#2 – 5: Use the quadratic formula to find the solutions to the following equations. Can any of these be

solved by factoring?

2) 2 5 14 0x x− − = 3) 2 3 2 0x x+ − =

4) 2 10 22x x+ = − 5) 22 9 7x x+ =

6) The length l (in feet) of runway needed for Air Bob to land his plane is given by 20.1 3 22l s s= − + , where s

is the plane’s speed (in feet per second). If Air Bob is landing his plane on a runway 2000 feet long, what is

the maximum speed at which the pilot can land? Round your answer to the nearest tenth. Graph it to verify

your answer.

7) An arrow is shot into the air. A function representing the relationship between the number of seconds it is in

the air, t, and the height of the arrow in meters, h, is given by: 2( ) 4.9 29.4 2.5h t t t= − + + How long does it

take for the arrow to hit the ground? Round your answer to the nearest tenth. Graph it to verify your answer.

5.2M Solving Quadratic Equations Using the Quadratic Formula to Find Real or

Complex Solutions



Answer Equation Answer Equation

An

swer

Eq

ua

tion

There are 12 cards included on these two

pages. Each card has two halves. One

half contains a quadratic equation; the

other half contains an answer for an

equation from a different card. Students will cut out the

cards along the bold border, solve each equation, and create a “snake”

pattern with an equation from one card being followed by its’ respective

answer on another card. The equation at the end of the snake will match with the answer at the

beginning of the snake.

An example of how the cards will be organized adjacent to each other is shown above and to the right.

Section

5.2M


Complex Solutions



5.2N Solving Quadratic Equations – Choosing the Best Method: Part I



Throughout unit 5, you have explored and worked with several methods for solving a

quadratic equation. Being able to identify which of those methods to use when working

with a situation that is modeled by a quadratic equation is an important part of

developing your mathematical knowledge of quadratic functions. The next few

examples will compare the various methods.

1) Solve the following equation using each method designated. Identify the solution to

the equation, and then answer the question in the end of the problem.

2 2 3 0x x+ − =

Graph:

Complete the Square:

Factor:

Quadratic Formula:

Solution(s):

Which method was the most efficient for this problem and why?

Section

5.2N



2) Solve the following equation choosing two different methods to use. Identify the solution(s) to the equation,

and then answer the question at the end of the problem.

23 8 3x x− =

Method 1: ________________________ Method 2: ________________________

Solution(s): Why did you choose the two methods that

you did, and which do you feel is more

efficient?

3) For the equation, determine an effective method for solving the quadratic equation and explain why you

chose that method. Solve the quadratic equation with the method of your choice, keeping the answer exact

(no decimal approximations). 27 350 0x x+ =




#4 – 5: Determine an answer for each situation. Be sure to clearly record your thinking.

4) The product of two consecutive integers is 156. Find the two numbers.

5) In the shot-put competition at an international track meet, the trajectory of Sharon Day’s best shot-put throw

is given by the function ( ) 20.0186 5h x x x= − + + , where � is the horizontal distance the shot travels, and

ℎ is the corresponding height of the shot above the ground

(both measured in feet).

a) How far did the shot go?

b) What was the greatest height obtained?

c) In the context of the situation explain the meaning of the 5 in the given function.

5.2O Solving Quadratic Equations – Choosing the Best Method: Part II


#1 – 5: Four equations and one situation are given. Solve one by the square root

property, one by factoring, one by completing the square, one using the

quadratic formula, and one by graphing. Express your answers in simplest

radical form. Verify your answer.

1) 2 2 15 0x x− − = 2) 2 6 9 8x x− + = 3) 22 7 15x x= − + 4) 2 12 20x x+ =

5) The Epic Toothpick Company can produce up to 700 cases of toothpicks per day. The company’s profit P,

in thousands of dollars, is given by the formula ( )2

3 5P x= − + , where x represents the number of hundreds

of cases produced. How many hundreds of toothpicks would result in a profit of $230,000?

Solve by: Square Root Property

#______ equation: ________________________

Solve:

Solution(s): ______________________

� Verify that your answer(s) are solution(s)

Solve by: Factoring

#______ equation: ________________________

Solve:

Solution(s): _____________________


Solve by: Completing the Square

#______ equation: ________________________

Solve:

Solution(s): ______________________


Solve by: Using the Quadratic Formula

#______ equation: ________________________

Solve:

Solution(s): _____________________


Section

5.2O




Solve by: Graphing

#______ equation: ________________________

Solution(s): ______________________

� Verify that your answer(s) are solution(s).

6) a) Solve �� − 2�� = 7.

b) Sketch the graphs of 2 = �� − 2�� − 5 and 2 = 2.

c) Describe the connections between the solution in

part a) and the graph in part b).

5.3A Number and Type of Solutions: Part I


#1 – 8: Use the quadratic formula to solve the following equations. State the type

and number of solutions. Look for patterns between the calculations and

the number and type of solutions.

1) 2 24 5 3 2 3x x x x+ + = − 2) 2 14 5c c− =

Number of solutions: _________ Number of solutions: _________

Type of solutions: ___________ Type of solutions: ___________

3) 2 2 1 0x x− + = 4) 23 4 8 0b b+ + =



5) 24 12 9 0x x+ + = 6) 2 2 1 0q q− − =



Section

5.3A


5.3 I can determine the number of real and non-real solutions for a quadratic equation. 221

#1 – 8 (continued): Use the quadratic formula to solve the following equations. State the type and number

of solutions. Look for patterns between the calculations and the number and type of solutions.

7) 2 4 3m + = 8) 23 7 4 0t t+ + =



9) What patterns did you notice in #1 – 8? How could you determine the number and type of solutions without

solving?

Value of the

Discriminant

POSITIVE

___________ 0>

ZERO

___________ 0=

NEGATIVE

___________ 0<

Example

showing number

and type of root

of

2 0ax bx c+ + =

2 6 5 0x x+ + =

If the discriminant is ___________________ , there are two real roots.

If the discriminant is

___________________ , the

two roots are ____________.


___________________ , the

two roots are ____________.

2 2 1 0x x− + =


___________________ ,

there is one real root.

This is also called a

________________ root.

2 3 10 0x x− + =


___________________ , the

two roots are

_______________.

Graph indicating

x-intercepts

2y ax bx c= + +



#10 – 13: Without solving (simplifying the equation is encouraged)…

• determine the number of solutions

• determine the type of solutions for each quadratic equation.

10) 2 7 33 8 3p p p+ + = − 11) 27 2 5 0x x+ + =

12) 2 22 10 4 3y y y y+ = + − 13) 24 9 13z z+ = −

14) Four possible graphs of f (x) are shown below. Consider the equation f (x) = 0. State whether the

discriminant of each quadratic equation is positive, negative, or equal to zero on the line below the graph.

a)

b) Identify which graph above matches with each discriminant below.

Discriminant A:

( ) ( ) ( )2

2 4 1 2− −

Graph #: ______

Discriminant B:

( ) ( ) ( )2

4 4 1 4− − − −

Graph #: ______

Discriminant C:

( ) ( ) ( )2

4 4 1 0− −

Graph #: ______

Discriminant D:

( ) ( ) ( )2

8 4 1 13− − − −

Graph #: ______

Graph #1 Graph #2 Graph #3 Graph #4

_____________________ ______________________ _______________________ ______________________

5.3B Number and Type of Solutions: Part II


From the work we did in section 5.3A, we found the following:

You can use the sign of the discriminant, 2 4b ac− , to determine the number of real

solutions to a quadratic equation in the form 2 0ax bx c+ + = , where 0a ≠ . If the

equation has a positive discriminant, there are two real solutions. If the equation has a

negative discriminant, there are no real solutions (two complex solutions). Finally, if the

equation has a discriminant that equals zero, then there is only one real solution.

1) Find a possible discriminant value for the following quadratic situations.

a) two real solutions

b) two real, rational solutions

c) one real solution

d) two complex

2) Given the following information about a quadratic function, determine the possible number and type of

solutions to the equation f(x) = 0.

a) y-intercept: 7 AND opens: up

b) maximum: (–1, 5) AND y = 3

c) minimum: (0, 3)

d) y-intercept: –3 AND opens: up

e) y-intercept: 5 AND x-intercept: 3

f) minimum: (–2, 0)

3) Three possible graphs of ( )f x are shown below. Consider the equation ( ) 0f x = . Identify whether the

discriminant would be positive, negative, or zero. a b c

a) ___________________

b) ___________________

c) ___________________

Section

5.3B



#4 – 6: The graph of (((( ))))f x is shown to the right.

4) What type of discriminant does the equation ( ) 2f x =

have?

a) Discriminant: positive, negative, or zero (circle one)

b) Number and type of solutions _______________

c) Can you determine the solutions from the graph? If

so, state the solutions ______________________

5) The graph of ( )f x is shown to the right. What type of

discriminant does the equation ( ) 2f x = − have? Write an equation to model the graph pictured.

a) Discriminant: positive, negative, or zero (circle one)


c) Can you determine the solutions from the graph? If so, state the solutions ________________________

6) The graph of ( )f x is shown to the right. What type of discriminant does the equation ( ) 4f x = − have?

a) Discriminant ____________________________


c) Can you determine the solutions from the graph? If so, state the solutions _______________________

7) A quadratic equation is given by 3x2 + bx + 12 = 0.

a) Choose a value for b such that the equation has two real solutions. Explain your reasoning.

b) Choose a value for b such that the equation has two imaginary solutions. Explain your reasoning.

c) Choose a value for b such that the equation has one real solution. Explain your reasoning.

d) Choose a value for b such that the equation has two rational solutions and could be solved by factoring.

5.4A Graphing Quadratic Inequalities

5.4 I can represent relationships involving quadratic inequalities in multiple ways, find solutions and 225 interpret these solutions to solve real-world situations.

To the right is the graph of the

equation 2 = −�� − 4� − 3

#1 – 4: Using the graph of the given equation, complete the graph for each of the following inequalities by

adding the appropriate boundary and shading.

1) 2 4 3y x x> − − − 2) 2 4 3y x x≥ − − −

2) 2 4 3y x x< − − − 4) 2 4 3y x x≤ − − −

x y

–4 –3

–3 0

–2 1

–1 0

0 –3

Section

5.4A



5) What features should be indicated in the graph of an inequality that do not appear in the graph of an

equation?

6) Graph the following inequalities.

( )2

2 1 4y x≥ + −

x-intercepts:____________

Vertex: _____________

Solid or Dashed Boundary Line?

Make a table of values:

x y

Test point for solutions.

( ) ( )2 2y x x> − + −

x-intercepts:____________

Vertex: _____________

Solid or Dashed Boundary Line?

Make a table of values:

x y

Test point for solutions.

-8 -6 -4 -2 2 4 6 8

-8

-6

-4

-2

2

4

6

8

x

y

-8 -6 -4 -2 2 4 6 8

-8

-6

-4

-2

2

4

6

8

x

y

5.4B Solving Quadratic Inequalities


-5 5

-5

5

x

y

-5 5

-5

5

x

y

Throughout the Functions unit, a strong emphasis was put upon determining and

investigating the value of a function. Throughout this section, the value of the function

becomes even more important as we solve quadratic inequalities.

1) Solve the quadratic equation ( )2

2 4x − = . Think: What will make this equation true?

Thinking about this equation graphically, graph the equation

( )2

2y x= − and on the same axes, graph 4y = . Notice, these are the

expressions from the two sides of the equation.

a) Upon graphing these two equations, what do you notice about the

graph?

b) Solution(s) identified by the graph: _____________________

2) Using the information from 1), solve the inequality ( )2

2x − < 4 .

Thinking about this inequality, we need to identify where the function

values of ( )2

2x − are less than 4?

a) Use a graphing strategy to solve. Graph the equation

( )2

2y x= − and on the same axes, graph 4y = .

b) On the graph, highlight the function values of ( )2

2x − that are

less than 4.

c) Think about the intersection points of the two graphs. Write one

or two sentences about the importance of the intersection point(s).

d) Graph on the number line, above and to the right, the values that satisfy the inequality. These are called

critical values.

e) Solve the inequality ( )2

2x − < 4 . _________________________________

Section

5.4B



-8 -6 -4 -2 2 4 6 8

-8

-6

-4

-2

2

4

6

8

x

y

-8 -6 -4 -2 2 4 6 8

-8

-6

-4

-2

2

4

6

8

x

y

3) Consider the inequality 2 4 3y x x≥ − + .

a) Write a sentence describing the meaning of the

inequality 2 4 3 0x x− + ≤ .

b) Find the solutions of 2 4 3 0x x− + ≤ .

Step 1: Graph the equation ( ) 2 4 3f x x x= − + .

Step 2: Look for the x-values for which ( ) 0f x ≤ .

c) State the solutions of 2 4 3 0x x− + ≥ .

(Keep in mind what the inequality means.)

d) Solve the quadratic inequality 2 4 3 0x x− + > .

4) Solve 2 8 12 0x x− + ≤

Locate the x-values for which ( ) 0f x ≤ .

a) What is the significance of the x-values of 2 and 6?

b) How could you have identified those two points

(critical values) without graphing?

c) Solve the inequality 2 8 12 0x x− + ≤



Many times, as inequalities become more complicated, it becomes easier to solve them algebraically.

5) Solve 2 5 2 4x x− + ≥ − .

In other words, locate the x-values for which ( ) 4f x ≥ − .

Graphical Solution Algebraic Solution

6) Solve the inequality both graphically and algebraically ( )2

2 1x − ≥ .

Graphical Solution Algebraic Solution



Part I: Solve each quadratic equation using the method listed in the first

column.

As a class, discuss strengths and weaknesses of each method and identify what

type of equation using this method of solving a quadratic would be most

efficient.

METHOD Example Strengths/Weaknesses When to Use

Graphing

20 8 12x x= − +

Factoring

2 7 10 0x x− + =

Square Root

Property

( )2

2 5x − =

Complete the

Square

2 4 18 0x x− + =

Quadratic

Formula

22 5 3 0x x− + =

-2 2 4 6

-4

-2

2

4

x

y

Unit 5

Review


Review Material 231

Part 2: Sort these equations into the category method you believe is best.

2( )1 8x − = 2 10 9 0x x− + = 25 8 1 0x x+ − = 2 10 8 0x x− + =

2 4 7 0x x+ + = 2 3 5 0x x+ + = 2 ( 2) 4x − = − 2 11 10 0x x− + =

2 6 3 0x x+ − = ( )2

4 25x + = 2 2 9 0x x− − = 24 4 1 0x x+ + =

2 6 8 0x x+ + = 22 4 1 0x x− + = 2 3 3 0x x+ + = ( )2

3 5x + =

Square

Root

Property:

1. 2.

3. 4.

Factoring:

1. 2.

3. 4.

Complete

the

Square:

1. 2.

3. 4.

Quadratic

Formula:

1. 2.

3. 4.



#1 – 3: The graph of a quadratic function is shown. Consider the equation f(x) = 0. Determine if the

discriminant is positive, zero, or negative.

1) 2) 3)

#4 – 9: Simplify these solutions:

4) 4 64

4x

− ±= 5)

6 24

4x

±= 6)

24 40

10x

± −=

7) 4 121

20x

±= 8)

9 27

21x

± −= 9)

12 20

10x

±=

#10 – 16: Use the given equation to answer the related questions.

10) 2 4 18 0x x− + =

a) Value of the discriminant: ___________________

b) Can this equation be factored? ______Explain._______________________________________

c) Number and type of solutions: (Multiple Choice) ________

[A] 2 Real & Rational Numbers

[B] 2 Real & Irrational Numbers

[C] 1 Real Number

[D] 2 Complex Numbers

d) Solve by graphing, using the square root property, factoring, completing the square, or by using

the quadratic formula.


Review Material 233

#10 – 16 (continued): Use the given equation to answer the related questions.

11) 23 7 6 0x x− − =



c) Number and type of solutions: (Multiple Choice)



[C] 1 Real Number


d) Solve by graphing, using the square root property, factoring, completing the square, or by using the

quadratic formula.

12) 2 10 9x x− = −






[C] 1 Real Number



quadratic formula.

13) 22 36 0x − =






[C] 1 Real Number



quadratic formula.



#10 – 16 (continued): Use the given equation to answer the related questions.

14) 2 2x x+ =






[C] 1 Real Number



quadratic formula.

15) 24 1 4x x+ =






[C] 1 Real Number



quadratic formula.

16) 2 5 14x x+ =






[C] 1 Real Number



quadratic formula.


Review Material 235

17) During a game of golf, Kayley hits her ball out of a sand trap. The height of the golf ball is modeled

by the equation ( ) 216 20 4h t t t= − + − , where h is the height in feet and t is the time in seconds since

the ball was hit.

Find how long it takes Kayley's golf ball to reach ground level.

a) Write the equation that you are trying to solve. _____________________________________________

b) Solve the equation by graphing, using the square root property, factoring, completing the square, or by

using the quadratic formula.

18) The height of a rock thrown off a cliff can be modeled by the equation ( ) 216 8 120h t t t= − − + ,

where h is the height in feet and t is the time (in seconds) since the rock was thrown. How long does

it take the rock to reach the ground?


b) Solve the equation by graphing, using the square root property, factoring, completing the square, or by

using the quadratic formula.



19) The height of an object launched vertically is given by 20 0( ) 16h t t v t h= − + ⋅ + where:

?@ = initial velocity ℎ@ = initial height

One of the games at a carnival involves trying to ring a bell with a ball by hitting a lever that propels

the ball into the air. The initial velocity is 40 ft/sec as the ball leaves the ground.

a) Write an equation that models the height of the ball at time t.

b) Find the height of the ball after 1 second.

c) What is the maximum height of the ball? After how many seconds does the ball reach this height?

d) The bell is 28 ft. above the ground. Will the ball ring the bell? Explain your reasoning.

e) Your mom took a photo of you at this game. When you looked at the photo later, the ball was at 16 ft high in the photo. How long after you hit the ball did your mom take the photo?

f) In part e), you should have gotten two answers. Do both of these answers make sense in the context of the situation? Explain your reasoning.

