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Chapter 4 Resource Masters
Bothell, WA • Chicago, IL • Columbus, OH • New York, NY
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Copyright © by The McGraw-Hill Companies, Inc.
All rights reserved. The contents, or parts thereof, may be reproduced in print form for non-profit educational use with Glencoe Algebra 2, provided such reproductions bear copyright notice, but may not be reproduced in any form for any other purpose without the prior written consent of The McGraw-Hill Companies, Inc., including, but not limited to, network storage or transmission, or broadcast for distance learning.
Send all inquiries to:McGraw-Hill Education8787 Orion PlaceColumbus, OH 43240
ISBN: 978-0-07-660785-3MHID: 0-07-660785-2
Printed in the United States of America.
1 2 3 4 5 6 7 8 9 DOH 16 15 14 13 12 11
CONSUMABLE WORKBOOKS Many of the worksheets contained in the Chapter Resource Masters booklets are available as consumable workbooks in both English and Spanish.
MHID ISBNStudy Guide and Intervention Workbook 0-07-660301-6 978-0-07-660301-5Homework Practice Workbook 0-07-660299-0 978-0-07-660299-5
Spanish VersionHomework Practice Workbook 0-07-660300-8 978-0-07-660300-8
Answers For Workbooks The answers for Chapter 4 of these workbooks can be found in the back of this Chapter Resource Masters booklet.
ConnectED All of the materials found in this booklet are included for viewing, printing, and editing at connected.mcgraw-hill.com.
Spanish Assessment Masters (MHID: 0-07-660298-2, ISBN: 978-0-07-660298-8) These masters contain a Spanish version of Chapter 4 Test Form 2A and Form 2C.
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Chapter 4 iii Glencoe Algebra 2
Teacher’s Guide to Using the Chapter 4 Resource Masters ......................................... iv
Chapter ResourcesStudent-Built Glossary ....................................... 1Anticipation Guide (English) .............................. 3Anticipation Guide (Spanish) ............................. 4
Lesson 4-1Graphing Quadratic FunctionsStudy Guide and Intervention ............................ 5Skills Practice ................................................... 7Practice ............................................................. 8Word Problem Practice ...................................... 9Enrichment ...................................................... 10
Lesson 4-2Solving Quadratic Equations by GraphingStudy Guide and Intervention ...........................11Skills Practice .................................................. 13Practice ............................................................ 14Word Problem Practice .................................... 15Enrichment ...................................................... 16
Lesson 4-3Solving Quadratic Equations by FactoringStudy Guide and Intervention .......................... 17Skills Practice .................................................. 19Practice ........................................................... 20Word Problem Practice .................................... 21Enrichment ...................................................... 22Graphing Calculator Activity ............................ 23
Lesson 4-4Complex NumbersStudy Guide and Intervention .......................... 24Skills Practice .................................................. 26Practice ........................................................... 27Word Problem Practice .................................... 28Enrichment ...................................................... 29
Lesson 4-5Completing the SquareStudy Guide and Intervention .......................... 30Skills Practice .................................................. 32Practice ........................................................... 33Word Problem Practice .................................... 34Enrichment ...................................................... 35
Lesson 4-6The Quadratic Formula and the DiscriminantStudy Guide and Intervention .......................... 36Skills Practice .................................................. 38Practice ........................................................... 39Word Problem Practice .................................... 40Enrichment ...................................................... 41Spreadsheet Activity ........................................ 42
Lesson 4-7Transformations of Quadratic GraphsStudy Guide and Intervention .......................... 43Skills Practice .................................................. 45Practice ........................................................... 46Word Problem Practice .................................... 47Enrichment ...................................................... 48
Lesson 4-8Quadratic InequalitiesStudy Guide and Intervention .......................... 49Skills Practice .................................................. 51Practice ........................................................... 52Word Problem Practice .................................... 53Enrichment ...................................................... 54Graphing Calculator Activity ............................ 55
AssessmentStudent Recording Sheet ................................ 57Rubric for Extended Response ....................... 58Chapter 4 Quizzes 1 and 2 ............................. 59Chapter 4 Quizzes 3 and 4 ............................. 60Chapter 4 Mid-Chapter Test ............................ 61Chapter 4 Vocabulary Test ............................... 62Chapter 4 Test, Form 1 .................................... 63Chapter 4 Test, Form 2A ................................. 65Chapter 4 Test, Form 2B ................................. 67Chapter 4 Test, Form 2C ................................. 69Chapter 4 Test, Form 2D ................................. 71Chapter 4 Test, Form 3 .................................... 73Chapter 4 Extended-Response Test ................ 75Standardized Test Practice .............................. 76
Answers ........................................... A1–A36
Contents
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Chapter 4 iv Glencoe Algebra 2
Teacher’s Guide to Using the Chapter 4 Resource Masters
The Chapter 4 Resource Masters includes the core materials needed for Chapter 4. These materials include worksheets, extensions, and assessment options. The answers for these pages appear at the back of this booklet.
All of the materials found in this booklet are included for viewing, printing, and editing at connectED.mcgraw-hill.com.
Chapter ResourcesStudent-Built Glossary (pages 1–2) These masters are a student study tool that presents up to twenty of the key vocabulary terms from the chapter. Students are to record definitions and/or examples for each term. You may suggest that students highlight or star the terms with which they are not familiar. Give this to students before beginning Lesson 4-1. Encourage them to add these pages to their mathematics study notebooks. Remind them to complete the appropriate words as they study each lesson.
Anticipation Guide (pages 3–4) This master, presented in both English and Spanish, is a survey used before beginning the chapter to pinpoint what students may or may not know about the concepts in the chapter. Students will revisit this survey after they complete the chapter to see if their perceptions have changed.
Lesson ResourcesStudy Guide and Intervention These masters provide vocabulary, key concepts, additional worked-out examples and Check Your Progress exercises to use as a reteaching activity. It can also be used in conjunction with the Student Edition as an instructional tool for students who have been absent.
Skills Practice This master focuses more on the computational nature of the lesson. Use as an additional practice option or as homework for second-day teaching of the lesson.
Practice This master closely follows the types of problems found in the Exercises section of the Student Edition and includes word problems. Use as an additional practice option or as homework for second-day teaching of the lesson.
Word Problem Practice This master includes additional practice in solving word problems that apply the concepts of the lesson. Use as an additional practice or as homework for second-day teaching of the lesson.
Enrichment These activities may extend the concepts of the lesson, offer an histori-cal or multicultural look at the concepts, or widen students’ perspectives on the mathematics they are learning. They are written for use with all levels of students.
Graphing Calculator, TI–Nspire, or Spreadsheet Activities These activities present ways in which technology can be used with the concepts in some lessons of this chapter. Use as an alternative approach to some concepts or as an integral part of your lesson presentation.
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Chapter 4 v Glencoe Algebra 2
Assessment OptionsThe assessment masters in the Chapter 4 Resource Masters offer a wide range of assessment tools for formative (monitoring) assessment and summative (final) assessment.
Student Recording Sheet This master corresponds with the standardized test practice at the end of the chapter.
Extended Response Rubric This master provides information for teachers and stu-dents on how to assess performance on open-ended questions.
Quizzes Four free-response quizzes offer assessment at appropriate intervals in the chapter.
Mid-Chapter Test This 1-page test provides an option to assess the first half of the chapter. It parallels the timing of the Mid-Chapter Quiz in the Student Edition and includes both multiple-choice and free-response questions.
Vocabulary Test This test is suitable for all students. It includes a list of vocabulary words and 10 questions to assess students’ knowledge of those words. This can also be used in conjunction with one of the leveled chapter tests.
Leveled Chapter Tests
• Form 1 contains multiple-choice questions and is intended for use with below grade level students.
• Forms 2A and 2B contain multiple-choice questions aimed at on grade level students. These tests are similar in format to offer comparable testing situations.
• Forms 2C and 2D contain free-response questions aimed at on grade level students. These tests are similar in format to offer comparable testing situations.
• Form 3 is a free-response test for use with above grade level students.
All of the above mentioned tests include a free-response Bonus question.
Extended-Response Test Performance assessment tasks are suitable for all students. Sample answers and a scoring rubric are included for evaluation.
Standardized Test Practice These three pages are cumulative in nature. It includes three parts: multiple-choice questions with bubble-in answer format, griddable questions with answer grids, and short-answer free-response questions.
Answers• The answers for the Anticipation Guide
and Lesson Resources are provided as reduced pages.
• Full-size answer keys are provided for the assessment masters.
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Chapter 4 1 Glencoe Algebra 2
This is an alphabetical list of the key vocabulary terms you will learn in Chapter 4. As you study the chapter, complete each term’s definition or description. Remember to add the page number where you found the term. Add these pages to your Algebra Study Notebook to review vocabulary at the end of the chapter.
Vocabulary Term
Found on Page
Defi nition/Description/Example
completing the square
complex conjugates
complex number
constant term
discriminant
(dihs·KRIH·muh·nuhnt)
imaginary unit
linear term
maximum value
minimum value
pure imaginary number
(continued on the next page)
Student-Built Glossary
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Chapter 4 2 Glencoe Algebra 2
Student-Built Glossary
Vocabulary TermFound
on PageDefi nition/Description/Example
quadratic equation
(kwah·DRA·tihk)
Quadratic Formula
quadratic inequality
quadratic term
root
standard form
vertex
vertex form
zero
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Chapter 4 3 Glencoe Algebra 2
Before you begin Chapter 4
• Read each statement.
• Decide whether you Agree (A) or Disagree (D) with the statement.
• Write A or D in the first column OR if you are not sure whether you agree or disagree, write NS (Not Sure).
After you complete Chapter 4
• Reread each statement and complete the last column by entering an A or a D.
• Did any of your opinions about the statements change from the first column?
• For those statements that you mark with a D, use a piece of paper to write an example of why you disagree.
STEP 1 A, D, or NS
StatementSTEP 2 A or D
1. All quadratic functions have a term with the variable to the second power.
2. If the graph of the quadratic function y = ax2 + c opens up then c < 0.
3. A quadratic equation whose graph does not intersect the x-axis has no real solution.
4. Since graphing shows the exact solutions to a quadratic equation, no other method is necessary for solving.
5. If (x - 3)(x + 4) = 0, then either x - 3 = 0 or x + 4 = 0.
6. An imaginary number contains i, which equals the square root of -1.
7. A method called completing the square can be used to rewrite a quadratic expression as a perfect square.
8. The quadratic formula can only be used for quadratic equations that cannot be solved by graphing or completing the square.
9. The discriminant of a quadratic equation can be used to determine the direction the graph will open.
10. The graph of y = 2x2 is a dilation of the graph of y = x2.
11. The graph of y = (x + 2)2 will be two units to the right of the graph of y = x2.
12. The graph of a quadratic inequality containing the symbol < will be a parabola opening downward.
Step 2
Step 1
Anticipation GuideQuadratic Functions and Relations
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Capítulo 4 4 Álgebra 2 de Glencoe
PASO 1 A, D o NS
EnunciadoPASO 2 A o D
1. Todas las funciones cuadráticas tienen un término con la variable elevada a la segunda potencia.
2. Si la gráfica de la función cuadrática y = ax2 + c se abre hacia arriba, entonces c > 0.
3. Una ecuación cuadrática cuya gráfica no interseca el eje x no tiene solución real.
4. Dado que graficar muestra las soluciones exactas de una ecuación cuadrática, no se necesita ningún otro método para resolverla.
5. Si (x - 3)(x + 4) = 0, entonces x - 3 = 0 ó x + 4 = 0.
6. Un número imaginario contiene i, la cual es igual a la raíz cuadrada de -1.
7. El método de completar el cuadrado se puede usar para volver a plantear una expresión cuadrática como un cuadrado perfecto.
8. La fórmula cuadrática se puede usar sólo para ecuaciones cuadráticas que no pueden resolverse mediante la completación del cuadrado o una gráfica.
9. Se puede usar el discriminante de una ecuación cuadrática para determinar la dirección en que se abrirá la gráfica.
10. La gráfica de y = 2x2 es una dilatación de la gráfica de y = x2.
11. La gráfica de y = (x + 2)2 estará dos unidades a la derecha de la gráfica de y = x2.
12. La gráfica de una desigualdad cuadrática con el símbolo < será una parábola que se abre hacia abajo.
Paso 2
Paso 1 Antes de comenzar el Capítulo 4
• Lee cada enunciado.
• Decide si estás de acuerdo (A) o en desacuerdo (D) con el enunciado.
• Escribe A o D en la primera columna O si no estás seguro(a) de la respuesta, escribe NS (No estoy seguro(a)).
Después de completar el Capítulo 4
• Vuelve a leer cada enunciado y completa la última columna con una A o una D.
• ¿Cambió cualquiera de tus opiniones sobre los enunciados de la primera columna?
• En una hoja de papel aparte, escribe un ejemplo de por qué estás en desacuerdo con los enunciados que marcaste con una D.
Ejercicios preparatoriosFunciones cuadráticas y las relaciones
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Chapter 4 5 Glencoe Algebra 2
4-1 Study Guide and InterventionGraphing Quadratic Functions
Graph Quadratic Functions
Quadratic Function A function defi ned by an equation of the form f(x) = ax2 + bx + c, where a ≠ 0
Graph of a
Quadratic Function
A parabola with these characteristics: y-intercept: c; axis of symmetry: x = -b − 2a
;
x- coordinate of vertex: -b − 2a
Find the y-intercept, the equation of the axis of symmetry, and the x-coordinate of the vertex for the graph of f(x) = x2 - 3x + 5. Use this information to graph the function.a = 1, b = -3, and c = 5, so the y-intercept is 5. The equation of the axis of symmetry is
x = -(-3) −
2(1) or 3 −
2 . The x-coordinate of the vertex is 3 −
2 .
Next make a table of values for x near 3 − 2 .
x x2 - 3x + 5 f(x) (x, f(x))
0 02 - 3(0) + 5 5 (0, 5)
1 12 - 3(1) + 5 3 (1, 3)
3 − 2 ( 3 −
2 )
2
- 3 ( 3 − 2 ) + 5 11
− 4 ( 3 −
2 , 11
− 4 )
2 22 - 3(2) + 5 3 (2, 3)
3 32 - 3(3) + 5 5 (3, 5)
ExercisesComplete parts a–c for each quadratic function.a. Find the y-intercept, the equation of the axis of symmetry, and the x-coordinate
of the vertex.b. Make a table of values that includes the vertex.c. Use this information to graph the function.
1. f(x) = x2 + 6x + 8 2. f(x) = -x2 - 2x + 2 3. f(x) = 2x2 - 4x + 3
Example
xO
f (x)
xO
f (x)
xO
f (x)
xO
f (x)
6
4
2
-2 2 4
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Chapter 4 6 Glencoe Algebra 2
4-1 Study Guide and Intervention (continued)
Graphing Quadratic Functions
Maximum and Minimum Values The y-coordinate of the vertex of a quadratic function is the maximum value or minimum value of the function.
Maximum or Minimum Value
of a Quadratic Function
The graph of f (x ) = ax 2 + bx + c, where a ≠ 0, opens up and has a minimum
when a > 0. The graph opens down and has a maximum when a < 0.
Determine whether each function has a maximum or minimum value, and find that value. Then state the domain and range of the function.
ExercisesDetermine whether each function has a maximum or minimum value, and find that value. Then state the domain and range of the function. 1. f(x) = 2x2 - x + 10 2. f(x) = x2 + 4x - 7 3. f(x) = 3x2 - 3x + 1
4. f(x) = x2 + 5x + 2 5. f(x) = 20 + 6x - x2 6. f(x) = 4x2 + x + 3
7. f(x) = -x2 - 4x + 10 8. f(x) = x2 - 10x + 5 9. f(x) = -6x2 + 12x + 21
Example
a. f(x) = 3x2 - 6x + 7
For this function, a = 3 and b = -6.Since a > 0, the graph opens up, and the function has a minimum value.The minimum value is the y-coordinate of the vertex. The x-coordinate of the
vertex is -b − 2a
= -(-6)
− 2(3)
= 1.
Evaluate the function at x = 1 to find the minimum value.f(1) = 3(1)2 - 6(1) + 7 = 4, so the minimum value of the function is 4. The domain is all real numbers. The range is all reals greater than or equal to the minimum value, that is {f(x) | f(x) ≥ 4}.
b. f(x) = 100 - 2x - x2
For this function, a = -1 and b = -2.Since a < 0, the graph opens down, and the function has a maximum value.The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is -b −
2a = - -2 −
2(-1) = -1.
Evaluate the function at x = -1 to find the maximum value.f (-1) = 100 - 2(-1) - (-1)2 = 101, so the maximum value of the function is 101. The domain is all real numbers. The range is all reals less than or equal to the maximum value, that is {f(x) ⎪ f(x) ≤ 101}.
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Chapter 4 7 Glencoe Algebra 2
4-1 Skills PracticeGraphing Quadratic Functions
Complete parts a–c for each quadratic function.a. Find the y-intercept, the equation of the axis of symmetry, and the x-coordinate
of the vertex.b. Make a table of values that includes the vertex.c. Use this information to graph the function.
1. f(x) = -2x2 2. f(x) = x2 - 4x + 4 3. f(x) = x2 - 6x + 8
Determine whether each function has a maximum or a minimum value, and find that value. Then state the domain and range of the function.
4. f(x) = 6x2 5. f(x) = -8x2 6. f(x) = x2 + 2x
7. f(x) = -2x2 + 4x - 3 8. f(x) = 3x2 + 12x + 3 9. f(x) = 2x2 + 4x + 1
10. f(x) = 3x2 11. f(x) = x2 + 1 12. f(x) = -x2 + 6x - 15
13. f(x) = 2x2 - 11 14. f(x) = x2 - 10x + 5 15. f(x) = -2x2 + 8x + 7
xO
f (x)
xO
f (x)
xO
f (x)
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Chapter 4 8 Glencoe Algebra 2
4-1 PracticeGraphing Quadratic Functions
Complete parts a–c for each quadratic function.a. Find the y-intercept, the equation of the axis of symmetry, and the x-coordinate
of the vertex.b. Make a table of values that includes the vertex.c. Use this information to graph the function.
1. f(x) = x2 - 8x + 15 2. f(x) = -x2 - 4x + 12 3. f(x) = 2x2 - 2x + 1
xO
f (x)
xO
f (x)
xO
f (x)
Determine whether each function has a maximum or minimum value, and find that value. Then state the domain and range of the function.
4. f (x) = x2 + 2x - 8 5. f (x) = x2 - 6x + 14 6. v(x) = -x2 + 14x - 57
7. f (x) = 2x2 + 4x - 6 8. f (x) = -x2 + 4x - 1 9. f (x) = - 2 − 3 x2 + 8x - 24
10. GRAVITATION From 4 feet above a swimming pool, Susan throws a ball upward with a velocity of 32 feet per second. The height h(t) of the ball t seconds after Susan throws it is given by h(t) = -16t2 + 32t + 4. For t ≥ 0, find the maximum height reached by the ball and the time that this height is reached.
11. HEALTH CLUBS Last year, the SportsTime Athletic Club charged $20 to participate in an aerobics class. Seventy people attended the classes. The club wants to increase the class price this year. They expect to lose one customer for each $1 increase in the price.
a. What price should the club charge to maximize the income from the aerobics classes?
b. What is the maximum income the SportsTime Athletic Club can expect to make?
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Chapter 4 9 Glencoe Algebra 2
4-1 Word Problem PracticeGraphing Quadratic Functions
1. TRAJECTORIES A cannonball is launched from a cannon on the wall of Fort Chambly, Quebec. If the path of the cannonball is traced on apiece of graph paper aligned so that the cannon is situated on the y-axis, the equation that describes the path is
y = - 1
−1600
x2+
1−2
x + 20,
where x is the horizontal distance from the cliff and y is the vertical distance above the ground in feet. How high above the ground is the cannon?
2. TICKETING The manager of a symphony computes that the symphony will earn -40P2
+ 1100P dollars per concert if they charge P dollars for tickets. What ticket price should the symphony charge in order to maximize its profits?
3. ARCHES An architect decides to use a parabolic arch for the main entrance of a science museum. In one of his plans, the top edge of the arch is described by the graph of y = -
1−4
x2+
5−2
x + 15. What are the coordinates of the vertex of this parabola?
4. FRAMING A frame company offers a line of square frames. If the side length of the frame is s, then the area of the opening in the frame is given by the function a(s) = s2
- 10s + 24. Graph a(s).
O
5. WALKING Canal Street and Walker Street are perpendicular to each other. Evita is driving south on Canal Street and is currently 5 miles north of the intersection with Walker Street. Jack is at the intersection of Canal and Walker Streets and heading east on Walker. Jack and Evita are both driving 30 miles per hour.
a. When Jack is x miles east of the intersection, where is Evita?
b. The distance between Jack and Evita is given by the formula √�����x2
+ (5 - x) 2 .For what value of x are Jack and Evita at their closest? (Hint: Minimize the square of the distance.)
c. What is the distance of closest approach?
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Chapter 4 10 Glencoe Algebra 2
4-1 Enrichment
Finding the x-intercepts of a Parabola
As you know, if f (x) = ax2 + bx + c is a quadratic function, the values of x
that make f (x) equal to zero are -b + √ ���� b2 - 4ac −
2a and -b - √ ���� b2 - 4ac
− 2a
.
The average of these two number values is - b − 2a
.
O
f(x)
x
– –, f( ( (( b––2a
b––2a
b––2ax = –
f(x) = ax2 + bx + c
The function f (x) has its maximum or minimum
value when x = - b − 2a
. The x-intercepts of the parabola,
when they exist, are √ ���� b2 - 4ac
− 2a
units to the left and right of the axis of symmetry.
Find the vertex, axis of symmetry, and x-intercepts for f(x) = 5x2 + 10x - 7.
Use x = - b − 2a
.
x = - 10 − 2(5)
= -1 The x-coordinate of the vertex is -1.
Substitute x = -1 in f (x) = 5x2 + 10x - 7.
f (-1) = 5(-1)2 + 10(-1) - 7 = -12. The vertex is (-1,-12).
The axis of symmetry is x = - b − 2a
, or x = -1.
The x-coordinates of the x-intercepts are –1 ± √ ���� b2 - 4ac
− 2a
= –1 ± √ ������� 102 - 4 � 5 � (-7)
−− 2 � 5
= –1 ± √ �� 240
− 10
. The x–intercepts are (–1 – 2 − 5 √ �� 15 , 0) and (–1 + 2 −
5 √ �� 15 , 0) .
ExercisesFind the vertex, axis of symmetry, and x-intercepts for the graph of each function using x = - b −
2a .
1. f (x) = x2 - 4x - 8 2. g(x) = -4x2 - 8x + 3
3. y = -x2 + 8x + 3 4. f (x) = 2x2 + 6x + 5
5. A(x) = x2 + 12x + 36 6. k(x) = -2x2 + 2x - 6
Example
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Chapter 4 11 Glencoe Algebra 2
Study Guide and InterventionSolving Quadratic Equations by Graphing
Solve Quadratic Equations
Quadratic Equation A quadratic equation has the form ax 2 + bx + c = 0, where a ≠ 0.
Roots of a Quadratic Equation solution(s) of the equation, or the zero(s) of the related quadratic function
The zeros of a quadratic function are the x-intercepts of its graph. Therefore, finding the x-intercepts is one way of solving the related quadratic equation.
Solve x2 + x - 6 = 0 by graphing.
Graph the related function f (x) = x2 + x - 6.
The x-coordinate of the vertex is -b − 2a
= - 1 − 2 , and the equation of the
axis of symmetry is x = - 1 − 2 .
Make a table of values using x-values around - 1 − 2 .
x -1 - 1 − 2 0 1 2
f (x) -6 -6 1 − 4 -6 -4 0
From the table and the graph, we can see that the zeros of the function are 2 and -3.
Exercises
Use the related graph of each equation to determine its solution.
1. x2 + 2x - 8 = 0 2. x2 - 4x - 5 = 0 3. x2 - 5x + 4 = 0
xO
f (x)
xO
f (x)
xO
f (x)
4. x2 - 10x + 21 = 0 5. x2 + 4x + 6 = 0 6. 4x2 + 4x + 1 = 0
xO
f (x)
xO
f (x)
xO
f (x)
xO
-6
-4
-2
-4 -2 2
f (x)Example
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Chapter 4 12 Glencoe Algebra 2
Study Guide and Intervention (continued)
Solving Quadratic Equations by Graphing
Estimate Solutions Often, you may not be able to find exact solutions to quadratic equations by graphing. But you can use the graph to estimate solutions.
Solve x2 - 2x - 2 = 0 by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are located.
The equation of the axis of symmetry of the related function is
xO-2
-2
-4
2
2 4
4f (x)
x = - -2 −
2(1) = 1, so the vertex has x-coordinate 1. Make a table of values.
x -1 0 1 2 3
f (x) 1 -2 -3 -2 1
The x-intercepts of the graph are between 2 and 3 and between 0 and -1. So one solution is between 2 and 3, and the other solution is between 0 and -1.
ExercisesSolve the equations. If exact roots cannot be found, state the consecutive integers between which the roots are located.
1. x2 - 4x + 2 = 0 2. x2 + 6x + 6 = 0 3. x2 + 4x + 2= 0
xO
f (x)
xO
f (x)
xO
f (x)
4. -x2 + 2x + 4 = 0 5. 2x2 - 12x + 17 = 0 6. - 1 −
2 x2 + x + 5 −
2 = 0
xO
f (x)
xO
f (x)
xO
f (x)
Example
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Chapter 4 13 Glencoe Algebra 2
Skills PracticeSolving Quadratic Equations By Graphing
Use the related graph of each equation to determine its solutions.
1. x2 + 2x - 3 = 0 2. -x2 - 6x - 9 = 0 3. 3x2 + 4x + 3 = 0
x
f (x)
O
-2
-4
2-2-4 4
2
f(x) = x2 + 2x - 3
x
f (x)
O
-2
-4
-6
-8
-2-6 -4
f(x) = -x2 - 6x - 9
xO
f(x) = 3x2 + 4x + 3
12
8
4
–2–4–6
f (x)
Solve each equation. If exact roots cannot be found, state the consecutive integers between which the roots are located.
4. x2 - 6x + 5 = 0 5. -x2 + 2x - 4 = 0 6. x2 - 6x + 4 = 0
xO
f (x)
xO
f (x)
xO
f (x)
7. -x2 - 4x = 0 8. -x2 + 36 = 0
xO
f (x)
xO
f (x)
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Chapter 4 14 Glencoe Algebra 2
PracticeSolving Quadratic Equations By Graphing
Use the related graph of each equation to determine its solutions.
1. -3x2 + 3 = 0 2. 3x2 + x + 3 = 0 3. x2 - 3x + 2 = 0
xO-2
-2
-4
-4 2
2
f (x)4
xO-2-4 2
2
4
6
8
4
f (x)
xO-2-4 2
2
4
6
8
4
f (x)
Solve each equation. If exact roots cannot be found, state the consecutive integers between which the roots are located. 4. -2x2 - 6x + 5 = 0 5. x2 + 10x + 24 = 0 6. 2x2 - x - 6 = 0
xO
f (x)
xO
f (x)
xO
f (x)
7. -x2 + x + 6 = 0 8. -x2 + 5x - 8 = 0
xO
f (x)
xO
f (x)
9. GRAVITY Use the formula h(t) = v0t - 16t2, where h(t) is the height of an object in feet, v0 is the object’s initial velocity in feet per second, and t is the time in seconds.
a. Marta throws a baseball with an initial upward velocity of 60 feet per second. Ignoring Marta’s height, how long after she releases the ball will it hit the ground?
b. A volcanic eruption blasts a boulder upward with an initial velocity of 240 feet per second. How long will it take the boulder to hit the ground if it lands at the same elevation from which it was ejected?
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Chapter 4 15 Glencoe Algebra 2
Word Problem PracticeSolving Quadratic Equations by Graphing
1. TRAJECTORIES David threw a baseball into the air. The function of the height of the baseball in feet is h = 80t -16t2, where t represents the time in seconds after the ball was thrown. Use this graph of the function to determine how long it took for the ball to fall back to the ground.
h
t1 2 3 4 5-1
-40
40
80
2. BRIDGES In 1895, a brick arch railway bridge was built on North Avenue in Baltimore, Maryland. The arch is described by the equation h = 9 – 1
−50
x2, where h is the height in yards and x is the distance in yards from the center of the bridge. Graph this equation and describe, to the nearest yard, where the bridge touches the ground.
3. LOGIC Wilma is thinking of two numbers. The sum is 2 and the product is -24. Use a quadratic equation to find the two numbers.
4. RADIO TELESCOPES The cross-section of a large radio telescope is a parabola. The dish is set into the ground. The equation that describes the cross-section is d =
2−75
x2-
4−3
x - 32−3
, where d givesthe depth of the dish below ground and x is the distance from the control center, both in meters. If the dish does not extend above the ground level, what is the diameter of the dish? Solve by graphing.
x
d
5. BOATS The distance between two boats is
d = √������t2- 10t + 35 ,
where d is distance in meters and t is time in seconds.
a. Make a graph of d2 versus t.d
tO
b. Do the boats ever collide?
x
h
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Chapter 4 16 Glencoe Algebra 2
Enrichment
Graphing Absolute Value Equations
You can solve absolute value equations in much the same way you solved quadratic equations. Graph the related absolute value function for each equation using a graphing calculator. Then use the ZERO feature in the CALC menu to find its real solutions, if any. Recall that solutions are points where the graph intersects the x-axis.