WORK SPACE

NO

GR

AP

HIN

G C

AL

CU

LA

TO

R N

O G

RA

PH

ING

CA

LC

UL

AT

OR

N

O G

RA

PH

ING

CA

LC

UL

AT

OR

Name ______________________________ Period __________


5.1 I can use tables and graphs to solve quadratic equations including real-world situations 237

and translate between representations.

1) What does “find the zeros of the function” mean?

2) When you are solving a quadratic equation f (x) = 0 by graphing, what do you look for on the graph?

#3 – 5: Determine for each quadratic function, (((( )))) 0f x ==== , the number and type of solutions. If possible, list

the zeros of the function.

3) 4) 5)

# of solutions: __________ # of solutions: __________ # of solutions: _____________

type of solutions: ________ type of solutions: _______ type of solutions: ___________

zeros: _________________ zeros: ________________ zeros: ____________________

#6 – 7: The graph of (((( ))))f x is provided. Find the solutions of the quadratic equation (((( )))) 0f x ==== . Verify the

solutions.

6) ( ) 2 2 15f x x x= + − 7) ( ) 22 8f x x x= − +

Solution(s): Solution(s):


x

y

x

y

x

y

-7 -6 -5 -4 -3 -2 -1 1 2 3 4 5

-18

-16

-14

-12

-10

-8

-6

-4

-2

2

4

6

8

x

f(x)

-3 -2 -1 1 2 3 4 5 6

-4

-3

-2

-1

1

2

3

4

5

6

7

8

x

f(x)

Name ______________________________ Period __________


238 UNIT 5 – SOLVING QUADRATIC EQUATIONS (PRACTICE)

#8 – 9: Graph the function (((( ))))f x . Find the solutions of the quadratic equation (((( )))) 0f x ==== . Verify the

solutions.

8) ( ) 2 4 12f x x x= − + + 9) ( ) 2 3 10f x x x= + −

Solution(s): Solution(s):


#10 – 15: Use a graphing utility to solve each equation by graphing. If needed, round your answer to the

nearest hundredth. Question #13 – 15, verify that the values truly are solutions.

10) 2 7 11x x− = 11) 26 19 15x x= − − 12) 25 7 3 8x x− − =


13) 218

2x x− = 14) 2 4 6x x+ = 15) 22 2 5 0x x− − =


� Verify: � Verify: � Verify:

Name ______________________________ Period __________




#16 – 21: Graph the function (((( ))))f x with your graphing utility. Find the solution(s) of the quadratic

equation (((( )))) 0f x ==== . Check the solutions.

16) ( ) 2 6 5f x x x= + + 17) ( ) 25 30 25f x x x= + + 18) ( ) 23 18 15f x x x= + +


19) ( ) 2 8f x x x= + 20) ( ) 214

2f x x x= + 21) ( ) 22 16f x x x= +


22) Investigation: ( ) 2 6 5f x x x= + +

a) Looking to Question #16 – 18, record the following:

� Function in #16 ____________________ Solutions in #16 _______________________



b) Looking to Question #19 – 21, record the following:

� Function in #19 _____________________ Solutions in #19 _______________________



c) Comparing the functions in questions 16, 17, and 18, and then again in 19, 20, and 21, write a conjecture

about the relationship of the functions within each set of questions and the solutions of those functions.

-6 -5 -4 -3 -2 -1 1

-20

-18

-16

-14

-12

-10

-8

-6

-4

-2

2

x

f(x)

-6 -5 -4 -3 -2 -1 1

-20

-18

-16

-14

-12

-10

-8

-6

-4

-2

2

x

f(x)

-10 -8 -6 -4 -2 2

-20

-16

-12

-8

-4

4

x

f(x)

-10 -8 -6 -4 -2 2

-20

-16

-12

-8

-4

4

x

f(x)

-6 -5 -4 -3 -2 -1 1

-20

-18

-16

-14

-12

-10

-8

-6

-4

-2

2

x

f(x)

-10 -8 -6 -4 -2 2

-20

-16

-12

-8

-4

4

x

f(x)

Name ______________________________ Period __________



23) A bottlenose dolphin jumps out of the water. The path the dolphin travels can be modeled by the function

( ) 20.2 2h d d d= − + , where h represents the height, in feet, of the dolphin and d represents the horizontal

distance, in feet, the dolphin traveled.

a) Sketch a graph of the quadratic equation.

b) What is the maximum height the dolphin reaches? Where is this represented on the graph of the

function?

c) What is the horizontal distance that the dolphin jumps? Where is this represented on the graph of the

function?

Section 5.1A

d

h(d)

Name ______________________________ Period __________




#1 – 10: Use your graphing utility to solve the following problems.

1) Phillip, Javier and Meg each throw a ball over a fence. The height of Phillip’s ball with respect to time can

be modeled by the equation ( ) 216 60P t t t= − + . The height of Javier’s ball with respect to time can be

modeled by the equation ( ) 216 50J t t t= − + . The height of Meg’s ball with respect to time can be modeled

by the equation ( ) 216 40M t t t= − + , where each function gives the height in feet at t seconds.

a) Phillip, Javier and Meg want to know whose ball hit the ground first. Javier thinks that they should find

the x-intercept(s) of the graphs to determine this. Phillip thinks that they should find the vertex of each

graph to find which ball hit the ground first. Which one is correct? Explain your answer.

b) Whose ball hit the ground first? How long did it take?

c) Whose ball hit the ground second? How long did it take?

2) A quarterback throws a football at an initial height of 5.5 feet with an initial upward velocity of 35 feet per

second. The height of a tossed ball with respect to time can be modeled by the quadratic function

( ) 20 016h t t v t h= − + ⋅ + where

0v is the initial upward velocity, 0h is the initial height and ( )h t is the

height of the ball after t seconds.

a) Write the function that models the height of the ball with respect to time.

b) How high will the football be after 1 second? (Consider what the 1 second represents.)

c) When will the football be 10 feet high? (Consider what the 10 feet represents.)

d) When will the football reach its maximum height? (When graphing the function, consider what

significant feature of the graph represents this concept.)

e) What is the maximum height of the football?

f) When will the football hit the ground if no one catches it? (When graphing the function, consider what

significant feature of the graph represents this concept.)

Name ______________________________ Period __________



#1 – 10 (continued): Use your graphing utility to solve the following problems.

3) Suppose a batter hits a baseball, and the height of the baseball above the ground can be modeled by the

function ( ) 216 50 2h t t t= − + + . Where is the vertex of the graph? Explain the meaning of the vertex in the

context of this situation.

4) A pool is treated with a chemical to reduce the amount of algae. The amount of algae in the pool t days after

the treatment begins can be approximated by the function ( ) 24 88 500A t t t= − + . How many days after

treatment begins will the pool have the least amount of algae?

5) The driver of a car traveling downhill on a road applied the brakes. The speed of the car, ( )s t , in kilometers

per hour t seconds after the driver started going downhill is modeled by the function ( ) 24 12 80s t t t= − + + .

a) After how many seconds did the car reach its maximum speed?

b) What was the maximum speed reached?

c) How long will it take the car to stop?

Name ______________________________ Period __________




#1 – 10 (continued): Use your graphing utility to answer the following problems.

6) Andrew has 100 feet of fence to enclose a rectangular tomato

patch. He wants to find the dimensions of the rectangle that

encloses the most area. The width of the rectangle can be found

by the expression 50 – L where L is the length of the rectangle.

a) In the expression representing the width of the rectangle (50 – L), what does the 50 represent? Explain

your thinking clearly.

b) Write a function to model the area of the rectangle. A(L) represents the area of the rectangular tomato

patch base on the length (L) of one side.

( ) _____________________A L =

c) Find the coordinate representing the maximum of the graph. Explain its meaning in the context of the

situation.

d) What size should Andrew make the tomato patch in order to enclose the most area within the fencing?

7) Sharon needs to create a fence for her new puppy. She purchased 40 feet of fencing to enclose the four sides

of a rectangular play area.

a) Write an expression (in terms of L) to represent the width of the rectangle.

b) Write a function to model the area of the play area.

c) What are the dimensions of the enclosure that will create the greatest area for her puppy to play?

????

50 – L

L

L

Name ______________________________ Period __________



#1 – 10 (continued): Use your graphing utility to answer the following problems.

8) Karen is throwing an orange to her brother Jim, who is standing on the balcony of their home. The height, h

(in feet), of the orange above the ground t seconds after Karen throws the orange is given by the function

( ) 216 32 3h t t t= − + + . If Jim's outstretched arms are 16 feet above the ground, will the orange ever be high

enough so that he can catch it? Explain your answer.

9) On wet concrete, the stopping distance, s (in feet), of a car traveling v miles per hour is given by

( ) 20.055 1.1s v v v= + . At what speed could a car be traveling and still stop at a stop sign 30 feet away?

10) The Buckingham Fountain in Chicago shoots water from a nozzle at the base of the fountain. The height, in

feet, of the water above the ground t seconds after it leaves the nozzle is given by ( ) 216 90 15h t t t= − + + .

a) What is the maximum height of the water spout to the nearest tenth of a foot?

b) How long does it take for the water to hit the ground?

Section 5.1B

Name ______________________________ Period __________


5.2 I can represent real-world situations with quadratic equations and solve using appropriate 245

methods, find real and non-real complex roots when they exist, and recognize that a


#1 – 12: Factor out the greatest common factor (GCF) for each polynomial and write in factored form.

1) 2 6x + 2) 3 9y − 3) 7 28a + 4) 36 12z −

Ans: _________________ Ans: _________________ Ans: _________________ Ans: _________________

5) 2b b+ 6) 22r r− 7) 29t t+ 8) 24 5n n−

Ans: _________________ Ans: _________________ Ans: _________________ Ans: _________________

9) 24 12h h+ 10) 29 27x x− 11) 22 4a a+ 12) 220 24d d−

Ans: _________________ Ans: _________________ Ans: _________________ Ans: _________________

#13 – 27: Factor each polynomial.

13) 2 13 42x x+ + 14) 2 6 9x x+ + 15) 2 12 32x x+ +

Ans: ____________________ Ans: ____________________ Ans: ____________________

16) 2 3 10x x+ − 17) 2 10 25x x− + 18) 2 12x x− −

Ans: ____________________ Ans: ____________________ Ans: ____________________

19) 23 4x x+ − 20) 22 5 12x x+ − 21) 24 12 9x x− +

Ans: ____________________ Ans: ____________________ Ans: ____________________

2x + 6 3y – 9 7a + 28 36z – 12

b2 + b 2r – r2 9t2 + t 4n2 – 5n

Name ______________________________ Period __________



Hei

gh

t

Time

Height of Tossed Ball

#13 – 27 (continued): Factor each polynomial.

22) 212 8 1x x− + 23) 22 8 10x x− − 24) 23 21 18x x+ +

Ans: ____________________ Ans: ____________________ Ans: ____________________

25) 2 9a − 26) 2 16x − 27) 225 36x −

Ans: ____________________ Ans: ____________________ Ans: ____________________

#28 – 29: The following quadratic equations are written in standard form. Convert them to factored form.

28) 2 3 2y x x= + + 29) 2 49y x= −

30) Why is factored form of a quadratic function also called intercept form?

#31 – 32: Convert the following quadratic equations to factored form and identify the x-intercepts.

31) 2 24 80y x x= − + 32) 2 9 10y x x= + −

factored form: _________________________ factored form: _________________________


33) The parabola graphed below shows the height of a ball tossed into the air.

a) Draw an arrow at the location of the graph that would

represent when the ball hits the ground.

b) Explain why you placed the arrow at that location.

Name ______________________________ Period __________





#1 – 3: Solve for x.

1) ( ) ( )4 9 0x x− + = 2) ( ) ( )2 3 6 0x x− − = 3) ( ) ( )4 3 2 5 0x x+ − =

#4 – 17: Solve the equation by factoring. Verify your solution(s).

4) 24 36 0x − = 5) 25 20 0x − = 6) 23 9 0x − =


7) 27 28 0x x− = 8) 2 8 9 0x x+ − = 9) 2 7 12 0x x+ + =


Name ______________________________ Period __________



#4 – 17 (continued): Solve the equation by factoring. Verify your solution(s).

10) 2 10 25 0x x− + = 11) 2 3 4x x− = 12) 2 4 5x x− =

�Verify: � Verify: � Verify:

13) 2 13 40x x− = − 14) 23 10 8 0x x+ + = 15) 28 6 5 0x x+ − =


16) 25 11 2x x+ = − 17) 22 15 8x x− =


Section 5.2B

Name ______________________________ Period __________





#1 – 9: Solve the following application problems.

1) One leg of a right triangle is 1 foot longer than the other leg. The hypotenuse is 5 feet. Draw a diagram. Find

the dimensions of the right triangle. (Hint: a2 + b2 = c2)

� Verify:

2) One leg of a right triangle is 7 feet longer than the other leg. The hypotenuse is 13. Draw a diagram. Find

the dimensions of the right triangle.

� Verify:

3) A rectangle has sides of 2x + and 1x − . What value of x gives an area of 108? Draw a diagram.

� Verify:

4) A rectangle has sides of 1x − and 1x + . What value of x gives an area of 120? Draw a diagram.

� Verify:

5) The product of two positive numbers is 120. Find the two numbers if one number is 7 more than the other.

� Verify:

Name ______________________________ Period __________



#1 – 9 (continued): Solve the following application problems.

6) A rectangle has a 50-foot diagonal. What are the dimensions of the rectangle if it is 34 feet longer than it is

wide? Draw a diagram.

� Verify:

7) Two positive numbers have a sum of 8, and their product is equal to the larger number plus 10. What are the

numbers?

� Verify:

8) The product of two negative integers is 24. The difference between the integers is 2. Find the integers.

� Verify:

9) Framing Warehouse offers a picture framing service. The cost for framing a picture is made up of two parts:

glass costs $20 per square foot and the frame costs $10 per linear foot. If the frame has to be a square, what

size picture can you frame, using a 6 inch wide frame, if you have $200 to spend?

� Verify:

Section 5.2C x

x

Name ______________________________ Period __________





1) Simplify each expression.

a) 40 b) 80 c) 200

d) 3 20 e) 5 24 f) 6 98

2) Add or subtract each expression.

a) ( ) ( )5 3 4 3− + + b) ( ) ( )4 5 2 2 6 2+ + + c) ( ) ( )6 8 7 4 2 7− − +

d) ( )8 3 5 2− + e) ( )2 5 3 5− + + f) ( ) ( )6 5 12 5 12+ + −

Name ______________________________ Period __________



3) Simplify each using two different methods.

a) 2 18⋅ b) 8 2⋅

1st method 2nd method 1st method 2nd method

2 18⋅ 2 18⋅

8 2⋅ 8 2⋅


a) 5 5⋅ b) 5 18⋅ c) 6 3⋅

d) 3 20 2 3⋅ e) 5 2 8 2− ⋅ f) 2 6 3 7− ⋅

Name ______________________________ Period __________






a) ( )5 3 7+ b) ( )3 2 3− − c) ( )2 1 4 3+

d) ( )5 2 3 2+ e) ( )2 3 18− + f) ( )3 5 8− −


a) ( ) ( )4 3 4 3+ + b) ( )2

2 3 2− c) ( ) ( )2

5 2 3 6 2 3− + −

d) ( ) ( )4 1 3 2 1 3 2+ + e) ( )2

3 4 2 7− f) ( ) ( )2

3 2 11 5 2 11 7− − − +

Name ______________________________ Period __________



7) Verify that the given answer (value of x) is a solution to the equation.

a) 2 7 25; 3 2x x+ = = b) ( )2 14 50 3; 6 2x x x− + = = + c) ( )22 20 40; 5 3 5x x x− = = −


a) 6 15

3

a + b)

6 15 2

3

+ c)

14 10 6

2

− − d)

25 15 40

5

− +


2 3 572 3 6 0;

4x x x

+− − = =

Section 5.2D

Name ______________________________ Period __________





#1 – 6: Solve each equation using the square root property and verify each answer.

1) 2 4x = 2) 22 32a = 3) 23 8 67m − =


4) ( )2

1 36x − = 5) ( )2

3 16 0x + − = 6) ( )2

2 2 3 21x − + =


Name ______________________________ Period __________



7) A physics teacher drops an object from an initial height of 64 feet. The height of the

ball (in feet) h at time t (in seconds) can be modeled by the equation ( ) 216 64h t t= − + .

How long does it take the ball to reach the ground?

�Verify:

8) The stopping distance d (in meters) that a car needs to come to a

complete stop when traveling at speed x (in km/h) can be modeled by

the equation ( ) ( )2

0.009 15 3xd x= + + . On a certain road, drivers

cannot see a stop sign until they are approximately 20 meters away.

What is the maximum speed that should be posted in order to allow

cars enough room to stop in time? Round your answer to the nearest

whole number and verify your solution.

9) A missing leg of a right triangle can be found using the Pythagorean Theorem: 2 2 2a b c+ = , where a and b

are the legs of the triangle and c is the hypotenuse of the triangle (the side directly across from the right

angle). Andy is trying to find the missing leg of the triangle below that represents the distance that the

person is from a flagpole. The flag pole is 12 feet tall and he knows that the distance from the person to the

top of the flagpole is 15 feet. Andy has started the problem by putting the values into the formula. Help him

find the solution.