For each equation, make a sketch of the related graph and find the solutions rounded to the nearest hundredth.
1. ⎪x + 5⎥ = 0 2. ⎪4x - 3⎥ + 5 = 0 3. ⎪x - 7⎥ = 0
4. ⎪x + 3⎥ - 8 = 0 5. - ⎪x + 3⎥ + 6 = 0 6. ⎪x - 2⎥ - 3 = 0
7. ⎪3x + 4⎥ = 2 8. ⎪x + 12⎥ = 10 9. ⎪x⎥ - 3 = 0
10. Explain how solving absolute value equations algebraically and finding zeros of absolute value functions graphically are related.
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Chapter 4 17 Glencoe Algebra 2
Study Guide and InterventionSolving Quadratic Equations by Factoring
Factored Form To write a quadratic equation with roots p and q, let (x - p)(x - q) = 0. Then multiply using FOIL.
Write a quadratic equation in standard form with the given roots.
a. 3, -5
(x - p)(x - q) = 0 Write the pattern.
(x - 3)[x - (-5)] = 0 Replace p with 3, q with -5.
(x - 3)(x + 5) = 0 Simplify.
x2 + 2x - 15 = 0 Use FOIL.
The equation x2 + 2x - 15 = 0 has roots 3 and -5.
b. - 7 −
8 , 1 −
3
(x - p)(x - q) = 0
[x - (- 7 − 8 ) ] (x - 1 −
3 ) = 0
(x + 7 − 8 ) (x - 1 −
3 ) = 0
(8x + 7) −
8 � (3x - 1)
− 3 = 0
24 � (8x + 7)(3x - 1) −−
24 = 24 � 0
24x2 + 13x - 7 = 0
The equation 24x2 + 13x - 7 = 0 has
roots - 7 − 8 and 1 −
3 .
ExercisesWrite a quadratic equation in standard form with the given root(s).
1. 3, -4 2. -8, -2 3. 1, 9
4. -5 5. 10, 7 6. -2, 15
7. - 1 − 3 , 5 8. 2, 2 −
3 9. -7, 3 −
4
10. 3, 2 − 5 11. - 4 −
9 , -1 12. 9, 1 −
6
13. 2 − 3 , - 2 −
3 14. 5 −
4 , - 1 −
2 15. 3 −
7 , 1 −
5
16. - 7 − 8 , 7 −
2 17. 1 −
2 , 3 −
4 18. 1 −
8 , 1 −
6
Example
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Chapter 4 18 Glencoe Algebra 2
Study Guide and Intervention (continued)
Solving Quadratic Equations by Factoring
Solve Equations by Factoring When you use factoring to solve a quadratic equation, you use the following property.
Zero Product Property For any real numbers a and b, if ab = 0, then either a = 0 or b =0, or both a and b = 0.
Solve each equation by factoring.
a. 3x2 = 15x
3x2 = 15x Original equation
3x2 - 15x = 0 Subtract 15x from both sides.
3x(x - 5) = 0 Factor the binomial.
3x = 0 or x - 5 = 0 Zero Product Property
x = 0 or x = 5 Solve each equation.
The solution set is {0, 5}.
b. 4x2 - 5x = 21 4x2 - 5x = 21 Original equation
4x2 - 5x - 21 = 0 Subtract 21 from both sides.
(4x + 7)(x - 3) = 0 Factor the trinomial.
4x + 7 = 0 or x - 3 = 0 Zero Product Property
x = - 7 − 4 or x = 3 Solve each equation.
The solution set is {- 7 − 4 , 3} .
ExercisesSolve each equation by factoring.
1. 6x2 - 2x = 0 2. x2 = 7x 3. 20x2 = -25x
4. 6x2 = 7x 5. 6x2 - 27x = 0 6. 12x2 - 8x = 0
7. x2 + x - 30 = 0 8. 2x2 - x - 3 = 0 9. x2 + 14x + 33 = 0
10. 4x2 + 27x - 7 = 0 11. 3x2 + 29x - 10 = 0 12. 6x2 - 5x - 4 = 0
13. 12x2 - 8x + 1 = 0 14. 5x2 + 28x - 12 = 0 15. 2x2 - 250x + 5000 = 0
16. 2x2 - 11x - 40 = 0 17. 2x2 + 21x - 11 = 0 18. 3x2 + 2x - 21 = 0
19. 8x2 - 14x + 3 = 0 20. 6x2 + 11x - 2 = 0 21. 5x2 + 17x - 12 = 0
22. 12x2 + 25x + 12 = 0 23. 12x2 + 18x + 6 = 0 24. 7x2 - 36x + 5 = 0
Example
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Chapter 4 19 Glencoe Algebra 2
Skills PracticeSolving Quadratic Equations by Factoring
Write a quadratic equation in standard form with the given root(s).
1. 1, 4 2. 6, -9
3. -2, -5 4. 0, 7
5. - 1 − 3 , -3 6. - 1 −
2 , 3 −
4
Factor each polynomial.
7. m2 + 7m - 18 8. 2x2 - 3x - 5
9. 4z2 + 4z - 15 10. 4p2 + 4p - 24
11. 3y2 + 21y + 36 12. c2 - 100
Solve each equation by factoring.
13. x2 = 64 14. x2 - 100 = 0
15. x2 - 3x + 2 = 0 16. x2 - 4x + 3 = 0
17. x2 + 2x - 3 = 0 18. x2 - 3x - 10 = 0
19. x2 - 6x + 5 = 0 20. x2 - 9x = 0
21. x2 - 4x = 21 22. 2x2 + 5x - 3 = 0
23. 4x2 + 5x - 6 = 0 24. 3x2 - 13x - 10 = 0
25. NUMBER THEORY Find two consecutive integers whose product is 272.
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Chapter 4 20 Glencoe Algebra 2
PracticeSolving Quadratic Equations by Factoring
Write a quadratic equation in standard form with the given root(s).
1. 7, 2 2. 0, 3 3. -5, 8
4. -7, -8 5. -6, -3 6. 3, -4
7. 1, 1 − 2 8. 1 −
3 , 2 9. 0, - 7 −
2
Factor each polynomial.
10. r3 + 3r2 - 54r 11. 8a2 + 2a - 6 12. c2 - 49
13. x3 + 8 14. 16r2 - 169 15. b4 - 81
Solve each equation by factoring.
16. x2 - 4x - 12 = 0 17. x2 - 16x + 64 = 0
18. x2 - 6x + 8 = 0 19. x2 + 3x + 2 = 0
20. x2 - 4x = 0 21. 7x2 = 4x
22. 10x2 = 9x 23. x2 = 2x + 99
24. x2 + 12x = -36 25. 5x2 - 35x + 60 = 0
26. 36x2 = 25 27. 2x2 - 8x - 90 = 0
28. NUMBER THEORY Find two consecutive even positive integers whose product is 624.
29. NUMBER THEORY Find two consecutive odd positive integers whose product is 323.
30. GEOMETRY The length of a rectangle is 2 feet more than its width. Find the dimensions of the rectangle if its area is 63 square feet.
31. PHOTOGRAPHY The length and width of a 6-inch by 8-inch photograph are reduced by the same amount to make a new photograph whose area is half that of the original. By how many inches will the dimensions of the photograph have to be reduced?
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Chapter 4 21 Glencoe Algebra 2
Word Problem PracticeSolving Quadratic Equations by Factoring
1. FLASHLIGHTS When Dora shines her flashlight on the wall at a certain angle, the edge of the lit area is in the shape of a parabola. The equation of the parabola is y = 2x2
+ 2x - 60. Factor this quadratic equation.
2. SIGNS David was looking through an old algebra book and came across this equation.
x2 6x + 8 = 0
The sign in front of the 6 was blotted out. How does the missing sign depend on the signs of the roots?
3. ART The area in square inches of the drawing Maisons prés de la mer by Claude Monet is approximated by the equation y = x2
– 23x + 130. Factor the equation to find the two roots, which are equal to the approximate length and width of the drawing.
4. PROGRAMMING Ray is a computer programmer. He needs to find the quadratic function of this graph for an algorithm related to a game involving dice. Provide such a function.
y
xO
-2
-4
2
2
4
4 6 8 10 12
5. ANIMATION A computer graphics animator would like to make a realistic simulation of a tossed ball. The animator wants the ball to follow the parabolic trajectory represented by the quadratic equation f(x) = -0.2(x + 5) (x - 5).
a. What are the solutions of f (x) = 0?
b. Write f (x) in standard form.
c. If the animator changes the equation to f(x) = -0.2x2 + 20, what are the solutions of f(x) = 0?
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Chapter 4 22 Glencoe Algebra 2
Enrichment
Using Patterns to Factor
Study the patterns below for factoring the sum and the difference of cubes.
a3 + b3 = (a + b)(a2 - ab + b2)a3 - b3 = (a - b)(a2 + ab + b2)
This pattern can be extended to other odd powers. Study these examples.
Factor a5 + b5.Extend the first pattern to obtain a5 + b5 = (a + b)(a4 - a3b + a2b2 - ab3 + b4).Check: (a + b)(a4 - a3b + a2b2 - ab3 + b4) = a5 - a4b + a3b2 - a2b3 + ab4 + a4b - a3b2 + a2b3 - ab4 + b5
= a5 + b5
Factor a5 - b5.
Extend the second pattern to obtain a5 - b5 = (a - b)(a4 + a3b + a2b2 + ab3 + b4).Check: (a - b) (a4 + a3b + a2 b2 + ab3 + b4) = a5 + a4b + a3b2 + a2b3 + ab4 - a4b - a3b2 - a2b3 - ab4 - b5
= a5 - b5
In general, if n is an odd integer, when you factor an + bn or an - bn, one factor will be either (a + b) or (a - b), depending on the sign of the original expression. The other factor will have the following properties:
• The first term will be an - 1 and the last term will be bn - 1.• The exponents of a will decrease by 1 as you go from left to right.• The exponents of b will increase by 1 as you go from left to right.• The degree of each term will be n - 1.• If the original expression was an + bn, the terms will alternately have + and - signs.• If the original expression was an - bn, the terms will all have + signs.
Use the patterns above to factor each expression.
1. a7 + b7
2. c9 - d9
3. f 11 + g11
To factor x10 - y10, change it to (x5 + y5)(x5 - y5) and factor each binomial. Use this approach to factor each expression.
4. x10 - y10
5. a14 - b14
Example 1
Example 2
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Chapter 4 23 Glencoe Algebra 2
Graphing Calculator ActivityUsing Tables to Factor by Grouping
The TABLE feature of a graphing calculator can be used to help factor a polynomial of the form ax2 + bx + c. (The same problems can be solved with the Lists and Spreadsheet application on the TI-Nspire.)
Factor 10x2 - 43x + 28 by grouping.
Make a table of the negative factors of 10 � 28 or 280. Look for a pair of factors whose sum is -43.
Enter the equation y = 280 − x in Y1 to find the factors of 280. Then,
find the sum of the factors using y = 280 − x + x in Y2. Set up the table
to display the negative factors of 280 by setting �Tbl = to -1. Examine the results.
Keystrokes: Y= 280 ÷
ENTER VARS ENTER ENTER + ENTER 2nd [TBLSET] (–) 1 ENTER (–) 1 ENTER 2nd
[TABLE].
The last line of the table shows that -43x may be replaced with -8x + (-35x).
10x2 - 43x + 28 = 10x2 - 8x + (-35x) + 28 = 2x(5x - 4) + (-7)(5x - 4) = (5x - 4)(2x - 7)Thus, 10x2 - 43x + 28 = (5x - 4)(2x - 7).
Factor each polynomial.
1. y2 - 20y - 96 2. 4z2 - 33z + 35 3. 4y2 + y -18 4. 6a2 + 2a - 15
5. 6m2 + 17m + 12 6. 24z2 - 46z + 15 7. 36y2 + 84y + 49 8. 4b2 + 36b - 403
Factor 12x2 - 7x - 12.
Look at the factors of 12(-12) or -144 for a pair with a sum of -7. Enter an equation to determine the factors in Y1 and an equation to find the sum of factors in Y2. Examine the table to find a sum of -7.Keystrokes: Y= (–) 144
÷
ENTER VARS ENTER ENTER + ENTER 2nd [TBLSET] 1 ENTER 1 ENTER 2nd
[TABLE].12x2 - 7x - 12 = 12x2 + 9x + (-16x) - 12 = 3x(4x + 3) - 4(4x + 3) = (4x + 3)(3x - 4) Thus, 12x2 - 7x - 12 = (4x + 3)(3x - 4).
Exercises
Example 1
Example 2
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Chapter 4 24 Glencoe Algebra 2
Study Guide and InterventionComplex Numbers
Solve x2 + 5 = 0.
x2 + 5 = 0 Original equation.
x2 = -5 Subtract 5 from each side.
x = ± √ � 5 i Square Root Property.
ExercisesSimplify.
1. √ �� -72 2. √ �� -24
3. √ �� -84 4. (2 + i) (2 - i)
Solve each equation.
5. 5x2 + 45 = 0 6. 4x2 + 24 = 0
7. -9x2 = 9 8. 7x2 + 84 = 0
Pure Imaginary Numbers A square root of a number n is a number whose square
is n. For nonnegative real numbers a and b, √ �� ab = √ � a � √ � b and √ � a − b =
√ � a −
√ � b , b ≠ 0.
• The imaginary unit i is defined to have the property that i2 = -1.
• Simplified square root expressions do not have radicals in the denominator, and any number remaining under the square root has no perfect square factor other than 1.
a. Simplify √ �� -48 . √ �� -48 = √ ���� 16 � (-3) = √ �� 16 � √ � 3 � √ �� -1 = 4i √ � 3
b. Simplify √ �� -63 . √ �� -63 = √ ���� -1 � 7 � 9 = √ �� -1 � √ � 7 . √ � 9 = 3i √ � 7
a. Simplify -3i � 4i.-3i � 4i = -12i2
= -12(-1) = 12
b. Simplify √ �� -3 � √ �� -15 . √ �� -3 � √ �� -15 = i √ � 3 � i √ �� 15 = i2 √ �� 45 = -1 � √ � 9 � √ � 5 = -3 √ � 5
Example 1 Example 2
Example 3
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Chapter 4 25 Glencoe Algebra 2
Study Guide and Intervention (continued)
Complex Numbers
Operations with Complex Numbers
Complex Number
A complex number is any number that can be written in the form a + bi, where a and b are
real numbers and i is the imaginary unit (i 2 = -1). a is called the real part, and b is called
the imaginary part.
Addition and
Subtraction of
Complex Numbers
Combine like terms.
(a + bi ) + (c + di ) = (a + c) + (b + d )i
(a + bi ) - (c + di ) = (a - c) + (b - d )i
Multiplication of
Complex NumbersUse the defi nition of i 2 and the FOIL method:
(a + bi )(c + di ) = (ac - bd ) + (ad + bc)i
Complex Conjugatea + bi and a - bi are complex conjugates. The product of complex conjugates is always a
real number.
To divide by a complex number, first multiply the dividend and divisor by the complex conjugate of the divisor.
ExercisesSimplify.
1. (-4 + 2i) + (6 - 3i) 2. (5 - i) - (3 - 2i) 3. (6 - 3i) + (4 - 2i)
4. (-11 + 4i) - (1 - 5i) 5. (8 + 4i) + (8 - 4i) 6. (5 + 2i) - (-6 - 3i)
7. (2 + i)(3 - i) 8. (5 - 2i)(4 - i) 9. (4 - 2i)(1 - 2i)
10. 5 − 3 + i
11. 7 - 13i − 2i
12. 6 - 5i − 3i
Simplify (6 + i) + (4 - 5i).
(6 + i) + (4 - 5i) = (6 + 4) + (1 - 5)i = 10 - 4i
Simplify (2 - 5i) � (-4 + 2i).
(2 - 5i) � (-4 + 2i) = 2(-4) + 2(2i) + (-5i)(-4) + (-5i)(2i) = -8 + 4i + 20i - 10i2
= -8 + 24i - 10(-1) = 2 + 24i
Simplify (8 + 3i) - (6 - 2i).(8 + 3i) - (6 - 2i) = (8 - 6) + [3 - (-2)]i = 2 + 5i
Simplify 3 - i − 2 + 3i
.
3 - i − 2 + 3i
= 3 - i − 2 + 3i
� 2 - 3i − 2 - 3i
= 6 - 9i - 2i + 3 i 2 −− 4 - 9 i 2
= 3 - 11i − 13
= 3 − 13
- 11 − 13
i
Example 1 Example 2
Example 3 Example 4
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Chapter 4 26 Glencoe Algebra 2
Skills PracticeComplex Numbers
Simplify.
1. √ �� 99 2. √ �� 27 −
49
3. √ ��� 52x3y5 4. √ ��� -108x7
5. √ ��� -81x6 6. √ �� -23 � √ �� -46
7. (3i)(-2i)(5i) 8. i11
9. i65 10. (7 - 8i) + (-12 - 4i)
11. (-3 + 5i) + (18 - 7i) 12. (10 - 4i) - (7 + 3i)
13. (7 - 6i)(2 - 3i) 14. (3 + 4i)(3 - 4i)
15. 8 - 6i − 3i
16. 3i − 4 + 2i
Solve each equation.
17. 3x2 + 3 = 0 18. 5x2 + 125 = 0
19. 4x2 + 20 = 0 20. -x2 - 16 = 0
21. x2 + 18 = 0 22. 8x2 + 96 = 0
Find the values of � and m that make each equation true.
23. 20 - 12i = 5� + (4m)i 24. � - 16i = 3 - (2m)i
25. (4 + �) + (2m)i = 9 + 14i 26. (3 - m) + (7� - 14)i = 1 + 7i
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Chapter 4 27 Glencoe Algebra 2
PracticeComplex Numbers
Simplify.
1. √ �� -36 2. √ �� -8 � √ �� -32 3. √ �� -15 � √ �� -25
4. (-3i) (4i)(-5i) 5. (7i)2(6i) 6. i42
7. i55 8. i89 9. (5 - 2i) + (-13 - 8i)
10. (7 - 6i) + (9 + 11i) 11. (-12 + 48i) + (15 + 21i) 12. (10 + 15i) - (48 - 30i)
13. (28 - 4i) - (10 - 30i) 14. (6 - 4i) (6 + 4i) 15. (8 - 11i) (8 - 11i)
16. (4 + 3i) (2 - 5i) 17. (7 + 2i) (9 - 6i) 18. 6 + 5i − -2i
19. 2 − 7 - 8i
20. 3 - i − 2 - i
21. 2 - 4i − 1 + 3i
Solve each equation.
22. 5n2 + 35 = 0 23. 2m2 + 10 = 0
24. 4m2 + 76 = 0 25. -2m2 - 6 = 0
26. -5m2 - 65 = 0 27. 3 − 4 x2 + 12 = 0
Find the values of ℓ and m that make each equation true.
28. 15 - 28i = 3ℓ + (4m)i 29. (6 - ℓ) + (3m)i = -12 + 27i
30. (3ℓ + 4) + (3 - m)i = 16 - 3i 31. (7 + m) + (4ℓ - 10)i = 3 - 6i
32. ELECTRICITY The impedance in one part of a series circuit is 1 + 3j ohms and the impedance in another part of the circuit is 7 - 5j ohms. Add these complex numbers to find the total impedance in the circuit.
33. ELECTRICITY Using the formula E = IZ, find the voltage E in a circuit when the current I is 3 - j amps and the impedance Z is 3 + 2j ohms.
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Chapter 4 28 Glencoe Algebra 2
Word Problem PracticeComplex Numbers
1. SIGN ERRORS Jennifer and Jessica come up with different answers to the same problem. They had to multiply (4 + i)(4 - i) and give their answer as a complex number. Jennifer claims that the answer is 15 and Jessica claims that the answer is 17. Who is correct? Explain.
2. COMPLEX CONJUGATES You have seen that the product of complex conjugates is always a real number. Show that the sum of complex conjugates is also always a real number.
3. PYTHAGOREAN TRIPLES If three integers a, b, and c satisfy a2
+ b2= c2,
then they are called a Pythagorean triple. Suppose that a, b, and c are a Pythagorean triple. Show that the real and imaginary parts of (a + bi)2, together with the number c2, form another Pythagorean triple.
4. ROTATIONS Complex numbers can be used to perform rotations in the plane. For example, if (x, y) are the coordinates of a point in the plane, then the real and imaginary parts of i(x + yi) are the horizontal and vertical coordinates of the 90° counterclockwise rotation of (x, y) about the origin. What are the real and imaginary parts of i(x + yi)?
5. ELECTRICAL ENGINEERING Alternating current (AC) in an electrical circuit can be described by complex numbers. In any electrical circuit, Z, the impedance in the circuit, is related to the voltage V and the current I by the formula Z = V −
I . The standard electrical
voltage in Europe is 220 volts, so in these problems use V = 220.
a. Find the impedance in a standard European circuit if the current is 22 – 11i amps.
b. Find the current in a standard European circuit if the impedance is 10 – 5i watts.
c. Find the impedance in a standard European circuit if the current is 20i amps.
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Chapter 4 29 Glencoe Algebra 2
Enrichment
Conjugates and Absolute Value
When studying complex numbers, it is often convenient to represent a complex number by a single variable. For example, we might let z = x + yi. We denote the conjugate of z by − z . Thus, − z = x - yi.
We can define the absolute value of a complex number as follows.
⎪z⎥ = ⎪x + yi⎥ = √ ��� x2 + y2
There are many important relationships involving conjugates and absolute values of complex numbers.
Show ⎪z⎥ 2 = z − z for any complex number z.
Let z = x + yi. Then, zz − z = (x + yi)(x - yi) = x2 + y2
= √ ���� (x2 + y2)2
= ⎪z⎥ 2
Show − z − ⎪z⎥ 2
is the multiplicative inverse for any nonzero
complex number z.
We know ⎪z⎥ 2 = z − z . If z ≠ 0, then we have z ( − z − ⎪z⎥ 2
) = 1.
Thus, − z − ⎪z⎥ 2
is the multiplicative inverse of z.
ExercisesFor each of the following complex numbers, find the absolute value and multiplicative inverse.
1. 2i 2. -4 - 3i 3. 12 - 5i
4. 5 - 12i 5. 1 + i 6. √ � 3 - i
7. √ � 3
− 3 +
√ � 3 −
3 i 8.
√ � 2 −
2 -
√ � 2 −
2 i 9. 1 −
2 -
√ � 3 −
2 i
Example 1
Example 2
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Chapter 4 30 Glencoe Algebra 2
Study Guide and InterventionCompleting the Square
Square Root Property Use the Square Root Property to solve a quadratic equation that is in the form “perfect square trinomial = constant.”
Solve each equation by using the Square Root Property. Round to the nearest hundredth if necessary.
a. x2 - 8x + 16 = 25 x2 - 8x + 16 = 25 (x - 4)2 = 25 x - 4 = √ �� 25 or x - 4 = - √ �� 25 x = 5 + 4 = 9 or x = -5 + 4 = -1
The solution set is {9, -1}.
b. 4x2 - 20x + 25 = 32 4x2 - 20x + 25 = 32 (2x - 5)2 = 322x - 5 = √ �� 32 or 2x - 5 = - √ �� 32 2x - 5 = 4 √ � 2 or 2x - 5 = -4 √ � 2
x = 5 ± 4 √ � 2 −
2
The solution set is {
5 ± 4 √ � 2 −
2 } .
ExercisesSolve each equation by using the Square Root Property. Round to the nearest hundredth if necessary.
1. x2 - 18x + 81 = 49 2. x2 + 20x + 100 = 64 3. 4x2 + 4x + 1 = 16
4. 36x2 + 12x + 1 = 18 5. 9x2 - 12x + 4 = 4 6. 25x2 + 40x + 16 = 28
7. 4x2 - 28x + 49 = 64 8. 16x2 + 24x + 9 = 81 9. 100x2 - 60x + 9 = 121
10. 25x2 + 20x + 4 = 75 11. 36x2 + 48x + 16 = 12 12. 25x2 - 30x + 9 = 96
Example
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Chapter 4 31 Glencoe Algebra 2
Study Guide and Intervention (continued)
Completing the Square
Complete the Square To complete the square for a quadratic expression of the form x2 + bx, follow these steps.
1. Find b − 2 . 2. Square b −
2 . 3. Add ( b −
2 )
2 to x2 + bx.
Find the value of c that makes x2 + 22x + c a perfect square trinomial. Then write the trinomial as the square of a binomial.
Step 1 b = 22; b − 2 = 11
Step 2 112 = 121Step 3 c = 121
The trinomial is x2 + 22x + 121, which can be written as (x + 11)2.
Solve 2x2 - 8x - 24 = 0 by completing the square.
2x2 - 8x - 24 = 0 Original equation
2x2 - 8x - 24 − 2 = 0 −
2 Divide each side by 2.
x2 - 4x - 12 = 0 x2 - 4x - 12 is not a perfect square.
x2 - 4x = 12 Add 12 to each side.
x2 - 4x + 4 = 12 + 4 Since ( 4 − 2 )
2
= 4, add 4 to each side.
(x - 2)2 = 16 Factor the square.
x - 2 = ±4 Square Root Property
x = 6 or x = - 2 Solve each equation.
The solution set is {6, -2}.
Exercises
Find the value of c that makes each trinomial a perfect square. Then write the trinomial as a perfect square. 1. x2 - 10x + c 2. x2 + 60x + c 3. x2 - 3x + c
4. x2 + 3.2x + c 5. x2 + 1 − 2 x + c 6. x2 - 2.5x + c
Solve each equation by completing the square. 7. y2 - 4y - 5 = 0 8. x2 - 8x - 65 = 0 9. w2 - 10w + 21 = 0
10. 2x2 - 3x + 1 = 0 11. 2x2 - 13x - 7 = 0 12. 25x2 + 40x - 9 = 0
13. x2 + 4x + 1 = 0 14. y2 + 12y + 4 = 0 15. t2 + 3t - 8 = 0
Example 1 Example 2
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Chapter 4 32 Glencoe Algebra 2
Skills PracticeCompleting the Square
Solve each equation by using the Square Root Property. Round to the nearest hundredth if necessary.
1. x2 - 8x + 16 = 1 2. x2 + 4x + 4 = 1
3. x2 + 12x + 36 = 25 4. 4x2 - 4x + 1 = 9
5. x2 + 4x + 4 = 2 6. x2 - 2x + 1 = 5
7. x2 - 6x + 9 = 7 8. x2 + 16x + 64 = 15
Find the value of c that makes each trinomial a perfect square. Then write the trinomial as a perfect square.
9. x2 + 10x + c 10. x2 - 14x + c
11. x2 + 24x + c 12. x2 + 5x + c
13. x2 - 9x + c 14. x2 - x + c
Solve each equation by completing the square.
15. x2 - 13x + 36 = 0 16. x2 + 3x = 0
17. x2 + x - 6 = 0 18. x2 - 4x - 13 = 0
19. 2x2 + 7x - 4 = 0 20. 3x2 + 2x - 1 = 0
21. x2 + 3x - 6 = 0 22. x2 - x - 3 = 0
23. x2 = -11 24. x2 - 2x + 4 = 0
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Chapter 4 33 Glencoe Algebra 2
PracticeCompleting the Square
Solve each equation by using the Square Root Property. Round to the nearest hundredth if necessary.
1. x2 + 8x + 16 = 1 2. x2 + 6x + 9 = 1 3. x2 + 10x + 25 = 16
4. x2 - 14x + 49 = 9 5. 4x2 + 12x + 9 = 4 6. x2 - 8x + 16 = 8
7. x2 - 6x + 9 = 5 8. x2 - 2x + 1 = 2 9. 9x2 - 6x + 1 = 2
Find the value of c that makes each trinomial a perfect square. Then write the trinomial as a perfect square.
10. x2 + 12x + c 11. x2 - 20x + c 12. x2 + 11x + c
13. x2 + 0.8x + c 14. x2 - 2.2x + c 15. x2 - 0.36x + c
16. x2 + 5 − 6 x + c 17. x2 - 1 −
4 x + c 18. x2 - 5 −
3 x + c
Solve each equation by completing the square.
19. x2 + 6x + 8 = 0 20. 3x2 + x - 2 = 0 21. 3x2 - 5x + 2 = 0
22. x2 + 18 = 9x 23. x2 - 14x + 19 = 0 24. x2 + 16x - 7 = 0
25. 2x2 + 8x - 3 = 0 26. x2 + x - 5 = 0 27. 2x2 - 10x + 5 = 0
28. x2 + 3x + 6 = 0 29. 2x2 + 5x + 6 = 0 30. 7x2 + 6x + 2 = 0
31. GEOMETRY When the dimensions of a cube are reduced by 4 inches on each side, the surface area of the new cube is 864 square inches. What were the dimensions of the original cube?
32. INVESTMENTS The amount of money A in an account in which P dollars are invested for 2 years is given by the formula A = P(1 + r)2, where r is the interest rate compounded annually. If an investment of $800 in the account grows to $882 in two years, at what interest rate was it invested?
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Chapter 4 34 Glencoe Algebra 2
Word Problem PracticeCompleting the Square
1. COMPLETING THE SQUARESamantha needs to solve the equation
x2- 12x = 40.
What must she do to each side of the equation to complete the square?
2. ART The area in square inches of the drawing Foliage by Paul Cézanne is approximated by the equation y = x2
– 40x + 396. Complete the square and find the two roots, which are equal to the approximate length and width of the drawing.
3. COMPOUND INTEREST Nikki invested $1000 in a savings account with interest compounded annually. After two years the balance in the account is $1210. Use the compound interest formula A = P(1 + r)t to find the annual interest rate.
4. REACTION TIME Lauren was eating lunch when she saw her friend Jason approach. The room was crowded and Jason had to lift his tray to avoid obstacles. Suddenly, a glass on Jason’s lunch tray tipped and fell off the tray. Lauren lunged forward and managed to catch the glass just before it hit the ground. The height h, in feet, of the glass t seconds after it was dropped is given by h = -16t2
+ 4.5. Lauren caught the glass when it was six inches off the ground. How long was the glass in the air before Lauren caught it?
5. PARABOLAS A parabola is modeled by y = x2
- 10x + 28. Jane’s homework problem requires that she find the vertex of the parabola. She uses the completing square method to express the function in the form y = (x - h)2
+ k, where (h, k) is the vertex of the parabola. Write the function in the form used by Jane.
6. AUDITORIUM SEATING The seats in an auditorium are arranged in a square grid pattern. There are 45 rows and 45 columns of chairs. For a special concert, organizers decide to increase seating by adding n rows and n columns to make a square pattern of seating 45 + n seats on a side.
a. How many seats are there after the expansion?
b. What is n if organizers wish to add 1000 seats?
c. If organizers do add 1000 seats, what is the seating capacity of the auditorium?
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Chapter 4 35 Glencoe Algebra 2
Enrichment
The Golden Quadratic Equations
A golden rectangle has the property that its length
a
a
a
b
b
a
can be written as a + b, where a is the width of the
rectangle and a + b − a = a − b . Any golden rectangle can be
divided into a square and a smaller golden rectangle, as shown.
The proportion used to define golden rectangles can be used to derive two quadratic equations. These are sometimes called golden quadratic equations.
Solve each problem.
1. In the proportion for the golden rectangle, let a equal 1. Write the resulting quadratic equation and solve for b.
2. In the proportion, let b equal 1. Write the resulting quadratic equation and solve for a.
3. Describe the difference between the two golden quadratic equations you found in exercises 1 and 2.
4. Show that the positive solutions of the two equations in exercises 1 and 2 are reciprocals.
5. Use the Pythagorean Theorem to find a radical expression for the diagonal of a golden rectangle when a = 1.
6. Find a radical expression for the diagonal of a golden rectangle when b = 1.
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Chapter 4 36 Glencoe Algebra 2
Study Guide and InterventionThe Quadratic Formula and the Discriminant
Quadratic Formula The Quadratic Formula can be used to solve any quadratic equation once it is written in the form ax2 + bx + c = 0.
Quadratic Formula The solutions of ax 2 + bx + c = 0, with a ≠ 0, are given by x = -b ± √ ���� b2 - 4ac
− 2a
.
Solve x2 - 5x = 14 by using the Quadratic Formula.
Rewrite the equation as x2 - 5x - 14 = 0.
x = -b ± √ ���� b2 - 4ac −
2a Quadratic Formula
= -(-5) ± √ ������� (-5)2 - 4(1)(-14)
−− 2(1)
Replace a with 1, b with -5, and c with -14.
= 5 ± √ �� 81 −
2 Simplify.
= 5 ± 9 − 2
= 7 or -2
The solutions are -2 and 7.
ExercisesSolve each equation by using the Quadratic Formula.
1. x2 + 2x - 35 = 0 2. x2 + 10x + 24 = 0 3. x2 - 11x + 24 = 0
4. 4x2 + 19x - 5 = 0 5. 14x2 + 9x + 1 = 0 6. 2x2 - x - 15 = 0
7. 3x2 + 5x = 2 8. 2y2 + y - 15 = 0 9. 3x2 - 16x + 16 = 0
10. 8x2 + 6x - 9 = 0 11. r2 - 3r − 5 + 2 −
25 = 0 12. x2 - 10x - 50 = 0
13. x2 + 6x - 23 = 0 14. 4x2 - 12x - 63 = 0 15. x2 - 6x + 21 = 0
Example
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Chapter 4 37 Glencoe Algebra 2
Roots and the Discriminant
DiscriminantThe expression under the radical sign, b2 - 4ac, in the Quadratic Formula is called
the discriminant.
Discriminant Type and Number of Roots
b2 - 4ac > 0 and a perfect square 2 rational roots
b2 - 4ac > 0, but not a perfect square 2 irrational roots
b2 - 4ac = 0 1 rational root
b2 - 4ac < 0 2 complex roots
Find the value of the discriminant for each equation. Then describe the number and type of roots for the equation.
a. 2x2 + 5x + 3The discriminant is b2 - 4ac = 52 - 4(2) (3) or 1.The discriminant is a perfect square, so the equation has 2 rational roots.
b. 3x2 - 2x + 5The discriminant is b2 - 4ac = (-2)2 - 4(3) (5) or -56.The discriminant is negative, so the equation has 2 complex roots.
ExercisesComplete parts a-c for each quadratic equation.a. Find the value of the discriminant.b. Describe the number and type of roots.c. Find the exact solutions by using the Quadratic Formula.
1. p2 + 12p = -4 2. 9x2 - 6x + 1 = 0 3. 2x2 - 7x - 4 = 0
4. x2 + 4x - 4 = 0 5. 5x2 - 36x + 7 = 0 6. 4x2 - 4x + 11 = 0
7. x2 - 7x + 6 = 0 8. m2 - 8m = -14 9. 25x2 - 40x = -16
10. 4x2 + 20x + 29 = 0 11. 6x2 + 26x + 8 = 0 12. 4x2 - 4x - 11 = 0
Example
Study Guide and Intervention (continued)
The Quadratic Formula and the Discriminant
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Chapter 4 38 Glencoe Algebra 2
Skills PracticeThe Quadratic Formula and the Discriminant
Complete parts a-c for each quadratic equation.a. Find the value of the discriminant.b. Describe the number and type of roots.c. Find the exact solutions by using the Quadratic Formula.
1. x2 - 8x + 16 = 0 2. x2 - 11x - 26 = 0
3. 3x2 - 2x = 0 4. 20x2 + 7x - 3 = 0
5. 5x2 - 6 = 0 6. x2 - 6 = 0
7. x2 + 8x + 13 = 0 8. 5x2 - x - 1 = 0
9. x2 - 2x - 17 = 0 10. x2 + 49 = 0
11. x2 - x + 1 = 0 12. 2x2 - 3x = -2
Solve each equation by using the Quadratic Formula.
13. x2 = 64 14. x2 - 30 = 0
15. x2 - x = 30 16. 16x2 - 24x - 27 = 0
17. x2 - 4x - 11 = 0 18. x2 - 8x - 17 = 0
19. x2 + 25 = 0 20. 3x2 + 36 = 0
21. 2x2 + 10x + 11 = 0 22. 2x2 - 7x + 4 = 0
23. 8x2 + 1 = 4x 24. 2x2 + 2x + 3 = 0
25. PARACHUTING Ignoring wind resistance, the distance d(t) in feet that a parachutist falls in t seconds can be estimated using the formula d(t) = 16t2. If a parachutist jumps from an airplane and falls for 1100 feet before opening her parachute, how many seconds pass before she opens the parachute?
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Chapter 4 39 Glencoe Algebra 2
PracticeThe Quadratic Formula and the Discriminant
Solve each equation by using the Quadratic Formula.
1. 7x2 - 5x = 0 2. 4x2 - 9 = 0
3. 3x2 + 8x = 3 4. x2 - 21 = 4x
5. 3x2 - 13x + 4 = 0 6. 15x2 + 22x = -8
7. x2 - 6x + 3 = 0 8. x2 - 14x + 53 = 0
9. 3x2 = -54 10. 25x2 - 20x - 6 = 0
11. 4x2 - 4x + 17 = 0 12. 8x - 1 = 4x2
13. x2 = 4x - 15 14. 4x2 - 12x + 7 = 0
Complete parts a-c for each quadratic equation.a. Find the value of the discriminant.b. Describe the number and type of roots.c. Find the exact solutions by using the Quadratic Formula.
15. x2 - 16x + 64 = 0 16. x2 = 3x 17. 9x2 - 24x + 16 = 0
18. x2 - 3x = 40 19. 3x2 + 9x - 2 = 0 20. 2x2 + 7x = 0
21. 5x2 - 2x + 4 = 0 22. 12x2 - x - 6 = 0 23. 7x2 + 6x + 2 = 0
24. 12x2 + 2x - 4 = 0 25. 6x2 - 2x - 1 = 0 26. x2 + 3x + 6 = 0
27. 4x2 - 3x2 - 6 = 0 28. 16x2 - 8x + 1 = 0 29. 2x2 - 5x - 6 = 0
30. GRAVITATION The height h(t) in feet of an object t seconds after it is propelled straight up from the ground with an initial velocity of 60 feet per second is modeled by the equation h(t) = -16t2 + 60t. At what times will the object be at a height of 56 feet?
31. STOPPING DISTANCE The formula d = 0.05s2 + 1.1s estimates the minimum stopping distance d in feet for a car traveling s miles per hour. If a car stops in 200 feet, what is the fastest it could have been traveling when the driver applied the brakes?
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Chapter 4 40 Glencoe Algebra 2
Word Problem PracticeThe Quadratic Formula and the Discriminant
1. PARABOLAS The graph of a quadratic equation of the form y = ax2
+ bx + c is shown below.
y
xO
5
-5
Is the discriminant b2- 4ac positive,
negative, or zero? Explain.
2. TANGENT Kathleen is trying to find bso that the x-axis is tangent to the parabola y = x2
+ bx + 4. She finds one value that works, b = 4. Is this the only value that works? Explain.
3. SPORTS In 1990, American Randy Barnes set the world record for the shot put. His throw can be described by the equation y = –16x2
+ 368x. Use the Quadratic Formula to find how far his throw was to the nearest foot.
4. EXAMPLES Give an example of a quadratic function f (x) that has the following properties.
I. The discriminant of f is zero.
II. There is no real solution of the equation f(x) = 10.
Sketch the graph of x = f(x).
yx
O
5. TANGENTS The graph of y = x2 is a parabola that passes through the point at (1, 1). The line y = mx - m + 1, where m is a constant, also passes through the point at (1, 1).
a. To find the points of intersection between the line y = mx - m + 1 and the parabola y = x2, set x2 = mx - m + 1 and then solve for x. Rearranging terms, this equation becomes x2 - mx + m - 1 = 0. What is the discriminant of this equation?
b. For what value of m is there only one point of intersection? Explain the meaning of this in terms of the corresponding line and the parabola.
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Chapter 4 41 Glencoe Algebra 2
Enrichment
Sum and Product of Roots
Sometimes you may know the roots of a quadratic equation without knowing the equation itself. Using your knowledge of factoring to solve an equation, you can work backward to find the quadratic equation. The rule for finding the sum and product of roots is as follows:
Sum and Product of RootsIf the roots of ax2 + bx + c = 0, with a ≠ 0, are s
1 and
s2, then s
1 + s
2 = - a −
b and s
1 � s
2 = c − a .
Write a quadratic equation that has the roots 3 and -8.
The roots are x = 3 and x = -8.
x
y
O
(–5–2, –30–1
4)
10
–10
–20
–30
2 4–2–4–6–8
3 + (-8) = -5 Add the roots.
3(-8) = -24 Multiply the roots. Equation: x2 + 5x - 24 = 0
ExercisesWrite a quadratic equation that has the given roots.
1. 6, -9 2. 5, -1 3. 6, 6
4. 4 ± √ � 3 5. - 2 − 5 , 2 −
7 6. -2 ± 3 √ � 5
− 7
Find k such that the number given is a root of the equation.
7. 7; 2x2 + kx - 21 = 0 8. -2; x2 - 13x + k = 0
Example
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Chapter 4 42 Glencoe Algebra 2
Spreadsheet ActivityApproximating the Real Zeros of Polynomials
You have learned the Location Principle, which can be used to approximate the real zeros of a polynomial.
A12345
C
xf(x)
B
5
G
0–2
D
–523
J
523
E
–3.33333339.1111111
H
1.66666670.7777778
I
3.33333339.1111111
F
–1.66666670.7777778
–5
Sheet 1 Sheet 2 Sheet 3
In the spreadsheet above, the positive real zero of ƒ(x) = x2 - 2 can be approximated in the following way. Set the spreadsheet preference to manual calculation. The values in A2 and B2 are the endpoints of a range of values. The values in D2 through J2 are values equally in the interval from A2 to B2. The formulas for these values are A2, A2 + (B2 - A2)/6, A2 + 2*(B2 - A2)/6, A2 + 3*(B2 - A2)/6, A2 + 4*(B2 - A2)/6, A2 + 5*(B2 - A2)/6, and B2, respectively.
Row 3 gives the function values at these points. The function ƒ(x) = x2 - 2 is entered into the spreadsheet in Cell D3 as D2^2 - 2. This function is then copied to the remaining cells in the row.
You can use this spreadsheet to study the function values at the points in cells D2 through J2. The value in cell F3 is positive and the value in cell G3 is negative, so there must be a zero between -1.6667 and 0. Enter these values in cells A2 and B2, respectively, and recalculate the spreadsheet. (You will have to recalculate a number of times.) The result is a new table from which you can see that there is a zero between 1.41414 and 1.414306. Because these values agree to three decimal places, the zero is about 1.414. This can be verified by using algebra.By solving x2 - 2 = 0, we obtain x = ± √ � 2 . The positive root is x = ± √ � 2 = 1.414213. . . , which verifies the result.
Exercises 1. Use a spreadsheet like the one above to approximate the zero of ƒ(x) = 3x - 2 to three
decimal places. Then verify your answer by using algebra to find the exact value of the root.
2. Use a spreadsheet like the one above to approximate the real zeros of f(x) = x2 + 2x + 0.5. Round your answer to four decimal places. Then, verify your answer by using the quadratic formula.
3. Use a spreadsheet like the one above to approximate the real zero of ƒ(x) = x3 - 3 −
2 x2 - 6x - 2 between -0.4 and -0.3.
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Chapter 4 43 Glencoe Algebra 2
Study Guide and InterventionTransformations of Quadratic Graphs
Write Quadratic Equations in Vertex Form A quadratic function is easier to graph when it is in vertex form. You can write a quadratic function of the form y = ax2 + bx + c in vertex from by completing the square.
Write y = 2x2 - 12x + 25 in vertex form. Then graph the function.
y = 2x2 - 12x + 25
x
y
O 2
2
4
6
8
4 6
y = 2(x2 - 6x) + 25y = 2(x2 - 6x + 9) + 25 - 18y = 2(x - 3)2 + 7
The vertex form of the equation is y = 2(x - 3)2 + 7.
Exercises
Write each equation in vertex form. Then graph the function.
1. y = x2 - 10x + 32 2. y = x2 + 6x 3. y = x2 - 8x + 6
x
y
O
x
y
O
x
y
O
4. y = -4x2 + 16x - 11 5. y = 3x2 - 12x + 5 6. y = 5x2 - 10x + 9
x
y
O
x
y
O
x
y
O
Example
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Chapter 4 44 Glencoe Algebra 2
Study Guide and Intervention (continued)
Transformations of Quadratic Graphs
Transformations of Quadratic Graphs Parabolas can be transformed by changing the values of the constants a, h, and k in the vertex form of a quadratic equation: y = a(x – h ) 2 + k.
• The sign of a determines whether the graph opens upward (a > 0) or downward (a < 0).
• The absolute value of a also causes a dilation (enlargement or reduction) of the parabola. The parabola becomes narrower if ⎪a⎥ >1 and wider if ⎪a⎥ < 1.
• The value of h translates the parabola horizontally. Positive values of h slide the graph to the right and negative values slide the graph to the left.
• The value of k translates the graph vertically. Positive values of k slide the graph upward and negative values slide the graph downward.
Graph y = (x + 7)2 + 3.
• Rewrite the equation as y = [x – (–7)]2 + 3.
• Because h = –7 and k = 3, the vertex is at (–7, 3). The axis of symmetry is x = –7. Because a = 1, we know that the graph opens up, and the graph is the same width as the graph of y = x2.
• Translate the graph of y = x2 seven units to the left and three units up.
Exercises
Graph each function.
1. y = –2x2 + 2 2. y = – 3(x – 1)2 3. y = 2(x + 2)2 + 3
Example
x
15
5
-15
-15
-55 15
y
-5
x
y
x
y
x
y
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Chapter 4 45 Glencoe Algebra 2
Skills PracticeTransformations of Quadratic Graphs
Write each quadratic function in vertex form. Then identify the vertex, axis of symmetry, and direction of opening.
1. y = (x - 2)2 2. y = -x2 + 4 3. y = x2 - 6
4. y = -3(x + 5)2 5. y = -5x2 + 9 6. y = (x - 2)2 - 18
7. y = x2 - 2x - 5 8. y = x2 + 6x + 2 9. y = -3x2 + 24x
Graph each function.
10. y = (x - 3)2 - 1 11. y = (x + 1)2 + 2 12. y = -(x - 4)2 - 4
x
y
O
x
y
O
x
y
O
13. y = - 1 − 2 (x + 2)2 14. y = -3x2 + 4 15. y = x2 + 6x + 4
x
y
O
x
y
O
x
y
O
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Chapter 4 46 Glencoe Algebra 2
PracticeTransformations of Quadratic Graphs
Write each equation in vertex form. Then identify the vertex, axis of symmetry, and direction of opening. 1. y = -6x2
- 24x - 25 2. y = 2x2 + 2 3. y = -4x2 + 8x
4. y = x2 + 10x + 20 5. y = 2x2 + 12x + 18 6. y = 3x2 - 6x + 5
7. y = -2x2 - 16x - 32 8. y = -3x2 + 18x - 21 9. y = 2x2 + 16x + 29
Graph each function.
10. y = (x + 3)2 - 1 11. y = -x2 + 6x - 5 12. y = 2x2 - 2x + 1
x
y
O
x
y
O
x
y
O
13. Write an equation for a parabola with vertex at (1, 3) that passes through (-2, -15).
14. Write an equation for a parabola with vertex at (-3, 0) that passes through (3, 18).
15. BASEBALL The height h of a baseball t seconds after being hit is given by h(t) = -16t2 + 80t + 3. What is the maximum height that the baseball reaches, and when does this occur?
16. SCULPTURE A modern sculpture in a park contains a parabolic arc that starts at the ground and reaches a maximum height of 10 feet after a horizontal distance of 4 feet. Write a quadratic function in vertex form that describes the shape of the outside of the arc, where y is the height of a point on the arc and x is its horizontal distance from the left-hand starting point of the arc. 10 ft
4 ft
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Chapter 4 47 Glencoe Algebra 2
Word Problem PracticeTransformations of Quadratic Graphs
1. ARCHES A parabolic arch is used as a bridge support. The graph of the arch is shown below.
y
xO 5
5
-5
If the equation that corresponds to this graph is written in the form y + a(x - h)2
+ k, what are h and k?
2. TRANSLATIONS For a computer animation, Barbara uses the quadratic function f(x) = -42(x - 20)2 + 16800 to help her simulate an object tossed on another planet. For one skit, she had to use the function f (x + 5) - 8000 instead of f (x). Where is the vertex of the graph of y = f (x + 5) - 8000?
3. BRIDGES The shape formed by the main cables of the Golden Gate Bridge approximately follows the equation y = 0.0002x2 - 0.23x + 227. Graph the parabola formed by one of the cables.
4. WATER JETS The graph shows the path of a jet of water.
y
xO
5
The equation corresponding to this graph is y = a(x - h) 2
+ k. What are a, h, and k?
5. PROFIT A theater operator predicts that the theater can make -4x2 + 160x dollars per show if tickets are priced at x dollars.
a. Rewrite the equation y = -4x2 + 160x in the form y = a(x - h) 2 + k.
b. What is the vertex of the parabola and what is its axis of symmetry?
c. Graph the parabola.
y
xO
Width
Heig
ht
x
y
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Chapter 4 48 Glencoe Algebra 2
Enrichment
A Shortcut to Complex RootsWhen graphing a quadratic function, the real roots are shown in the graph. You have learned that quadratic functions can also have imaginary roots that cannot be seen on the graph of the function. However, there is a way to graphically represent the complex roots of a quadratic function.
Find the complex roots of the quadratic function y = x2 - 4x + 5.
Step 1 Graph the function.
y
xO
6
5
Step 2 Reflect the graph over the horizontal line containing the vertex. In this example, the vertex is (2, 1).
y
xO
Step 3 The real part of the complex root is the point halfway between the x-intercepts of the reflected graph and the imaginary part of the complex roots are + and - half the distance between the x-intercepts of the reflected graph. So, in this example, the complex roots are 2 + 1i and 2 - 1i.
ExercisesUsing this method, find the complex roots of the following quadratic functions.
1. y = x2 + 2x + 5 2. y = x2 + 4x + 8
3. y = x2 + 6x + 13 4. y = x2 + 2x + 17
Example
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Chapter 4 49 Glencoe Algebra 2
Study Guide and InterventionQuadratic Inequalities
Graph Quadratic Inequalities To graph a quadratic inequality in two variables, use the following steps:
1. Graph the related quadratic equation, y = ax2 + bx + c. Use a dashed line for < or >; use a solid line for ≤ or ≥.
2. Test a point inside the parabola. If it satisfies the inequality, shade the region inside the parabola; otherwise, shade the region outside the parabola.
Graph the inequality y > x2 + 6x + 7.
First graph the equation y = x2 + 6x + 7. By completing the
x
y
O 2
-2
2
4
-4 -2
square, you get the vertex form of the equation y = (x + 3)2 - 2, so the vertex is (-3, -2). Make a table of values around x = -3, and graph. Since the inequality includes >, use a dashed line.Test the point (-3, 0), which is inside the parabola. Since (-3)2 + 6(-3) + 7 = -2, and 0 > -2, (-3, 0) satisfies the inequality. Therefore, shade the region inside the parabola.
ExercisesGraph each inequality.
1. y > x2 - 8x + 17 2. y ≤ x2 + 6x + 4 3. y ≥ x2 + 2x + 2
x
y
O
x
y
O
x
y
O
4. y < -x2 + 4x - 6 5. y ≥ 2x2 + 4x 6. y > -2x2 - 4x + 2
x
y
O
x
y
O
x
y
O
Example
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Chapter 4 50 Glencoe Algebra 2
Study Guide and Intervention (continued)
Quadratic Inequalities
Solve Quadratic Inequalities Quadratic inequalities in one variable can be solved graphically or algebraically.
Graphical Method
To solve ax 2 + bx + c < 0:
First graph y = ax 2 + bx + c. The solution consists of the x-values
for which the graph is below the x-axis.
To solve ax 2 + bx + c > 0:
First graph y = ax 2 + bx + c. The solution consists of the x-values
for which the graph is above the x-axis.
Algebraic Method
Find the roots of the related quadratic equation by factoring,
completing the square; or using the Quadratic Formula.
2 roots divide the number line into 3 intervals.
Test a value in each interval to see which intervals are solutions.
If the inequality involves ≤ or ≥, the roots of the related equation are included in the solution set.
Solve the inequality x2 - x - 6 ≤ 0.
First find the roots of the related equation x2 - x - 6 = 0. The
x
y
O
equation factors as (x - 3)(x + 2) = 0, so the roots are 3 and -2. The graph opens up with x-intercepts 3 and -2, so it must be on or below the x-axis for -2 ≤ x ≤ 3. Therefore the solution set is {x|-2 ≤ x ≤ 3}.
ExercisesSolve each inequality.
1. x2 + 2x < 0 2. x2 - 16 < 0 3. 0 < 6x - x2 - 5
4. c2 ≤ 4 5. 2m2 - m < 1 6. y2 < -8
7. x2 - 4x - 12 < 0 8. x2 + 9x + 14 > 0 9. -x2 + 7x - 10 ≥ 0
10. 2x2 + 5x- 3 ≤ 0 11. 4x2 - 23x + 15 > 0 12. -6x2 - 11x + 2 < 0
13. 2x2 - 11x + 12 ≥ 0 14. x2 - 4x + 5 < 0 15. 3x2 - 16x + 5 < 0
Example
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Chapter 4 51 Glencoe Algebra 2
Skills PracticeQuadratic Inequalities
Graph each inequality.
1. y ≥ x2 - 4x + 4 2. y ≤ x2 - 4 3. y > x2 + 2x - 5
x
y
O
x
y
O
x
y
O
Solve each inequality by graphing.
4. x2 - 6x + 9 ≤ 0 5. -x2 - 4x + 32 ≥ 0 6. x2 + x - 10 > 10
xO
y
x
y
O
Solve each inequality algebraically.
7. x2 - 3x - 10 < 0 8. x2 + 2x - 35 ≥ 0
9. x2 - 18x + 81 ≤ 0 10. x2 ≤ 36
11. x2 - 7x > 0 12. x2 + 7x + 6 < 0
13. x2 + x - 12 > 0 14. x2 + 9x + 18 ≤ 0
15. x2 - 10x + 25 ≥ 0 16. -x2 - 2x + 15 ≥ 0
17. x2 + 3x > 0 18. 2x2 + 2x > 4
19. -x2 - 64 ≤ -16x 20. 9x2 + 12x + 9 < 0
x
y
O
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Chapter 4 52 Glencoe Algebra 2
PracticeQuadratic Inequalities
Graph each inequality.
1. y ≤ x2 + 4 2. y > x2 + 6x + 6 3. y < 2x2 - 4x - 2
x
y
O
x
y
O
x
y
O
Solve each inequality.
4. x2 + 2x + 1 > 0 5. x2 - 3x + 2 ≤ 0 6. x2 + 10x + 7 ≥ 0
7. x2 - x - 20 > 0 8. x2 - 10x + 16 < 0 9. x2 + 4x + 5 ≤ 0
10. x2 + 14x + 49 ≥ 0 11. x2 - 5x > 14 12. -x2 - 15 ≤ 8x
13. -x2 + 5x - 7 ≤ 0 14. 9x2 + 36x + 36 ≤ 0 15. 9x ≤ 12x2
16. 4x2 + 4x + 1 > 0 17. 5x2 + 10 ≥ 27x 18. 9x2 + 31x + 12 ≤ 0
19. FENCING Vanessa has 180 feet of fencing that she intends to use to build a rectangular play area for her dog. She wants the play area to enclose at least 1800 square feet. What are the possible widths of the play area?
20. BUSINESS A bicycle maker sold 300 bicycles last year at a profit of $300 each. The maker wants to increase the profit margin this year, but predicts that each $20 increase in profit will reduce the number of bicycles sold by 10. How many $20 increases in profit can the maker add in and expect to make a total profit of at least $100,000?
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Chapter 4 53 Glencoe Algebra 2
Word Problem PracticeQuadratic Inequalities
1. HUTS The space inside a hut is shaded in the graph. The parabola is described
by the equation y = - 4−5
(x - 1)2+ 4.
y
xO
Write an inequality that describes the shaded region.
2. DISCRIMINANTS Consider the equation ax2 + bx + c = 0. Assume that the discriminant is zero and that a is positive. What are the solutions of the inequality ax2 + bx + c ≤ 0?
3. KIOSKS Caleb is designing a kiosk by wrapping a piece of sheet metal with dimensions x + 5 inches by 4x + 8 inches into a cylindrical shape. Ignoring cost, Caleb would like a kiosk that has a surface area of at least 4480 square inches. What values of x satisfy this condition?
4. DAMS The Hoover Dam is a concrete arch dam designed to hold the water of Lake Mead. At its center, the dam’s height is approximately 725 feet, and the dam varies from 45 to 660 feet thick. The dark line on this sketch of the cross-section of the dam is a parabola.
y
xThickness
Heig
ht
Cross-Section ofHoover Dam
Dam
Lake Mead
a. Write an equation for the Hoover Dam parabola. Let the height be the y-value of the parabola and the thickness be the x-value of the parabola. (Hint: the equation will be in the form: y = k(x – maximum thickness) + maximum height.)
b. Using your equation, graph the parabola of the Hoover Dam for 45 ≤ x ≤ 660.
Thickness
Heig
ht
x
y
c. Estimate to the nearest foot the thickness of the dam when the height is 200 feet.
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Chapter 4 54 Glencoe Algebra 2
Enrichment
Graphing Absolute Value Inequalities
You can solve absolute value inequalities by graphing in much the same manner you graphed quadratic inequalities. Graph the related absolute function for each inequality by using a graphing calculator. For > and ≥, identify the x-values, if any, for which the graph lies below the x-axis. For < and ≤, identify the x values, if any, for which the graph lies above the x-axis.
For each inequality, make a sketch of the related graph and find the solutions rounded to the nearest hundredth.
1. ⎪x - 3⎥ > 0 2. ⎪x⎥ - 6 < 0 3. - ⎪x + 4⎥ + 8 < 0
4. 2 ⎪x + 6⎥ - 2 ≥ 0 5. ⎪3x - 3⎥ ≥ 0 6. ⎪x - 7⎥ < 5
7. ⎪7x - 1⎥ > 13 8. ⎪x - 3.6⎥ ≤ 4.2 9. ⎪2x + 5⎥ ≤ 7
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Chapter 4 55 Glencoe Algebra 2
Graphing Calculator ActivityQuadratic Inequalities and the Test Menu
The inequality symbols, called relational operators, in the TEST menu can be used to display the solution of a quadratic inequality. Another method that can be used to find the solution set of a quadratic inequality is to graph each side of an inequality separately. Examine the graphs and use the intersect function to determine the range of values for which the inequality is true.