2 2 2a b c+ =

( ) ( )2 2212 15b+ =

Section 5.2E

Name ______________________________ Period __________





1) On a quiz, Omar solved a quadratic equation and got the answer wrong. His work is shown below. Identify

his mistake and then solve the equation correctly to find the real solution.

#2 – 5: Verify that each of the following values are solutions to the given equation. Show ALL of your

work.

2) 22 3 21; 3, 3 x x x+ = = = − 3) ( )2

5 1 17; 9, 1 x x x− + = = =

4) 2 7 35; 2 7, 2 7 x x x+ = = = − 5) ( )2

3 5 70; 3 5, 3 5 x x x+ − = = − + = − −

Name ______________________________ Period __________



6) Samantha solved the following problem on a test and got the right answer. Unfortunately, she doesn’t know

which answer is the actual solution. Explain to her which solution is correct and why.

#7 – 10: Solve each equation for real solutions and simplify your answers. Verify your solutions.

7) 2 3 21 x + = 8) ( )2

1 32 x − =


Not sure which is

right???

HELP!

5) The height h of a water balloon (in feet) at time x (in seconds) is given by the equation.

( ) ( )2

16 0.65 10h x x= − − + . If a student throws the balloon and it hits a student who is 6

feet tall in the head, how long was the balloon in the air?

Name ______________________________ Period __________





#7 – 10 (continued): Solve each equation for real solutions and simplify your answers. Verify your

solutions.

9) 22 8 0 x − = 10) ( )2

5 1 3 42 x − − =


#11 – 14: Find the roots of each function and simplify your answers. Verify your solutions.

11) ( ) 2 75f x x= − 12) ( ) ( )2

2f x x= +


Name ______________________________ Period __________



#11 – 14 (continued): Find the roots of each function and simplify your answers. Verify your solutions.

13) ( ) ( )2

2 1 18f x x= − − 14) ( ) 23 24f x x −=


15) The height of a ball in the air, h, at time t can be modeled by the equation

( ) ( )2

16 1 32h t t= − − + . How long does it take for the ball to reach the

ground? Round answers to the nearest hundredth.

16) Big Bertha, a cannon used in WW1, could fire shells incredibly long distances. The path of a shell could be

modeled by ( )2

0.5 3 12y x= − − + where x was the horizontal distance traveled (in miles), and y was the

height (in miles). How far could Big Bertha fire a shell? Round your answer to the nearest mile.

Section 5.2F Big Bertha (Paris Gun), courtesy

http://www.militaryimages.net

Name ______________________________ Period __________






a) 2i b) 4i c) 17i

d) 10i− e) 101i f) 87i

g) 25− h) 4 9− i) 5 28−

j) 7 8− k) 4 10− − l) 5 100−

2) Add or subtract each expression.

a) ( ) ( )5 3 4 7i i− + + b) ( ) ( )4 8 9 2i i− + + c) ( ) ( )3 2 5 4i i− − −

d) ( )6 5 9− − − − e) ( )8 9 11i i+ − f) ( ) ( )7 81 5 100− − + − −


a) 3 2i i⋅ b) 6 2i i− ⋅ c) 4 6 2 3− ⋅

d) 3 20 2 5− − ⋅ e) 8 2 3 2− ⋅ f) 2 6 3 3− − ⋅ −

Name ______________________________ Period __________



4) Kasem simplified the expression 4 9− ⋅ − using various methods.

1 WRONG 2 CORRECT 3 CORRECT

Line 1 4 9

Line 2 4 9

Line 3 36

Line 4 6

− ⋅ −

− ⋅ −

( )

2

Line 1 4 9

Line 2 1 4 1 9

Line 3 1 4 1 9

Line 4 1 1 4 9

Line 5 4 9

Line 6 36

Line 7 1 6

Line 8 6

i i

i

− ⋅ −

− ⋅ ⋅ − ⋅

− ⋅ ⋅ − ⋅

− ⋅ − ⋅ ⋅

⋅ ⋅ ⋅

⋅

−

−

( )

2

Line 1 4 9

Line 2 1 4 1 9

Line 3 1 4 1 9

Line 4 2 3

Line 5 2 3

Line 6 6

Line 7 6 1

Line 8 6

i i

i i

i

− ⋅ −

− ⋅ ⋅ − ⋅

− ⋅ ⋅ − ⋅

⋅ ⋅ ⋅

⋅ ⋅ ⋅

⋅

⋅ −

−

a) Other than simplifying to the different numeric values, identify differences in the process shown used in

methods 1 and 2.

b) Although the methods have created the same numeric value, identify differences in the process shown

through methods 2 and 3.

c) When simplifying the multiplication of radicals that contain negative values under each radical, write a

summary statement identifying important points to consider when simplifying.


a) 6 2− ⋅ − b) 3 5 2 8− ⋅ − c) 3 2 7 9− − ⋅ −

Name ______________________________ Period __________






a) ( )5 1 3i− b) ( )3 3 i+ c) ( )3 2 4 25− −

d) ( )3 2 4i i− + e) ( )5 8 2i i− − + f) ( )10 2 2− − −


a) ( ) ( )3 5 3 5i i+ + b) ( )2

3 4 2i− c) ( )2

5 3 2i−


a) ( ) ( )4 1 3 2 1 3 2i i+ + b) ( )2

3 4 2 7i− c) ( ) ( )2

3 2 11 5 2 11 7i i− − − +

Name ______________________________ Period __________




a) 22 6 56; 5x x i− + = = b) ( )2 8 30 5; 4 3x x x i− + = = + c) ( )2 2 48 2; 1 3 5x x x i− + = = −

10) Simplify each expression (no decimal values allowed).

a) 6 8

2

i+ b)

6 10 2

2

i+ c)

15 21 6

3

i− − d)

30 75

5

− + −


2 32 6 8 3;

2

ix x x

+− + = =

Section 5.2G

Name ______________________________ Period __________


Solutions




#1 – 6: Review - Simplify the following radicals.

1) 9− 2) 7 50− 3) 2 3−

4) 3 100− 5) 11 18− 6) 121−

7) What is the value of i2?

#8 – 13: Review - Simplify the following complex expressions.

8) ( )2

3i 9) ( )2

5i− 10) ( )2

i−

11) ( )2

3 5i− 12) ( )2

2 3 4i+ 13) ( )2

2 7 5i− − +

#14 – 17: Verify that each of the following values are solutions to the given equation. Show all of your

work.

14) 22 3 21; 3 , 3x x i x i− + = = = − 15) ( )2

5 1 17; 5 4 , 5 4x x i x i− − = − = + = −

Name ______________________________ Period __________


Solutions


#14 – 17 (continued): Verify that each of the following values are solutions to the given equation. Show all

of your work.

16) 2 11 7; 2 , 2x x i x i+ = = = − 17) ( )2

2 25; 2 5 , 2 5x x i x i+ = − = − + = − −

#18 – 21: Solve each equation for real or complex solutions. Verify your solutions.

18) 2 3 51x + = 19) ( )2

1 24x − = −


Name ______________________________ Period __________


Solutions




#18 – 21 (continued): Solve each equation for real or complex solutions. Verify your solutions.

20) 23 27 0x − = 21) ( )2

5 1 3 48x + − = −


22) The height, h, of a water balloon (in feet) at time t (in seconds) is given by the equation

( ) ( )2

16 0.45 32h t t= − − + . If a student throws the balloon and it lands on the ground, how long is the

balloon in the air? Verify your solution(s).

#23 – 26: Find the real or complex roots of each function. Verify your solutions.

23) ( ) 2 125f x x= − 24) ( ) ( )2

7f x x= +


Name ______________________________ Period __________


Solutions


#23 – 26 (continued): Find the real or complex roots of each function. Verify your solutions.

25) ( ) ( )2

2 2 18f x x= − − − 26) ( ) 24 24f x x= +


27) The area of a square can be found using the formula 2A s= , where A is the area and s is the length of one

side. If the area of a square is 50 square inches, what is the length of one side? Round your answer to the

nearest thousandth and verify your solution(s).

28) The function ( ) 2 4f x x= + has no x-intercepts, as shown in

the graph to the right. Use algebra to show that no real roots

exist for this function.

Section 5.2H

Name ______________________________ Period __________

5.2I (ext) Solving Quadratic Equations by Completing the Square to Find Rational

Solutions




#1 – 3: Solve using square roots.

1) ( )2

2 25x − = 2) ( )2

4 16x + = 3) ( )2

121 7x= −

4) Find and explain the error made when solving the following equation, then solve it correctly.

2 10 24x x+ =

( )

( )

2

2

2

2

Line 1 10 24

Line 2 10 25 24

Line 3 5 24

Line 4 5 24

Line 5 5 24

Line 6 5 2 6

Line 7 5 2 6 and 5 2 6

Line 8 5 2 6 5 2 6

Line 9 5 2 6

x x

x x

x

x

x

x

x x

x x

x

+ =

+ + =

+ =

+ =

+ =

+ = ±

+ = + = −

= − + = − −

= − ±

#5 – 6: Fill in the missing value to create a perfect square trinomial. Then solve by completing the square.

5) 240 ______ 6 ______x x+ = + + 6) 2 18 ______ 88 ______x x− + = +

#7 – 10: Solve by completing the square. Then verify your solutions.

7) 229 28x x= + 8) 2 10 56 0x x− − =


Name ______________________________ Period __________

5.2I (ext) Solving Quadratic Equations by Completing the Square to Find Rational

Solutions


#7 – 10 (continued): Solve by completing the square. Then verify your solutions.

9) 223 12 13x x− = − + 10) 2 7 30 74x x+ = −


11) The product of two consecutive positive even integers is 528. What are the numbers?

Solve by completing the square.

12) A square garden is altered so that one dimension is decreased by 3 yards, while the other dimension is

increased by 5 yards. The area of the resulting rectangle is 20 square yards. Find the length of each side of

the original garden and its area. Solve by completing the square.

Section 5.2I

Name ______________________________ Period __________

5.2J (ext) Solving Quadratic Equations by Completing the Square to Find Real Solutions





1) ( )2

7 15x + = 2) ( )2

9 12x − = 3) ( )2

32 5x −


4) 27 ______ 5 ______x x+ = + + 5) 2 7 ______ 3 ______x x− + = − +


6) 288 28x x− = + 7) 22 8 13 21x x− − = −


8) 28 4 4 13x x= + − 9) 29 1 6x x− =


Name ______________________________ Period __________

5.2J (ext) Solving Quadratic Equations by Completing the Square to Find Real Solutions


10) For which values of y, given 2

2

2

by x bx

= + + , will you find 2 real solutions? 1 real solution?

11) A pool measuring 12 meters by 16 meters is to have a sidewalk installed

all around it, increasing the total area to 285 square meters. What will be

the width of the sidewalk? Draw a diagram. Solve by completing the

square and round your solution to the nearest hundredth.

12) The height h in feet of an arrow shot upward from the top of a 96-foot tall tower when

time t = 0 is given by ( ) 216 80 96h t t t= − + + . How long will it take the arrow to

strike the ground? Solve by completing the square and round your solution to the

nearest hundredth.

13) The height h in feet of a bottle rocket launched from a deck 8 feet above the ground is given by

( ) 216 240 8h t t t= − + + , where t is the time in seconds.

a) What is the height after 2 seconds?

b) At what times will the rocket be at a height of 400 feet? Solve by

completing the square and round your solution to the nearest hundredth.

Section 5.2J

Name ______________________________ Period __________

5.2K (ext) Solving Quadratic Equations by Completing the Square to Find Real or

Complex Solutions





1) ( )2

5 144x − = − 2) ( )2

7 24x − = − 3) ( )2

128 13x− = +


4) 240 ______ 12 ______x x− + = + + 5) 2 24 ______ 216 ______x x− + = − +

Name ______________________________ Period __________


Complex Solutions



6) 24 5 4x x x= + + 7) 2 4 20 0x x− + =


8) 26 23 10 26x x+ = + 9) 25 8 6x x− = +


Name ______________________________ Period __________


Complex Solutions




10) Emma hits a golf ball off the tee. The height of the ball is given by ( ) 20.02 5h x x x= − + where h is the

height in yards above the ground and x is the horizontal distance from the tee in yards. How far does Emma

hit the ball? Solve by completing the square and round your solution to the nearest hundredth.

11) Gail and Veronica are fixing a leak in a roof. Gail is working on the roof and Veronica is tossing up supplies

to Gail. When Veronica tosses up a tape measure, the height h, in feet, of the object above the ground t

seconds after Veronica tosses it is ( ) 216 32 5h t t t= − + + . Gail can catch the object any time it is above 17

feet. When does Gail catch the tape measure? Solve by completing the square and round your solution to

the nearest hundredth. Explain how both of these solutions could make sense.

12) For which values of y, given

2

2

2

by x bx

= + +

, will you find 2 complex solutions?

13) Solve the following quadratic for x by completing the square.

What is the result? 2 0ax bx c+ + =

Section 5.2K

Name ______________________________ Period __________


Complex Solutions



Name ______________________________ Period __________


Solutions




1) Describe a situation where you would have to use the quadratic formula to solve a quadratic equation if you

did not want to graph it.

#2 – 7: Determine a, b, and c and then solve using the quadratic formula. Remember to show all work.

2) 22 5 3 0x x− − = 3) 2 7 9 0x x− + =

a: b: c: a: b: c:

Could you have solved by factoring? Explain. Could you have solved by factoring? Explain.

4) 25 3 1x x+ = 5) 2 1 0x x+ − =

a: b: c: a: b: c:

Could you have solved by factoring? Explain. Could you have solved by factoring? Explain.

Name ______________________________ Period __________


Solutions


#2 – 7 (continued): Determine a, b, and c and then solve using the quadratic formula. Remember to show

all work.

6) 29 6 1 0x x+ − = 7) 22 3 2 3x x+ + =

a: b: c: a: b: c:

8) A cliff diver jumps up and away from the cliff as he jumps. His path can be modeled

by the equation ( ) 216 12 25h t t t= − + + , where h is the height, in feet, of the diver at

a specific time, t, in seconds. How long will it take for the diver to reach the water

below? Solve using the quadratic formula. Round answers to the nearest hundredth.

9) The volcanic cinder cone Puu Puai in Hawaii was formed in 1959 when a massive “lava fountain” erupted at

Kilauea Iki Crater, shooting lava hundreds of feet into the air. When the eruption was most intense, the

height h (in feet) of the lava t seconds after being ejected from the ground could be modeled by

( ) 216 352h t t t= − + . Solve using any method you have learned. Round your answers to the nearest

hundredth.

a) How long was the lava in the air?

b) How long did it take the lava to reach its maximum height of 1936 feet?

Section 5.2L

Name ______________________________ Period __________


Complex Solutions




#1 – 3: Review: Simplify the following radicals.

1) 12− 2) 24 3) 16−

#4 – 9: The following solutions were found using the quadratic formula but are not simplified completely.

Put the solutions in simplest form. No decimals allowed!

4) 2 4

2x

− ±= 5)

6 0

4x

− ±= 6)

3 4

2x

± −=

7) ( )

6 11

2 2x

− ± −= 8)

( )2 20

2 1x

− ± −= 9)

( )2 64

2 1x

− ±=

10) Carter solved the following quadratic equation using the quadratic formula – his work is shown below.

However, he did not simplify his answer correctly. Find his mistake(s) and then simplify the solution

correctly.

( ) ( ) ( )

( )

+ + =

= = =

− ± −=

− ± −=

− ±=

= − ±

2

2

Line 1 6 10 0

Line 2 1, 6, 10

6 6 4 1 10Line 3

2 1

6 4Line 4

2

6 2Line 5

2

Line 6 3 2

x x

a b c

x

x

ix

x i

Name ______________________________ Period __________


Complex Solutions


#11 – 13: Solve using the quadratic formula. Be sure to simplify your answer, keeping your answers exact

(no decimals approximations). Remember to show all work.

11) 2 6 2x x+ = − 12) 22 8 8x x− = − 13) 25 13 9 0x x− + =

14) The path of an object thrown straight up in the air with an initial velocity of 40 feet per second and from an

initial height of 4 feet can be modeled by the equation ( ) 216 40 4,h t t t= − + + where h is the height of the

object at time t. Use the quadratic formula and show your work.

a) How long does the object remain in the air before landing? Put your answer in decimal form, rounded to

the nearest tenth of a second.

b) From part a) of this problem, the quadratic formula gives you two answers. How did you know which

answer to choose?

c) How long is the object in the air when it reaches a height of 25 feet?

Section 5.2M

Name ______________________________ Period __________





1) List 5 ways to solve a quadratic equation.

�

�

�

�

�

#2 – 3: Solve the following quadratic equations using the indicated methods.

2) 2 6x x− =

Graph:

Factor:

Quadratic Formula:

Solution(s):


Name ______________________________ Period __________



#2 – 3 (continued): Solve the following quadratic equations using the indicated methods.

3) 2 3 0x − =

Graph:

Square Root:

Quadratic Formula:

Solution(s):


4) List the method(s) for solving quadratic equations that:

a) You can always use if the equation has real solutions, but it often only gives approximate answers.

b) You can always use to solve quadratic equations, regardless of the number and type of solutions.

c) You can use sometimes to solve quadratic equations, if it has rational solutions.

Name ______________________________ Period __________





#5 – 7: For each equation, determine an effective method for solving the quadratic equation and explain

why you chose that method. Solve the quadratic equation with the method of your choice, keeping

the answer exact (no decimal approximations).

5) 22 5 3 0x x+ − =

6) 24 4000 0x x− + =

7) 2 7 18 0x x+ − =

#8 – 11: Determine an answer for each situation. Be sure to clearly record your thinking.

8) The product of two consecutive integers is 72. Find the two numbers.

9) The length of a rectangle exceeds its width by 3 inches. The area of the rectangle is 70 square inches. Find its

dimensions.