ExercisesSolve each inequality.
1. -x2 - 10x - 21 < 0 2. x2 - 9 < 0 3. x2 + 10x + 25 ≤ 0
4. x2 + 3x ≤ 28 5. 2x2 + x ≥ 3 6. 4x2 + 12x + 9 > 0
7. 23 > -x2 + 10x 8. x2 - 4x - 13 ≤ 0 9. (x + 1)(x -3) > 0
Solve x2 + x ≥ 6.
Place the calculator in Dot mode. Enter the inequality into Y1. Then trace the graph and describe the solution as an inequality.
Keystrokes: Y= x2 + 2nd [TEST] 4 6 ZOOM 4.
Use TRACE to determine the endpoints of the segments. Theses values are used to express the solution of the inequality, { x | x ≥ - 3 or x ≥ 2 }.
Solve 2x2 + 4x - 5 ≤ 3.
Place the left side of the inequality in Y1 and the right side in Y2. Determine the points of intersection. Use the intersection points to express the solution set of the inequality. Be sure to set the calculator to Connected mode.
Keystrokes: Y= 2 x2 + 4 — 5 ENTER 3 ENTER ZOOM 6.
Press 2nd [CALC] 5 and use the key to move the cursor to the left of the first intersection point. Press ENTER . Then move
the cursor to the right of the intersection point and press ENTER
ENTER . One of the values used in the solution set is displayed. Repeat the procedure on the other intersection point.
The solution is { x | -3.24 ≤ x ≤ 1.24}.
Example 1
Example 2
[-4.7, 4.7] scl:1 by [-3.1, 3.1] scl:1
[-10, 10] scl:1 by [-10, 10] scl:1
[-10, 10] scl:1 by [-10, 10] scl:1
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Use this recording sheet with pages 298–299 of the Student Edition.
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Chapter 4 57 Glencoe Algebra 2
4 Student Recording Sheet
Read each question. Then fill in the correct answer.
1. A B C D
2. F G H J
3. A B C D
4. F G H J
5. A B C D
6. F G H J
7. A B C D
8. F G H J
Multiple Choice
Short Response/Gridded Response
Record your answer in the blank.
For gridded response, also enter your answer in the grid by writing each number or symbol in a box. Then fill in the corresponding circle for that number or symbol.
9. ———————— (grid in)
10. ———————— (grid in)
11. a. ———————
b. ———————
c. ———————
12. ————————
13. ————————
9. 10.
Extended Response
Record your answers for Questions 14–16 on the back of this paper.
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
9
8
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5
4
3
2
1
0
9
8
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5
4
3
2
1
0
9
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6
5
4
3
2
1
0
9
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7
6
5
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3
2
1
0
9
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5
4
3
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Chapter 4 58 Glencoe Algebra 2
4
General Scoring Guidelines
• If a student gives only a correct numerical answer to a problem but does not show how he or she arrived at the answer, the student will be awarded only 1 credit. All extended response questions require the student to show work.
• A fully correct answer for a multiple-part question requires correct responses for all parts of the question. For example, if a question has three parts, the correct response to one or two parts of the question that required work to be shown is not considered a fully correct response.
• Students who use trial and error to solve a problem must show their method. Merely showing that the answer checks or is correct is not considered a complete response for full credit.
Exercises 14–16 Rubric
Score Specifi c Criteria
4 For Exercise 14, the three possible results of the discriminant (positive, negative, or zero) are explained. Positive has two real roots, zero has one real root, and negative has no real roots.For Exercise 15, a system of equation is written in part a. One equation should be based on the cost of the printers; the second equation will be based on the number of each that exist. In part b, the system of equations is set up in a matrix and used in part c to find the inverse to solve the system of equations.For Exercise 16, the quadratic equation is graphed in part a. The vertex of this graph will be used in parts b and c to solve for maximum height and time.
3 A generally correct solution, but may contain minor flaws in reasoning or computation.
2 A partially correct interpretation and/or solution to the problem.
1 A correct solution with no evidence or explanation.
0 An incorrect solution indicating no mathematical understanding of the concept or task, or no solution given.
Rubric for Scoring Extended Response
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Chapter 4 59 Glencoe Algebra 2
For Questions 1 and 2, consider f(x) = x2 + 2x - 3. 1. Find the y-intercept, the equation of the axis of symmetry,
and the x-coordinate of the vertex. 1.
2. Graph the function, labeling the y-intercept, vertex, and 2. axis of symmetry.
3. MULTIPLE CHOICE Determine the maximum or minimum value of f (x) = 2x2 - 8x + 9.
A min. 1 B max. 1 C min. 2 D max. 2 3.
Solve each equation. If exact roots cannot be found, state the consecutive integers between which the roots are located.
4. x2 - 2x = 3 5. x2 + 4x - 7 = 0 4.
5.
For Questions 1 and 2, solve each equation by factoring.
1. 3x2 = 10 - 13x 2. x2 + 4x = 45
3. MULTIPLE CHOICE Solve 5x2 + 100 = 0.
A ±2 √ � 5 B ±10 √ � 5 C ±2i √ � 5 D ±10i √ � 5
Write a quadratic equation with the given roots. Write the equation in the form ax2 + bx + c = 0, where a, b, and c are integers.
4. -6 and 2 5. 2 − 3 and -4
Simplify.
6. √ �� -80 7. √ �� -6 � √ �� -12
8. (6 - 9i) - (17 - 12i) 9. (7 - 3i)(8 + 4i)
10. 2 + i − 3 - i
xO
f(x)
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
4
Chapter 4 Quiz 1(Lessons 4-1 and 4-2)
Chapter 4 Quiz 2(Lessons 4-3 and 4-4)
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Chapter 4 60 Glencoe Algebra 2
Solve each equation by using the Square Root Property.
1. x2 + 8x + 16 = 36 2. x2 - 2x + 1 = 45 1.
2.
3. MULTIPLE CHOICE What is the solution of x2 - 10x = 11?
A {-11, 1} C -5 ± √ �� 14
B {-1, 11} D 5 ± √ �� 14 3.
4. Solve x2 - 4x = 1 by using the Quadratic Formula. Find exact solutions. 4.
5. Find the value of the discriminant for 3x2 = 6x - 11. Then describe the number and type of roots for the equation. 5.
1. Graph y = -(x - 2)2 - 1. Show and label the vertex and 1. axis of symmetry.
2. Write an equation for the parabola whose vertex is at (-5, 0) and passes through (0, 50). 2.
3. Graph y ≤ - 1 − 3 (x + 2)2 + 3. 3.
4. Use the graph of its related function to write the solutions of -x2 + 6x - 5 > 0. 4.
5. MULTIPLE CHOICE What is the solution of 4x2 + 1 ≥ 4x?
A all reals B empty set C {x ⎪x ≥ 1 − 2 } D {x ⎪x ≤ 1 −
2 } 5.
xO
y
Chapter 4 Quiz 3(Lessons 4-5 and 4-6)
y
xO
Chapter 4 Quiz 4(Lessons 4-7 and 4-8)
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Chapter 4 61 Glencoe Algebra 2
4
Part I Write the letter for the correct answer in the blank at the right of each question.
1. Which function is graphed?
A f (x) = x2- 2x - 3
B f (x) = x2+ 2x - 3
C f (x) = x2+ x - 3
D f (x) = (x - 3)2 1.
2. By the Zero Product Property, if (2x - 1)(x - 5) = 0, then .
F x = 1 or x = 5 H x =1−2
or x = 5
G x = - 1−2
or x = -5 J x = -1 or x = -5 2.
3. Write a quadratic equation with 7 and 2−5
as its roots. Write the equation in the form ax2
+ bx + c = 0, where a, b, and c are integers.
A 5x2- 37x + 14 = 0 C 5x2
+ 37x + 14 = 0 B 2x2
+ 9x - 35 = 0 D 2x2- 9x - 35 = 0 3.
4. The current in one part of a series circuit is 3 - 2j amps. The current in another part of the circuit is 2 + 4j amps. Find the total amps in the circuit.
F 5 + 2j H 1 + 2j G 6 - 8j J 7j 4.
5. Solve x2+ 6x = -6. If exact roots cannot be found, state the
consecutive integers between which the roots are located.
A -2, -3 C between -4 and -3; between -2 and -1 B -3 D between -5 and -4; between -2 and -1 5.
Part II
6. Solve x2- 4x + 3 = 0 by graphing. 6.
7. Determine whether f (x) = 1−2
x2- x - 9 7.
has a maximum or a minimum value and find that value.
For Questions 8 and 9, solve each equation by factoring.
8. x2- 7x = 18 9. 4x2
= x 9.
10. Simplify 5i−3 - 5i
. 10.
y
xO
Chapter 4 Mid-Chapter Test(Lessons 4-1 through 4-4)
8.
O
f (x)4
2
2 4
–2
–4
-4 -2 x
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Chapter 4 62 Glencoe Algebra 2
4
Write whether each sentence is true or false. If false, replace the underlined word or words to make a true sentence.
1. Two complex numbers of the form a + bi and a - bi are called imaginary units. 1.
2. In f(x) = 3x2 - 2x + 5, the linear term is 5. 2.
3. 2x2 + 3x - 4 ≤ 0 is an example of a quadratic equation. 3.
4. The solutions of a quadratic equation are called its zeros. 4.
5. The quadratic equation y = 2(x + 3)2 - 1 is written in vertex form. 5.
6. If a parabola opens upward, the y-coordinate of the vertex is the maximum value. 6.
7. In f(x) = -x2 + 2x - 1, the constant term is -x2. 7.
8. Pure imaginary numbers are square roots of negative real numbers. 8.
9. The highest or lowest point on a parabola is called the vertex. 9.
10. In the Quadratic Formula, the expression b2 - 4ac is called the quadratic term. 10.
Define each term in your own words.
11. discriminant 11.
12. standard form 12.
completing the square
complex conjugates
complex number
constant term
discriminant
imaginary unit
linear term
maximum value
minimum value
pure imaginary number
quadratic equation
Quadratic Formula
quadratic inequality
quadratic term
root
standard form
vertex
vertex form
zero
Chapter 4 Vocabulary Test
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Chapter 4 63 Glencoe Algebra 2
4
Write the letter for the correct answer in the blank at the right of each question.
1. Find the y-intercept for f (x) = -(x + 1)2. A 1 B -1 C x D 0 1.
2. What is the equation of the axis of symmetry of y = -3(x + 6)2 + 12? F x = 2 G x = -6 H x = 6 J x = -18 2.
3. The graph of f (x) = -2x2 + x opens and has a value. A down; maximum C up; maximum B down; minimum D up; minimum 3.
4. The related graph of a quadratic equation is shown at the right. Use the graph to determine the solutions of the equation.
F -2, 3 H -3, 2 G 0, -6 J 0, 2 4.
5. The quadratic function f (x) = x2 has . A no zeros C exactly two zeros B exactly one zero D more than two zeros 5.
6. Solve x2 - 3x - 10 = 0 by factoring. F {-5, 2} G (-2, -5) H {-2, 5} J {-10, 1} 6.
7. Which quadratic equation has roots -2 and 3? A x2 + x + 6 = 0 C x2 - 6x + 1 = 0 B x2 - x - 6 = 0 D x2 + x - 6 = 0 7.
8. Simplify (5 + 2i)(1 + 3i). F 5 + 6i G -1 H -1 + 17i J 11 + 17i 8.
9. ELECTRICITY The total impedance of a series circuit is the sum of the impedances of all parts of the circuit. A technician determined that the impedance of the first part of a particular circuit was 2 + 5j ohms. The impedance of the remaining part of the circuit was 3 - 2j ohms. What was the total impedance of the circuit?
A 5 + 3j ohms C -1 + 7j ohms B 5 + 7j ohms D 16 + 11j ohms 9.
10. To solve x2 + 8x + 16 = 25 by using the Square Root Property, you would first rewrite the equation as .
F (x + 4)2 = 25 H x2 + 8x - 9 = 0 G (x + 4)2 = 5 J x2 + 8x = 9 10.
xO
f (x)2
2 4
–2
–4
–6
-4 -2
Chapter 4 Test, Form 1
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Chapter 4 64 Glencoe Algebra 2
4
11. Find the value of c that makes x2 + 10x + c a perfect square. A 100 B 25 C 10 D 50 11.
12. The quadratic equation x2 + 6x = 1 is to be solved by completing the square. Which equation would be the first step in that solution?
F x2 + 6x - 1 = 0 H x2 + 6x + 36 = 1 + 36 G x(x + 6) = 1 J x2 + 6x + 9 = 1 + 9 12.
13. Find the exact solutions to x2 - 3x + 1 = 0 by using the Quadratic Formula.
A -3 ± √ � 5 −
2 B 3 ± √ �� 13
− 2 C -3 ± √ �� 13
− 2 D 3 ± √ � 5
− 2 13.
For Questions 14 and 15, use the value of the discriminant to determine the number and type of roots for each equation.
14. x2 - 3x + 7 = 0 F 2 complex roots H 2 real, irrational roots G 2 real, rational roots J 1 real, rational root 14.
15. x2 = 4x - 4 A 2 real, rational roots C 1 real, rational root B 2 real, irrational roots D no real roots
16. What is the vertex of y = 2(x - 3)2 + 6? F (-3, -6) G (3, -6) H (-3, 6) J (3, 6)
17. What is the equation of the axis of symmetry of y = -3(x + 6)2 + 1? A x = 2 B x = -6 C x = -3 D x = 6 17.
18. Which quadratic function has its vertex at (2, 3) and passes through (1, 0)? F y = 2(x - 2)2 + 3 H y = -3(x + 2)2 + 3
G y = -3(x - 2)2 + 3 J y = 2(x - 2)2 - 3 18.
19. Which quadratic inequality is graphed at the right?
A y ≥ (x + 1)2 + 4 B y ≤ -(x + 1)2 + 4 C y ≤ -(x - 1)2 + 4
D y ≤ -(x - 1)2 - 4 19.
20. Solve (x - 4)(x + 2) ≤ 0. F {x ⎪x ≤ -2 or x ≥ 4} H {x ⎪-4 ≤ x ≤ 2} G {x ⎪-2 ≤ x ≤ 4} J {x ⎪x = -2 or x = 4} 20.
Bonus Find the x-intercepts and the y-intercept of the graph of y = 2(x - 4)2 - 18. B:
y
xO
4
2
2 4
–2
–4
-4 -2
Chapter 4 Test, Form 1 (continued)
15.
16.
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Chapter 4 65 Glencoe Algebra 2
4
Write the letter for the correct answer in the blank at the right of each question.
1. Identify the y-intercept and the axis of symmetry for the graph of f (x) = 10x2 + 40x + 42.
A 42; x = 4 B 0; x = -4 C 42; x = -2 D -42; x = 2 1.
2. Identify the quadratic function graphed at the right. F f (x) = -x2 - 2x G f (x) = -x2 + 2x H f (x) = x2 - 2x
J f (x) = -(x + 2)2 2.
3. Determine whether f (x) = 4x2 - 16x + 6 has a maximum or a minimum value and find that value.
A minimum; -10 B minimum; 2 C maximum; -10 D maximum; 2 3.
4. Solve -x2 = 4x by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are located.
F 4, 0 H -4, 0 G between -4 and 4 J -2, 4 4.
5. Solve x2 - 3x = 18 by factoring.
A {6} B {-6, 3} C {-9, 2} D {-3, 6} 5.
6. Which quadratic equation has roots -2 and 1 − 5 ?
F x2 + 4x + 4 = 0 H 5x2 - 9x - 2 = 0 G 5x2 + 9x - 2 = 0 J 5x2 - 11x + 2 = 0 6.
7. Simplify (4 - 12i) - (-8 + 4i). A 12 - 8 B 28 C 12 - 16i D 12 + 16i 7.
8. Simplify 4 - 2i − 7 + 3i
.
F 11 − 29
- 13 − 29
i G 11 − 29
- 14 − 29
i H 13 − 29
- 17 − 29
i J 11 − 29
- 13 − 29
i 8.
9. To solve 9x2 - 12x + 4 = 49 by using the Square Root Property, you would first rewrite the equation as .
A 9x2 - 12x - 45 = 0 C (3x - 2)2 = 7 B (3x - 2)2 = ±49 D (3x - 2)2 = 49 9.
10. Find the value of c that makes x2 - 9x + c a perfect square.
F 81 − 4 G 9 −
2 H - 81 −
4 J 81 10.
xO
f(x)2
2 4
–2
–4
-4 -2
Chapter 4 Test, Form 2A
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Chapter 4 66 Glencoe Algebra 2
4
11. The quadratic equation x2 - 8x = -20 is to be solved by completing the square. Which equation would be a step in that solution? A (x - 4)2 = 4 C x2 - 8x + 20 = 0
B x - 4 = ± 2i D x2 - 8x + 16 = -20 11.
12. Find the exact solutions to 3x2 = 5x - 1 by using the Quadratic Formula.
F -5 ± √ �� 13 −
6 G 5 ± √ �� 13
− 6 H 5 ± √ �� 37
− 6 J 5 ±
√ �� 13 −
6
For Questions 13 and 14, use the value of the discriminant to determine the number and type of roots for each equation.
13. 2x2 - 7x + 9 = 0 A 2 real, rational C 2 complex B 2 real, irrational D 1 real, rational 13.
14. x2 + 20 = 12x - 16 F 1 real, irrational H 2 real, rational G no real J 1 real, rational 14.
15. Identify the vertex, axis of symmetry, and direction of opening for y = 1 −
2 (x - 8)2 + 2.
A (-8, 2); x = -8; up C (8, -2); x = 8; up B (-8, -2); x = -8; down D (8, 2); x = 8; up 15.
16. Which quadratic function has its vertex at (-2, 7) and opens down? F y = -3(x + 2)2 + 7 H y = (x - 2)2 + 7 G y = -12(x + 2)2 - 7 J y = -2(x - 2)2 + 7 16.
17. Write y = x2 + 4x - 1 in vertex form. A y = (x - 2)2 + 5 C y = (x + 2)2 - 1 B y = (x + 2)2 - 5 D y = (x + 2)2 + 3 17.
18. Write an equation for the parabola whose vertex is at (-8, 4) and passes through (-6, -2).
F y = - 3 − 2 (x + 8)2 + 4 H y = - 1 −
4 (x + 8)2 + 4
G y = 3 − 2 (x + 6)2 - 2 J y = - 3 −
2 (x - 8)2 + 4 18.
19. Which quadratic inequality is graphed at the right?A y ≥ (x - 2)(x + 3) C y > (x + 2)(x - 3)B y > (x - 2)(x + 3) D y < (x + 2)(x - 3) 19.
20. Solve x2 ≥ 2x + 24. F {x ⎪-4 ≤ x ≤ 6} H {x ⎪-6 ≤ x ≤ 4} G {x ⎪x ≤ -6 or x ≥ 4} J {x ⎪x ≤ -4 or x ≥ 6} 20.
Bonus Write a quadratic equation with roots ± i √ � 3 −
4 . B:
yxO 2 4
–2
–4
–6
-4 -2
Chapter 4 Test, Form 2A (continued)
12.
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Chapter 4 67 Glencoe Algebra 2
4
Write the letter for the correct answer in the blank at the right of each question.
1. Identify the y-intercept and the axis of symmetry for the graph of f (x) = -3x2 + 6x + 12.
A 2; x = -12 B 12; x = 1 C -2; x = 0 D -12; x = -1 1.
2. Identify the quadratic function graphed at the right. F f (x) = x2 - 4xG f (x) = -x2 + 4x H f (x) = -x2 - 4xJ f (x) = -(x + 4)2 2.
3. Determine whether f (x) = -5x2 - 10x + 6 has a maximum or a minimum value and find that value.
A minimum; -1 B maximum; 11 C maximum; -1 D minimum; 11 3.
4. Solve x2 = 4x by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are located.
F -4, 0 H between -4 and 4 G 2, -4 J 0, 4 4.
5. Solve x2 - 3x = 28 by factoring. A {-4, 7} B {-14, 2} C {-7, 4} D {-2, 14} 5.
6. Which quadratic equation has roots 7 and - 2 − 3 ?
F 2x2 - 11x - 21 = 0 H 3x2 - 19x - 14 = 0 G 3x2 + 23x + 14 = 0 J 2x2 + 11x - 21 = 0 6.
7. Simplify (15 - 13i) - (-1 + 17i). A 16 - 30i B 16 + 4i C 16 + 30i D 46 7.
8. Simplify 1 + 2i − 2 - 3i
.
F 8 − 7 + 1 −
7 i G 8 −
7 + i H -4 + 7i J - 4 −
13 + 7 −
13 i 8.
9. To solve 4x2 - 28x + 49 = 25 by using the Square Root Property, you would first rewrite the equation as .
A (2x - 7)2 = 25 C (2x - 7)2 = ±5 B (2x - 7)2 = 5 D 4x2 - 28x + 24 = 0 9.
10. Find the value of c that makes x2 + 5x + c a perfect square trinomial.
F 25 − 16
G 5 − 4 H 25 −
4 J 25 −
4
xO
f(x)4
2
2-4 -2
Chapter 4 Test, Form 2B
10.
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Chapter 4 68 Glencoe Algebra 2
4
11. The quadratic equation x2 - 18x = -106 is to be solved by completing the square. Which equation would be a step in that solution? A (x - 9)2 = 25 C x - 9 = ±5i B x2 - 18x + 106 = 0 D x2 - 18x + 81 = -106 11.
12. Find the exact solutions to 2x2 = 5x - 1 by using the Quadratic Formula.
F -5 ± √ �� 17 −
4 G 5 ± √ �� 17
− 4 H 5 ± √ �� 33
− 4 J 5 ± √ �� 17
− 2
For Questions 13 and 14, use the value of the discriminant to determine the number and type of roots for each equation.
13. 3x2 - x - 12 = 0 A 2 complex roots C 2 real, rational roots B 1 real, rational root D 2 real, irrational roots 13.
14. x2 + 10 = 3x - 3 F 2 complex roots H 2 real, irrational roots G 1 real, rational root J 2 real, rational roots 14.
15. Identify the vertex, axis of symmetry, and direction of opening for y = -8(x + 2)2.
A (-8, -2); x = -8 up C (2, 0); x = 2; down B (-2, 0); x = -2; down D (-2, -8); x = -2; down 15.
16. Which quadratic function has its vertex at (-3, 5) and opens down? F y = (x - 3)2 + 5 H y = (x + 3)2 - 5 G y = -(x + 3)2 + 5 J y = -(x - 3)2 + 5 16.
17. Write y = x2 - 18x + 52 in vertex form. A y = (x - 9)2 + 113 C y = (x - 9)2 + 52 B y = (x + 9)2 - 29 D y = (x - 9)2 - 29 17.
18. Write an equation for the parabola whose vertex is at (-5, 7) and passes through (-3, -1).
F y = - 1 − 11
(x + 5)2 + 7 H y = -2(x + 5)2 + 7
G y = - 1 − 2 (x + 5)2 + 7 J y = - 1 −
2 (x - 5)2 + 7 18.
19. Which quadratic inequality is graphed at the right? A y ≤ (x - 3)(x + 1) C y ≥ (x + 3)(x - 1) B y > (x - 3)(x + 1) D y > (x + 3)(x - 1) 19.
20. Solve 2x + 3 ≥ x2. F {x ⎪-1 ≤ x ≤ 3} H {x ⎪-3 ≤ x ≤ 1} G {x ⎪x ≤ -1 or x ≥ 3} J {x ⎪x ≤ -3 or x ≥ 1} 20.
Bonus Write a quadratic equation with roots ± i √ � 2 −
3 . B:
y
xO 2 4
–2
–4
-4 -2
Chapter 4 Test, Form 2B (continued)
12.
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Chapter 4 69 Glencoe Algebra 2
4
1. Graph f (x) = -5x2 + 10x, labeling the y-intercept, vertex, 1. and axis of symmetry.
2. Determine whether f (x) = -3x2 + 6x + 1 has a maximum or a minimum value and find that value. 2.
3. Solve x2 = 6x - 8 by graphing. If exact roots cannot befound, state the consecutive integers between which the roots are located. 3.
4. Solve 5x2 + 13x = 6 by factoring. 4.
5. GEOMETRY The length of a rectangle is 7 inches longer than its width. If the area of the rectangle is 144 square inches, what are its dimensions? 5.
6. ELECTRICITY The total impedance of a series circuit is the sum of the impedances of all parts of the circuit. Suppose that the first part of a circuit has an impedance of 6 - 5j ohms and that the total impedance of the circuit was 12 + 7j ohms. What is the impedance of the remainder of the circuit? 6.
7. ELECTRICITY In an AC circuit, the voltage E (in volts), current I (in amps), and impedance Z (in ohms) are related by the formula E = I � Z. Find the current in a circuit with voltage 10 - 3j volts and impedance 4 + j ohms. 7.
8. Write a quadratic equation with -6 and 3 − 4 as its roots.
Write the equation in the form ax2 + bx + c = 0, where a, b, and c are integers. 8.
9. Solve x2 + 6x + 9 = 25 by using the Square Root Property. 9.
xO
f (x)
xO
y
Chapter 4 Test, Form 2C
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Chapter 4 70 Glencoe Algebra 2
4
For Questions 10 and 11, solve each equation by completing the square.
10. x2 + 4x - 9 = 0 10.
11. 2x2 + 3x - 2 = 0 11.
12. Find the exact solutions to 5x2 = 3x - 2 by using the Quadratic Formula. 12.
For Questions 13 and 14, find the value of the discriminant for each quadratic equation. Then describe the number and type of roots for the equation.
13. 9x2 - 12x + 4 = 0 13.
14. 4x2 + 1 = 9x - 2 14.
15. Identify the vertex, axis of symmetry, and direction of
opening for y = - 2 − 3 (x + 5)2 - 7. 15.
16. Write an equation for the parabola with vertex at (2, -1) and y-intercept 5. 16.
17. Write y = x2 - 6x + 8 in vertex form. 17.
18. PHYSICS The height h (in feet) of a certain rocket t seconds after it leaves the ground is modeled by h(t) = -16t2 + 48t + 15. Write the function in vertex form and find the maximum height reached by the rocket. 18.
19. Graph y < x2 + 6x + 9. 19.
20. Solve 2x2 - 5x - 3 ≥ 0 algebraically. 20.
Bonus Write a quadratic equation with roots ± √ � 7
− 3 . Write the
equation in the form ax2 + bx + c = 0, where a, b, and c are integers. B:
y
xO
Chapter 4 Test, Form 2C (continued)
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Chapter 4 71 Glencoe Algebra 2
4
1. Graph f (x) = x2 - 4x + 3, labeling the y-intercept, vertex, 1. and axis of symmetry.
2. Determine whether f (x) = 5x2 - 20x + 3 has a maximum or a minimum value and find that value. 2.
3. Solve x2 + 2x - 3 = 0 by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are located. 3.
4. Solve 3x2 - x = 4 by factoring. 4.
5. GEOMETRY The length of a rectangle is 10 inches longer than its width. If the area of the rectangle is 144 square inches, what are its dimensions? 5.
6. ELECTRICITY The total impedance of a series circuit is the sum of the impedances of all parts of the circuit. Suppose that the first part of a circuit has an impedance of 7 + 4j ohms and that the total impedance of the circuit was 16 - 2j ohms. What is the impedance of the remainder of the circuit? 6.
7. ELECTRICITY In an AC circuit, the voltage E (in volts), current I (in amps), and impedance Z (in ohms) are related by the formula E = I � Z. Find the impedance in a circuit with voltage 12 + 2j volts and current 3 + 5j amps. 7.
8. Write a quadratic equation with -4 and 3 − 2 as its roots.
Write the equation in the form ax2 + bx + c = 0, where a, b, and c are integers. 8.
y
xO
xO
f(x)
Chapter 4 Test, Form 2D
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Chapter 4 72 Glencoe Algebra 2
4
9. Solve 9x2 + 12x + 4 = 6 by using the Square Root Property. 9.
Solve each equation by completing the square.
10. x2 - 8x + 14 = 0 10.
11. 3x2 + x - 2 = 0 11.
12. Find the exact solutions to 2x2 = 9x - 5 by using the Quadratic Formula. 12.
For Questions 13 and 14, find the value of the discriminant for each quadratic equation. Then describe the number and type of roots for the equation.
13. 25x2 - 20x + 4 = 0 13.
14. 2x2 + 10x + 9 = 2x 14.
15. Identify the vertex, axis of symmetry, and direction of opening for y = -(x - 6)2 - 5. 15.
16. Write an equation for the parabola with vertex at (-4, 2) and y-intercept -2. 16.
17. Write y = x2 + 4x + 8 in vertex form. 17.
18. PHYSICS The height h (in feet) of a certain rocket t seconds after it leaves the ground is modeled by h(t) = -16t2 + 64t + 12. Write the function in vertex form and find the maximum height reached by the rocket. 18.
19. Graph y ≥ x2 - 4x + 4. 19.
20. Solve 2x2 - 7x - 15 < 0 algebraically. 20.
Bonus Write a quadratic equation with roots ± √ � 5
− 4 .
Write the equation in the form ax2 + bx + c = 0, where a, b, and c are integers. B:
y
xO
Chapter 4 Test, Form 2D (continued)
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Chapter 4 73 Glencoe Algebra 2
4
1. Graph f (x) = 3 + 3x2 + 2x, labeling the y-intercept, vertex, 1. and axis of symmetry.
2. Determine whether f (x) = 1 - 3 − 5 x + 3 −
4 x2
has a maximum or a minimum value and find that value.
3. BUSINESS Khalid charges $10 for a one-year subscription to his on-line newsletter. Khalid currently has 600 subscribers and he estimates that for each $1 decrease in the subscription price, he would gain 100 new subscribers. What subscription price will maximize Khalid’s 2. income? If he charges this price, how much income should Khalid expect? 3.
For Questions 4 and 5, solve each equation by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are located.
4. 0.5x2 + 9 = 4.5x 4.
5. 2 − 3 x + 3 = 1 −
3 x2 5.
6. Solve 18x2 + 15 = 39x by factoring. 6.
7. Write a quadratic equation with - 2 − 3 and 1.75 as its roots.
Write the equation in the form ax2 + bx + c = 0, where a, b, and c are integers. 7.
8. Simplify (5 - i) + (2 - 4i) - (3 + i). 8.
9. Simplify 2 - i √ � 5 −
2 + i √ � 5 . 9.
y
xO
y
xO
xO
f (x)
Chapter 4 Test, Form 3
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Chapter 4 74 Glencoe Algebra 2
4
10. Solve 4x2 - 2x + 0.25 = 1.44 by using the Square Root Property. 10.
For Questions 11 and 12, solve each equation by completing the square.
11. 2x2 - 5 − 2 x + 2 = 0 11.
12. x2 + 2.5x - 3 = 0.5 12.
13. Find the exact solutions to 1 − 4 x2 - 3x + 1 = 0 by using the
Quadratic Formula. 13.
14. Find the value of the discriminant for 3x(0.2x - 0.4) + 1 = 0.9. Then describe the number and type of roots for the equation. 14.
15. Find all values of k such that x2 + kx + 1 = 0 has two complex roots. 15.
16. Write an equation of the parabola with equation
y = - 3 − 5 (x - 1 −
2 )
2
- 5 − 2 , translated 4 units left and 2 units up.
Then identify the vertex, axis of symmetry, and direction of opening of your function. 16.
17. PHYSICS The height h (in feet) of a certain aircraft t seconds after it leaves the ground is modeled by h(t) = -9.1t2 + 591.5t + 20,388.125. Write the function in vertex form and find the maximum height reached by the aircraft. 17.
18. Write an equation for the parabola that has the same
vertex as y = 1 − 3 x2 + 6x + 83 −
2 and x-intercept 1. 18.
19. Graph y < -(x2 + 2x) + 5.25. 19.
20. Solve (x + 7 − 2 ) (x - 1) 2 ≤ 0. 20.
Bonus Write a quadratic equation with roots -3 ± 2i √ � 5 −
4 .
Write the equation in the form ax2 + bx + c = 0, where a, b, and c are integers. B:
y
xO
Chapter 4 Test, Form 3 (continued)
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Chapter 4 75 Glencoe Algebra 2
4
Demonstrate your knowledge by giving a clear, concise solution to each problem. Be sure to include all relevant drawings and justify your answers. You may show your solution in more than one way or investigate beyond the requirements of the problem.
1. Mr. Moseley asked the students in his Algebra class to work in groups to solve (x - 3)2 = 25, stating that each student in the first group to solve the equation correctly would earn five bonus points on the next quiz. Mi-Ling’s group solved the equation using the Square Root Property. Emilia’s group used the Quadratic Formula to find the solutions. In which group would you prefer to be? Explain your reasoning.
2. The next day, Mr. Moseley had his students work in pairs to review for their chapter exam. He asked each student to write a practice problem for his or her partner. Len wrote the following problem for his partner, Jocelyn: Write an equation for the parabola whose vertex is (-3, -4), that passes through (-1, 0), and opens down.
a. Jocelyn had trouble solving Len’s problem. Explain why.b. How would you change Len’s problem?c. Make the change you suggested in part b and complete the
problem.
3. a. Write a quadratic function in vertex form whose maximum value is 8.
b. Write a quadratic function that transforms the graph of your function from part a so that it is shifted horizontally. Explain the change you made and describe the transformation that results from this change.
4. When asked to write f (x) = 2x2 + 12x - 5 in vertex form, Joseph wrote:
f (x) = 2x2 + 12x - 5 Step 1 f (x) = 2(x2 + 6x) - 5 Step 2 f (x) = 2(x2 + 6x + 9) - 5 + 9 Step 3 f (x) = 2(x + 3)2 + 4 Is Joseph’s answer correct? Explain your reasoning.
5. The graph of y = x2 + 4x + 4 is shown. Susan used this graph to solve three quadratic inequalities. Her three solutions are given below. Replace each ● with an inequality symbol (<, >, ≤, ≥) so that each solution is correct. Explain your reasoning for each.
a. The solution of x2 + 4x + 4 ● 0 is {x ⎪x < -2 or x > -2}.
b. The solution of x2 + 4x + 4 ● 0 is ∅.c. The solution of x2 + 4x + 4 ● 0 is all real numbers.
y
xO
4
2
2 4
–2
–4
-4 -2
Chapter 4 Extended-Response Test
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Chapter 4 76 Glencoe Algebra 2
4
Part 1: Multiple Choice
Instructions: Fill in the appropriate circle for the best answer.
1. If a − b = 3 −
2 ,then 8a equals which of the following?
A 16b B 12b C 3b − 2 D 8 −
3 b 1. A B C D
2. 20% of 3 yards is how many fifths of 9 feet?
F 1 G 6 H 10 J 15 2. F G H J
3. If u > v and t > 0, which of the following are true? I. ut > vt II. u + t > v + t III. u - t > v - t A I only C I and II only B III only D I, II, and III 3. A B C D
4. Which of the following is the greatest?
F 2 − 3 G 7 −
9 H 10 −
15 J 8 −
11 4. F G H J
5. If 2a + 3b represents the perimeter of a rectangle and a - 2b represents its width, the length is .
A 7b B b C 7b − 2 D 14b 5. A B C D
6. In the figure, what is the area of the shaded region?
F 30 units2 H 36 units2
G 54 units2 J 27 units2 6. F G H J
7. Mr. Salazár rented a car for d days. The rental agency charged x dollars per day plus c cents per mile for the model he selected. When Mr. Salazár returned the car, he paid a total of T dollars. In terms of d, x, c, and T, how many miles did he drive?
A T - (xd + c) B T - xd − c C T − xd + c
D T - xd − c 7. A B C D
8. If P(3, 2) and Q(7, 10) are the endpoints of the diameter of a circle, what is the area of the circle in square units?
F 2 √ � 5 π G 80π H 4 √ � 5 π J 20π 8. F G H J
9. If (x - y)2 = 100 and xy = 20, what is the value of x2 + y2?
A 120 B 140 C 80 D 60 9. A B C D
10. The tenth term in the sequence 7, 12, 19, 28, . . . is . F 124 G 103 H 57 J 147 10. F G H J
6
3
15
Standardized Test Practice(Chapters 1–4)
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Chapter 4 77 Glencoe Algebra 2
4
Part 2: Gridded Response
Instructions: Enter your answer by writing each digit of the answer in a column box
and then shading in the appropriate circle that corresponds to that entry.
15. The bar graph shows the distribution of votes 15. among the candidates for senior class president. If 220 seniors voted in all, how many students voted for either Theo or Pam?
16. Find the median of x, 2x + 1, x − 2 - 13, 45, and 16.
x + 22 if the mean of this set of numbers is 83.
JoeyAnaPamTheo
20
30
10
0
40
Perc
ent
of
vote
s re
ceiv
ed 50
Candidates
11. If t2 + 6t = -9, what is the value of (t - 1 − 2 )
2?
A -3 B 12 1 − 4 C 6 1 −
4 D -12 1 −
4 11. A B C D
12. All four walls of a rectangular room that is 14 feet wide, 20 feet long, and 8 feet high, are to be painted. What is the minimum cost of paint if one gallon covers at most 130 square feet and the paint costs $22 per gallon?
F $92 G $102 H $110 J $190 12. F G H J
13. If i 2 = -1, then what is the value of i32? A -1 B 1 C -i D i 13. A B C D
14. Which of the following is the sum of both solutions of the equation x2 + x - 42 = 0?
F 13 G -1 H -13 J 1 14. F G H J
Standardized Test Practice (continued)
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
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Chapter 4 78 Glencoe Algebra 2
4
17. Find the value of 12 + 36 ÷ 4 - (5 - 7)2. 17.
18. Find the slope of the line that is parallel to the line with equation 3x + 4y = 10. 18.
19. Describe the system 2x - 3y = 21 and y - 5 = 2 − 3 x as
consistent and independent, consistent and dependent, or inconsistent. 19.
20. Find the coordinates of the vertices of the figure formed by the system of inequalities. x ≥ -2 x + y ≤ 6y ≥ -2 x + y ≥ -2 20.
21. Find the value of ⎪ 5 12 -6 4
⎥ . 21.
22. Solve [ 4 -1 -2 3
] � [a b
] = [ 11 -13
] by using inverse matrices. 22.
23. Solve 2x2 + 40 = 0. 23.
24. PHYSICS An object is thrown straight up from the top of a 100-foot platform at a velocity of 48 feet per second. The height h(t) of the object t seconds after being thrown is given by h(t) = -16t2 + 48t + 100. Find the maximum height reached by the object and the time it takes to achieve this height. 24.
25. Solve x2 = 2x + 3 by graphing. 25.
26. Solve 4x2 - 4x = 24 by factoring. 26.
27. Find the value of the discriminant for 7x2 + 5x + 1 = 0. Then describe the number and type of roots for the equation. 27.
28. Use y = x2 - 7x + 5 for parts a – c.a. Write the equation in vertex form. 28a.
b. Identify the vertex. 28b.
c. Identify the axis of symmetry. 28c.
y
xO
Standardized Test Practice (continued)
Part 3: Short Response
Instructions: Write your answer in the space provided.
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Chapter 4 A1 Glencoe Algebra 2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Lesson 4-1
NA
ME
DAT
E
P
ER
IOD
Cha
pter
4
5 G
lenc
oe A
lgeb
ra 2
4-1
Stud
y Gu
ide
and
Inte
rven
tion
Gra
ph
ing
Qu
ad
rati
c F
un
cti
on
s
Gra
ph Q
uadr
atic
Fun
ctio
ns
Qua
drat
ic F
unct
ion
A fu
nctio
n de
fi ned
by
an e
quat
ion
of th
e fo
rm f(
x) =
ax2 +
bx
+ c
, whe
re a
≠ 0
Gra
ph o
f a
Qua
drat
ic F
unct
ion
A p
arab
ola
with
thes
e ch
arac
teris
tics:
y-in
terc
ept:
c; a
xis
of s
ymm
etry
: x =
-b
−
2a ;
x- co
ordi
nate
of v
erte
x: -
b −
2a
F
ind
the
y-in
terc
ept,
the
equa
tion
of
the
axis
of
sym
met
ry, a
nd t
he
x-co
ordi
nate
of
the
vert
ex f
or t
he g
raph
of
f(x)
= x
2 - 3
x +
5. U
se t
his
info
rmat
ion
to g
raph
the
fun
ctio
n.a
= 1
, b =
-3,
and
c =
5, s
o th
e y-
inte
rcep
t is
5. T
he e
quat
ion
of t
he a
xis
of s
ymm
etry
is
x =
-(-
3)
−
2(1)
or
3 −
2 . The
x-c
oord
inat
e of
the
ver
tex
is 3 −
2 .
Nex
t m
ake
a ta
ble
of v
alue
s fo
r x
near
3 −
2 .
xx2 -
3x
+ 5
f(x)
(x, f
(x))
002 -
3(0
) + 5
5(0
, 5)
112 -
3(1
) + 5
3(1
, 3)
3 −
2 ( 3 −
2 ) 2 - 3 (
3 −
2 ) +
5 11
−
4 ( 3 −
2 , 11
−
4 )
222 -
3(2
) + 5
3(2
, 3)
332 -
3(3
) + 5
5(3
, 5)
Exer
cise
sC
ompl
ete
part
s a–
c fo
r ea
ch q
uadr
atic
fun
ctio
n.a.
Fin
d th
e y-
inte
rcep
t, th
e eq
uati
on o
f th
e ax
is o
f sy
mm
etry
, and
the
x-c
oord
inat
e of
the
ver
tex.
b. M
ake
a ta
ble
of v
alue
s th
at in
clud
es t
he v
erte
x.c.
Use
thi
s in
form
atio
n to
gra
ph t
he f
unct
ion.
1. f
(x) =
x2 +
6x
+ 8
2.
f(x)
= -
x2 - 2x
+ 2
3.
f(x)
= 2
x2 - 4
x +
3
Exam
ple
xO12
-4
f(x)
( -3,
-1)
4-
8-
4
8 4x
O4
8-
8-
4
( -1,
3)
f(x)
4
-4
-8
xO
48
-4
( 1, 1
)
f(x)
12 8 4
xO
f(x)
6 4 2
-2
24
8
, x =
-3,
-3
2, x
= -
1, -
1 3
, x =
1, 1
x-
3-
2-
1-
4f(
x)-
10
30
x-
10
-2
1f(
x)3
22
-1
x1
02
3f(
x)1
33
9
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Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Chapter Resources
4
Cha
pter
4
3 G
lenc
oe A
lgeb
ra 2
B
efor
e yo
u be
gin
Cha
pter
4
• R
ead
each
sta
tem
ent.
• D
ecid
e w
heth
er y
ou A
gree
(A) o
r D
isag
ree
(D) w
ith
the
stat
emen
t.
• W
rite
A o
r D
in t
he fi
rst
colu
mn
OR
if y
ou a
re n
ot s
ure
whe
ther
you
agr
ee o
r di
sagr
ee,
wri
te N
S (N
ot S
ure)
.
A
fter
you
com
plet
e C
hapt
er 4
• R
erea
d ea
ch s
tate
men
t an
d co
mpl
ete
the
last
col
umn
by e
nter
ing
an A
or
a D
.
• D
id a
ny o
f you
r op
inio
ns a
bout
the
sta
tem
ents
cha
nge
from
the
firs
t co
lum
n?
• Fo
r th
ose
stat
emen
ts t
hat
you
mar
k w
ith
a D
, use
a p
iece
of p
aper
to
wri
te a
n ex
ampl
e of
why
you
dis
agre
e.
STE
P 1
A
, D, o
r N
SS
tate
men
tST
EP
2
A o
r D
1.
All
quad
rati
c fu
ncti
ons
have
a t
erm
wit
h th
e va
riab
le t
o th
e se
cond
pow
er.
2.
If t
he g
raph
of t
he q
uadr
atic
func
tion
y =
ax2 +
c o
pens
up
then
c <
0.
3.
A q
uadr
atic
equ
atio
n w
hose
gra
ph d
oes
not
inte
rsec
t th
e x-
axis
has
no
real
sol
utio
n.
4.
Sinc
e gr
aphi
ng s
how
s th
e ex
act
solu
tion
s to
a q
uadr
atic
eq
uati
on, n
o ot
her
met
hod
is n
eces
sary
for
solv
ing.
5.
If (x
- 3
)(x +
4) =
0, t
hen
eith
er x
- 3
= 0
or
x +
4 =
0.
6.
An
imag
inar
y nu
mbe
r co
ntai
ns i,
whi
ch e
qual
s th
e sq
uare
roo
t of
-1.
7.
A m
etho
d ca
lled
com
plet
ing
the
squa
re c
an b
e us
ed t
o re
wri
te a
qua
drat
ic e
xpre
ssio
n as
a p
erfe
ct s
quar
e.
8.
The
quad
rati
c fo
rmul
a ca
n on
ly b
e us
ed fo
r qu
adra
tic
equa
tion
s th
at c
anno
t be
sol
ved
by g
raph
ing
or
com
plet
ing
the
squa
re.
9.
The
disc
rim
inan
t of
a q
uadr
atic
equ
atio
n ca
n be
use
d to
de
term
ine
the
dire
ctio
n th
e gr
aph
will
ope
n.
10.
The
grap
h of
y =
2x2 i
s a
dila
tion
of t
he g
raph
of y
= x
2 .
11.
The
grap
h of
y =
(x +
2)2 w
ill b
e tw
o un
its
to t
he r
ight
of
the
grap
h of
y =
x2 .
12.
The
grap
h of
a q
uadr
atic
ineq
ualit
y co
ntai
ning
the
sy
mbo
l < w
ill b
e a
para
bola
ope
ning
dow
nwar
d.
Step
2
Step
1
Antic
ipat
ion
Guid
eQ
uad
rati
c F
un
cti
on
s an
d R
ela
tio
ns
A D A A A A AD D D D D
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PM
Answers (Anticipation Guide and Lesson 4-1)
A01_A13_ALG2_A_CRM_C04_AN_660785.indd A1A01_A13_ALG2_A_CRM_C04_AN_660785.indd A1 12/20/10 10:57 PM12/20/10 10:57 PM
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pyrig
ht ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
pa
nie
s, In
c.
PDF Pass
Chapter 4 A2 Glencoe Algebra 2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Cha
pter
4
6 G
lenc
oe A
lgeb
ra 2
4-1
Stud
y Gu
ide
and
Inte
rven
tion
(con
tinu
ed)
Gra
ph
ing
Qu
ad
rati
c F
un
cti
on
s
Max
imum
and
Min
imum
Val
ues
The
y-co
ordi
nate
of t
he v
erte
x of
a q
uadr
atic
fu
ncti
on is
the
max
imum
val
ue o
r m
inim
um v
alue
of t
he fu
ncti
on.
Max
imum
or
Min
imum
Val
ue
of a
Qua
drat
ic F
unct
ion
The
grap
h of
f(x
) = a
x2 + b
x +
c, w
here
a ≠
0, o
pens
up
and
has
a m
inim
um
whe
n a
> 0
. The
gra
ph o
pens
dow
n an
d ha
s a
max
imum
whe
n a
< 0
.
D
eter
min
e w
heth
er e
ach
func
tion
has
a m
axim
um o
r m
inim
um
valu
e, a
nd f
ind
that
val
ue. T
hen
stat
e th
e do
mai
n an
d ra
nge
of t
he f
unct
ion.
Exer
cise
sD
eter
min
e w
heth
er e
ach
func
tion
has
a m
axim
um o
r m
inim
um v
alue
, and
fin
d th
at v
alue
. The
n st
ate
the
dom
ain
and
rang
e of
the
fun
ctio
n. 1
. f(x
) = 2
x2 - x
+ 1
0 2.
f(x)
= x
2 + 4
x -
7
3. f(
x) =
3x2 -
3x
+ 1
4. f
(x) =
x2 +
5x
+ 2
5.
f(x)
= 2
0 +
6x
- x
2 6.
f(x)
= 4
x2 + x
+ 3
7. f
(x) =
-x2 -
4x
+ 1
0 8.
f(x)
= x
2 - 1
0x +
5
9. f(
x) =
-6x
2 + 1
2x +
21
Exam
ple
a. f
(x)
= 3
x2 - 6
x +
7
For
this
func
tion
, a =
3 a
nd b
= -
6.Si
nce
a >
0, t
he g
raph
ope
ns u
p, a
nd t
he
func
tion
has
a m
inim
um v
alue
.Th
e m
inim
um v
alue
is t
he y
-coo
rdin
ate
of
the
vert
ex. T
he x
-coo
rdin
ate
of t
he
vert
ex is
-b
−
2a =
-
(-
6)
−
2(3)
=
1.
Eva
luat
e th
e fu
ncti
on a
t x
= 1
to
find
the
min
imum
val
ue.
f(1)
= 3
(1)2 -
6(1
) + 7
= 4
, so
the
min
imum
val
ue o
f the
func
tion
is 4
. The
do
mai
n is
all
real
num
bers
. The
ran
ge is
al
l rea
ls g
reat
er t
han
or e
qual
to
the
min
imum
val
ue, t
hat
is {f
(x) |
f(x)
≥ 4
}.
b. f
(x)
= 1
00 -
2x
- x
2
For
this
func
tion
, a =
-1
and
b =
-2.
Sinc
e a
< 0
, the
gra
ph o
pens
dow
n, a
nd
the
func
tion
has
a m
axim
um v
alue
.Th
e m
axim
um v
alue
is t
he y
-coo
rdin
ate
of
the
vert
ex. T
he x
-coo
rdin
ate
of t
he v
erte
x is
-b
−
2a =
-
-2
−
2(-
1) =
-1.
Eva
luat
e th
e fu
ncti
on a
t x
= -
1 to
find
th
e m
axim
um v
alue
.f(
-1)
= 1
00 -
2(-
1) -
(-1)
2 = 1
01, s
o th
e m
axim
um v
alue
of t
he fu
ncti
on is
101
. The
do
mai
n is
all
real
num
bers
. The
ran
ge is
al
l rea
ls le
ss t
han
or e
qual
to
the
max
imum
val
ue, t
hat
is {f
(x) ⎪
f(x)
≤ 1
01}.
m
in.,
9 7 −
8 ; al
l rea
ls;
min
., -
11; a
ll re
als;
m
in.,
1 −
4 ; al
l rea
ls;
{ f
(x)⎥
f(x)
≥ 9
7 −
8 }
{f(x
)⎥ f(
x) ≥
-11
}
{f(
x)⎥ f(
x) ≥
1 −
4 }
m
in.,
- 17
−
4 ;
all
real
s;
max
., 29
; all
real
s;
min
., 2
15
−
16 ;
all r
eals
;
{
f(x)
⎥ f(
x) ≥
- 17
−
4 }
{
f(x)
⎥ f(
x) ≤
29 }
{
f(x)
⎥ f(
x) ≥
2 15
−
16
}
m
ax.,
14; a
ll re
als;
m
in.,
-20
; all
real
s;
max
., 27
; all
real
s;
{f(
x)⎥
f(x
) ≤ 1
4 }
{f(
x)⎥
f(x
) ≥ -
20}
{f(
x)⎥
f(x
) ≤ 2
7 }
001_
010_
ALG
2_A
_CR
M_C
04_C
R_6
6078
5.in
dd
612
/20/
10
9:05
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Lesson 4-1
NA
ME
DAT
E
P
ER
IOD
Cha
pter
4
7 G
lenc
oe A
lgeb
ra 2
4-1
Skill
s Pr
actic
eG
rap
hin
g Q
uad
rati
c F
un
cti
on
s
Com
plet
e pa
rts
a–c
for
each
qua
drat
ic f
unct
ion.
a. F
ind
the
y-in
terc
ept,
the
equa
tion
of
the
axis
of
sym
met
ry, a
nd t
he x
-coo
rdin
ate
of t
he v
erte
x.b.
Mak
e a
tabl
e of
val
ues
that
incl
udes
the
ver
tex.
c. U
se t
his
info
rmat
ion
to g
raph
the
fun
ctio
n.
1. f
(x) =
-2x
2 2.
f(x)
= x
2 - 4
x +
4
3. f(
x) =
x2 -
6x
+ 8
Det
erm
ine
whe
ther
eac
h fu
ncti
on h
as a
max
imum
or
a m
inim
um v
alue
, and
fi
nd t
hat
valu
e. T
hen
stat
e th
e do
mai
n an
d ra
nge
of t
he f
unct
ion.
4. f
(x) =
6x2
5. f(
x) =
-8x
2 6.
f(x)
= x
2 + 2
x
7. f
(x) =
-2x
2 + 4
x -
3
8. f(
x) =
3x2 +
12x
+ 3
9.
f(x)
= 2
x2 + 4
x +
1
10. f
(x) =
3x2
11. f
(x) =
x2 +
1
12. f
(x) =
-x2 +
6x
- 1
5
13. f
(x) =
2x2 -
11
14. f
(x) =
x2 -
10x
+ 5
15
. f(x
) = -
2x2 +
8x
+ 7
xO
f(x)
( 0, 0
)
-2
-4
-6
-8
2-
2-
44
2
xO16 12 8 4
2–
24
6
f(x) ( 2
, 0)
xO
( 3, –
1)
f(x)
8 6 4 2
2
–2
46
0
; x =
0; 0
4
; x =
2; 2
8
; x =
3; 3
x-
2-
10
12
f(x)
-8
-2
0-
2-
2x
-2
02
46
f(x)
164
04
16x
02
34
6f(
x)8
0-
10
8
m
ax.;
-1;
m
in.;
-9;
m
in.;
-1;
D
= {
all r
eal n
umbe
rs};
D
= {
all r
eal n
umbe
rs};
D
= {
all r
eal n
umbe
rs};
R
= {
f(x) ⎥
f(x
) ≤ -
1}
R =
{f(
x) ⎥
f(x
) ≥ -
9}
R =
{f(
x) ⎥
f(x
) ≥ -
1}
m
in.;
0;
max
.; 0;
m
in.;
-1;
D =
{al
l rea
l num
bers
};
D =
{al
l rea
l num
bers
};
D =
{al
l rea
l num
bers
};
R =
{f(
x) ⎥
f(x
) ≥ 0
} R
= {
f(x) ⎥
f(x
) ≤ 0
} R
= {
f(x) ⎥
f(x
) ≥ -
1}
m
in.;
0;
min
.; 1;
m
ax.;
-6;
D =
{al
l rea
l num
bers
};
D =
{al
l rea
l num
bers
};
D =
{al
l rea
l num
bers
};
R =
{f(
x) ⎥
f(x
) ≥ 0
} R
= {
f(x) ⎥
f(x
) ≥ 1
} R
= {
f(x) ⎥
f(x
) ≤ -
6}
m
in.;
-11
; m
in.;
-20
; m
ax.;
15;
D
= {
all r
eal n
umbe
rs};
D
= {
all r
eal n
umbe
rs};
D
= {
all r
eal n
umbe
rs};
R
= {
f(x) ⎥
f(x
) ≥ -
11}
R =
{f(
x) ⎥
f(x
) ≥ -
20}
R =
{f(
x) ⎥
f(x
) ≤ 1
5}
001_
010_
ALG
2_A
_CR
M_C
04_C
R_6
6078
5.in
dd
712
/20/
10
9:06
PM
Answers (Lesson 4-1)
A01_A13_ALG2_A_CRM_C04_AN_660785.indd A2A01_A13_ALG2_A_CRM_C04_AN_660785.indd A2 12/20/10 10:57 PM12/20/10 10:57 PM
Copyri
ght
© G
lencoe/M
cG
raw
-Hill
, a
div
isio
n o
f T
he
McG
raw
-Hill
Co
mp
an
ies,
Inc.
PDF Pass
Chapter 4 A3 Glencoe Algebra 2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Cha
pter
4
8 G
lenc
oe A
lgeb
ra 2
4-1
Prac
tice
Gra
ph
ing
Qu
ad
rati
c F
un
cti
on
sC
ompl
ete
part
s a–
c fo
r ea
ch q
uadr
atic
fun
ctio
n.a.
Fin
d th
e y-
inte
rcep
t, th
e eq
uati
on o
f th
e ax
is o
f sy
mm
etry
, and
the
x-c
oord
inat
e of
the
ver
tex.
b. M
ake
a ta
ble
of v
alue
s th
at in
clud
es t
he v
erte
x.c.
Use
thi
s in
form
atio
n to
gra
ph t
he f
unct
ion.
1. f
(x) =
x2 -
8x
+ 1
5 2.
f(x)
= -
x2 - 4
x +
12
3. f(
x) =
2x2 -
2x
+ 1
xO16 12 8 4
28
f(x)
( 4, -
1)64
16 12 8 4
xO
-2
-4
-6
( -2,
16)
f(x)
2
xO
f(x)
( 0.5
, 0.5
)2
2468
-2
-4
4
Det
erm
ine
whe
ther
eac
h fu
ncti
on h
as a
max
imum
or
min
imum
val
ue, a
nd f
ind
that
val
ue. T
hen
stat
e th
e do
mai
n an
d ra
nge
of t
he f
unct
ion.
4. f
(x) =
x2 +
2x
- 8
5.
f(x
) = x
2 - 6
x +
14
6. v
(x) =
-x2 +
14x
- 5
7
7. f
(x) =
2x2 +
4x
- 6
8.
f(x
) = -
x2 + 4
x -
1
9. f
(x) =
- 2 −
3 x2 + 8
x -
24
10. G
RAV
ITAT
ION
Fro
m 4
feet
abo
ve a
sw
imm
ing
pool
, Sus
an t
hrow
s a
ball
upw
ard
wit
h a
velo
city
of 3
2 fe
et p
er s
econ
d. T
he h
eigh
t h(
t) o
f the
bal
l t s
econ
ds a
fter
Sus
an t
hrow
s it
is
giv
en b
y h(
t) =
-16
t2 + 3
2t +
4. F
or t
≥ 0
, fin
d th
e m
axim
um h
eigh
t re
ache
d by
the
ba
ll an
d th
e ti
me
that
thi
s he
ight
is r
each
ed.