Name ______________________________ Period __________



#8 – 11 (continued): Determine an answer for each situation. Be sure to clearly record your thinking.

10) Suzie wants to build a garden that has three separate rectangular sections. She wants to fence around the

whole garden and between each section as shown. The plot is twice as long as it is wide and the total area is

200 square feet. How much fencing does Suzie need?

11) Mike wants to fence three sides of a rectangular patio that is adjacent to the back of his house. The area of

the patio is 192 ft2 and the length is 4 feet longer than the width. Find how much fencing Mike will need.

#12 – 13: Solve each quadratic equation 2 different ways. Record the method that you are using for each.

(Square Roots – Factoring – Completing the Square –Quadratic Formula)

12) 22 3 39x + =

Method 1: ________________________ Method 2: ________________________


you did and which do you feel is more

efficient?

x

2x

Name ______________________________ Period __________





#12 – 13 (continued): Solve each quadratic equation 2 different ways. Record the method that you are

using for each. (Square Roots – Factoring – Completing the Square –Quadratic Formula)

13) The length of a rectangular pool is 10 meters more than its width. The area of the pool is 875 square meters.

Find the dimensions of the pool.

Method 1: ________________________ Method 2: ________________________


you did and which do you feel is more

efficient?

#14 – 17: Solve the following quadratic equations with the method of your choice. Verify that the answer

is correct.

14) 23 6 10x x+ = −

� Verify:

Name ______________________________ Period __________



#14 – 17 (continued): Solve the following quadratic equations with the method of your choice. Verify that

the answer is correct.

15) 23 12 1 0x x− + + =

� Verify:

16) 2 6 9 0x x+ + =

� Verify:

17) 281 1 0x + =

� Verify:

Name ______________________________ Period __________





18) The height h (in feet) above the ground of a baseball depends upon the time t (in seconds) it has been in

flight. Joe takes a mighty swing and hits a bloop single whose height is described approximately by the

equation ( ) 280 16h t t t= − .

a) How long is the ball in the air?

b) When does the ball reach its maximum height?

c) What is the maximum height?

d) It takes approximately 0.92 seconds for the ball to reach a height of 60 feet. On its way back down, the

ball is again 60 feet above the ground. What is the value of ! when this happens?

Section 5.2N

Name ______________________________ Period __________




Name ______________________________ Period __________





#1 – 5: Four equations and one situation are given. Solve one by the square root property, one by

factoring, one by completing the square, one using the quadratic formula, and one by graphing.

Express your answers in simplest radical form. Verify your answers.

1) 22 7 15x x= − + 2) 2 2 15 0x x− − = 3) 2 12 20x x+ = 4) ( )2

3 8x − =

5) A group of friends hiked to Havasupai Point in Grand Canyon National Park. The Colorado River was 4755

feet below them. A rock was thrown upward at an initial velocity of 24 feet per second. The rock's height

t seconds after it was thrown upward is given by the function ( ) 216 24 4755h t t t= − + + . How long did it

take for the rock to hit the river?

Solve by: Square Root Property

#______ equation: ________________________

Solve:

Solution(s): ______________________


Solve by: Factoring

#______ equation: ________________________

Solve:

Solution(s): ______________________


Solve by: Completing the Square

#______ equation: ________________________

Solve:

Solution(s): ______________________


Solve by: Using the Quadratic Formula

#______ equation: ________________________

Solve:

Solution(s): ______________________


Name ______________________________ Period __________



#1 – 5 (continued): Four equations and one situation are given. Solve one by the square root property, one

by factoring, one by completing the square, one using the quadratic formula, and one by graphing.

Express your answers in simplest radical form. Verify your answers.

Solve by: Graphing

#______ equation: ________________________

Solution(s): ______________________


6) a) Solve ( )2

4 9x − = .

b) Sketch the graphs of ( )2

4y x= − and 9y = .

c) Describe the connection between the solution

in part a) and the graph in part b).

Name ______________________________ Period __________





7) A hose used by the fire department shoots water out in a parabolic arc. Let " be the horizontal distance from

the hose’s nozzle, and # be the corresponding height of the stream of water, both in feet. The quadratic

equation is 20.016 0.5 4.5y x x= − + + .

a) Explain the meaning in the context of the situation of the 4.5 that appears in the

equation.

b) What is the horizontal distance from the nozzle to where the stream hits the ground?

c) Will the stream go over a 6-foot high fence that is located 28 feet from the nozzle? Explain your

reasoning.

8) The graph of 2 400y x= − is shown at right. Notice that no

coordinates appear in the diagram. Without using your graphing

calculator, figure out the actual window that was used for this graph.

Find the high and low values for both the "- and #-axis. After you

get your answer check it on your calculator.

Xmin: _________ Xmax: _________

Ymin: _________ Ymax: _________

Section 5.2O

Name ______________________________ Period __________




Name ______________________________ Period __________



1) What is the discriminant? What information does it provide?

#2 – 9: Find the discriminant, the number of solutions and the type of solutions for each equation.

2) 2 6 10 0x x+ + = 3) 23 2 1x x+ =

discriminant: ___________________ discriminant: _______________________

number of solutions: _____________ number of solutions: _________________

type of solutions: ________________ type of solutions: ___________________

4) 20 4 4x x= − + 5) 212 11 2x x= +

discriminant: ___________________ discriminant: _______________________

number of solutions: _____________ number of solutions: _________________

type of solutions: ________________ type of solutions: ___________________

Name ______________________________ Period __________



#2 – 9 (continued): Find the discriminant, the number of solutions and the type of solutions for each

equation.

6) 28 1 16x x+ = − 7) 27 16 11 0x x+ + =

discriminant: ___________________ discriminant: _____________________

number of solutions: _____________ number of solutions: _______________

type of solutions: ________________ type of solutions: __________________

8) 25 11 6 0x x− + = 9) 20 4 5 2x x= + +

discriminant: ___________________ discriminant: _____________________

number of solutions: _____________ number of solutions: _______________

type of solutions: ________________ type of solutions: __________________

10) On a quiz, Brittani used the discriminant to find the number and type of solutions. Find her mistake and find

the correct solution.

( ) ( )

2

2

2

0 6 5

4

6 4 1 5

36 20

discriminant: 56

solutions: 2 imaginary solutions

x x

b ac

= − +

−

− −

− −

−

Name ______________________________ Period __________



11) Emma and Brandon own a factory that produces bike helmets. Their accountant says that their profit per year

is given by the function P(x) = 0.003x2 + 12x + 27760, where x represents the number of helmets produced.

Their goal is to make a profit of $40,000 this year.

a) Write the function that would represent a $40,000 profit.

b) Write the equation in standard form.

c) Find the discriminant.

d) Based on the discriminant, is the profit possible? Explain your thinking. If this is possible, find the

number of helmets that need to be sold to make a profit of $40,000.

12) Marty is outside his apartment building, and he borrowed his friend Yolanda’s phone. He needs to return it.

Yolanda lives on the third floor, and he is in a hurry, so he tosses her phone up to her. He throws it straight

up with a vertical velocity of 55 feet/second. The equation for the height of the phone above the ground at a

certain time t is given by the equation 216 55 4y t t= − + + (he released the phone when it was 4 feet above

the ground). Will the phone reach her if she is 36 feet up?

13) Bryson owns a business that manufactures and sells tires. The revenue from selling the tires in the month of

July is given by the function ( ) ( )200 4R x x x= − where " is the number of tires sold. Can Bryson’s

business generate revenue, R, of $20,000 in the month of July?

Section 5.3A

Name ______________________________ Period __________




Name ______________________________ Period __________



#1 – 2: Consider the equation (((( )))) 0f x ==== .

1) ( ) 2 2 8f x x x= − −

a) What is the discriminant?

b) Number of solutions?

c) Type of solutions?

d) What are the zeros (roots)?


x

y

f) What is the vertex?

g) Is the vertex a minimum or maximum?

h) What is the y-intercept?

i) What is the domain?

j) What is the range?

2) ( )24 x f x− =

a) What is the discriminant?

b) Number of solutions?

c) Type of solutions?

d) What are the zeros (roots)?


x

y

f) What is the vertex?

g) Is the vertex a minimum or maximum?

h) What is the y-intercept?

i) What is the domain?

j) What is the range?

Name ______________________________ Period __________



3) Which equation could model the graph to the right?

[A] ( ) ( )4 1y a x x= − − −

[B] ( ) ( )4 1y a x x= − + +

[C] ( ) ( )4 1y a x x= − −

[D] ( ) ( )4 1y a x x= + +

#4 – 6: Find the discriminant of each equation and then state the

number and type of solutions.

4) 2 6 2x x+ = − 5) 22 8 8x x− = − 6) 25 13 9 0x x− + =

discriminant: _____________ discriminant: _____________ discriminant: _______________

number of solutions: _______ number of solutions: _______ number of solutions: _________

real or imaginary: _________ real or imaginary: _________ real or imaginary: ___________

7) Three possible graphs of ( )f x are shown below. Consider the equation ( ) 0f x = . Label each graph below

as having a positive, negative, or zero discriminant.

8) The graph of ( )f x is shown. Consider the equation ( ) 0f x = . What type of discriminant does the

equation have? How many solutions does the equation have? Write a

function to model the graph pictured to the right.

Type of Discriminant (+), (–), (0) __________________________

Type and number of solutions ____________________________

Function _____________________________________________

Section 5.3B

a) b) c)

Name ______________________________ Period __________


5.4 I can represent relationships involving quadratic inequalities in multiple ways, find solutions and 299

interpret these solutions to solve real-world situations.

#1 – 3: Determine whether each of the given points is a solution to the given quadratic inequality.

1) 2 3 3y x x≥ − + 2) 216

2y x x< − − + 3) 2 4y x x> 2 − +

a) ( )0, 0 a) ( )0, 0 a) ( )0, 0

b) ( )1,1 b) ( )3, 3 b) ( )1, 5

#4 – 6: For each inequality and graph (the points plotted are points that exist on the boundary):

� Determine whether the boundary is included as a solution (solid) or not included as part of the

solution (dashed).

� Use a test point to determine the solution region.

Graph the solution to each inequality.

4) 2 4y x≤ − 5) 22 4 3y x x≥ − + − 6) 2 20 6y x x> −5 − +

-5 -4 -3 -2 -1 1 2 3

-8

-4

4

8

12

16

20

24

28

x

y

-4 -3 -2 -1 1 2 3 4

-9

-8

-7

-6

-5

-4

-3

-2

-1

1

2

3

x

y

-4 -3 -2 -1 1 2 3 4

-5

-4

-3

-2

-1

1

2

3

4

5

x

y

Name ______________________________ Period __________



#7 – 9: Fill in the blank with the appropriate inequality sign.

7) 21 5 1

4 4 y x x− − 8) 21 5

14 4

y x x− + + 9) 2 5 4 y x +

#10 – 15: Match the inequality with its graph.

______ 10) 2 4 3y x x≥ − + − ______ 11) 2 4 3y x x≤ − − − ______ 12) 2 2 3y x x≤ + −

______ 13) 2 4 3y x x< − + ______ 14) 2 4 3y x x> − + − ______ 15) 2 2 3y x x> − −

[A] [B] [C]

[D] [E] [F]

-2 2 4 6

-4

-2

2

4

x

y

-2 2 4 6

-4

-2

2

4

x

y

-4 -2 2 4

-2

2

4

6

8

10

x

y

Name ______________________________ Period __________




#16 – 23: Draw the graph of each quadratic inequality. When graphing the boundary, consider the

various forms of a quadratic and the significant features that are identified from each form.

16) 2 2 1y x x< − − 17) 2 9y x≥ −

18) 2 7 12y x x> − + − 19) 22 7 1y x x≤ − − +

-4 -2 2 4

-10

-8

-6

-4

-2

2

4

x

y

-4 -2 2 4

-4

-2

2

4

x

y

-2 2 4 6

-4

-2

2

4

x

y

-6 -4 -2 2

-2

2

4

6

8

x

y

Name ______________________________ Period __________



#16 – 23 (continued): Draw the graph of each quadratic inequality. When graphing the boundary,

consider the various forms of a quadratic and the significant features that are identified from

each form.

20) 2 5 1y x x≤ − − + 21) ( )21

6 32

y x> + −

22) ( ) ( )1

1 52

y x x> + − 23) ( )2

2 1 6y x≤ − + +

-6 -4 -2 2

-2

2

4

6

8

x

y

-10 -8 -6 -4 -2 2

-6

-4

-2

2

4

x

y

Section 5.4A

Name ______________________________ Period __________




1) Graph ( ) 2 2 8.f x x x= + −

a) On what interval(s) is 2 2 8 0x x+ − > ? (What are the

critical values?)

b) On what interval(s) is2 2 8 0x x+ − < ?

2) Graph ( ) 2 6.f x x x= − − +

a) On what interval(s) is 2 6 0x x− − + > ? (What are the

critical values?)

b) On what interval(s) is 2 6 0x x− − + < ?

3) Draw a quadratic function that is:

� Positive when 1 and 5x x< − >

� Negative when 1 5x− < <

4) Draw a quadratic function that is:

� Positive when 4 0x− < <

� Negative when 4 and 0x x< − >

-6 -4 -2 2 4 6

-10

-8

-6

-4

-2

2

-6 -4 -2 2 4 6

-2

2

4

6

-6 -4 -2 2 4 6

-6

-4

-2

2

4

6

Name ______________________________ Period __________



#5 – 12: Use a graph to find the solutions for the following inequalities.

5) 2 25 0x − > 6) 2 4 0x x− ≥

Solution: _________________ Solution: _________________

7) 2 7 10 0x x− + < 8) 2 4 5 0x x− + + ≤


x

y

x

y

Name ______________________________ Period __________




#5 – 12 (continued): Use a graph to find the solutions for the following inequalities.

9) 2 8 20x x> + 10) 2 27 12x x+ <


11) 22 11 5 0x x− + ≥ 12) 23 17 10 0x x− + <


x

y

x

y

Name ______________________________ Period __________



#13 – 17: Use a graphing utility to solve the following problems. Sketch the graph and label the

x-intercepts and vertex.

13) The profit a coat manufacturer makes each day is modeled by the

equation ( ) 2 120 2000P x x x= − + − , where P is the profit and x is

the price for each coat sold. For what values of x does the company

make a profit?

14) When a baseball is hit by a batter, the height of the ball, h, at time t,

is determined by the equation ( ) 216 64 4h t t t= − + + . For which

interval of time is the height of the ball greater than or equal to 52

feet?

15) The path of a soccer ball kicked from the ground can be modeled

by 20.0540 1.43y x x= − + where x is the horizontal distance (in

feet) from where the ball was kicked and y is the corresponding

height (in feet).

a) A soccer goal is 8 feet high. Write and solve an inequality to

find at what values of x the ball is low enough to go into the

goal.

b) A soccer player kicks the ball toward the goal from a distance of 15 feet away. No one is blocking the

goal. Will the player score a goal? Explain your reasoning.

Name ______________________________ Period __________




#13 – 17 (continued): Use a graphing utility to solve the following problems. Sketch the graph and label

the x-intercepts and vertex.

16) A manilla rope used for rappelling down a cliff can safely support a

weight W (in pounds) modeled by the inequality 2

1480W d≤

where d is the rope’s diameter (in inches). What diameter of rope

would be needed to support a weight of at least 5920 pounds?

17) For a driver aged x years, a study found that the driver’s reaction time R (in milliseconds) to a visual

stimulus such as a traffic light can be modeled by 20.005 0.23 22T x x= − + when 16 70x≤ ≤ . At what age

does a driver’s reaction time tend to be greater than 25 milliseconds??

Section 5.4B

Name ______________________________ Period __________




Name ______________________________ Period __________


Review Material 309

1) Describe what you look for in determining which method is best to solve a quadratic equation:

a) Square Root:

b) Completing the Square:

c) Factoring:

d) Quadratic Formula:

2) Choose the most efficient method for each equation. You must select 3 equations for each method.

� Place a circle around the letter of each equation you would solve using the square root method.

� Place a square around the letter of each equation you would solve by completing the square.

� Place a triangle around the letter of each equation you would solve by factoring.

� The three equations with no mark would represent the equations you would solve using the

quadratic formula.

[A] 22 5 41x + = [B] 2 173 10x x+ = − [C] 2 20 104x x+ = −

[D] 2 6 15 0xx − − = [E] 2 4 12x x+ = [F] 2 288 60 0x x− − =

[G] 23 6 2 0x x+ + = [H] 2 8 36 xx − = − [I] 2 129 4 0x x+ + =

[J] ( )2

1 363 x − =− [K] ( )2

6 45 532 x − − = [L] 2 135 6 0x x− + =

3) The letters you placed a circle around represent the equations you would solve using the square root method.

Write one equation in each blank below (3 blanks, 3 equations). Solve each equation using the square root

method to find the real or imaginary solutions.

[ __ ] ___________________ [ __ ] ____________________ [ __ ] ____________________

4) The letters you placed a square around represent the equations you would solve by completing the square.

Write one equation in each blank below (3 blanks, 3 equations). Solve each equation by completing the

square to find the real or imaginary solutions.

[ __ ] ___________________ [ __ ] ____________________ [ __ ] ____________________

Name ______________________________ Period __________



5) The letters you placed a triangle around represent the equations you would solve by factoring. Write one

equation in each blank below (3 blanks, 3 equations). Solve each equation by factoring to find the real or

imaginary solutions.

[ __ ] ___________________ [ __ ] ____________________ [ __ ] ____________________

6) The letters with no mark around them, you would solve by using the quadratic formula. Write one equation

in each blank below (3 blanks, 3 equations). Solve each equation using the quadratic formula to find the real

or imaginary solutions.

[ __ ] ___________________ [ __ ] ____________________ [ __ ] ____________________

7) Simplify.

a) 4 50− ± b) 3 81

12

± c)

6 45

3

− ± − d)

7 12

14

± −

8) A ball is thrown off of a rooftop 200 feet high with an initial velocity of 40 feet per second. The function

( ) 216 40 200h t t t= − + + represents the height of the ball h after t seconds. Write and solve the function

you would use to determine when the ball would hit the ground.

Name ______________________________ Period __________


Review Material 311

9) A water balloon is catapulted into the air. The height h of the balloon in meters is represented by the function

( ) 24.9 27 2.4h t t t= − + + where t represents the time in seconds. Write and solve the equation you would

use to determine when the balloon would hit the ground.

10) The graph of ( )f x is provided. Consider the equation ( ) 0f x = . Determine if the discriminant is positive,

negative, or zero. State the type of solutions the parabola has and how many solutions there are.

Pos/Neg/Zero: ____________ Pos/Neg/Zero: _______________ Pos/Neg/Zero: ___________________

# of Solutions: ____________ # of Solutions: _______________ # of Solutions: ___________________

Type of solutions: _________ Type of solutions: _____________ Type of solutions: ________________

11) Determine the discriminant of each quadratic function. State how many solutions the equation will have and

what type of solutions they will be.

a) 20 2 5x x= + + b) 20 5 13x x= − − c) 20 4 4 1x x= − +

Discriminant: ____________ Discriminant: ________________ Discriminant: ____________________

# of Solutions: ____________ # of Solutions: _______________ # of Solutions: ___________________

Type of solutions: _________ Type of solutions: _____________ Type of solutions: ________________

Name ______________________________ Period __________



12) Which is the correct graph of 2 6 5y x x> + + ?

[a]

[b]

[c]

[d]

13) Use the graph above to write the solution to the inequality: 2 6 5 0x x+ + < .

14) List below the 4 steps that you should follow to solve a quadratic inequality graphically.

��

��

��

��

15) Solve each inequality by completing the 4 steps stated above.

a) 2 2 0x x+ − < b) 2 12 32x x+ ≥ −

Name ______________________________ Period __________


Review Material 313

16) Which is the graph that represents the solutions of ( ) ( )4 1y x x≤ − + ? Graph ___________________

[a]

[b]

[c]

[d]

17) Use the graph above to write the solution to the inequality: ( ) ( )4 1 0x x− + ≥ .

Name ______________________________ Period __________



18) The height of an object launched vertically is given by 2

0 0( ) 16h t t v t h= − + ⋅ + where:

$% = initial velocity ℎ% = initial height

A trebuchet launches a watermelon on a parabolic arc at an initial velocity of 48 ft/s from a height

of 32 ft.

a) Write a function that models the height of the

watermelon at time t.

b) Find the height of the watermelon after

1 second.

c) What is the maximum height of the watermelon? After

how many seconds does the watermelon reach this

height?

d) When will the watermelon return to its original height of

32 ft?

e) When will the watermelon land on the ground?

f) In part e), you should have gotten two answers. Do both

of these answers make sense in the context of the

situation? Explain your reasoning.

Unit 5 Review

WORK SPACE

NO

GR

AP

HIN

G C

AL

CU

LA

TO

R N

O G

RA

PH

ING

CA

LC

UL

AT

OR

Appendix A ~ Glossary ~ Units 4 – 5 315

Appendix A ~ Glossary ~ Units 4 – 5

Axis of symmetry – p. 3 – resources, section 4.1

All parabolic functions (quadratic functions) have an imaginary line that runs through the vertex of the

parabola and divides it into two sides that are mirror images of one another. Any two points with the same

y-value will be the same distance from this axis. The axis of symmetry will be the vertical line 2

bx

a= − .

Binomial – p. 11 – resources, section 4.2

A polynomial with two terms.

Coefficient – p. 2 – resources, section 4.1

The number which is multiplied by one or more variables or powers of variables in the term. There are two

coefficients for the standard quadratic form: 2y ax bx c= + + . The coefficient of the

quadratic term (x2-term) is a and the coefficient of the linear term (x-term) is b. The value c is called a

constant term.

• The coefficient of the x2-term is called a.

• If a is positive, the parabola will open upward. The vertex will be a minimum.

• If a is negative, the parabola will open downward. The vertex will be a maximum.

• If a > 1 or a < –1, the parabola will be stretched vertically about the axis of symmetry.

• If –1 < a < 1 and 0a ≠ , the parabola will be compressed vertically about the axis of

symmetry.

• The value of c is the y-intercept, or where the parabola will intersect the y-axis.

Complete the square – p. 159 – resources, section 5.2

An algorithm performed on an expression that is in the form �� + ��, add ��. Then the expression

becomes �� + �� + �� which is a perfect square trinomial that can be factored as �� + �

��.

Complex numbers – p. 155 – resources, section 5.2

Numbers that have the form � + �, where a and b are real numbers. a is the real part of the complex

number and b is the coefficient of the imaginary part.

Difference of squares – p. 21 – resources, section 4.3

A quadratic expression in the form: 2 2

a b− .

Discriminant – p. 166 – resources, section 5.3

The value under the radical in the Quadratic Formula, �� − 4�� . The discriminant gives us information to

determine the number and type of solution(s) a quadratic equation has.

• If �� − 4�� > 0, the equation has two separate real solutions.

• If �� − 4�� = 0, the equation has one real solution, a double root.

• If �� − 4�� < 0, the equation has only non-real solutions (2 imaginary solutions).

Distributive property – p. 11 – resources, section 4.2

Used to rewrite a polynomial as the product of the greatest common factor (GCF) and the other parts of the

polynomial. It says that ( )a b c a b a c+ = ⋅ + ⋅ which is the factored form of a polynomial.

316 APPENDICES

Domain (of a function or relation) – p. 2 – resources, section 4.1

The set of all possible independent values (input values, x-coordinates) the relation can take. It is the

collection of all possible inputs. The domain of every quadratic equation is all real numbers.

Double root – p. 147 – resources, section 5.2

A solution that is repeated twice.

Factors – p. 16 – resources, section 4.3

Expressions that multiply together to produce another expression.

Factoring – p. 16 – resources, section 4.3

The process of breaking an expression down into its multiplicative factors.

Greatest Common Factor (GCF) – p. 15 – resources, section 4.3

The greatest common factor of a polynomial can be a number, a variable, or a combination of numbers and

variables that appear in every term of the polynomial.

Imaginary number – p. 155 – resources, section 5.2

A number that has the imaginary unit, to represent √−1 and does not have a real portion as part of the

number.

Intercept (or factored) form – p. 6 – resources, section 4.1

A quadratic equation in the form: ( ) ( )y a x m x n= − − . Translating a quadratic equation to this form

allows easier identification of the point(s) where the parabola crosses the x-axis (the x-intercept(s)) of the

parabola. They are at points (m, 0) and (n, 0) because the x-axis is where y = 0

Monomial – p. 11 – resources, section 4.2

A polynomial with one term.

Parabola – p. 2 – resources, section 4.1

The U-shaped graph of a quadratic function.

Perfect square trinomial – p. 20 – resources, section 4.3

A quadratic expression in the form: 2 22a ab b+ + or

2 22a ab b− + .

Polynomial – p. 11 – resources, section 4.2

An expression made with constants, variables, and positive integer exponents of the variables.

Quadratic Formula – p. 163 – resources, section 5.2

The solutions to any quadratic equation in standard form�� + �� + � = 0 can be found by using the

quadratic formula: 2 4

2

b b acx

a

− ± −= .

Quadratic Inequality – p. 173 – resources, section 5.4

A term describing a squared function that is specified to be smaller or larger than a given value.

Appendix A ~ Glossary ~ Units 4 – 5 317

Range – p. 2 – resources, section 4.1

The set of all possible dependent values (output values, y-coordinates) the relation can produce from the

domain values. It is the collection of all possible outputs. The range of a parabola depends upon whether the

parabola opens up or down. If k is the y-coordinate of the vertex:

• If a is positive, the range will be y ≥ k.

• If a is negative, the range will be y ≤ k.

Solution – p. 146 – resources, section 5.2

A value that makes an equation true. Solutions can also be called zeros or roots.

Solutions of the quadratic equation (((( )))) 0f x ==== – p. 143 – resources, section 5.1

The x-intercepts of the parabola. The x-coordinate of these points are often called the roots and zeros.

Standard form – p. 2 – resources, section 4.1

The standard form is 2y ax bx c= + + with a, b, and c being constants and a ≠ 0.

Vertex – p. 2 – resources, section 4.1

The highest or lowest point on a parabola. For a parabola that opens upward, the vertex is the lowest point

(minimum); for a parabola that opens downward, the vertex is the highest point (maximum). When the

quadratic function is given in standard form, 2y ax bx c= + + , the x-coordinate of the vertex is always

2

b

a− . The vertex is always half-way between the two x-intercepts of the graph of a quadratic function. If

the equation is written in vertex form, the vertex is ( ),h k .

Vertex form – p. 8 – resources, section 4.1

A quadratic equation in the form: ( )2

y a x h k= − + .

Unit 4 – Practice Problem ANSWERS (4th Edition)

318 APPENDICES

SECTION 4.1A

1) Parabola

a. Root, Zeros

b. Axis of symmetry

c. Substitute x = 0 into the

equation to find the value of

y.

2)

3) Maximum

4) Minimum

5)

axis of symmetry: � = 1

y-intercept: (0, –3)

vertex: (1, –4)

domain: All Real Numbers

range: y ≥ –4

opens: up

No vertical stretch nor

compression

6)


y-intercept: (0, 5)

vertex: (1, 7)


range: y ≤ 7

opens: down

Vertical stretch

7)

axis of symmetry: x = 1


vertex: (1, –4.5)


range: y ≥ –4.5

opens: up

Vertical compression

8)

axis of symmetry: x = –1

y-intercept: (0, 1)

vertex: (−1,−3) domain: All Real Numbers

range: y ≥ –3

opens: up

Vertical stretch

9)


y-intercept: (0, 0)

vertex: (1, 1)


range: y ≤ 1

opens: down

No vertical stretch nor

compression

-5 -4 -3 -2 -1 1 2 3 4 5

-8

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

8

x

y

-5 -4 -3 -2 -1 1 2 3 4 5

-8

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

8

x

y

x-int

or

y-int

vertex

x-int or

zero

Axis of

symmetry


Appendix B ~ Selected Answers ~ Unit 4 319

SECTION 4.1A (continued)

10)

axis of symmetry: x = –1

y-intercept: (0, 4)

vertex: (–1, 6)


range: y ≤ 6

opens: down

Vertical stretch

11) a. (0, −8), 1

8x = , down

b. (0,−1), 3

4x

−= , up

12) a. (0, 7), � = −3, up

b. (0,−1), � = ��, down

13) a.

b. 10 feet

c. 0.5 seconds, at the vertex

(maximum)

13) d. 1 second if the juggler

catches it; at approximately

1.291 seconds if it hits the

ground, this is an x-intercept

e. (0, 6); The height of the

jugglers hand is 6 feet when

he lets go of the ring.

14) a.

b. 784 ft

c. 7 seconds, at the vertex

d. 14 sec, the positive x-int

15) a.

b. 144 feet, at the vertex

c. 2 seconds

d. 5 seconds

16) a. � provides information

whether the graph opens up

of down

� provides information

whether the graph is a

vertical stretch or vertical

compression of the parent

graph.

b. Use b to find the axis of

symmetry; x = − ��. Then

use this x-value to find the

y-value of the vertex.

c. c is the y-intercept where

x = 0.

d. x = − ��

e. Once you know the x

coordinate from − ��, you

can substitute the value into

the original function and

solve for y

17) a. y-intercept: (0, −8) axis of symmetry: � = 0

vertex: (0, −8) opens: up

Vertical compression

domain: All real numbers

range: � ≥ −8

b. y-intercept: (0, −3) axis of symmetry: x = –2

vertex: (–2, –7)

opens: up

Vertical: same


range: y ≥ –7

c. y-intercept: (0, −2) axis of symmetry: � = 1

vertex: (1, 1)

opens: down

Vertical stretch


range: � ≤ 1

-5 -4 -3 -2 -1 1 2 3 4 5

-8

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

8

x

y

Time (seconds)

Hei

ght

of

fire

work

s (f

t)

Time (seconds)

Hei

gh

t of

frogsa

y (

ft)


320 APPENDICES

SECTION 4.1B

1)

2) a. Answers will vary. Ex: Start

by using � = − ��, then

substitute the x-value into the original function to solve for y and find the vertex. From there, use a table to find 2 other points on the parabola. Use the c-value of the equation to graph the y-intercept at (0,−15).

b. Answers will vary. Ex: Set each of the factors equal to zero to find the two x-intercepts. Then, graph the points (3, 0) and (–5, 0) and find the x-coordinate halfway between them to use as the x-coordinate of the vertex. Then, substitute that x-value to find the y-value of the vertex. Because there is no negative in front of either factor, the parabola opens up.

c. Answers will vary. Ex: With factored form you start with the x-intercepts, then find the vertex. With standard form you start by finding the

vertex using − ��. You use

the vertex to graph with both forms.

3) a. x-intercepts are (−9,0) and (25,0). The x-intercepts represent the number $2 decreases the company should have in order to increase profit.

b. 8 $2 decreases. $16 off $50, so $34. Income at this price will be $28,900

c.

d. Domain: −9 ≤ � ≤ 25

Range: 0 ≤ � ≤ 28,900

4) a.

b. 3 seconds

c. 1.5 seconds

d. 11.025 meters

e. Domain: all real numbers

Range: � ≤ 11.025

4) f. * The domain represents time, and because time is negative it must be greater than or equal to 0. The object hits the ground after 3 seconds, so any time after 3 seconds implies the object is below the ground which is not possible.

* The range represents the height of the object. The object starts at the ground and goes to a maximum height of 11 meters so the height will always remain between 0 and 11.

5)

x-intercept(s): (2, 0), (6, 0)

Axis of Symmetry: � = 4

Minimum at (4, –4) No Max

Domain: All real numbers

Range: y ≥ –4

6)

x-intercept(s): (–1, 0), (1, 0)

Axis of Symmetry: � = 0

Minimum at (0, –4) No Max


Range: y ≥ –4

-10 -5 5 10 15 20 25

5000

10000

15000

20000

25000

30000

x

y

-2 -1 1 2 3 4 5

2

4

6

8

10

12

14

x

y

x-int

y-int

vertex

x-int

Axis of

symmetry



-2 -1 1 2 3 4 5 6 7

-20

-15

-10

-5

x

y

-7 -6 -5 -4 -3 -2 -1 1 2

-5

5

10

15

x

y

-3 -2 -1 1 2 3 4

-8

-6

-4

-2

2

4

6

8

10

x

y

-7 -6 -5 -4 -3 -2 -1 1 2 3

-8

-6

-4

-2

2

4

6

8

10

x

y

-4 -3 -2 -1 1 2 3 4 5 6

-6

-4

-2

2

4

6

8

10

12

x

y

SECTION 4.1B (continued)

7)

x-intercept(s): (−3,0)(7,0) Maximum at (2, 12.5) Axis of symmetry: � = 2


Range: � ≤ 12.5

8)

x-intercept(s): (0,0)(5,0) Axis of Symmetry: � = 2.5

Minimum at (2.5, −18.75)


Range: � ≥ −18.75

9)

x-intercept(s): (−5,0)(−1,0) Minimum at (−3,−4) Axis of symmetry: � = −3


Range: � ≥ −4

10)

x-intercept(s): (3,0)(−1,0) Maximum at (1, 8) Axis of symmetry: � = 1


Range: � ≤ 8

11)

x-intercept(s): (−5,0)(1,0) Maximum at (−2,9) Axis of symmetry: � = −2


Range: � ≤ 9

12)

x-intercept(s): (3,0)(−1,0) Minimum at (1,−2) Axis of symmetry: � = 1


Range: � ≥ −2

13) x-int(s): (–6,0), ( –2, 0)

Axis of symmetry: � = −4

Vertex: (–4, –4)

Opens: Up

14) x-int(s): (−3,0)(−5,0) Axis of symmetry: � = −4

Vertex: (–4, 1)

Opens: Down

15) x-int(s): (1,0)(−7,0) Axis of symmetry: � = −3

Vertex: (−3,−32) Opens: Up

16) x-int(s): (3,0)(−5,0) Axis of symmetry: � = −1

Vertex: (−1,8) Opens: Down

17) a. The ‘a’ value determines

whether the parabola opens

up or down and whether the

function is vertically

compressed or stretched.

b. Set each factor equal to zero

and solve. These values are

the x-intercepts.

c. The x-coordinate of the

vertex can be found by

going half way between the

x-intercepts.

d. After finding the

x-coordinate of the vertex,

substitute it into the function

for x to solve for y.

e. The appropriate x values are

near the vertex.

18) � = (� − 1)(� − 5) Domain: all real numbers

Range: y ≥ –4

Minimum at (3, −4) 19) � = (� − 1)(� + 6) Domain: all real numbers

Range: y ≥ –12.25

Minimum at (−2.5,−12.25)


322 APPENDICES

SECTION 4.1C

1)

2) a. a gives the direction of the

opening, opening up or

down, and if there is a

vertical stretch or

compression of the parent

graph.

b. You have to change the sign

on h, and leave k as is; the

vertex is located at (h, k)

c. Answers will vary. Ex: you

can use 2-3 numbers on one

side of the vertex, then

reflect those points over the

axis of symmetry.