11. H
EALT
H C
LUBS
Las
t ye
ar, t
he S
port
sTim
e A
thle
tic
Clu
b ch
arge
d $2
0 to
par
tici
pate
in
an a
erob
ics
clas
s. S
even
ty p
eopl
e at
tend
ed t
he c
lass
es. T
he c
lub
wan
ts t
o in
crea
se t
he
clas
s pr
ice
this
yea
r. Th
ey e
xpec
t to
lose
one
cus
tom
er fo
r ea
ch $
1 in
crea
se in
the
pri
ce.
a. W
hat
pric
e sh
ould
the
clu
b ch
arge
to
max
imiz
e th
e in
com
e fr
om t
he a
erob
ics
clas
ses?
b. W
hat
is t
he m
axim
um in
com
e th
e Sp
orts
Tim
e A
thle
tic
Clu
b ca
n ex
pect
to
mak
e?
1
5; x
= 4
; 4
12;
x =
-2;
-2
1; x
= 0
.5; 0
.5
m
in.;
-9;
all
real
s;
min
.; 5;
all
real
s;
max
.; -
8; a
ll re
als;
{f
(x) ⎥
f(x
) ≥ -
9}
{f(
x) ⎥
f(x
) ≥ 5
} {
f(x) ⎥
f(x
) ≤ -
8}
m
in.;
-8;
all
real
s;
max
.; 3;
all
real
s;
max
.; 0;
all
real
s;
{f(x
) ⎥ f(x
) ≥ -
8}
{f(
x) ⎥
f(x
) ≤ 3
} {
f(x) ⎥
f(x
) ≤ 0
}
20 ft
; 1 s
$202
5
$45
x0
24
68
f(x)
153
-1
315
x-
6-
4-
20
2f(
x)0
1216
120
x-
10
0.5
12
f(x)
51
0.5
15
001_
010_
ALG
2_A
_CR
M_C
04_C
R_6
6078
5.in
dd
812
/20/
10
9:06
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Lesson 4-1
NA
ME
DAT
E
P
ER
IOD
Cha
pter
4
9 G
lenc
oe A
lgeb
ra 2
4-1
Wor
d Pr
oble
m P
ract
ice
Gra
ph
ing
Qu
ad
rati
c F
un
cti
on
s
1. T
RAJE
CTO
RIES
A c
anno
nbal
l is
laun
ched
from
a c
anno
n on
the
wal
l of
Fort
Cha
mbl
y, Q
uebe
c. If
the
pat
h of
the
ca
nnon
ball
is
trac
ed o
n a
piec
e of
gra
ph
pape
r al
igne
d so
th
at t
he c
anno
n is
sit
uate
d on
th
e y-
axis
, the
eq
uati
on t
hat
desc
ribe
s th
e pa
th is
y
=-
1−
1600
x2+
1−
2x +
20,
whe
re x
is t
he h
oriz
onta
l dis
tanc
e fr
om
the
cliff
and
y is
the
ver
tica
l dis
tanc
e ab
ove
the
grou
nd in
feet
. How
hig
h ab
ove
the
grou
nd is
the
can
non?
2. T
ICK
ETIN
G T
he m
anag
er o
f a
sym
phon
y co
mpu
tes
that
the
sym
phon
y w
ill e
arn
-40
P2+
110
0P d
olla
rs p
er
conc
ert
if th
ey c
harg
e P
dolla
rs fo
r ti
cket
s. W
hat
tick
et p
rice
sho
uld
the
sym
phon
y ch
arge
in o
rder
to
max
imiz
e it
s pr
ofit
s?
3. A
RCH
ES A
n ar
chit
ect
deci
des
to u
se a
pa
rabo
lic a
rch
for
the
mai
n en
tran
ce o
f a
scie
nce
mus
eum
. In
one
of h
is p
lans
, the
to
p ed
ge o
f the
arc
h is
des
crib
ed b
y th
e gr
aph
of y
=-
1−
4x2
+
5−
2x
+ 1
5. W
hat
are
the
coor
dina
tes
of t
he v
erte
x of
thi
s pa
rabo
la?
4. F
RAM
ING
A fr
ame
com
pany
offe
rs a
lin
e of
squ
are
fram
es. I
f the
sid
e le
ngth
of
the
fram
e is
s, t
hen
the
area
of t
he
open
ing
in t
he fr
ame
is g
iven
by
the
func
tion
a(s
) =s2
- 1
0s+
24.
G
raph
a(s
).
a
sO5
5
5. W
ALK
ING
Can
al S
tree
t an
d W
alke
r St
reet
are
per
pend
icul
ar t
o ea
ch o
ther
. E
vita
is d
rivi
ng s
outh
on
Can
al S
tree
t an
d is
cur
rent
ly 5
mile
s no
rth
of t
he
inte
rsec
tion
wit
h W
alke
r St
reet
. Jac
k is
at
the
inte
rsec
tion
of C
anal
and
Wal
ker
Stre
ets
and
head
ing
east
on
Wal
ker.
Jack
and
Evi
ta a
re b
oth
driv
ing
30 m
iles
per
hour
.
a. W
hen
Jack
is x
mile
s ea
st o
f the
in
ters
ecti
on, w
here
is E
vita
?
b. T
he d
ista
nce
betw
een
Jack
and
Evi
ta is
gi
ven
by t
he fo
rmul
a √
��
��
�
x2+
(5 -
x) 2 .
For
wha
t va
lue
of x
are
Jac
k an
d E
vita
at
thei
r cl
oses
t?
(Hin
t: M
inim
ize
the
squa
re o
f the
di
stan
ce.)
c. W
hat
is t
he d
ista
nce
of c
lose
st
appr
oach
?
20 ft
$13.
75
(5, 2
1.25
)
5
- x
mi n
orth
of t
he
inte
rsec
tion
5 √
�
2 −
2
mi
x =
2.5
001_
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Answers (Lesson 4-1)
A01_A13_ALG2_A_CRM_C04_AN_660785.indd A3A01_A13_ALG2_A_CRM_C04_AN_660785.indd A3 12/20/10 10:57 PM12/20/10 10:57 PM
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pyrig
ht ©
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nco
e/M
cG
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-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
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nie
s, In
c.
PDF Pass
Chapter 4 A4 Glencoe Algebra 2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Cha
pter
4
10
Gle
ncoe
Alg
ebra
2
4-1
Enri
chm
ent
Fin
din
g t
he x
-inte
rcep
ts o
f a P
ara
bo
la
As
you
know
, if f
(x) =
ax2 +
bx
+ c
is a
qua
drat
ic fu
ncti
on, t
he v
alue
s of
x
that
mak
e f(
x) e
qual
to
zero
are
-
b +
√
��
��
b2 -
4ac
−
2a
and
-
b -
√
��
��
b2 -
4ac
−
2a
.
The
aver
age
of t
hese
tw
o nu
mbe
r va
lues
is -
b −
2a .
Of(x)
x
––
,f(
(
((
b –– 2a b –– 2a
b –– 2ax
= –
f(x) =
ax2 +
bx
+ c
The
func
tion
f(x
) has
its
max
imum
or
min
imum
valu
e w
hen
x =
- b −
2a . T
he x
-inte
rcep
ts o
f the
par
abol
a,
whe
n th
ey e
xist
, are
√
��
��
b2 -
4ac
−
2a
uni
ts t
o th
e le
ft a
nd
righ
t of
the
axi
s of
sym
met
ry.
F
ind
the
vert
ex, a
xis
of s
ymm
etry
, and
x-i
nter
cept
s fo
r f(
x) =
5x2 +
10x
- 7
.
Use
x =
- b −
2a .
x =
- 10
−
2(5)
= -
1
The
x-c
oord
inat
e of
the
ver
tex
is -
1.
Subs
titu
te x
= -
1 in
f(x
) = 5
x2 + 1
0x -
7.
f(-
1) =
5(-
1)2 +
10(
-1)
- 7
= -
12. T
he v
erte
x is
(-1,
-12
).
The
axis
of s
ymm
etry
is x
= -
b −
2a , o
r x
= -
1.
The
x-co
ordi
nate
s of
the
x-in
terc
epts
are
–1
± √
��
��
b2 -
4ac
−
2a
= –
1 ±
√
��
��
��
�
102 -
4 �
5 � (-
7)
−
−
2 � 5
= –
1 ±
√
��
24
0 −
10
. The
x–i
nter
cept
s ar
e (–
1 –
2 −
5 √
��
15
, 0)
and
(–1
+ 2
− 5 √
��
15
, 0)
.
Exer
cise
sF
ind
the
vert
ex, a
xis
of s
ymm
etry
, and
x-i
nter
cept
s fo
r th
e gr
aph
of e
ach
func
tion
us
ing
x =
- b −
2a .
1. f
(x) =
x2 -
4x
- 8
2. g
(x)
= -
4x2 -
8x
+ 3
3. y
= -
x2 + 8
x +
34.
f(x
) = 2
x2 + 6
x +
5
5. A
(x) =
x2 +
12x
+ 3
66.
k(x
) = -
2x2 +
2x
- 6
Exam
ple
(4, 1
9); x
= 4
;4
± √
��
19
(2, -
12);
x =
2;
2 ±
2 √
� 3
(-1,
7);
x =
-1;
-1
± 2
√ �
7
(-6,
0);
x =
-6;
-6
( 1 −
2 , -5
1 −
2 ) ; x
= 1 −
2 ;
no x
-inte
rcep
ts
(- 3 −
2 , 1 −
2 ) ; x
= -
3 −
2 ;
no x
-inte
rcep
ts
001_
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Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 4-2
4-2
Cha
pter
4
11
Gle
ncoe
Alg
ebra
2
Stud
y Gu
ide
and
Inte
rven
tion
So
lvin
g Q
uad
rati
c E
qu
ati
on
s b
y G
rap
hin
g
Solv
e Q
uadr
atic
Equ
atio
ns
Qua
drat
ic E
quat
ion
A q
uadr
atic
equ
atio
n ha
s th
e fo
rm a
x2 + b
x +
c =
0, w
here
a ≠
0.
Roo
ts o
f a Q
uadr
atic
Equ
atio
nso
lutio
n(s)
of t
he e
quat
ion,
or t
he z
ero(
s) o
f the
rela
ted
quad
ratic
func
tion
The
zero
s of
a q
uadr
atic
func
tion
are
the
x-in
terc
epts
of i
ts g
raph
. The
refo
re, f
indi
ng t
he
x-in
terc
epts
is o
ne w
ay o
f sol
ving
the
rel
ated
qua
drat
ic e
quat
ion.
So
lve
x2 + x
- 6
= 0
by
grap
hing
.
Gra
ph t
he r
elat
ed fu
ncti
on f
(x) =
x2 +
x -
6.
The
x-co
ordi
nate
of t
he v
erte
x is
-b
−
2a =
- 1 −
2 , and
the
equ
atio
n of
the
axis
of s
ymm
etry
is x
= -
1 −
2 .
Mak
e a
tabl
e of
val
ues
usin
g x-
valu
es a
roun
d -
1 −
2 .
x-
1-
1 −
2 0
12
f(x)
-6
-6
1 −
4 -
6-
40
Fro
m t
he t
able
and
the
gra
ph, w
e ca
n se
e th
at t
he z
eros
of t
he fu
ncti
on a
re 2
and
-3.
Exer
cise
s
Use
the
rel
ated
gra
ph o
f ea
ch e
quat
ion
to d
eter
min
e it
s so
luti
on.
1. x
2 + 2
x -
8 =
0
2. x
2 - 4
x -
5 =
0
3. x
2 - 5
x +
4 =
0
xO
f(x)
-2
-2
-4
-6
-8
-4
2
xO
-2
-4
-6
-8
-2
24
f(x)
xO
f(x)
-2
-2
-4
2
246
4
4. x
2 - 1
0x +
21
= 0
5.
x2 +
4x
+ 6
= 0
6.
4x2 +
4x
+ 1
= 0
xO
-2
-4
2
2
46
f(x)
xO
-2
-4
2
246
f(x)
xO
-2
-4
2
2468
4
f(x)
xO
-6
-4
-2
-4
-2
2
f(x)
Exam
ple
3
, 7
no
real
sol
utio
ns
- 1 −
2
2, -
45,
-1
1, 4
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Answers (Lesson 4-1 and Lesson 4-2)
A01_A13_ALG2_A_CRM_C04_AN_660785.indd A4A01_A13_ALG2_A_CRM_C04_AN_660785.indd A4 12/20/10 10:57 PM12/20/10 10:57 PM
Copyri
ght
© G
lencoe/M
cG
raw
-Hill
, a
div
isio
n o
f T
he
McG
raw
-Hill
Co
mp
an
ies,
Inc.
PDF Pass
Chapter 4 A5 Glencoe Algebra 2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
4-2
Cha
pter
4
12
Gle
ncoe
Alg
ebra
2
Stud
y Gu
ide
and
Inte
rven
tion
(con
tinu
ed)
So
lvin
g Q
uad
rati
c E
qu
ati
on
s b
y G
rap
hin
g
Esti
mat
e So
luti
ons
Oft
en, y
ou m
ay n
ot b
e ab
le t
o fin
d ex
act
solu
tion
s to
qua
drat
ic
equa
tion
s by
gra
phin
g. B
ut y
ou c
an u
se t
he g
raph
to
esti
mat
e so
luti
ons.
So
lve
x2 - 2
x -
2 =
0 b
y gr
aphi
ng. I
f ex
act
root
s ca
nnot
be
foun
d,
stat
e th
e co
nsec
utiv
e in
tege
rs b
etw
een
whi
ch t
he r
oots
are
loca
ted.
The
equa
tion
of t
he a
xis
of s
ymm
etry
of t
he r
elat
ed fu
ncti
on is
xO
-2
-2
-42
24
4f(x
)
x =
- -
2 −
2(1)
= 1
, so
the
vert
ex h
as x
-coo
rdin
ate
1. M
ake
a ta
ble
of v
alue
s.
x-
10
12
3
f(x)
1-
2-
3-
21
The
x-in
terc
epts
of t
he g
raph
are
bet
wee
n 2
and
3 an
d be
twee
n 0
and
-1.
So
one
solu
tion
is b
etw
een
2 an
d 3,
and
the
oth
er s
olut
ion
is
betw
een
0 an
d -
1.
Exer
cise
sSo
lve
the
equa
tion
s. I
f ex
act
root
s ca
nnot
be
foun
d, s
tate
the
con
secu
tive
inte
gers
be
twee
n w
hich
the
roo
ts a
re lo
cate
d.
1. x
2 - 4
x +
2 =
0
2. x
2 + 6
x +
6 =
0
3. x
2 + 4
x +
2=
0
xO
-2
-2
-4
-4
2
24
4
f(x)
xO
-2
-2
-4
-4
2
2
f(x)
xO
-2
-2
-4
-4
2
24f(x
)
4. -
x2 + 2
x +
4 =
0
5. 2
x2 - 1
2x +
17
= 0
6.
- 1 −
2 x2 + x
+ 5 −
2 = 0
xO
-2
2
246
-2
4
f(x)
xO
2
246
46
f(x)
xO
-2
-2
-4
2
24
4
f(x)
Exam
ple
b
etw
een
0 an
d 1;
b
etw
een
-2
and
-1;
b
etw
een
-1
and
0;be
twee
n 3
and
4 b
etw
een
-5
and
-4
bet
wee
n -
4 an
d -
3
b
etw
een
3 an
d 4;
b
etw
een
2 an
d 3;
b
etw
een
-2
and
-1;
betw
een
-2
and
-1
bet
wee
n 3
and
4 b
etw
een
3 an
d 4
011_
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PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 4-2
4-2
Cha
pter
4
13
Gle
ncoe
Alg
ebra
2
Skill
s Pr
actic
eS
olv
ing
Qu
ad
rati
c E
qu
ati
on
s B
y G
rap
hin
g
Use
the
rel
ated
gra
ph o
f ea
ch e
quat
ion
to d
eter
min
e it
s so
luti
ons.
1. x
2 + 2
x -
3 =
0
2. -
x2 - 6
x -
9 =
0
3. 3
x2 + 4
x +
3 =
0
x
f(x)
O
-2
-4
2-
2-
44
2
f(x) =
x2 +
2x
- 3
x
f(x)
O
-2
-4
-6
-8
-2
-6
-4
f(x) =
-x2 -
6x
- 9
xO
f(x) =
3x2 +
4x +
3
12 8 4
–2
–4
–6
f(x)
Solv
e ea
ch e
quat
ion.
If
exac
t ro
ots
cann
ot b
e fo
und,
sta
te t
he c
onse
cuti
ve in
tege
rs
betw
een
whi
ch t
he r
oots
are
loca
ted.
4. x
2 - 6
x +
5 =
0
5. -
x2 + 2
x -
4 =
0
6. x
2 - 6
x +
4 =
0
xO
f(x)
4 2
2
–2
–4
46
f(x) =
x2 -
6x
+ 5
xO
f(x)
2
–2
–2
–4
–6
–8
46
f(x) =
-x2 +
2x
- 4
xO
f(x)
4 2
2
–2
–4
46
f(x) =
x2 -
6x
+ 4
7. -
x2 - 4
x =
0
8. -
x2 + 3
6 =
0
xO
f(x)
4 2
2
–2
–4
-4
-2
-6f(x
) = -
x2 - 4
x
xO
12–1
2
36 24 12
f(x)
–66f(x
) = -
x2 + 3
6
0
, -4
-
6, 6
1
, 5
no
real
sol
utio
ns
bet
wee
n 0
and
1;
betw
een
5 an
d 6
-
3, 1
-
3 n
o re
al s
olut
ions
011_
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ALG
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5.in
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10
9:06
PM
Answers (Lesson 4-2)
A01_A13_ALG2_A_CRM_C04_AN_660785.indd A5A01_A13_ALG2_A_CRM_C04_AN_660785.indd A5 12/20/10 10:57 PM12/20/10 10:57 PM
Co
pyrig
ht ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
pa
nie
s, In
c.
PDF Pass
Chapter 4 A6 Glencoe Algebra 2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
4-2
Cha
pter
4
14
Gle
ncoe
Alg
ebra
2
Prac
tice
So
lvin
g Q
uad
rati
c E
qu
ati
on
s B
y G
rap
hin
g
3, -
2no
suc
h re
al
num
bers
exi
st
Use
the
rel
ated
gra
ph o
f ea
ch e
quat
ion
to d
eter
min
e it
s so
luti
ons.
1. -
3x2 +
3 =
0
2. 3
x2 + x
+ 3
= 0
3.
x2 -
3x
+ 2
= 0
xO
-2
-2
-4
-4
2
2
f(x)
4
xO
-2
-4
2
2468
4
f(x)
xO
-2
-4
2
2468
4
f(x)
Solv
e ea
ch e
quat
ion.
If
exac
t ro
ots
cann
ot b
e fo
und,
sta
te t
he c
onse
cuti
ve in
tege
rs
betw
een
whi
ch t
he r
oots
are
loca
ted.
4. -
2x2 -
6x
+ 5
= 0
5.
x2 +
10x
+ 2
4 =
0
6. 2
x2 - x
- 6
= 0
xO
-2
-6
f(x)
12 8 4
-4
xO
f(x)
246
-2
-4
-6
xO
f(x)
-2
-2
-4
-6
24
6
7. -
x2 + x
+ 6
= 0
8.
-x2 +
5x
- 8
= 0
x
O
f(x)
-2
2
246
46
xO
f(x)
-2
2
-2
-4
-6
-8
46
9. G
RAV
ITY
Use
the
form
ula
h(t)
= v
0t -
16t
2 , w
here
h(t
) is
the
heig
ht o
f an
obje
ct in
feet
, v 0 i
s th
e ob
ject
’s in
itia
l vel
ocit
y in
feet
per
sec
ond,
and
t is
the
tim
e in
sec
onds
.
a. M
arta
thr
ows
a ba
seba
ll w
ith
an in
itia
l upw
ard
velo
city
of 6
0 fe
et p
er s
econ
d.
Igno
ring
Mar
ta’s
heig
ht, h
ow lo
ng a
fter
she
rel
ease
s th
e ba
ll w
ill it
hit
the
gro
und?
b. A
vol
cani
c er
upti
on b
last
s a
boul
der
upw
ard
wit
h an
init
ial v
eloc
ity
of
240
feet
per
sec
ond.
How
long
will
it t
ake
the
boul
der
to h
it t
he g
roun
d if
it la
nds
at
the
sam
e el
evat
ion
from
whi
ch it
was
eje
cted
?
-
1, 1
n
o re
al s
olut
ions
1
, 2
b
etw
een
0 an
d 1;
-
6, -
4 -
1.5,
2be
twee
n -
4 an
d -
3
15 s
3.75
s
011_
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9:06
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 4-2
4-2
Cha
pter
4
15
Gle
ncoe
Alg
ebra
2
Wor
d Pr
oble
m P
ract
ice
So
lvin
g Q
uad
rati
c E
qu
ati
on
s b
y G
rap
hin
g
1. T
RAJE
CTO
RIES
Dav
id t
hrew
a b
aseb
all
into
the
air.
The
func
tion
of t
he h
eigh
t of
the
bas
ebal
l in
feet
is h
= 8
0t-
16t2 ,
whe
re t
repr
esen
ts t
he t
ime
in s
econ
ds
afte
r th
e ba
ll w
as t
hrow
n. U
se t
his
grap
h of
the
func
tion
to
dete
rmin
e ho
w
long
it t
ook
for
the
ball
to fa
ll ba
ck t
o th
e gr
ound
. h
t1
23
45
-1
-40
4080
2. B
RID
GES
In
1895
, a b
rick
arc
h ra
ilway
br
idge
was
bui
lt o
n N
orth
Ave
nue
in
Bal
tim
ore,
Mar
ylan
d. T
he a
rch
is
desc
ribe
d by
the
equ
atio
n h
= 9
–1−
50x2 ,
whe
re h
is t
he h
eigh
t in
yar
ds a
nd x
is
the
dist
ance
in y
ards
from
the
cen
ter
of
the
brid
ge. G
raph
thi
s eq
uati
on a
nd
desc
ribe
, to
the
near
est
yard
, whe
re t
he
brid
ge t
ouch
es t
he g
roun
d.
3. L
OG
IC W
ilma
is t
hink
ing
of t
wo
num
bers
. The
sum
is 2
and
the
pro
duct
is
-24
. Use
a q
uadr
atic
equ
atio
n to
find
th
e tw
o nu
mbe
rs.
4. R
AD
IO T
ELES
COPE
S Th
e cr
oss-
sect
ion
of a
larg
e ra
dio
tele
scop
e is
a p
arab
ola.
Th
e di
sh is
set
into
the
gro
und.
The
eq
uati
on t
hat
desc
ribe
s th
e cr
oss-
sect
ion
is d
=
2−
75x2
-
4−
3x
-
32−
3 , w
here
d g
ives
the
dept
h of
the
dis
h be
low
gro
und
and
x is
the
dis
tanc
e fr
om t
he c
ontr
ol c
ente
r, bo
th in
met
ers.
If t
he d
ish
does
not
ex
tend
abo
ve t
he g
roun
d le
vel,
wha
t is
th
e di
amet
er o
f the
dis
h? S
olve
by
grap
hing
.
x40
6080
20-
20
-2020406080
d
5. B
OAT
S Th
e di
stan
ce b
etw
een
two
boat
s is
d
=√
��
��
��
t2-
10t
+ 3
5 ,w
here
d is
dis
tanc
e in
met
ers
and
t is
tim
e in
sec
onds
.
a. M
ake
a gr
aph
of d
2 ver
sus
t.d
tO
1015
2025
5
102030
b. D
o th
e bo
ats
ever
col
lide?
N
o
5
sec
onds
6 an
d -
4
21 y
ards
from
the
cent
er o
f the
br
idge
on
eith
er s
ide
at (-
21, 0
) an
d (2
1, 0
)
x
30 10
-20
-40
-30
-10
2040
h
6
4 m
011_
029_
ALG
2_A
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M_C
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5.in
dd
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10
9:06
PM
Answers (Lesson 4-2)
A01_A13_ALG2_A_CRM_C04_AN_660785.indd A6A01_A13_ALG2_A_CRM_C04_AN_660785.indd A6 12/20/10 10:57 PM12/20/10 10:57 PM
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cG
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-Hill
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isio
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he
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raw
-Hill
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mp
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ies,
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Chapter 4 A7 Glencoe Algebra 2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
4-2
Cha
pter
4
16
Gle
ncoe
Alg
ebra
2
Enri
chm
ent
Gra
ph
ing
Ab
solu
te V
alu
e E
qu
ati
on
s
You
can
solv
e ab
solu
te v
alue
equ
atio
ns in
muc
h th
e sa
me
way
you
sol
ved
quad
rati
c eq
uati
ons.
Gra
ph t
he r
elat
ed a
bsol
ute
valu
e fu
ncti
on fo
r ea
ch
equa
tion
usi
ng a
gra
phin
g ca
lcul
ator
. The
n us
e th
e ZE
RO fe
atur
e in
the
CA
LC m
enu
to fi
nd it
s re
al s
olut
ions
, if a
ny. R
ecal
l tha
t so
luti
ons
are
poin
ts
whe
re t
he g
raph
inte
rsec
ts t
he x
-axi
s.
For
each
equ
atio
n, m
ake
a sk
etch
of
the
rela
ted
grap
h an
d fi
nd t
he
solu
tion
s ro
unde
d to
the
nea
rest
hun
dred
th.
1. ⎪
x +
5⎥
= 0
2.
⎪ 4x
- 3
⎥
+ 5
= 0
3.
⎪ x -
7⎥
= 0
4. ⎪
x +
3⎥
- 8
= 0
5.
- ⎪ x
+ 3
⎥
+ 6
= 0
6.
⎪ x -
2⎥
- 3
= 0
7. ⎪
3x +
4⎥
= 2
8.
⎪ x +
12⎥
= 1
0 9.
⎪ x⎥
- 3
= 0
10. E
xpla
in h
ow s
olvi
ng a
bsol
ute
valu
e eq
uati
ons
alge
brai
cally
and
find
ing
zero
s of
abs
olut
e va
lue
func
tion
s gr
aphi
cally
are
rel
ated
.
-
5 N
o so
lutio
ns
7
Sam
ple
answ
er: v
alue
s of
x w
hen
solv
ing
alge
brai
cally
are
the
x-in
terc
epts
(or
zero
s) o
f the
func
tion
whe
n gr
aphe
d.
-
11, 5
-
9, 3
-
1, 5
-
2, -
2 −
3 -
22, -
2 -
3, 3
011_
029_
ALG
2_A
_CR
M_C
04_C
R_6
6078
5.in
dd
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/20/
10
9:06
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 4-3
4-3
Cha
pter
4
17
Gle
ncoe
Alg
ebra
2
Stud
y Gu
ide
and
Inte
rven
tion
So
lvin
g Q
uad
rati
c E
qu
ati
on
s b
y Fa
cto
rin
g
Fact
ored
For
m T
o w
rite
a q
uadr
atic
equ
atio
n w
ith
root
s p
and
q, le
t (x
- p
)(x -
q) =
0.
Then
mul
tipl
y us
ing
FOIL
.
W
rite
a q
uadr
atic
equ
atio
n in
sta
ndar
d fo
rm w
ith
the
give
n ro
ots.
a. 3
, -5
(x
- p
)(x -
q) =
0
Writ
e th
e pa
ttern
.
(x
- 3
)[x -
(-5)
] = 0
R
epla
ce p
with
3, q
with
-5.
(x
- 3
)(x +
5) =
0
Sim
plify
.
x2 +
2x
- 1
5 =
0
Use
FO
IL.
The
equa
tion
x2 +
2x
- 1
5 =
0 h
as r
oots
3
and
-5.
b. -
7 −
8 , 1 −
3
(x
- p
)(x -
q) =
0
[x -
(-
7 −
8 ) ] (
x -
1 −
3 ) =
0
(
x +
7 −
8 ) (x
- 1 −
3 ) =
0
(8
x +
7)
−
8 �
(3x
- 1
) −
3 =
0
24
� (8
x +
7)(3
x -
1)
−
−
24
=
24
� 0
24
x2 + 1
3x -
7 =
0
The
equa
tion
24x
2 + 1
3x -
7 =
0 h
as
root
s -
7 −
8 and
1 −
3 .
Exer
cise
sW
rite
a q
uadr
atic
equ
atio
n in
sta
ndar
d fo
rm w
ith
the
give
n ro
ot(s
).
1. 3
, -4
2. -
8, -
2 3.
1, 9
4. -
5 5.
10,
7
6. -
2, 1
5
7. -
1 −
3 , 5
8. 2
, 2 −
3 9.
-7,
3 −
4
10. 3
, 2 −
5 11
. - 4 −
9 , -1
12. 9
, 1 −
6
13. 2
− 3 , -
2 −
3 14
. 5 −
4 , - 1 −
2 15
. 3 −
7 , 1 −
5
16. -
7 −
8 , 7 −
2 17
. 1 −
2 , 3 −
4 18
. 1 −
8 , 1 −
6
Exam
ple
1
6x2 -
42x
- 4
9 =
0
8x
2 - 1
0x +
3 =
0
48x
2 - 1
4x +
1 =
0
9
x2 -
4 =
0
8x
2 - 6
x -
5 =
0
35x
2 - 2
2x +
3 =
0
x
2 + x
- 1
2 =
0
x2 +
10x
+ 1
6 =
0
x2 -
10x
+ 9
= 0
x
2 + 1
0x +
25
= 0
x
2 - 1
7x +
70
= 0
x
2 - 1
3x -
30
= 0
5
x2 -
17x
+ 6
= 0
9
x2 +
13x
+ 4
= 0
6
x2 -
55x
+ 9
= 0
3
x2 -
14x
- 5
= 0
3
x2 -
8x
+ 4
= 0
4
x2 +
25x
- 2
1 =
0
011_
029_
ALG
2_A
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M_C
04_C
R_6
6078
5.in
dd
1712
/20/
10
9:06
PM
Answers (Lesson 4-2 and Lesson 4-3)
A01_A13_ALG2_A_CRM_C04_AN_660785.indd A7A01_A13_ALG2_A_CRM_C04_AN_660785.indd A7 12/20/10 10:57 PM12/20/10 10:57 PM
Co
pyrig
ht ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
pa
nie
s, In
c.