3) a. a = 3, h = 1, k = –2

b. a= –0.5, h= –3, k = 1

4)

Vertex: (2, 1)

Minimum at (2, 1)

Opens: Up


Range: y ≥ 1

5)

Vertex: (–1, 2)

Minimum at (–1, 2)

Opens: Up


Range: y ≥ 2

6)

Vertex: (–3, –1)

Maximum at (–3, –1)

Opens: down


Range: y ≤ –1

7)

7) Vertex: (–2, 1)

Maximum at (–2, 1)

Opens: down


Range: y ≤ 1

8)

Vertex: (2, –5)

Minimum at (2, –5)

Opens: Up


Range: y ≥ –5

9)

Vertex: (5, –3)

Minimum at (5, –3)

Opens: Up


Range: y ≥ –3

-5 5

-5

5

x

y

-5 5

-5

5

x

y

-5 5

-5

5

x

y

x-int

y-int

vertex

x-int

Axis of

symmetry



SECTION 4.1C (continued)

10)

Form Equation Significant Features Vertex

Vertex � = �(� − ℎ)� + � Vertex opens up or down.

Vertical stretch or

compression of graph in

relation to parent graph � =��.

The vertex is (h, k). Remember to

set the quantity inside the

parentheses equal to zero to identify

the value of h.

Intercept � = �(� − �)(� − �) x-intercepts, opens up or

down, Vertical stretch or

compression of graph in

reference to the parent graph

� = ��.

The x-coordinate is found half-way

between the x-intercepts. Substitute

that number in for x in the original

equation to find the y-coordinate of

the vertex.

Standard � = �� + � + ! Opens up or down, Vertical

stretch or compression of

graph in relation to the parent

graph � = ��.

� = − �� gives the x-coordinate of

the vertex. Substitute this number

into the original equation and solve

for y to get the y-coordinate.

11) Answers will vary. Possible

choices include: Vertex form

because I get the vertex right

away. Intercept form because I

get the x-intercepts right away.

Standard form because I am

used to it and I can use the

x = − ��to find the x-intercepts.

12) a.

b. (4, 14)

c. x = 4 feet, so at his highest

point in the air he is 4 feet

away (horizontally) from

where he started on the

diving board;

y = 14 feet, his highest point

is 14 feet in the air.

12) d. (9.916, 0) The diver is 9.916

feet out from the diving

board horizontally and is

entering the water (height

above the water is 0).

13) a. 2.225 seconds

b. 24 feet

c. Domain: 0 < x < 2.225

Range: 0 < y < 24

13) d. Domain has to be positive

because it represents time,

and the ball is no longer in

the air after 2.225 seconds,

Range has to be positive

because the ball can’t go

below ground level and

doesn’t reach a height above

24 feet.

14) Up, x = 2, (2, 4)

15) Down, x = –5, (–5, –1)

16) Up, x = –10, (–10, 0)

17) Up, x = –1, (–1, –12)

18) Down, x = 3, (3, 1)

19) Down, x = 14, (14, –74)

-2 2 4 6 8 10 12

-4

4

8

12

16

x

y


324 APPENDICES

-6 -4 -2 2 4

-25

-20

-15

-10

-5

5

10

15

x

y

-4 -2 2 4 6

-25

-20

-15

-10

-5

5

10

x

y

SECTION 4.2A

1) 3x + 12

2) 4x + 6

3) 4x + 8

4) 6x + 3

5) 3x – 12

6) –5a – 10

7) 2b2 + 6b

8) –k2 + k

9) 3x2 + 12x – 9

10) –2x2 +10x – 2

11) 10x2 +15x +5

12) 6n3 – 2n2 – 8n

13) –32x2 + 28x +12

14) –10x2 + 40x – 20

15) 32x2 – 32x +8

16) –6x3 + 21x2 – 6x

17) x2 + 19x + 60

18) 2x2 + 14x + 24

19) x2 + 7x +10

20) 2x2 + 9x + 4

21) x2 + 8x +15

22) x2 – 8x + 15

23) x2 – 9

24) x2 – 6x – 16

25) x2 + 10x + 25

26) x2 – 8x + 16

27) 4x2 + 7x – 2

28) 3x2 – 18x – 21

29) –x2 + 7x –6

30) 2x2 + 2x – 40

31) 2x2 + 3x – 14

32) 2x2 + 3x – 2

33) 6x2 + 38x + 60

34) 24x2 – 5x – 14

SECTION 4.2B

1) y = 3x2 + 6x – 18

2) y = 2x2 – 8x – 10

3) y = –x2 – 5x +8

4) y = –5x2 + 25x – 5

5) y = x2 – 4x – 12

6) y = x2 – 4x + 3

7) y = 3x2 +18x + 24

8) y = –2x2 – 2x + 40

9) a. x-int: (–2, 0), (–1, 0)

b. y = x2 + 3x + 2

c. (0, 2)

d. graph

10) a. x-int: (–4, 0), (3, 0)

b. y = 2x2 + 2x – 24

c. (0, –24)

d.

11) a. x-int: (1, 0), (4, 0)

b. y = –2x2 + 10x – 8

c. (0, –8)

d.

12) a. x-int: (–7, 0), (3, 0)

b. y = –x2 – 4x +21

c. (0, 21)

d.

13) a. 250,000

people

b. P(20) = 1050

{1,050,000 people}

c.

t 0 5 10 15 20 30

P(t) 250 412.5 600 812.5 1050 1600

-6 -4 -2 2 4

-4

4

8

12

16

x

y




14) a. at 7 seconds b. 68.6 meters c.

x 0 1 2 3 4 5 6 7

y 68.6 88.2 98 98 88.2 68.6 39.2 0

15) a. y = 2x2 + 2x – 24

b.

16) a. y = –3x2 – 9x – 6

b.

SECTION 4.2C

1) y = x2 – 6x + 10

2) y = –2x2 + 12x – 24

3) y = –x2 – 8x – 24

4) y = x2 – 16x + 67

5) y = –2x2 – 28x – 108

6) y = 2.4x 2 – 24.48x + 65.424

7) Equivalent

8) No

9) a. (1, –16)

b. y = x2 – 2x – 15

c. (0, –15)

d.

x –4 –3 –2 –1 0 1

y 9 0 –7 –12 –15 –16

x 2 3 4 5 6

y –15 –12 –7 0 9

10) a. (–3, 7)

b. y = 2x2 + 12x + 25

c. (0, 25)

d.

x –5 –4 –3 –2 –1

y 15 9 7 9 15

11) C

12) A

13) D

14) B

15) a. y = –0.5x2 + 6.8x + 2.88

b. 26 meters

c. 2.88 meters

16) a. y = –x2 + 12x – 34

16) b.

17) a. y = 2x2 + 4x – 2 b.

18) a. 130 meters

b. 4 seconds

c. 51.6 meters

d.

19) C


326 APPENDICES

-4 -2 2 4 6 8 10

-30

-24

-18

-12

-6

6

12

x

f(x)

2 4 6 8 10

-8

-4

4

8

12

16

x

f(x)

SECTION 4.3A

1) 3(x + 2)

2) 5(x – 4)

3) 8(2b + 1)

4) 6(4n – 1)

5) 4(6w – 7)

6) 5(6p + 5q)

7) 3(a2 + 3)

8) x(x – 2)

9) 7y(2y + 1)

10) t(5t + 7)

11) 3(9p2 – 3p + 1)

12) 5(9b2 – b + 8)

13) 3(2x2 – 3x + 12)

14) 4(3w2 + 4w – 1)

15) –5(5w2 – 2w – 6)

16) –7(5m2 – 2m + 10)

17) 4v(4v2 – 4v – 7)

18) –x(5x2 – 4x – 3)

19) –5m2(4m4 – 9m + 10)

20) –10n2(5n5 – n2 – 5)

21) Not equivalent

22) Equivalent

23) y = 2(x2 – 8x – 4),


vertex: (4, –40)

24) y = 3(x2 – 7x + 4)

y-intercept: (0, 12)

vertex: (3.5, –24.75)

25) y = –2(x2 – 6x + 2)


vertex: (3, 14)

26) y = –6(x2 – 4x – 1)

y-int: (0, 6),

vertex: (2, 30)

27) 12xy

28) 6pr2

29) 5q

30) 9xy2z

31) 10(9b5 + 6b4 + 7)

32) –3(4b3 – 7b + 7)

33) 7(4r6 + 10r + 9)

34) –n(41n4 – 10n – 100)

35) x2 (5x2 + 7x + 8)

36) 6n5(7n2 + 8n + 3)

37) –7(3a3 – 7a2 – 6a – 5)

38) 4x3( x3 + 20x2 – 8x – 14)

39) 3d 2(13d 4 – 5d 3 + 2d 2 – 3)

40) 7b2 (4b5 + 5b3 – 10b2 + 2)

SECTION 4.3B

1) +, +

2) –, –

3) +, –

4) –, +

5) (b – 8)(b + 3)

6) (m + 7)(m + 10)

7) (n – 5)(n – 7))

8) (a – 9)(a + 6)

9) (n + 3)(n – 2)

10) (v – 6)(v – 2)

11) y = (x + 6)(x + 10)

x-int: (–10,0), ( –6,0)

12) y = (x + 2)(x – 1)

x-int. (–2, 0), (1, 0)

13) y = (x + 8)(x + 5)

x-int: (–8, 0), (–5, 0)

14) y = (x – 9)(x – 5)

x-int: (9, 0), (5, 0)

15) (k + 2)(k + 8)

16) (n + 4)(n + 1)

17) (x – 1)(x – 5)

18) (a – 6)(a + 8)

19) (v – 2)(v – 3)

20) (x – 7)(x + 2)

21) (x + 10)(x – 2)

22) (a + 7)(a + 8)

23) (x + 1)(x + 3)

24) (x + 8)(x – 2)

25) Prime (not factorable)

26) (x + 13)(x + 2)

27) a. x = –5

b. f(x) = (x + 9)(x + 1)

c. x-int: (–9, 0), (–1, 0)

d. x = –5

e. yes

f.

28) 3(b – 7)2

29) 5(x + 9)(x – 5)

30) f(x) = (x – 8)(x – 3)

x-int: (8, 0), (3, 0)

31) ( ) ( ) ( )8 2f x x x= − +

x-int: (8, 0), (–2, 0)



-1 1 2 3 4

-12

12

24

36

48

t

h(t)


32) y = 4(x – 10)(x + 4)

x-int: (10, 0), (–4, 0)

33) y = 3(x – 2)(x + 5)

x-int: (2, 0), (–5, 0)

34) a. 7

b. 2(x – 4)(x – 10)

c. (4, 0), (10, 0)

d. 7

e. yes

34) f.

35) a. (0, 45), (1, 80), (2, 105) 0 seconds � 45 meters 1 second � 80 meters 2 seconds � 105 meters

b. 45 meters

c. h(t)= –5(t – 9)(t + 1), 9 sec

d. 125 meters at 4 seconds

e.

SECTION 4.3C

1) +, +

2) –, –

3) –, +

4) +, –

5) (3n – 4)(n – 8)

6) (k – 5)(5k – 7)

7) (7x + 9)(x + 6)

8) (b – 4)(2b + 3)

9) (5x + 4)(x – 7)

10) (p + 8)(3p – 7)

11) (3a – 5)(2a – 5)

12) (4x + 5)(2x + 1)

13) not equivalent

14) Equivalent

15) y = (2x + 1)(x – 3)

vertex:(1.25, –6.125)

x-int: (–0.5, 0), (3, 0)

16) y = (3x + 2)(x – 2)

vertex: "�� , −5#�$

x-int: "− �� , 0$ , (2, 0)

17) y = (8x – 3)(x + 1)

vertex: (–0.31, –3.8)

x-int:(0.375, 0),(–1, 0)

18) y = (3x + 8)(2x + 3)

vertex: (–2.08, –2.04)

x-int:

"− %� , 0$ , "−

�� , 0$

19) 5(2x – 1)(x – 10)

20) 2(7v + 5)(v + 3)

21) 2(5r + 7)(r – 1)

22) 4(3n – 8)(n + 5)

23) D

24) a. 24 feet

b. –8(2t + 1)(t – 3),

The ball hits the

ground at t = 3

seconds; t = –0.5

does not make

sense in this

context.

24) c. 49 feet at 1.25

seconds

d.

25) (3x + 2)(x + 1)

26) (2x + 1)(x + 3)

27) (7x – 2)(x – 1)

28) (3x – 1)(x – 5)

29) (3x + 1)(2x + 3)

30) (4b + 1)(2b + 1)

31) (8b – 1)(b – 1)

32) (6b + 1)(b + 2)

33) (4b – 3)(3b – 1)

34) (3b + 2)(b + 4)

35) (5k – 1)(k + 1)

36) (2k + 1)(k – 3)

37) (3k + 5)(k – 1)

38) (7k + 1)(k – 2)

39) (5k + 1)(2k – 1)

40) (3v + 1)(2v – 3)

41) (5v – 2)(v + 4)

42) (4v – 3)(v – 3)

43) (3v – 2)(v + 6)

44) –1(4v + 3)(2v – 5)

45) (3x + 4)(3x + 4)

or (3x + 4)2

2 4 6 8 10

20

40

60

80

100

120

140

0 t

h(t)


328 APPENDICES

SECTION 4.3D

1) (x + 2)(x – 2)

2) (2x + 3)(2x – 3)

3) (x – 11)(x – 11)

4) (2x + 3)(2x + 3)

5) (x – 10)(x – 10) or (x – 10)2

6) (x + 4)(x – 4)

7) (3x – 5)(3x + 5)

8) (x + 7)(x + 7) or (x + 7)2

9) (8x – 13)(8x + 13)

10) (3x +4)(3x + 4) or (3x + 4)2

11) (x – 5)(x + 5)

12) (4x – 9)(4x + 9)

13) Equivalent

14) No

15) y = (x + 12)(x – 12)

vertex: (0, –144)

x-int: (12,0), (–12,0)

16) y = (x +4)2

vertex: (–4, 0)

x-intercept: (–4, 0)

17) y = (x + 6)2

vertex and x-int: (–6,0)

18) (5x + 2)(5x – 2)

vertex: (0, –4)

x-intercept: "�& , 0$ , "−�& , 0$

19) 2(7x + 12)(7x – 12)

20) 2(x + 8)2 or 2(x + 8)(x + 8)

21) (4x – 3y)(4x + 3y)

22) (xy – 10)(xy + 10)

23) (2a + b)(2a + b) or (2a + b)2

24) (2x + 5y)2 or (2x + 5y)(2x + 5y)

25) (x2 + 3y2)2 or (x2 + 3y2)(x2 + 3y2)

26) (2k2 + 5w2)2 or (2k2 + 5w2) (2k2 + 5w2)

SECTION 4.3E

1) 16, 4,

? in grid is 4 • 4 = 16

2) 25, 5

? in grid is 5 • 5 = 25

3) 4, 2

x 2

x x2 2x

2 2x 4

4) 100, 10

x 10

x x2 10x

10 10x 100

5) 1, 1

6) 9, 3

7) 81, 9

8) 64, 8

9) 36, (x – 6)2

10) 144, 12

11) 49, (x + 7)2

12) 36, (x – 6)2

13) %#' , "� − (

�$�

14) �&' , "� + &

�$�

15) a. x = –4

b. y = (x + 4)2 – 21

c. (–4, –21)

d. yes

-16 16

-150

-100

-50

x

y

-8 1

-2

12

x

y

-12 -6

-2

12

x

y

-0.8 -0.4 0.4 0.8

-5

10

x

y



SECTION 4.3E (continued)

15) e.

x –10 –9 –8 –7 –6 –5 –4

f (x) 15 4 –5 –12 –17 –20 –21

x –3 –2 –1 0 1 2 3

f (x) –20 –17 –12 –5 4 15 28

16) f (x) = (x + 3)2 – 11

x –6.3 –4 –3 –2 0.32

f (x) 0 –10 –11 –10 0

Tables will vary, but the graphs should match.

17) f (x) = (x – 7)2 – 36

18) f (x) = (x + 4)2 – 31

19) f (x) = (x – 1)2 – 7

20) a. –2 b. f (x) = (x + 2)2 – 16 c. (–2, –16) d. yes e.

x –6 –4 –2 0 2

f (x) 0 –12 –16 –12 0

Tables will vary, but the graphs should match.

SECTION 4.3ext

1) a. x = –4

b. f (x) = 3(x + 4)2 – 53

c. (–4, –53)

d. yes

e.

x –8.2 –6 –4 –3 0.2

f (x) 0 –41 –53 –50 0

Tables will vary, but the graphs

should match.

1) e. (con’t) 1) f. x = –4

g. Domain: All real numbers

Range: y ≥ –53

-7 3

-14

10

x

y

4 8 12 16

-40

-32

-24

-16

-8

8

x

y

-12 -8 -4 4

-40

-32

-24

-16

-8

8

x

y

-4 4 8

-8

8

16

24

x

y

-8 -4 4

-16

-8

8

x

y


330 APPENDICES

SECTION 4.3ext (continued)

2) f (x) = –1(x + 1)2 + 5

Possible table points (one of

which should be the vertex):

(–3, 1), (–1, 5), (1, 1)

3) f (x) = 5(x + 3)2 – 40

Possible table points (one of


(–6, 5), (–3, –40), (0, 5)

4) f (x) = –3(x + 1)2 + 7 Possible table points (one of


(–3, –5), (–1, 7), (1, –5)

5) f (x) = –2(x + 2)2 + 5 Possible table points (one of

which should be the vertex): (–4, –3), (–2, 5), (0, –3)

6) )(�) = 2 "� + &'$

�− '#

%

7) )(�) = 3 "� + ��$

�+ �

�

8) )(�) = 5 "� + �&$

�− *'

&

9) )(�) = 2 "� + ('$

�− %#

%

10) a. f (x) = 2(x + 5)(x – 3)

b. (–5, 0) and (3, 0)

c. f (x) = 2(x + 1)2 – 32

d. Vertex: (–1, –32)

e.