PDF Pass
Chapter 4 A8 Glencoe Algebra 2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
4-3
Cha
pter
4
18
Gle
ncoe
Alg
ebra
2
Stud
y Gu
ide
and
Inte
rven
tion
(con
tinu
ed)
So
lvin
g Q
uad
rati
c E
qu
ati
on
s b
y Fa
cto
rin
g
Solv
e Eq
uati
ons
by F
acto
ring
Whe
n yo
u us
e fa
ctor
ing
to s
olve
a q
uadr
atic
equ
atio
n,
you
use
the
follo
win
g pr
oper
ty.
Zero
Pro
duct
Pro
pert
yFo
r any
real
num
bers
a a
nd b
, if a
b =
0, t
hen
eith
er a
= 0
or b
=0,
or b
oth
a an
d b
= 0
.
So
lve
each
equ
atio
n by
fac
tori
ng.
a. 3
x2 = 1
5x
3x2 =
15x
O
rigin
al e
quat
ion
3x
2 - 1
5x =
0
Sub
tract
15x
from
bot
h si
des.
3x
(x -
5) =
0
Fact
or th
e bi
nom
ial.
3x =
0 o
r x
- 5
= 0
Ze
ro P
rodu
ct P
rope
rty
x
= 0
or
x
= 5
S
olve
eac
h eq
uatio
n.
Th
e so
luti
on s
et is
{0, 5
}.
b. 4
x2 - 5
x =
21
4x2 -
5x
= 2
1 O
rigin
al e
quat
ion
4x
2 - 5
x -
21
= 0
S
ubtra
ct 2
1 fro
m b
oth
side
s.
(4x
+ 7
)(x -
3) =
0
Fact
or th
e tri
nom
ial.
4x +
7 =
0
or x
- 3
= 0
Ze
ro P
rodu
ct P
rope
rty
x
= -
7 −
4 or
x =
3
Sol
ve e
ach
equa
tion.
The
solu
tion
set
is {
- 7 −
4 , 3} .
Exer
cise
sSo
lve
each
equ
atio
n by
fac
tori
ng.
1. 6
x2 - 2
x =
0
2. x
2 = 7
x 3.
20x
2 = -
25x
4. 6
x2 = 7
x 5.
6x2 -
27x
= 0
6.
12x
2 - 8
x =
0
7. x
2 + x
- 3
0 =
0
8. 2
x2 - x
- 3
= 0
9.
x2 +
14x
+ 3
3 =
0
10. 4
x2 + 2
7x -
7 =
0
11. 3
x2 + 2
9x -
10
= 0
12
. 6x2 -
5x
- 4
= 0
13. 1
2x2 -
8x
+ 1
= 0
14
. 5x2 +
28x
- 1
2 =
0
15.
2x2 -
250
x +
500
0 =
0
16. 2
x2 - 1
1x -
40
= 0
17
. 2x2 +
21x
- 1
1 =
0
18. 3
x2 + 2
x -
21
= 0
19. 8
x2 - 1
4x +
3 =
0
20. 6
x2 + 1
1x -
2 =
0
21. 5
x2 + 1
7x -
12
= 0
22. 1
2x2 +
25x
+ 1
2 =
0
23. 1
2x2 +
18x
+ 6
= 0
24
. 7x2 -
36x
+ 5
= 0
Exam
ple
{
5, -
6}
{ 3 −
2 , -
1 }
{-
11, -
3}
{
1 −
6 , 1 −
2 }
{ 2 −
5 , -6 }
{
100,
25}
{
- 4 −
3 , - 3 −
4 }
{-
1 −
2 , -1 }
{
1 −
7 , 5}
{
0, 7 −
6 }
{0,
9 −
2 }
{0,
2 −
3 }
{
0, 1 −
3 }
{0,
7}
{0,
- 5 −
4 }
{
8, -
5 −
2 }
{-
11, 1
− 2 }
{
7 −
3 , -
3 }
{
3 −
2 , 1 −
4 }
{-
2, 1 −
6 }
{ 3 −
5 , -
4 }
{
1 −
4 , -
7 }
{-
10, 1
− 3 }
{
- 1 −
2 , 4 −
3 }
011_
029_
ALG
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5.in
dd
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/20/
10
9:06
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 4-3
4-3
Cha
pter
4
19
Gle
ncoe
Alg
ebra
2
Skill
s Pr
actic
eS
olv
ing
Qu
ad
rati
c E
qu
ati
on
s b
y Fa
cto
rin
g
Wri
te a
qua
drat
ic e
quat
ion
in s
tand
ard
form
wit
h th
e gi
ven
root
(s).
1. 1
, 4
2. 6
, -9
3. -
2, -
5 4.
0, 7
5. -
1 −
3 , -3
6. -
1 −
2 , 3 −
4
Fact
or e
ach
poly
nom
ial.
7. m
2 + 7
m -
18
8.
2x2 -
3x
- 5
9. 4
z2 + 4
z -
15
10
. 4p2 +
4p
- 2
4
11. 3
y2 + 2
1y +
36
12
. c2 -
100
Solv
e ea
ch e
quat
ion
by f
acto
ring
.
13. x
2 = 6
4
14. x
2 - 1
00 =
0
15. x
2 - 3
x +
2 =
0
16. x
2 - 4
x +
3 =
0
17. x
2 + 2
x -
3 =
0
18. x
2 - 3
x -
10
= 0
19. x
2 - 6
x +
5 =
0
20. x
2 - 9
x =
0
21. x
2 - 4
x =
21
22
. 2x2 +
5x
- 3
= 0
23. 4
x2 + 5
x -
6 =
0
24. 3
x2 - 1
3x -
10
= 0
25. N
UM
BER
THEO
RY F
ind
two
cons
ecut
ive
inte
gers
who
se p
rodu
ct is
272
.
x2 -
5x
+ 4
= 0
x2 +
3x
- 5
4 =
0
x2 +
7x
+ 1
0 =
0x
2 - 7
x =
0
3x2 +
10x
+ 3
= 0
8x2 -
2x
- 3
= 0
(
m -
2)(
m +
9)
(2x
- 5
)(x
+ 1
)
(
2z +
5)(
2z -
3)
4(p
- 2
)(p
+ 3
)
3
(y +
4)(
y +
3)
(c +
10)
(c -
10)
{-8,
8} {1
, 2}
{1, 3
}
{5, -
2}
{1, 5
}{0
, 9}
{-3,
7}
{
1 −
2 , -3 }
{
3 −
4 , -2 }
{
- 2 −
3 , 5}
16, 1
7 or
-16
, -17
{10,
-10
}
{1, -
3}
011_
029_
ALG
2_A
_CR
M_C
04_C
R_6
6078
5.in
dd
1912
/20/
10
9:06
PM
Answers (Lesson 4-3)
A01_A13_ALG2_A_CRM_C04_AN_660785.indd A8A01_A13_ALG2_A_CRM_C04_AN_660785.indd A8 12/20/10 10:57 PM12/20/10 10:57 PM
Copyri
ght
© G
lencoe/M
cG
raw
-Hill
, a
div
isio
n o
f T
he
McG
raw
-Hill
Co
mp
an
ies,
Inc.
PDF Pass
Chapter 4 A9 Glencoe Algebra 2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
4-3
Cha
pter
4
20
Gle
ncoe
Alg
ebra
2
Prac
tice
So
lvin
g Q
uad
rati
c E
qu
ati
on
s b
y Fa
cto
rin
g
Wri
te a
qua
drat
ic e
quat
ion
in s
tand
ard
form
wit
h th
e gi
ven
root
(s).
1. 7
, 2
2. 0
, 3
3. -
5, 8
4. -
7, -
8 5.
-6,
-3
6. 3
, -4
7. 1
, 1 −
2 8.
1 −
3 , 2
9. 0
, - 7 −
2
Fact
or e
ach
poly
nom
ial.
10. r
3 + 3
r2 - 5
4r
11. 8
a2 + 2
a -
6
12. c
2 - 4
9
13. x
3 + 8
14
. 16r
2 - 1
69
15. b
4 - 8
1
Solv
e ea
ch e
quat
ion
by f
acto
ring
.
16. x
2 - 4
x -
12
= 0
17
. x2 -
16x
+ 6
4 =
0
18. x
2 - 6
x +
8 =
0
19. x
2 + 3
x +
2 =
0
20. x
2 - 4
x =
0
21. 7
x2 = 4
x
22. 1
0x2 =
9x
23. x
2 = 2
x +
99
24. x
2 + 1
2x =
-36
25
. 5x2 -
35x
+ 6
0 =
0
26. 3
6x2 =
25
27. 2
x2 - 8
x -
90
= 0
28. N
UM
BER
THEO
RY F
ind
two
cons
ecut
ive
even
pos
itiv
e in
tege
rs w
hose
pro
duct
is 6
24.
29. N
UM
BER
THEO
RY F
ind
two
cons
ecut
ive
odd
posi
tive
inte
gers
who
se p
rodu
ct is
323
.
30. G
EOM
ETRY
The
leng
th o
f a r
ecta
ngle
is 2
feet
mor
e th
an it
s w
idth
. Fin
d th
e di
men
sion
s of
the
rec
tang
le if
its
area
is 6
3 sq
uare
feet
.
31. P
HO
TOG
RAPH
Y T
he le
ngth
and
wid
th o
f a 6
-inch
by
8-in
ch p
hoto
grap
h ar
e re
duce
d by
th
e sa
me
amou
nt t
o m
ake
a ne
w p
hoto
grap
h w
hose
are
a is
hal
f tha
t of
the
ori
gina
l. B
y ho
w m
any
inch
es w
ill t
he d
imen
sion
s of
the
pho
togr
aph
have
to
be r
educ
ed?
x
2 - 9
x +
14
= 0
x
2 - 3
x =
0
x2 -
3x
- 4
0 =
0
x
2 + 1
5x +
56
= 0
x
2 + 9
x +
18
= 0
x
2 + x
- 1
2 =
0
r
(r +
9)(
r -
6)
2
(4a -
3)(
a +
1)
(c
- 7
)(c
+ 7
)
(
x +
2)(
x2 -
2x
+ 4
) (
4r +
13)
(4r
- 1
3)
(b
2 + 9
)(b
+ 3
)(b
- 3
)
{8}
{-2,
-1}
{
0, 4 −
7 }
{-9,
11}
{-6}
{3, 4
}
{9, -
5}
7 ft
by 9
ft
2
x2 -
3x
+ 1
= 0
3
x2 -
7x
+ 2
= 0
2
x2 +
7x
= 0 2
in.
{6, -
2}
{2, 4
}
{0, 4
}
{
0, 9 −
10
}
{
5 −
6 , - 5 −
6 }
24, 2
6
17, 1
9
011_
029_
ALG
2_A
_CR
M_C
04_C
R_6
6078
5.in
dd
2012
/20/
10
9:06
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 4-3
4-3
Cha
pter
4
21
Gle
ncoe
Alg
ebra
2
Wor
d Pr
oble
m P
ract
ice
So
lvin
g Q
uad
rati
c E
qu
ati
on
s b
y Fa
cto
rin
g
1. F
LASH
LIG
HTS
Whe
n D
ora
shin
es h
er
flash
light
on
the
wal
l at
a ce
rtai
n an
gle,
th
e ed
ge o
f the
lit
area
is in
the
sha
pe o
f a
para
bola
. The
equ
atio
n of
the
par
abol
a is
y=
2x2
+ 2
x-
60.
Fac
tor
this
qu
adra
tic
equa
tion
.
2. S
IGN
S D
avid
was
look
ing
thro
ugh
an
old
alge
bra
book
and
cam
e ac
ross
thi
s eq
uati
on.
x
26x
+ 8
= 0
The
sign
in fr
ont
of t
he 6
was
blo
tted
ou
t. H
ow d
oes
the
mis
sing
sig
n de
pend
on
the
sig
ns o
f the
roo
ts?
3. A
RT T
he a
rea
in s
quar
e in
ches
of t
he
draw
ing
Mai
sons
pré
s de
la m
er b
y C
laud
e M
onet
is a
ppro
xim
ated
by
the
equa
tion
y=
x2– 2
3x+
130
. Fa
ctor
the
eq
uati
on t
o fin
d th
e tw
o ro
ots,
whi
ch a
re
equa
l to
the
appr
oxim
ate
leng
th a
nd
wid
th o
f the
dra
win
g.
4. P
ROG
RAM
MIN
G R
ay is
a c
ompu
ter
prog
ram
mer
. He
need
s to
find
the
qu
adra
tic
func
tion
of t
his
grap
h fo
r an
al
gori
thm
rel
ated
to
a ga
me
invo
lvin
g di
ce. P
rovi
de s
uch
a fu
ncti
on.
y
xO
-2
-4
2
24
46
810
12
5. A
NIM
ATIO
N A
com
pute
r gr
aphi
cs
anim
ator
wou
ld li
ke t
o m
ake
a re
alis
tic
sim
ulat
ion
of a
tos
sed
ball.
The
ani
mat
or
wan
ts t
he b
all t
o fo
llow
the
par
abol
ic
traj
ecto
ry r
epre
sent
ed b
y th
e qu
adra
tic
equa
tion
f(x
) = -
0.2(
x +
5) (
x -
5).
a. W
hat
are
the
solu
tion
s of
f(x
) = 0
?
b. W
rite
f(x
) in
stan
dard
form
.
c. I
f the
ani
mat
or c
hang
es t
he e
quat
ion
to f
(x) =
-0.
2x2 +
20,
wha
t ar
e th
e so
luti
ons
of f(
x) =
0?
2(x
- 5
)(x
+ 6
)
The
mis
sing
sig
n is
the
oppo
site
of
the
sign
of t
he tw
o ro
ots,
be
caus
e th
eir
prod
uct i
s a
posi
tive
num
ber,
8.
10 in
ches
by
13 in
ches
f
(x) =
x2 -
18x
+ 7
7
x
= -
5 or
x =
5
f
(x) =
-0.
2x2 +
5
x
= -
10 o
r x
= 1
0
011_
029_
ALG
2_A
_CR
M_C
04_C
R_6
6078
5.in
dd
2112
/20/
10
9:06
PM
Answers (Lesson 4-3)
A01_A13_ALG2_A_CRM_C04_AN_660785.indd A9A01_A13_ALG2_A_CRM_C04_AN_660785.indd A9 12/20/10 10:57 PM12/20/10 10:57 PM
Co
pyrig
ht ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
pan
ies, In
c.
PDF Pass
Chapter 4 A10 Glencoe Algebra 2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
4-3
Ch
ap
ter
4
22
Gle
ncoe
Alg
ebra
2
Enri
chm
ent
Usi
ng
Patt
ern
s to
Facto
r
Stud
y th
e pa
tter
ns b
elow
for
fact
orin
g th
e su
m a
nd t
he d
iffer
ence
of c
ubes
.
a3 + b
3 = (a
+ b
)(a2 -
ab
+ b
2 )a3 -
b3 =
(a -
b)(a
2 + a
b +
b2 )
This
pat
tern
can
be
exte
nded
to
othe
r od
d po
wer
s. S
tudy
the
se e
xam
ples
.
Fa
ctor
a5 +
b5 .
Ext
end
the
first
pat
tern
to
obta
in a
5 + b
5 = (a
+ b
)(a4 -
a3 b
+ a
2 b2 -
ab3 +
b4 ).
Che
ck: (
a +
b)(a
4 - a
3 b +
a2 b
2 - a
b3 + b
4 ) =
a5 -
a4 b
+ a
3 b2 -
a2 b
3 + a
b4
+ a
4 b -
a3 b
2 + a
2 b3 -
ab4 +
b5
= a
5 +
b5
Fa
ctor
a5 -
b5 .
Ext
end
the
seco
nd p
atte
rn t
o ob
tain
a5 -
b5 =
(a -
b)(a
4 + a
3 b +
a2 b
2 + a
b3 + b
4 ).C
heck
: (a
- b
) (a4 +
a3 b
+ a
2 b2 +
ab3 +
b4 )
= a
5 + a
4 b +
a3 b
2 + a
2 b3 +
ab4
-
a4 b
- a
3 b2 -
a2 b
3 - a
b4 - b
5
= a
5 -
b5
In g
ener
al, i
f n is
an
odd
inte
ger,
whe
n yo
u fa
ctor
an +
bn o
r an -
bn ,
one
fact
or w
ill b
e ei
ther
(a
+ b
) or
(a -
b),
depe
ndin
g on
the
sig
n of
the
ori
gina
l exp
ress
ion.
The
oth
er fa
ctor
will
ha
ve t
he fo
llow
ing
prop
erti
es:
• Th
e fir
st t
erm
will
be
an -
1 a
nd t
he la
st t
erm
will
be
bn -
1.
• Th
e ex
pone
nts
of a
will
dec
reas
e by
1 a
s yo
u go
from
left
to
righ
t.•
The
expo
nent
s of
b w
ill in
crea
se b
y 1
as y
ou g
o fr
om le
ft t
o ri
ght.
• Th
e de
gree
of e
ach
term
will
be
n -
1.
• If
the
ori
gina
l exp
ress
ion
was
an +
bn ,
the
term
s w
ill a
lter
nate
ly h
ave
+ a
nd -
sig
ns.
• If
the
ori
gina
l exp
ress
ion
was
an -
bn ,
the
term
s w
ill a
ll ha
ve +
sig
ns.
Use
the
pat
tern
s ab
ove
to f
acto
r ea
ch e
xpre
ssio
n.
1. a
7 + b
7
2. c
9 - d
9
3. f
11 +
g11
To f
acto
r x10
- y
10, c
hang
e it
to
(x5 +
y5 )
(x5 -
y5 )
and
fac
tor
each
bin
omia
l. U
se t
his
appr
oach
to
fact
or e
ach
expr
essi
on.
4. x
10 -
y10
5. a
14 -
b14
Exam
ple
1
Exam
ple
2
(a +
b)(
a6 -
a5 b
+ a
4 b2 -
a3 b
3 + a
2 b4 -
ab
5 + b
6 ) (a
- b
)
(a
6 + a
5 b +
a4 b
2 + a
3 b3 +
a2 b
4 + a
b5 +
b6 )
(c -
d)(
c8 +
c7 d
+ c
6 d2 +
c5 d
3 + c
4 d4 +
c3 d
5 + c
2 d6 +
cd
7 + d
8 )
(f +
g)(f
10 -
f 9 g
+ f
8 g2 -
f 7 g
3 + f
6 g4 -
f 5 g
5 + f
4 g6 -
f 3 g
7 + f
2 g8 -
fg9 +
g10
)
(x +
y)(x
4 - x
3 y +
x2 y
2 - x
y3 +
y4 )(
x -
y)(x
4 + x
3 y +
x2 y
2 + x
y3 +
y4 )
(a +
b)(
a6 -
a5 b
+ a
4 b2 -
a3 b
3 + a
2 b4 -
ab
5 + b
6 )
011_
029_
ALG
2_A
_CR
M_C
04_C
R_6
6078
5.in
dd
2212
/20/
10
9:06
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 4-3
4-3
Ch
ap
ter
4
23
Gle
ncoe
Alg
ebra
2
Grap
hing
Cal
cula
tor
Activ
ityU
sin
g T
ab
les
to F
acto
r b
y G
rou
pin
g
The
TAB
LE
feat
ure
of a
gra
phin
g ca
lcul
ator
can
be
used
to
help
fact
or a
pol
ynom
ial o
f the
fo
rm a
x2 + b
x +
c. (
The
sam
e pr
oble
ms
can
be s
olve
d w
ith
the
List
s an
d Sp
read
shee
t ap
plic
atio
n on
the
TI-
Nsp
ire.
)
Fa
ctor
10x
2 - 4
3x +
28
by g
roup
ing.
Mak
e a
tabl
e of
the
neg
ativ
e fa
ctor
s of
10
� 28
or 2
80. L
ook
for
a pa
ir
of fa
ctor
s w
hose
sum
is -
43.
Ent
er t
he e
quat
ion
y =
280
−
x
in Y
1 to
find
the
fact
ors
of 2
80. T
hen,
find
the
sum
of t
he fa
ctor
s us
ing
y =
280
−
x
+ x
in Y
2. S
et u
p th
e ta
ble
to d
ispl
ay t
he n
egat
ive
fact
ors
of 2
80 b
y se
ttin
g �
Tbl
= t
o -
1.
Exa
min
e th
e re
sult
s.
Key
stro
kes:
Y=
280
÷
E
NT
ER
V
AR
S
EN
TE
R E
NT
ER
+
E
NT
ER
2
nd
[TB
LSE
T] (
–) 1
EN
TE
R (
–) 1
EN
TE
R
2n
d
[TA
BLE
].
The
last
line
of t
he t
able
sho
ws
that
-43
x m
ay b
e re
plac
ed w
ith
-8x
+ (-
35x)
.
10x2 -
43x
+ 2
8 =
10x
2 - 8
x +
(-35
x) +
28
=
2x(
5x -
4) +
(-7)
(5x
- 4
)
= (5
x -
4)(2
x -
7)
Thus
, 10x
2 - 4
3x +
28
= (5
x -
4)(2
x -
7).
Fact
or e
ach
poly
nom
ial.
1. y
2 - 2
0y -
96
2. 4
z2 - 3
3z +
35
3. 4
y2 + y
-18
4.
6a2 +
2a
- 1
5
5. 6
m2 +
17m
+ 1
2 6.
24z
2 - 4
6z +
15
7. 3
6y2 +
84y
+ 4
9 8.
4b2 +
36b
- 4
03
Fa
ctor
12x
2 - 7
x -
12.
Look
at
the
fact
ors
of 1
2(-
12) o
r -
144
for
a pa
ir w
ith
a su
m o
f -7.
E
nter
an
equa
tion
to
dete
rmin
e th
e fa
ctor
s in
Y1
and
an e
quat
ion
to
find
the
sum
of f
acto
rs in
Y2.
Exa
min
e th
e ta
ble
to fi
nd a
sum
of -
7.K
eyst
roke
s: Y
= (
–) 1
44
÷
EN
TE
R
VA
RS
E
NT
ER
EN
TE
R
+
EN
TE
R
2n
d [T
BLS
ET]
1 E
NT
ER
1 E
NT
ER
2
nd
[T
AB
LE].
12x2 -
7x
- 1
2 =
12x
2 + 9
x +
(-16
x) -
12
=
3x(
4x +
3) -
4(4
x +
3)
=
(4x
+ 3
)(3x
- 4
) Th
us, 1
2x2 -
7x
- 1
2 =
(4x
+ 3
)(3x
- 4
).
Exer
cise
s
Exam
ple
1
Exam
ple
2
(
y +
4)(
y -
24)
(
4z -
5)(
z -
7)
(4y +
9)(
y -
2)
pr
ime
(
2m +
3)(
3m +
4)
(1
2z -
5)(
2z -
3)
(
6y +
7)2
(
2b +
31)
(2b
- 1
3)
011_
029_
ALG
2_A
_CR
M_C
04_C
R_6
6078
5.in
dd
2312
/20/
10
9:06
PM
Answers (Lesson 4-3)
A01_A13_ALG2_A_CRM_C04_AN_660785.indd A10A01_A13_ALG2_A_CRM_C04_AN_660785.indd A10 12/20/10 10:57 PM12/20/10 10:57 PM
Copyri
ght
© G
lencoe/M
cG
raw
-Hill
, a
div
isio
n o
f T
he
McG
raw
-Hill
Co
mp
an
ies,
Inc.
PDF Pass
Chapter 4 A11 Glencoe Algebra 2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
4-4
Cha
pter
4
24
Gle
ncoe
Alg
ebra
2
Stud
y Gu
ide
and
Inte
rven
tion
Co
mp
lex
Nu
mb
ers
So
lve
x2 + 5
= 0
.
x2 + 5
= 0
O
rigin
al e
quat
ion.
x
2 = -
5
Sub
tract
5 fr
om e
ach
side
.
x
= ±
√
�
5 i
Squ
are
Roo
t Pro
pert
y.
Exer
cise
sSi
mpl
ify.
1. √
��
-
72
2.
√
��
-
24
3. √
��
-
84
4.
(2 +
i) (
2 -
i)
Solv
e ea
ch e
quat
ion.
5.
5x2 +
45
= 0
6.
4x2 +
24
= 0
7. -
9x2 =
9
8. 7
x2 + 8
4 =
0
Pure
Imag
inar
y N
umbe
rs A
squ
are
root
of a
num
ber
n is
a n
umbe
r w
hose
squ
are
is n
. For
non
nega
tive
rea
l num
bers
a a
nd b
, √
��
ab
= √
�
a �
√
�
b an
d √
�
a −
b =
√
�
a −
√
�
b , b ≠
0.
•
The
imag
inar
y un
it i
is d
efin
ed t
o ha
ve t
he p
rope
rty
that
i2 =
-1.
•
Sim
plifi
ed s
quar
e ro
ot e
xpre
ssio
ns d
o no
t ha
ve r
adic
als
in t
he d
enom
inat
or, a
nd a
ny
num
ber
rem
aini
ng u
nder
the
squ
are
root
has
no
perf
ect
squa
re fa
ctor
oth
er t
han
1.
a. S
impl
ify
√
��
-
48 .
√
��
-
48 =
√
��
��
16
� (-
3)
=
√
��
16
� √
�
3 � √
��
-
1
= 4
i √
�
3
b. S
impl
ify
√
��
-
63 .
√
��
-
63 =
√
��
��
-
1 � 7
� 9
=
√
��
-
1 � √
�
7 .
√
�
9
= 3
i √
�
7
a. S
impl
ify
-3i
� 4i
.-
3i �
4i =
-12
i2
=
-12
(-1)
=
12
b. S
impl
ify
√ �
�
-3
� √
��
-15
.
√
��
-
3 � √
��
-
15 =
i √
�
3 � i
√
��
15
= i
2 √
��
45
= -
1 � √
�
9 � √
�
5
= -
3 √
�
5
Exam
ple
1Ex
ampl
e 2
Exam
ple
3 6i √
� 2
2i √
� 6
2i √
��
21
5
±3i
±i √
� 6
± i
±2i
√ �
3
011_
029_
ALG
2_A
_CR
M_C
04_C
R_6
6078
5.in
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/20/
10
9:06
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 4-4
4-4
Cha
pter
4
25
Gle
ncoe
Alg
ebra
2
Stud
y Gu
ide
and
Inte
rven
tion
(con
tinu
ed)
Co
mp
lex
Nu
mb
ers
Ope
rati
ons
wit
h Co
mpl
ex N
umbe
rs
Com
plex
Num
ber
A c
ompl
ex n
umbe
r is
any
num
ber t
hat c
an b
e w
ritte
n in
the
form
a +
bi,
whe
re a
and
b a
re
real
num
bers
and
i is
the
imag
inar
y un
it (i
2 = -
1). a
is c
alle
d th
e re
al p
art,
and
b is
cal
led
the
imag
inar
y pa
rt.
Add
ition
and
S
ubtr
actio
n of
C
ompl
ex N
umbe
rs
Com
bine
like
term
s.
(a +
bi)
+ (c
+ d
i) =
(a +
c) +
(b +
d)i
(a +
bi)
- (c
+ d
i) =
(a -
c) +
(b -
d)i
Mul
tiplic
atio
n of
C
ompl
ex N
umbe
rsU
se th
e de
fi niti
on o
f i2 a
nd th
e FO
IL m
etho
d:
(a +
bi)
(c +
di)
= (a
c -
bd
) + (a
d +
bc)
i
Com
plex
Con
juga
tea
+ b
i an
d a
- b
i ar
e co
mpl
ex c
onju
gate
s. T
he p
rodu
ct o
f com
plex
con
juga
tes
is a
lway
s a
real
num
ber.
To d
ivid
e by
a c
ompl
ex n
umbe
r, fir
st m
ulti
ply
the
divi
dend
and
div
isor
by
the
com
plex
co
njug
ate
of t
he d
ivis
or.
Exer
cise
sSi
mpl
ify.
1. (
-4
+ 2
i) +
(6 -
3i)
2.
(5 -
i) -
(3 -
2i)
3.
(6 -
3i)
+ (4
- 2
i)
4. (
-11
+ 4
i) -
(1 -
5i)
5.
(8 +
4i)
+ (8
- 4
i)
6. (5
+ 2
i) -
(-6
- 3
i)
7. (
2 +
i)(3
- i
) 8.
(5 -
2i)
(4 -
i)
9. (4
- 2
i)(1
- 2
i)
10.
5
−
3 +
i
11. 7
- 1
3i
−
2i
12
. 6 -
5i
−
3i
Si
mpl
ify
(6 +
i) +
(4 -
5i)
.
(6 +
i) +
(4 -
5i)
=
(6 +
4) +
(1 -
5)i
=
10
- 4
i
Si
mpl
ify
(2 -
5i)
� (-
4 +
2i)
.