11) a. (4, 0), (–2, 0)

b. f (x) = 3x2 – 6x – 24

c. f (x) = 3(x – 1)2 – 27

d. (1, –27)

e.

12) a. A(x) = –1(x – 15)2 + 225

b. 15 feet by 15 feet

c. 225 sq. feet

UNIT 4 REVIEW

1) Up or down: Up

Vertical str/comp: stretch

vertex: (1, 0)

Axis of Sym: x = 1

y-intercept: (0, 2)

1)

1) x-intercept(s): (1,0)

Domain: all real numbers

Range: y ≥ 0

-5 5

-10

10

x

f(x)



UNIT 4 REVIEW (continued)

2) Up or down: Up

vertical str/comp: compression

vertex: (1, –2)

Axis of Symmetry: x = 1

y-intercept: (0, –1.5)

x-intercept: (3, 0) and (–1, 0),


Range: y ≥ –2

3) Up or down: Down

vertical str/comp: stretch

vertex: (1, –3)

Axis of Symmetry: x = 1

y-intercept: (0, –5),

x-intercept: none


Range: y ≤ –3

4) y = 2x2 + 9x – 5

5) y = –2x2 + 20x – 48

6) y = 9x2 + 72x + 132

7) y = –2x2 + 4x + 1

8) y = (x – 3)(x – 8)

9) y = (x + 3)(5x – 2)

10) y = 5(2x – 3)(x – 1)

11) y = (2x + 5)2

12) y = (3x – 2)2

13) y = 4(2x + 5)(2x – 5)

14) y = (2x + 1)(5x – 4)

15) y = (x – 7)2 – 41

16) � = "� − ��$

�− #*

'

17) y = 4(x + 3)2 + 4

18) � = 3 "� − #�$

�+ #

'

19) 64 feet

20) 2 seconds

21) 4 seconds

22) 0 ≤ t ≤ 4, the time has to be

positive and stops in the

context of this problem

when the ball comes back to

the ground.

23) 0 ≤ h(t) ≤ 64, the height

cannot be negative because

that would be below ground

and the highest the ball gets

it 64 feet.

24) 12.25 feet

25) 0.625 seconds

26) 10 feet

27) 1.5 seconds, the positive x-int.

28) 6 feet, that is the y-intercept

29) a. ℎ�+ = −16+� + 315+ + 4

b.

t h (t)

0 4

3 805

3.957 1000

3.957 1000

9.84 1554.4

15.75 1000

19.7 0

b.

c. (19.7, 0); the rocket returns

to the ground at 19.7

seconds.

d. (4, 0); the rocket was 4feet

above ground when

launched.

e. (9.84, 1554.4); at 9.84

seconds the rocket reached

its maximum height of

1554.4 feet.

f. 805 feet

g. 3.957 ≤ � ≤ 15.73

h. Domain: 0 ≤ � ≤ 19.7; the

rocket is in the air during

these times.

Range: 0 ≤ � ≤ 1554.4;

this represents the possible

heights of the rocket at any

given time.


332 APPENDICES

SECTION 5.1A

1) “Find the zeros of the function” means to find the x-value(s) that make the function f(x) = 0. (answers will vary)

2) The x-intercepts (answers will vary)

3) 1 solution real roots (rational) zeros: x = 3

4) 2 solutions roots are complex-no real roots zeros: none

5) 2 solutions real roots (rational) zeros: x = 0 or x = –2

6) solutions: x = 3 or x = –5 check: f(3) = (3)2 + 2(3) – 15 = 0 f(–5) = (–5)2 + 2(–5) – 15 = 0 both solutions create a zero

7) solutions: x = 0 or x = 4 check: f(0) = –2(0)2 + 8(0) = 0 f(4) = –2(4)2 + 8(4) = 0

both solutions create a zero

8)

solutions: x = –2 or x = 6 check: f(–2) = –(–2)2 + 4(–2) + 12 = 0 f(6) = –(6)2 + 4(6) + 12 = 0

9)

solutions: x = –5 or x = 2 check: f(–5) = (–5)2 + 3(–5) – 10 = 0 f(2) = (2)2 + 3(2) – 10 = 0

10) solutions: x ≈ –1.32 or x ≈ 8.32 11) solutions: x ≈ –1.67or x = –1.50 12) solutions: x ≈ –0.94 or x ≈ 2.34 13) solutions: x ≈ –3.12 or x ≈ 5.12

0.5��3.12� � ��3.12 � 8 , 0

0.5�5.12� � �5.12 � 8 , 0

14) solutions: x ≈ –5.16 or x ≈ 1.16 ��5.16� � 4��5.16 , 6

�1.16� � 4�1.16 , 6

15) solutions: x ≈ –1.16 or x ≈ 2.16 2��1.16� � 2��1.16 � 5 , 0

2�2.16� � 2�2.16 � 5 , 0

16)

solutions: x = –5 or x = –1

17)


18)


19)

solutions: x = 0 or x = –8



SECTION 5.1A (continued)

20)


21)


22) a. (16) f(x) = x2 + 6x + 5 x = –5, x = –1

(17) f(x) = 5x2 + 30x + 25

x = –5, x = –1

(18) f(x) = 3x2 + 18x + 15

x = –5, x = –1

b. (19) f(x) = x2 + 8x

x = 0, x = –8

(20) f(x) = (½)x2 + 4x

x = 0, x = –8

(21) f(x) = 2x2 + 16x

x = 0, x = –8

c. (answers will vary) If one

function is a multiple of

another function, then they

will have the same

solutions.

23) a.

b. 5 feet; the point (5, 5) is the

vertex of the graph of the

equation.

c. (answers will vary) 10 feet;

this is the x-intercept of the

graph of the equation (that is

greater than zero).

SECTION 5.1B

1) a. Javier is correct. The x-intercept gives the time (t) when y = 0, which is where the ball has a height of 0 on the ground. (explanations will vary)

b. Meg’s ball hit the ground

first. It took 2.5 seconds

for it to hit the ground.

c. Javier’s ball hit the ground

second. It took 3.125

seconds for it to hit the

ground.

2) a. h(t) = –16t2 + 35t + 5.5 b. 24.5 feet

c. After 0.14 seconds and

then again after 2.05

seconds.

d. After 1.09 seconds.

2) e. 24.64 feet is the maximum

that the football ever gets

into the air.

f. 2.33 seconds is when the

ball hits the ground

(the x-intercept).

3) Vertex: (1.56, 41.06). The vertex reveals how long it takes (1.56 seconds) for the ball to reach the maximum height (41.06 ft). (explanations will vary)

4) t = 11 days, which is the x-coordinate of the vertex.

5) a. 1.5 sections is the maximum speed (the x-coordinate of the vertex of the graph)

b. 89 km/h (the y-coordinate

of the vertex of the graph)

5) c. 6.22 seconds (the

x-coordinate of the

x-intercept of the graph.

6) a. 50 is half the perimeter, so one length plus one width equals 50 feet of fencing. (explanations will vary)

b. A(L) = L(50 – L)

A(L) = –L2 + 50L

c. (L, A(L)) = (25, 625)

when the length is 25, the

width is 50 – L = 25. So,

Area = L × W

25 ft × 25 ft

625 ft2

Area of the tomato patch is

625 square feet.

d. The tomato patch is a

square shape (25ft × 25ft).


334 APPENDICES


7) a. According to #6d, Sharon should make a square with her 40 feet of fencing. So, 40 ÷ 4 = 10 feet on each side.

b. A(L) = L(20 – L)

A(L) = –L2 + 20L

calc the max, shows a

vertex at (10, 100)

7) c. (L, A(L)) = (10, 100)

when the length is 10, the

width is 20 – L = 10. So,

Area = L × W

10 ft × 10 ft = 100 ft2

The area of the fenced play

area is 100 square feet.

8) Yes, the maximum height of the orange is 19 feet (which is the y-coordinate of the vertex of the graph). Jim can grab the orange either on the way up for the orange or on its way down (explanations will vary).

9) At approximately 15 miles per hour will the car be able to stop for a sign 30 feet away.

10) a. 141.56 feet b. 5.79 seconds

SECTION 5.2A

1) 2(x + 3)

2) 3(y – 3)

3) 7(a + 4)

4) 12(3z – 1)

5) b(b + 1)

6) r(2 – r)

7) t(9t + 1)

8) n(4n – 5)

9) 4h(h + 3)

10) 9x(1 – 3x)

11) 2a(a + 2)

12) 4d(5d – 6)

13) (x + 7)(x + 6)

14) (x + 3)(x + 3) or (x + 3)2

15) (x + 8)(x + 4)

16) (x + 5)(x – 2)

17) (x – 5)(x – 5) or (x – 5)2

18) (x – 4)(x + 3)

19) (3x + 4)(x – 1)

20) (2x – 3)(x + 4)

21) (2x – 3)( 2x – 3) or (2x – 3)2

22) (6x – 1)(2x – 1)

23) 2(x + 1)(x – 5)

24) 3(x + 1)(x + 6)

25) (a – 3)(a + 3)

26) (x – 4)(x + 4)

27) (5x – 6)(5x + 6)

28) y = (x + 1)(x + 2)

29) y = (x + 7)(x – 7)

30) The x-intercepts have a y-coordinate of zero. Setting y = 0 and solving for x gives the solution(s), which are the x-intercept(s). (answers will vary)

31) y = (x – 4)(x – 20) (4, 0) and (20, 0)

32) y = (x + 10)(x – 1) (1, 0) and (–10, 0)

33) a. An arrow should be drawn to the location that the graph of the height of the ball in relationship to time touches the x-axis.

b. y = 0 means the place where

the ball hits the ground

(where the height is zero).

SECTION 5.2B

1) x = 4, x = –9 2) x = 2 (double root)

3) 3

4x = − ,

5

2x =

4) x = 3, x = –3 5) x = 2, x = –2 6) x = 0, x = 3 7) x = 0, x = 4

8) x = 1, x = –9 9) x = –4, x = –3 10) x = 5 (double root) 11) x = –1, x = 4 12) x = –1, x = 5 13) x = 5, x = 8

14) 4

3x = − , x = –2

15) 5

4x = − ,

1

2x =

16) 1

5x = − , x = –2

17) 1

2x = − , x = 8

SECTION 5.2C

1) The legs of the triangle have lengths of 3 units and 4 units.

2) The legs of the triangle have lengths of 5 units and 12 units.

3) The value of x is 10 (side lengths are 9 units and 12 units).

4) The value of x is 11 (side lengths are 10 units and 12 units).

5) The two numbers that satisfy the situation are 8 and 15.

6) The dimensions are 14 feet by 48 feet.

7) The two positive numbers are 5 and 3.

8) The two negative integers are –4 and –6.

9) (x = 2) You can frame a 2ft × 2ft square picture.



SECTION 5.2D

1) a. 2√10

b. 4√5

c. 10 2

d. 6 5

e. 10 6

f. 42 2

2) a. 9

b. 6 11 2+

c. 2 10 7−

d. 5 5 2−

e. 3 5−

f. 11 8 3+

3) a.

b.

4) a. 5

b. 3 10

c. 3 2

d. 12 15

e. –80

f. 6 42−

5) a. 15 5 7+

b. 6 3 3− +

c. 2 8 3+

d. 2 5 3 10+

e. 6 3 2− −

f. 5 3 2 6− +

6) a. 19 8 3+

b. 22 12 2−

c. 47 26 3−

d. 76 24 2+

e. 132 48 7−

f. 42 7 11−

7) a. yes, this x-value is a solution

b. no (x = 6 2+ ) is not a

solution to the equation.

c. yes, (x = 5 3 5− ) is a

solution to the equation.

8) a. 2a + 5

b. 2 5 2+

c. 7 5 6− −

d. 5 6 10− +

9) 2

?

?

?

?

?

?

?

?

3 57 3 572 3 6 0

4 4

9 6 57 57 3 572 3 6 0

16 4

66 6 57 9 3 572 6 0

16 4

33 3 57 9 3 572 6 0

8 4

33 3 57 9 3 576 0

4 4

33 9 3 57 3 576 0

4

246 0

4

6 6 0

0 0

+ +− − =

+ + +− − =

+ − −+ − =

+ − −+ − =

+ − −+ − =

− + −− =

− =

− =

=

SECTION 5.2E

1) x = –2, x = 2 ( 2x = ± )

2) x = –4, x = 4 ( 4x = ± )

3) x = –5, x = 5 ( 5x = ± )

4) x = –5, x = 7

5) x = –7, x = 1

6) x = –1, x = 5

7) It takes 2 seconds to reach the ground.

8) A sign should be posted stating the maximum speed is 28 km per hour.

9) The other leg of the triangle Andy created is 9 feet long.


336 APPENDICES

SECTION 5.2F

1) The error occurs when Omar tried to add 3 to both sides before he takes the square root of both sides. The correct solutions are x = 6 or x = 0

2) Yes, both of the values given are solutions.



5) No, neither value given is a solution.

6) Both answers are correct. The balloon could have hit a 6 foot tall student on the way up after 0.15 seconds, or it could have hit the student on its descending path after 1.15 seconds.

(explanations will vary)

7) 3 2x = ±

8) 1 4 2x = ±

9) 2x = ±

10) x = –2, x = 4

11) 5 3x = ±

12) x = –2

13) x = –2, x = 4

14) 2 2x = ±

15) 2.41t ≈ seconds

16) 3 ± 2√6 miles, but only 3 �

2√6 , 7.9miles makes sense.

SECTION 5.2G

1) a. 1− b. 1

c. i d. 1

e. i f. i−

g. 5i h. 12i

i. 10 7i j. 14 2i

k. 4 10i− l. 50i

2) a. 9 4i+ b. 13 6i−

c. 2 2i− + d. 11 3i+

e. 9 3i− f. 12 19i−

3) a. 6− b. 12

c. 24 2i d. 60i−

e. 48i f. 18 2

4) (all answers in #4 may vary) a. In method 1, Kasem never

used the definition on

1i = − to rewrite either

radical. Method 2 uses the definition correctly.

b. (1) Method 2 combines

4 9 36⋅ = instead of

simplifying each separately like in method 3.

4) c. If there is a negative number under a radical, you must simplify it to include the i’s before combining with other radicals.

5) a. 2 3−

b. 12 10−

c. 63 2

6) a. 5 15i−

b. 9 3i+

c. 6 60i−

d. 12 6i−

e. 10 40i+

f. 2 5 2 10i+

7) a. 16 30i− +

b. 36 48i−

c. 7 30 2i−

8) a. 68 24 2i− +

b. 36 48 7i− −

c. 24 7 11i− − 9) a. Yes

( )2

2 5 6 56

56 56

i− + =

=

b. Yes

( ) ( )2

4 3 8 4 3 30 5

5 5

i i+ − + + =

=

9) c. Yes

( ) ( )

( ) ( )( ) ( ) ( )

2

1 3 5 2 1 3 5 48 2

1 3 5 1 3 5 2 6 5 48 2

1 6 5 9 1 5 2 6 5 48 2

2 2

i i

i i i

i i

− − − + =

− − − + + =

− + − − + + =

=

10) a. 3 4i+

b. 3 5 2i+

c. 5 7 6i− −

d. 6 3i− +

11)

( )

( )

2?

2 ?

?

?

?

?

3 32 6 8 3

2 2

9 62 3 3 8 3

4

9 6 12 9 3 8 3

4

8 62 1 3 3

4

4 32 1 3 3

2

4 3 1 3 3

3 3

i i

i ii

ii

ii

ii

i i

+ + − + =

+ +− + + =

+ + −− − + =

+ − − =

+ − − =

+ − − =

=



SECTION 5.2H

1) 3i

2) 35 2i

3) 2 3i

4) 30i

5) 33 2i

6) 11i

7) –1

8) –9

9) –25

10) –1

11) 16 30i− −

12) 14 48i− +

13) 40 28i− +

14) Yes

−2(3/)� + 3 = 21 21 = 21

15) Yes

0�5 + 4/ − 51�

− 1 = −17 −17 = −17

16) Yes �2/� + 11 = 7

7 = 7

17) Yes

0�−2 + 5/ + 21�

= −25 −25 = −25

18) � = ±4√3

19) � = 1 ± 2/√6 20) � = ±3 21) � = −1 ± 3/ 22) + ≈ 1.86 seconds. The balloon

lands on the ground after approximately 1.86 seconds. (–0.96 does not work when verified).

(#23-26, verification of solutions should

be shown by student.)

23) � = ±5√5 24) � = −7 25) � = 2 ± 3/ 26) � = ±/√6

27) 7.071s ≈ inches; The length of one side of the square is approximately 7.071 inches.

28) � = ±2/. With these solutions being imaginary, it reinforces the reason why the graph does not cross the x-axis. There are no real solutions, there are two complex solutions.

SECTION 5.2I

1) 7x = , 3x = −

2) 0x = , 8x = −

3) 18x = , 4x = −

4) From line 1 to line 2, the person forgot to add 25 to both sides of the equation.

5) 9; 4x = , 10x = −

6) 81; 22x = , 4x = −

7) 1x = , 29x = −

8) 14x = , 4x = −

9) 6x =

10) � = 27 , � = 3

11) � = 22 The two numbers that satisfy the conditions are 22 and 24.

12) � = 5 The length of each side of the original garden is 5 yards and the area of the garden is 25 square yards.

SECTION 5.2J

1) 7 15x = − ±

2) 9 2 3x = ±

3) 5 4 2x = ±

4) 5 53

2x

− ±=

5) 7 37

2x

±=

6) 14 6 3x = − ±

7) 2x =

8) 1 22

2x

− ±=

9) 1 2

3x

±=

10) When y > 0, there will be 2 real solutions.

When y = 0 there will be 1

real solution.

11) 1.5x = ; The width of the

pool is 15 meters and the

length is 19 meters.

12) 6t = ; The arrow will strike

the ground after 6 seconds.

13) a. ( )2 424h = ; The bottle

rocket is 424 feet high after

2 seconds.

b. 13.13t ≈ or 1.87t ≈ ; The

rocket will be at 400 feet in

the air at two different

times; once on the way

‘up’ and once on the way

‘down’ during its flight.

SECTION 5.2K

1) 5 12x i= ±

2) 7 2 6x i= ±

3) 13 8 2x i= − ±

4) 6 2x i= − ±

5) 12 6 2x i= ±

6) 1 15

2

ix

− ±=

7) 2 4x i= ±

8) 3 21

10

ix

±=

9) 3 31

8

ix

− ±=

10) 250x = ; Emma hits the golf

ball 250 yards.


338 APPENDICES

SECTION 5.2K (continued)

11) 1

2x = or

11

2x = ; At exactly

½ second and 1½ seconds, the tape measure is at exactly 17 feet above the ground. Gail can catch the tape measure anywhere between ½ second and 1½ seconds after Veronica tosses the tool.

12) When y < 0 there will be 2 imaginary solutions for a quadratic equation.

13) Create the quadratic formula: 2

2

2

2

2 22

2

2 2

2 2

0

0

0

2 4

4

2 4 4

ax bx c

ax bx c

a a

bx cx

a a

bx cx

a a

bx b c bx

a a a a

b b acx

a a a

+ + =

+ +=

+ + =

+ = −

+ + = − +

+ = −

13) (continued)

2 2

2

2

2 2

2 2

2

2

4

2 4

4

2 2

4 4 or

2 2 2 2

4 4 or

2 2 2 2

4

2 2

4

2

b b acx

a a

b b acx

a a

b b ac b b acx x

a a a a

b b ac b b acx x

a a a a

b b acx

a a

b b acx

a

− + =

−+ =

− −+ = − + = +

− −= − − = − +

−= − ±

− ± −=

SECTION 5.2L

1) If the quadratic equation does not factor, the roots are either irrational or imaginary. (Answers will vary)

2) a: 2 b: –5 c: –3

x = 3 or 1

2x = −

solve by factoring? Yes, 2 4b ac− (49) is a perfect

square number.

3) a: 1 b: –7 c: 9

7 13

2x

±=

solve by factoring? No, 2 4b ac− (13) is not a perfect

square number. 4) a: 5 b: 3 c: –1

3 29

10x

− ±=


square number.

5) a: 1 b: 1 c: –1

� =−1 ± √5

2


square number 6) a: 9 b: 6 c: –1

1 2

3x

− ±=

7) a: 2 b: 3 c: –1

3 17

4x

− ±=

8) 1.68 seconds 9) a. 22 seconds

b. 11 seconds

SECTION 5.2M

1) 2 3i

2) 2 6

3) 4i

4) 1x = , 3x = −

5) 3

2x = −

6) 3 2

2

ix

±=

7) 6 11

4

ix

− ±=

8) 1 5x i= − ±

9) 3x = , 5x = −

10) Error on line 6, need to divide

both terms of the numerator

by the denominator value of 2;

3x i= − ±

11) 3 7x = − ±

12) 2x =

13) 13 11

10

ix

±=

14) a. 2.6 seconds

b. Since t = 0 means time

when object is initially

thrown, positive time

means time after the throw

was started. Negative time

implies time before the

object was thrown, which

doesn’t make sense here.

(Answers will vary)

c. The object was at a height

of 25 feet after 0.75

seconds and again at 1.75

seconds.



SECTION 5.2N

1) » Use a graphing utility to find real zeros.

» Factor and use the zero product property.

» Square root property » Complete the square » Quadratic formula

2) Graph: The solutions are the x-intercepts of x = 3, x = –2

Factor:

( 3)( 2) 03 or 2

x x

x x

− + =

= = −

Quadratic Formula:

1 25

2

1 5

2

3 or 2

x

x

x x

±=

±=

= = −

Solutions:

3 or 2x x= = − Method:: The factoring method

is the fastest and yields rational solutions with fewer possibilities of careless errors.

(explanations may vary) 3) Graph: The solutions are the

x-intercepts of x ≈ ±1.73

3) (continued)

Square Root:

2

2

3 0

3

3

3

x

x

x

x

− =

=

=

= ±

Quadratic Formula:

0 12

2

2 3

2

3

x

x

x

±=

±=

= ±

Solutions: 3x = ±

Method: The square root method. There is no linear term (the x term). The b-value is equal to zero. (Answers will vary)

4) a. Graphing calculator (but it can’t be used to find complex/imaginary solutions)

b. Completing the Square and Quadratic Formula

c. Factoring or Square Root property

5) Factoring – it’s a trinomial that factors nicely. (method and explanation may vary)

1

2x = , 3x = −

6) Factoring – it is a binomial with a common factor. (method and explanation may vary)

0x = , 1000x =

7) Factoring – it is a trinomial with a leading coefficient of 1. (method and explanation may vary)

9x = − , 2x =

8) There are two scenarios for this situation:

(1) Numbers 8 and 9 (2) Numbers –8 and –9

9) width: 7 in. length: 10 in. 10) Suzie needs 80 feet of fencing. 11) Mike needs 40 feet of fencing.

12) Solutions: � = ±3√2 Methods included here may

vary

Method 1: Square Root 2

2

2

2 3 39

2 36

18

3 2

x

x

x

x

+ =

=

=

= ±

Method 2: Quadratic Formula

0 288

4

12 2

4

3 2

x

x

x

±=

±=

= ±

Why? Using square roots has the fewest number of steps and is the easiest. The function is not also factorable, so the quadratic formula was the only other method to use.

13) Solutions: 25x = ; so the width is 25 m and length is 35 m Method 1:Complete the

Square

( )

( )

2

2

2

10 875

10 ____ 875 ____

10 25 900

5 900

continue solving from here

x x

x x

x x

x

+ =

+ + = +

+ + =

+ =

…

Note that –35 is extraneous. Method 2: Factoring

( )

( ) ( )

2

10 875

10 875 0

25 35 0

continue solving from here

x x

x x

x x

+ =

+ − =

− + =

…

Why? Complete the Square may be more efficient since

"��$� was a whole number, and

the square root property is well engrained. Factoring might have taken longer to find 2 integers with a product of –875. (answers/methods may vary)

-5 -4 -3 -2 -1 1 2 3 4 5

-7-6-5-4-3-2-1

12345

x

y


340 APPENDICES

SECTION 5.2N (continued)

14) 3 21

3

ix

− ±= ; solutions

should be verified

15) 6 39

3x

±= ; solutions should

be verified

16) 3x = −

( ) ( )2

3 6 3 9 0

0 0

− + − + =

=

17) 9

ix = ±

2

2

81 1 09

0 0 and

81 1 09

0 0

i

i

+ =

=

− + =

=

18) a. The ball was in the air 5 sec.

b. The ball reaches the max

height at 2.5 seconds.

c. The max height is 100 feet.

d. Graph a second line at y2 =

60 and use “calculate

intersect” to get the x-

coordinate that creates a

function value of 60. The

ball is at 60 feet again at

approximately 4.08 seconds.

SECTION 5.2O

(1) – (5) (answers will vary)

Square Root Property:

Eq’n #4: ( )2

3 8x − =

Solutions: 3 2 2x = ±

( )( )( )

( )( )( )

2

2

2

2

3 2 2 3 8

2 2 8

8 8 and

3 2 2 3 8

2 2 8

8 8

+ − =

=

=

− − =

− =

=

Factoring:

Eq’n #2: 2 2 15 0x x− − =

Solutions: 5 or 3x x= = −

( ) ( )

( ) ( )

2

2

5 2 5 15 0

25 10 15 0

0 0 and

3 2 3 15 0

9 6 15 0

0 0

− − =

− − =

=

− − − − =

+ − =

=

Factoring:

Eq’n #3: 2 12 20x x+ =

Solutions: 6 2 14x = − ±

( ) ( )

( ) ( )

2

2

6 2 14 12 6 2 14 20

20 20

and

6 2 14 12 6 2 14 20

20 20

− + + − + =

=

− − + − − =

=

Quadratic Formula:

Eq’n #1: 22 7 15 0x x+ − =

Solution(s): 3

or 52

x x= = −

( ) ( )

2

2

3 32 7 15 0

2 20 0 and

2 5 7 5 15 0

0 0

+ − =

=

− + − − =

=

Graphing:

Eq’n #5:

( ) 216 24 4755h t t t= − + +

Solution: 18.005t ≈

It took approximately 18 seconds for the rock to hit the ground.

( ) ( )2

16 18.005 24 18.005 4755 0

0 0

− + + =

=

6) a. 7 or 1x x= =

b. graph:

c. The intersection points of

the 2 graphs in part b) are the solutions for x in part a).

9 = (7 − 4)� 9 = 9 and

9 = �1 − 4� 9 = 9

7) a. The 4.5 feet is how far the

hose’s nozzle is above the

ground where the water

begins to shoot out.

b. 38.55 feet

c. No. At 27.89 feet from the

nozzle, the stream would

hit the top of the 6 foot

fence. However, at the 28

foot distance, the water

height is lower, reaching

only 5.96 feet high.

8) (answers will vary)

Xmin: –30 Xmax: 30

Ymin: –400 Ymax: 1000

-25 25-500

5000

t

h(t)



SECTION 5.3A

1) In the quadratic formula, 2 4

2

b b acx

a

− ± −= , the

discriminant is the value of the

expression 2 4b ac− that is

under the radical. This number is used to determine the number and type of solutions of a quadratic equation.

2) discriminant: –4 number of solutions: 2 type: complex

3) discriminant: 16 number of solutions: 2 type: real, rational


5) discriminant: 217 number of solutions: 2 type: real, irrational





10) the second line of ‘work’, the negative 6 must be in

parentheses ( ) ( ) ( )2

6 4 1 5− −

discriminant: 16

# of solutions 2

type: real, rational

11) a. 240000 0.003 12 27760x x= + +

b. 20.003 12 12240 0x x+ − =

c. 290.88

d. Yes, the discriminant > 0,

which yields two real

solutions. Only the positive

solution would pertain to

this story. 843 helmets.

12) D = 977. Yes, the discriminant is greater than zero and yields two real solutions.

13) D = –280,000. No, since the discriminant is less than 0 (negative) there are no real solutions.

SECTION 5.3B

1) a. 36

b. 2

c. real, rational

d. 2, 4x x= − =

e.

f. (1, –9)

g. minimum

h. (0, –8)

i. all real numbers

j. � ≥ −9

2) a. 16

b. 2

c. real, rational

d. 2x = ±

e.

f. (0, 4)

g. maximum

h. (0, 4)

i. all reals

j. � ≤ 4

3) D 4) discriminant = 28 2 solutions, real, irrational

5) discriminant = 0 1 solution, real, rational

6) discriminant = –11 2 complex solutions

7) a. negative discriminant b. zero discriminant

c. positive discriminant

8) Discriminant is a positive perfect square number

2 real, rational solutions

Possible equation:

( ) ( )2

3 7

10 21

y x x

y x x

= − − −

= − + −

(Answers may vary)

-10 10

-10

10

x

y

-10 10

-10

10

x

y


342 APPENDICES

-4 -2 2 4

-4

-2

2

4

x

y

SECTION 5.4A

1) a. Not a sol. b. Yes, a sol. 2) a. Yes, a sol. b. Not a sol. 3) a. Not a sol. b. Not a sol. 4) (test points may vary) Use (0, 0)

Is0 ≤ −4? No, so points

outside the parabola are

solutions.

5) (test points may vary) Use (0, 0)

Is0 ≥ −3? Yes, so points

outside the parabola are

solutions.

6) (test points may vary) Use (0, 0)

Is0 > 6? No, so graph

outside the parabola.

7) < 8) ≤ 9) > 10) C 11) A 12) F 13) E 14) B 15) D 16)

17)

18)

19)

20)

21)

22)

23)

-2 2 4 6

-4

-2

2

4

x

y

-6 -4 -2 2

-2

2

4

6

8

x

y

-10 -8 -6 -4 -2 2

-6

-4

-2

2

4

x

y

-2 2 4 6

-4

-2

2

4

x

y

-6 -4 -2 2

-2

2

4

6

8

x

y



SECTION 5.4B

1) a. When 4 or 2x x< − >

b. 4 2x− < <

2) a. 3 2x− < < b. 3 or 2x x< − >

3) Answers will vary, but

x-intercepts and concavity must match. Example:

4) Answers will vary, but

x-intercepts and concavity must match. Example:

5) 5 or 5x x< − >

6) � ≤ 0or� ≥ 4

7) 2 < � < 5

8) 2 4 5 0

1 or 5

x x

x x

− + + ≤

≤ − ≥

9) 2 8 20 0x x− − >

2 or 10x x< − >

10) 2 12 27 0x x− + < 3 < � < 9

11) � ≤ #

� or� ≥ 5

12) 2

53

x< <

Function values are

MORE than 0

above x-axis

Function values are

MORE than 0

above x-axis

-6 6

-5

5

x

y

Function values are

LESS than 0

below x-axis

-5 5 10

-5

5

x

y

Function values are

MORE than 0

above x-axis

Function values are

LESS than 0

below x-axis

Function values are

MORE than 0

above x-axis

Function values are

LESS than 0 below x-axis

-5 8

-5

5

x

y

Function values are

LESS than 0

below x-axis


344 APPENDICES


13) $20 $100x< < The x-intercepts of the graph

are at (20, 0) and (100, 0) with

a vertex is at (60, 1600).

The company will make a

profit if they sell the coats for a

selling price between $20 and

$100.

14) 1 3t≤ ≤ seconds

The curve intercepts the horizontal line h(t) = 52 are at t = 1 and t = 3, vertex is at (2, 68)

The ball will be greater than or

equal to 52 feet above the

ground from 1 second after

hitting the ball to 3 seconds

after hitting the ball.

15) a. 0 8.03x≤ ≤ ft. and

18.45x ≥ ft. The

x-intercepts are at (8.03, 0)

and (18.45, 0) and the

vertex is at (13.24, 1.47)

b. No. From 15 feet away,

the height of the ball would be 9.3 feet, so the ball would go over the top of the net. Also, from the graph and algebraic solution inequalities above, 15 is not in the correct ranges (explanations may vary).

16) 2d ≥ inches. Values in the

domain 0d ≤ are extraneous (a rope cannot have a negative diameter). The x-intercepts of the graph

are at (2, 0) and (–2, 0) with

the vertex is at (0, 0). The

graph of 21480W d≤

intercepts the horizontal line

of y = 5920 at (2, 5920).

The diameter of the rope

would need to be 2 inches or

more to support 5920 pounds.

17) Drivers with ages between 56.6 years and 70 years (inclusive on 70) have a reaction time slower than 25 milliseconds.

10 20 30 40 50 60 70

10

20

30

x

R



UNIT 5 REVIEW

1) Answers will vary

a. There is no x-term, such as 2 4x = or if it is in vertex

form, like ( )2

3 2 5 0x − + =

.

b. If it is not factorable and

a = 1 and b is even

c. a, b, and c are all integers

and fairly small numbers

d. If it is not factorable and

a, b, and c are larger

numbers and/or decimals.

2) Answers may vary

Suggest A, J and K for square

root method

Suggest C, D, and E for

completing the square

Suggest B, F and I for

factoring

Suggest G, H and L for

quadratic formula

3) Square Root Method

[A] 3 2x = ±

[J] 1 2 3x i= ±

[K] 13 or 1x x= = −

4) Completing the Square

[C] 10 2x i= − ±

[D] 3 2 6x = ±

[E] 2 or 6x x= = −

5) Factoring

[B] 2

or 53

x x= − = −

[F] 3

or 52

x x= − =

[I] 2

3x = −

6) Quadratic Formula

[G] 3 3 3

or 13 3

x− ±

= − ±

[H] 4 2

6

ix

±=

[L] 3

2 or 5

x x= =

7) a. 4 5 2x = − ±

b. 1

1 or 2

x x= = −

c. 2 5x i= − ±

d. 7 2 3

14

ix

±=

8) t = 5 seconds

9) 5.6 seconds

10) a. Discrim. is negative.

There are two complex

solutions.

b. Discrim. is zero. There is

one real, rational solution.

c. Discrim. is positive. There

are two real solutions.

11) a. Discrim. is –16. There are

two complex solutions.

b. Discrim. is 261. There are

two real, irrational

solutions.

c. Discrim. is 0. There is one

real, rational solution.

12) B

13) 5 1x− < < −

14)

• Algebraically find the

x-intercepts.

• Sketch the graph of a

parabola that has these

x-intercepts and opens up if

a > 0 or down if a < 0. Also

determine dashed or solid

line.

• Identify the x-values

(critical values) for which

the graph lies below the x-

axis (#15a) or above (or on)

the x-axis (#15b).

• For ≤ or ≥ include the

x-intercepts in the solution.

15) a. 2 1x− < < Function values less than 0 (below x-axis)

b. 8 or 4x x≤ − ≥ − Function

values are ≥ 0 (on or

above the x-axis)

16) Graph B 17) 1 or 4x x≤ − ≥

18) a. ( ) 216 48 32h t t t= − + +

b. 64 feet c. 68 feet d. 3 seconds e. 3.56 seconds f. No – doesn’t make sense to

talk about negative time

INTERMEDIATE A UNITS 4-5 · 2016-11-08 · Intermediate Algebra iii Intermediate Algebra Unit 4:...

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Transcript of INTERMEDIATE A UNITS 4-5 · 2016-11-08 · Intermediate Algebra iii Intermediate Algebra Unit 4:...