(2 -
5i)
� (-
4 +
2i)
=
2(-
4) +
2(2
i) +
(-5i
)(-4)
+ (-
5i)(2
i)
= -
8 +
4i
+ 2
0i -
10i
2
=
-8
+ 2
4i -
10(
-1)
=
2 +
24i
Si
mpl
ify
(8 +
3i)
- (6
- 2
i).
(8 +
3i)
- (6
- 2
i)
= (8
- 6
) + [3
- (-
2)]i
=
2 +
5i
Si
mpl
ify
3 -
i
−
2 +
3i .
3 -
i
−
2 +
3i =
3
- i
−
2 +
3i �
2 -
3i
−
2 -
3i
=
6 -
9i
- 2
i +
3 i 2
−
−
4 -
9 i 2
=
3 -
11i
−
13
=
3 −
13 -
11
−
13 i
Exam
ple
1Ex
ampl
e 2
Exam
ple
3Ex
ampl
e 4
2
- i
2
+ i
1
0 -
5i
-
12 +
9i
16
11
+ 5
i
7 +
i18
- 1
3i-
10i
- 13
−
2 -
7 −
2 i-
5 −
3 - 2
i 3 −
2 - 1 −
2 i
011_
029_
ALG
2_A
_CR
M_C
04_C
R_6
6078
5.in
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/20/
10
9:06
PM
Answers (Lesson 4-4)
A01_A13_ALG2_A_CRM_C04_AN_660785.indd A11A01_A13_ALG2_A_CRM_C04_AN_660785.indd A11 12/20/10 10:57 PM12/20/10 10:57 PM
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pyrig
ht ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
pa
nie
s, In
c.
PDF Pass
Chapter 4 A12 Glencoe Algebra 2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 4-4
4-4
Cha
pter
4
27
Gle
ncoe
Alg
ebra
2
Prac
tice
Co
mp
lex
Nu
mb
ers
Sim
plif
y.
1. √
��
-
36
2. √
��
-
8 �
√
��
-
32
3. √
��
-
15 �
√
��
-
25
4. (
-3i
) (4i
)(-5i
) 5.
(7i)
2 (6i)
6.
i42
7. i
55
8. i
89
9. (5
- 2
i) +
(-13
- 8
i)
10. (
7 -
6i)
+ (9
+ 1
1i)
11. (
-12
+ 4
8i) +
(15
+ 2
1i)
12. (
10 +
15i
) - (4
8 -
30i
)
13. (
28 -
4i)
- (1
0 -
30i
) 14
. (6
- 4
i) (6
+ 4
i)
15. (
8 -
11i
) (8
- 1
1i)
16. (
4 +
3i)
(2 -
5i)
17
. (7
+ 2
i) (9
- 6
i)
18. 6
+ 5
i −
-2i
-
5 −
2 + 3
i
19.
2 −
7 -
8i
20
. 3 -
i
−
2 -
i
21
. 2 -
4i
−
1 +
3i
Solv
e ea
ch e
quat
ion.
22. 5
n2 + 3
5 =
0
23. 2
m2 +
10
= 0
24. 4
m2 +
76
= 0
25
. -2m
2 - 6
= 0
26. -
5m2 -
65
= 0
27
. 3 −
4 x2 +
12
= 0
Fin
d th
e va
lues
of
ℓ an
d m
tha
t m
ake
each
equ
atio
n tr
ue.
28. 1
5 -
28i
= 3
ℓ +
(4m
)i
29. (
6 -
ℓ) +
(3m
)i =
-12
+ 2
7i
30. (
3ℓ +
4) +
(3 -
m)i
= 1
6 -
3i
31. (
7 +
m) +
(4ℓ
- 1
0)i
= 3
- 6
i
32. E
LECT
RICI
TY T
he im
peda
nce
in o
ne p
art
of a
ser
ies
circ
uit
is 1
+ 3
j ohm
s an
d th
e im
peda
nce
in a
noth
er p
art
of t
he c
ircu
it is
7 -
5j o
hms.
Add
the
se c
ompl
ex n
umbe
rs t
o fin
d th
e to
tal i
mpe
danc
e in
the
cir
cuit
.
33. E
LECT
RICI
TY U
sing
the
form
ula
E =
IZ,
find
the
vol
tage
E in
a c
ircu
it w
hen
the
curr
ent
I is
3 -
j am
ps a
nd t
he im
peda
nce
Z is
3 +
2j o
hms.
2
3 -
14i
7
5 -
24i
-1
- i
±i √
��
19
±i √
� 3
±i √
� 5
±i √
��
13
±4i
5, -
7
4, 6
8 -
2j o
hms 11
+ 3
j vol
ts
6
i -
16
-5 √
��
15
-
60i
-29
4i
-1
-
i i
-
8 -
10i
1
6 +
5i
3 +
69i
-
38 +
45i
1
8 +
26i
5
2 -
57 -
176
i
±i √
� 7
14
−
113 +
16
−
113 i
7 −
5 + 1 −
5 i
1, -
4
18, 9
011_
029_
ALG
2_A
_CR
M_C
04_C
R_6
6078
5.in
dd
2712
/20/
10
9:06
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
4-4
Cha
pter
4
26
Gle
ncoe
Alg
ebra
2
Skill
s Pr
actic
eC
om
ple
x N
um
bers
Sim
plif
y.
1. √
��
99
2.
√
��
27
−
49
3. √
��
�
52
x3 y5
4.
√
��
�
-
108x
7
5. √
��
�
-
81x6
6. √
��
-
23 �
√
��
-
46
7. (
3i)(-
2i)(5
i)
8. i
11
9. i
65
10. (
7 -
8i)
+ (-
12 -
4i)
11. (
-3
+ 5
i) +
(18
- 7
i)
12. (
10 -
4i)
- (7
+ 3
i)
13. (
7 -
6i)
(2 -
3i)
14
. (3
+ 4
i)(3
- 4
i)
15. 8
- 6
i −
3i
16
.
3i
−
4 +
2i
Solv
e ea
ch e
quat
ion.
17. 3
x2 + 3
= 0
18
. 5x2 +
125
= 0
19. 4
x2 + 2
0 =
0
20. -
x2 - 1
6 =
0
21. x
2 + 1
8 =
0
22. 8
x2 + 9
6 =
0
Fin
d th
e va
lues
of
� a
nd m
tha
t m
ake
each
equ
atio
n tr
ue.
23. 2
0 -
12i
= 5
� +
(4m
)i
24. �
- 1
6i =
3 -
(2m
)i
25. (
4 +
�) +
(2m
)i =
9 +
14i
26
. (3
- m
) + (7
� -
14)
i =
1 +
7i
3 √ �
11
2⎪x
⎪y2 √
��
�
13x
y
6i ⎪ x
3 ⎥ √
��
3x
9 ⎪x
3 ⎥ i
-23
√ �
2
30i
-i
i-
5 -
12i
15 -
2i
3 -
7i
-4
- 3
3i25
±i
±5i
±i √
� 5
±4i
±3i
√ �
2 ±
2i √
� 3
4, -
33,
8
5, 7
3, 2
3 −
10 +
3 −
5 i-
2 -
8 −
3 i
3 √ �
3 −
7
011_
029_
ALG
2_A
_CR
M_C
04_C
R_6
6078
5.in
dd
2612
/20/
10
9:06
PM
Answers (Lesson 4-4)
A01_A13_ALG2_A_CRM_C04_AN_660785.indd A12A01_A13_ALG2_A_CRM_C04_AN_660785.indd A12 12/20/10 10:57 PM12/20/10 10:57 PM
Copyri
ght
© G
lencoe/M
cG
raw
-Hill
, a
div
isio
n o
f T
he
McG
raw
-Hill
Co
mp
an
ies,
Inc.
PDF Pass
Chapter 4 A13 Glencoe Algebra 2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 4-4
4-4
Cha
pter
4
29
Gle
ncoe
Alg
ebra
2
Enri
chm
ent
Co
nju
gate
s an
d A
bso
lute
Valu
e
Whe
n st
udyi
ng c
ompl
ex n
umbe
rs, i
t is
oft
en c
onve
nien
t to
rep
rese
nt a
com
plex
nu
mbe
r by
a s
ingl
e va
riab
le. F
or e
xam
ple,
we
mig
ht le
t z
= x
+ y
i. W
e de
note
th
e co
njug
ate
of z
by
−
z . T
hus,
−
z =
x -
yi.
We
can
defin
e th
e ab
solu
te v
alue
of a
com
plex
num
ber
as fo
llow
s.
⎪ z⎥
= ⎪
x +
yi ⎥
= √
��
�
x2 +
y2
Ther
e ar
e m
any
impo
rtan
t re
lati
onsh
ips
invo
lvin
g co
njug
ates
and
abs
olut
e va
lues
of c
ompl
ex n
umbe
rs.
Sh
ow ⎪
z⎥ 2 =
z −
z fo
r an
y co
mpl
ex n
umbe
r z.
Let
z =
x +
yi.
Then
,
zz −
z =
(x +
yi)
(x -
yi)
=
x2 +
y2
=
√
��
��
(x
2 + y
2 )2
=
⎪z⎥
2
Sh
ow
−
z −
⎪ z⎥ 2
is t
he m
ulti
plic
ativ
e in
vers
e fo
r an
y no
nzer
o
com
plex
num
ber
z.
We
know
⎪z⎥
2 = z
−
z . I
f z ≠
0, t
hen
we
have
z (
−
z −
⎪ z⎥
2 )
= 1
.
Thus
, −
z −
⎪ z⎥
2 i
s th
e m
ulti
plic
ativ
e in
vers
e of
z.
Exer
cise
sFo
r ea
ch o
f th
e fo
llow
ing
com
plex
num
bers
, fin
d th
e ab
solu
te v
alue
and
m
ulti
plic
ativ
e in
vers
e.
1. 2
i
2. -
4 -
3i
3.
12
- 5
i
4. 5
- 1
2i
5. 1
+ i
6.
√
�
3 -
i
7. √
�
3 −
3 +
√
�
3 −
3 i
8. √
�
2 −
2 -
√
�
2 −
2 i
9. 1 −
2 - √
�
3 −
2 i
Exam
ple
1
Exam
ple
2
2;
-i
−
2 5
; -
4 +
3i
−
25
1
3; 12
+ 5
i −
16
9
13;
5 +
12i
−
169
√
� 2 ; 1
- i
−
2
2;
√ �
3 +
i −
4
√
� 6
−
3 ; √
� 3 -
i √
� 3
−
2
1;
√ �
2 −
2
+ √
� 2
−
2 i
1;
1 −
2 + √
� 3
−
2 i
011_
029_
ALG
2_A
_CR
M_C
04_C
R_6
6078
5.in
dd
2912
/20/
10
9:06
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
4-4
Cha
pter
4
28
Gle
ncoe
Alg
ebra
2
Wor
d Pr
oble
m P
ract
ice
Co
mp
lex
Nu
mb
ers
1. S
IGN
ERR
ORS
Jen
nife
r an
d Je
ssic
a co
me
up w
ith
diffe
rent
ans
wer
s to
the
sa
me
prob
lem
. The
y ha
d to
mul
tipl
y (4
+i)
(4 -
i) a
nd g
ive
thei
r an
swer
as
a co
mpl
ex n
umbe
r. Je
nnife
r cl
aim
s th
at
the
answ
er is
15
and
Jess
ica
clai
ms
that
the
ans
wer
is 1
7. W
ho is
cor
rect
? E
xpla
in.
2. C
OM
PLEX
CO
NJU
GAT
ES Y
ou h
ave
seen
tha
t th
e pr
oduc
t of
com
plex
co
njug
ates
is a
lway
s a
real
num
ber.
Show
tha
t th
e su
m o
f com
plex
co
njug
ates
is a
lso
alw
ays
a re
al n
umbe
r.
3. P
YTH
AG
ORE
AN
TRI
PLES
If t
hree
in
tege
rs a
, b, a
nd c
sat
isfy
a2
+b2
=c2 ,
then
the
y ar
e ca
lled
a Py
thag
orea
n tr
iple
. Sup
pose
tha
t a,
b, a
nd c
are
a
Pyth
agor
ean
trip
le. S
how
tha
t th
e re
al
and
imag
inar
y pa
rts
of (a
+bi
)2 , to
geth
er w
ith
the
num
ber
c2 , fo
rm
anot
her
Pyth
agor
ean
trip
le.
4. R
OTA
TIO
NS
Com
plex
num
bers
can
be
used
to
perf
orm
rot
atio
ns in
the
pla
ne.
For
exam
ple,
if (x
, y) a
re t
he c
oord
inat
es
of a
poi
nt in
the
pla
ne, t
hen
the
real
an
d im
agin
ary
part
s of
i(x
+ y
i) a
re
the
hori
zont
al a
nd v
erti
cal c
oord
inat
es
of t
he 9
0° c
ount
ercl
ockw
ise
rota
tion
of
(x, y
) abo
ut t
he o
rigi
n. W
hat
are
the
real
an
d im
agin
ary
part
s of
i(x
+ y
i)?
5. E
LECT
RICA
L EN
GIN
EERI
NG
A
lter
nati
ng c
urre
nt (A
C) i
n an
ele
ctri
cal
circ
uit
can
be d
escr
ibed
by
com
plex
nu
mbe
rs. I
n an
y el
ectr
ical
cir
cuit
, Z, t
he
impe
danc
e in
the
cir
cuit
, is
rela
ted
to
the
volt
age
V a
nd t
he c
urre
nt I
by
the
form
ula
Z =
V
− I . Th
e st
anda
rd e
lect
rica
l vo
ltag
e in
Eur
ope
is 2
20 v
olts
, so
in
thes
e pr
oble
ms
use
V =
220
.
a. F
ind
the
impe
danc
e in
a s
tand
ard
Eur
opea
n ci
rcui
t if
the
curr
ent
is
22 –
11i
am
ps.
b. F
ind
the
curr
ent
in a
sta
ndar
d E
urop
ean
circ
uit
if th
e im
peda
nce
is
10 –
5i
wat
ts.
c. F
ind
the
impe
danc
e in
a s
tand
ard
Eur
opea
n ci
rcui
t if
the
curr
ent
is
20i
amps
.
a
+ b
i and
a -
bi a
re c
ompl
ex
conj
ugat
es a
nd th
eir
sum
is 2
a,
whi
ch is
rea
l.
(
a +
bi)
2 = a
2 - b
2 + 2
ab
i; a
2 - b
2 and
2a
b a
re in
tege
rs a
nd
(a2 -
b2 )2 +
(2a
b)2 =
a
4 - 2
a2 b
2 + b
4 + 4
a2 b
2 =
a4 +
2a
2 b2 +
b4 =
(a2 +
b2 )2 =
(c
2 )2 , so
a2 +
b2 =
c2 a
s de
sire
d.
T
he r
eal p
art i
s -
y an
d im
agin
ary
part
is x
.
18 –
8i a
mps
J
essi
ca is
cor
rect
; (4
+ i
)(4 -
i) =
1
6 +
4i -
4i -
i2
=
16
- (-
1) =
16
+ 1
= 1
7.
8 +
4i
–11
i am
ps
011_
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Answers (Lesson 4-4)
A01_A13_ALG2_A_CRM_C04_AN_660785.indd A13A01_A13_ALG2_A_CRM_C04_AN_660785.indd A13 12/20/10 10:57 PM12/20/10 10:57 PM
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pyrig
ht ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
pa
nie
s, In
c.
PDF Pass
Chapter 4 A14 Glencoe Algebra 2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 4-5
4-5
Cha
pter
4
31
Gle
ncoe
Alg
ebra
2
Stud
y Gu
ide
and
Inte
rven
tion
(con
tinu
ed)
Co
mp
leti
ng
th
e S
qu
are
Com
plet
e th
e Sq
uare
To
com
plet
e th
e sq
uare
for
a qu
adra
tic
expr
essi
on o
f the
form
x2 +
bx,
fol
low
the
se s
teps
.
1. F
ind
b −
2 .
2.
Squ
are
b −
2 .
3.
Add
( b −
2 ) 2 to
x2 + b
x.
F
ind
the
valu
e of
c t
hat
mak
es x
2 + 2
2x +
c a
pe
rfec
t sq
uare
tri
nom
ial.
The
n w
rite
the
tri
nom
ial a
s th
e sq
uare
of
a bi
nom
ial.
Step
1
b =
22
; b −
2 = 1
1
Step
2 1
12 = 1
21St
ep 3
c
= 1
21
The
trin
omia
l is
x2 + 2
2x +
121
, w
hich
can
be
wri
tten
as
(x +
11)
2 .
So
lve
2x2 -
8x
- 2
4 =
0 b
y co
mpl
etin
g th
e sq
uare
.
2x2 -
8x
- 2
4 =
0
Orig
inal
equ
atio
n
2x
2 - 8
x -
24
−
2
= 0 −
2 D
ivid
e ea
ch s
ide
by 2
.
x2 -
4x
- 1
2 =
0
x2 - 4
x -
12
is n
ot a
per
fect
squ
are.
x2 -
4x
= 1
2 A
dd 1
2 to
eac
h si
de.
x2 -
4x
+ 4
= 1
2 +
4
Since
( 4 −
2 ) 2 = 4
, add
4 to
eac
h si
de.
(x
- 2
)2 =
16
Fact
or th
e sq
uare
.
x
- 2
= ±
4 S
quar
e R
oot P
rope
rty
x =
6 o
r x
= -
2
Sol
ve e
ach
equa
tion.
The
solu
tion
set
is {6
, -2}
.
Exer
cise
s
Fin
d th
e va
lue
of c
tha
t m
akes
eac
h tr
inom
ial a
per
fect
squ
are.
The
n w
rite
the
tr
inom
ial a
s a
perf
ect
squa
re.
1. x
2 - 1
0x +
c
2. x
2 + 6
0x +
c
3. x
2 - 3
x +
c
4. x
2 + 3
.2x
+ c
5.
x2 +
1 −
2 x +
c
6. x
2 - 2
.5x
+ c
Solv
e ea
ch e
quat
ion
by c
ompl
etin
g th
e sq
uare
. 7
. y2 -
4y
- 5
= 0
8.
x2 -
8x
- 6
5 =
0
9. w
2 - 1
0w +
21
= 0
10. 2
x2 - 3
x +
1 =
0
11. 2
x2 - 1
3x -
7 =
0
12. 2
5x2 +
40x
- 9
= 0
13. x
2 + 4
x +
1 =
0
14. y
2 + 1
2y +
4 =
0
15. t
2 + 3
t - 8
= 0
Exam
ple
1Ex
ampl
e 2
2
5; (x
- 5
)2 9
00; (
x +
30)
2 9
− 4 ;
(x -
3 −
2 ) 2
2
.56;
(x +
1.6
)2
1 −
16
; (x
+ 1 −
4 ) 2 1
.562
5; (x
- 1
.25)
2
-
1, 5
-
5, 1
3 3
, 7
1
, 1 −
2 -
1 −
2 , 7
1 −
5 , -
9 −
5
-
2 ±
√ �
3 -
6 ±
4 √
� 2
-3
± √
��
41
−
2
030_
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Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
4-5
Cha
pter
4
30
Gle
ncoe
Alg
ebra
2
Stud
y Gu
ide
and
Inte
rven
tion
Co
mp
leti
ng
th
e S
qu
are
Squa
re R
oot
Prop
erty
Use
the
Squ
are
Roo
t Pr
oper
ty t
o so
lve
a qu
adra
tic
equa
tion
th
at is
in t
he fo
rm “p
erfe
ct s
quar
e tr
inom
ial =
con
stan
t.”
So
lve
each
equ
atio
n by
usi
ng t
he S
quar
e R
oot
Pro
pert
y. R
ound
to
the
near
est
hund
redt
h if
nec
essa
ry.
a. x
2 - 8
x +
16
= 2
5
x2 - 8
x +
16
= 2
5
(x -
4)2 =
25
x
- 4
= √
��
25
or
x
- 4
= -
√
��
25
x =
5 +
4 =
9 o
r
x =
-5
+ 4
= -
1
Th
e so
luti
on s
et is
{9, -
1}.
b. 4
x2 - 2
0x +
25
= 3
2 4
x2 - 2
0x +
25
= 3
2
(2x
- 5
)2 = 3
22x
- 5
= √
��
32
or
2x -
5 =
- √
��
32
2x
- 5
= 4
√
�
2 or
2x
- 5
= -
4 √
�
2
x
= 5
± 4
√
�
2 −
2
The
solu
tion
set
is {
5 ±
4 √
�
2 −
2
}
.
Exer
cise
sSo
lve
each
equ
atio
n by
usi
ng t
he S
quar
e R
oot
Pro
pert
y. R
ound
to
the
near
est
hund
redt
h if
nec
essa
ry.
1. x
2 - 1
8x +
81
= 4
9 2.
x2 +
20x
+ 1
00 =
64
3. 4
x2 + 4
x +
1 =
16
4. 3
6x2 +
12x
+ 1
= 1
8 5.
9x2 -
12x
+ 4
= 4
6.
25x
2 + 4
0x +
16
= 2
8
7. 4
x2 - 2
8x +
49
= 6
4 8.
16x
2 + 2
4x +
9 =
81
9. 1
00x2 -
60x
+ 9
= 1
21
10. 2
5x2 +
20x
+ 4
= 7
5 11
. 36x
2 + 4
8x +
16
= 1
2 12
. 25x
2 - 3
0x +
9 =
96
Exam
ple
{
-2
± 5
√ �
3 −
5
} {
-2
± √
� 3
−
3 }
{ 3
± 4
√ �
6 −
5
}
{
2, 1
6}
{-
2, -
18}
{ 3 −
2 , -
5 −
2 }
{
-1
± 3
√ �
2 −
6
} {
0, 4 −
3 } {
-4
± 2
√ �
7 −
5
}
{
15
−
2 , -
1 −
2 } {
3 −
2 , -3 }
-
{0.8
, 1.4
}
030_
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Answers (Lesson 4-5)
A14_A26_ALG2_A_CRM_C04_AN_660785.indd A14A14_A26_ALG2_A_CRM_C04_AN_660785.indd A14 12/20/10 10:56 PM12/20/10 10:56 PM
Copyri
ght
© G
lencoe/M
cG
raw
-Hill
, a
div
isio
n o
f T
he
McG
raw
-Hill
Co
mp
an
ies,
Inc.
PDF Pass
Chapter 4 A15 Glencoe Algebra 2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
4-5
Cha
pter
4
32
Gle
ncoe
Alg
ebra
2
Skill
s Pr
actic
eC
om
ple
tin
g t
he S
qu
are
Solv
e ea
ch e
quat
ion
by u
sing
the
Squ
are
Roo
t P
rope
rty.
Rou
nd t
o th
e ne
ares
t hu
ndre
dth
if n
eces
sary
.
1. x
2 - 8
x +
16
= 1
2.
x2 +
4x
+ 4
= 1
3. x
2 + 1
2x +
36
= 2
5
4. 4
x2 - 4
x +
1 =
9
5. x
2 + 4
x +
4 =
2
6. x
2 - 2
x +
1 =
5
7. x
2 - 6
x +
9 =
7
8. x
2 + 1
6x +
64
= 1
5
Fin
d th
e va
lue
of c
tha
t m
akes
eac
h tr
inom
ial a
per
fect
squ
are.
The
n w
rite
the
tr
inom
ial a
s a
perf
ect
squa
re.
9. x
2 + 1
0x +
c
10. x
2 - 1
4x +
c
11. x
2 + 2
4x +
c
12. x
2 + 5
x +
c
13. x
2 - 9
x +
c
14. x
2 - x
+ c
Solv
e ea
ch e
quat
ion
by c
ompl
etin
g th
e sq
uare
.
15. x
2 - 1
3x +
36
= 0
16
. x2 +
3x
= 0
17. x
2 + x
- 6
= 0
18
. x2 -
4x
- 1
3 =
0
19. 2
x2 + 7
x -
4 =
0
20. 3
x2 + 2
x -
1 =
0
21. x
2 + 3
x -
6 =
0
22. x
2 - x
- 3
= 0
23. x
2 = -
11
24. x
2 - 2
x +
4 =
0
-1,
-3
-1,
2
-11
.87,
-4.
13
2 ±
√ �
�
17
1 −
3 , -1
± i
√ �
11
3, 5 -
1, -
11
25; (
x +
5)2
49; (
x -
7)2
144;
(x +
12)
2
81
−
4 ; (
x -
9 −
2 ) 2
4, 9
0, -
3
2, -
3
-4,
1 −
2
1 ±
i √
� 3
1 ±
√ �
�
13
−
2
-3
± √
��
33
−
2
25
−
4 ; (
x +
5 −
2 ) 2
1 −
4 ; (x
- 1 −
2 ) 2
-3.
41, -
0.59
0.35
, 2.6
5
-1.
24, 3
.24
030_
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5.in
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10
9:06
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 4-5
4-5
Cha
pter
4
33
Gle
ncoe
Alg
ebra
2
Prac
tice
Co
mp
leti
ng
th
e S
qu
are
Solv
e ea
ch e
quat
ion
by u
sing
the
Squ
are
Roo
t P
rope
rty.
Rou
nd t
o th
e ne
ares
t hu
ndre
dth
if n
eces
sary
.
1. x
2 + 8
x +
16
= 1
2.
x2 +
6x
+ 9
= 1
3.
x2 +
10x
+ 2
5 =
16
4. x
2 - 1
4x +
49
= 9
5.
4x2 +
12x
+ 9
= 4
6.
x2 -
8x
+ 1
6 =
8
7. x
2 - 6
x +
9 =
5
8. x
2 - 2
x +
1 =
2
9. 9
x2 - 6
x +
1 =
2
Fin
d th
e va
lue
of c
tha
t m
akes
eac
h tr
inom
ial a
per
fect
squ
are.
The
n w
rite
the
tr
inom
ial a
s a
perf
ect
squa
re.
10. x
2 + 1
2x +
c
11. x
2 - 2
0x +
c
12. x
2 + 1
1x +
c
13. x
2 + 0
.8x
+ c
14
. x2 -
2.2
x +
c
15. x
2 - 0
.36x
+ c
16. x
2 + 5 −
6 x +
c
17. x
2 - 1 −
4 x +
c
18. x
2 - 5 −
3 x +
c
Solv
e ea
ch e
quat
ion
by c
ompl
etin
g th
e sq
uare
.
19. x
2 + 6
x +
8 =
0
20. 3
x2 + x
- 2
= 0
21
. 3x2 -
5x
+ 2
= 0
22. x
2 + 1
8 =
9x
23. x
2 - 1
4x +
19
= 0
24
. x2 +
16x
- 7
= 0
25. 2
x2 + 8
x -
3 =
0
26. x
2 + x
- 5
= 0
27
. 2x2 -
10x
+ 5
= 0
28. x
2 + 3
x +
6 =
0
29. 2
x2 + 5
x +
6 =
0
30. 7
x2 + 6
x +
2 =
0
31. G
EOM
ETRY
Whe
n th
e di
men
sion
s of
a c
ube
are
redu
ced
by 4
inch
es o
n ea
ch s
ide,
the
su
rfac
e ar
ea o
f the
new
cub
e is
864
squ
are
inch
es. W
hat
wer
e th
e di
men
sion
s of
the
or
igin
al c
ube?
32. I
NV
ESTM
ENTS
The
am
ount
of m
oney
A in
an
acco
unt
in w
hich
P d
olla
rs a
re in
vest
ed
for
2 ye
ars
is g
iven
by
the
form
ula
A =
P(1
+ r
)2 , w
here
r is
the
inte
rest
rat
e co
mpo
unde
d an
nual
ly. I
f an
inve
stm
ent
of $
800
in t
he a
ccou
nt g
row
s to
$88
2 in
tw
o ye
ars,
at
wha
t in
tere
st r
ate
was
it in
vest
ed?
-
5, -
3 -
4, -
2 -
9, -
1
-
4, -
2 2
− 3 ,
-1
1, 2
− 3
4
, 10
- 1 −
2 , -
5 −
2 4
± 2
√ �
2
3
± √
� 5
1 ±
√ �
2 1
± √
� 2
−
3
3
6; (x
+ 6
)2 1
00; (
x -
10)
2 12
1 −
4
; (x
+ 11
−
2 )
2
0
.16;
(x +
0.4
)2 1
.21;
(x -
1.1
)2 0
.032
4; (x
- 0
.18)
2
25
−
144 ;
(x +
5 −
12
) 2
1 −
64
; (x
- 1 −
8 ) 2 25
−
36
; (x
- 5 −
6 ) 2
6
, 3
7 ±
√ �
�
30
-8
± √
��
71
-
4 ±
√ �
�
22
−
2
-1
± √
��
21
−
2
5 ±
√ �
�
15
−
2
16 in
. by
16 in
. by
16 in
.
5%
-
3 ±
i √
��
15
−
2
-5
± i
√ �
�
23
−
4
-3
± i
√ �
5 −
7
030_
041_
ALG
2_A
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M_C
04_C
R_6
6078
5.in
dd
3312
/20/
10
9:06
PM
Answers (Lesson 4-5)
A14_A26_ALG2_A_CRM_C04_AN_660785.indd A15A14_A26_ALG2_A_CRM_C04_AN_660785.indd A15 12/20/10 10:56 PM12/20/10 10:56 PM
Co
pyrig
ht ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
pa
nie
s, In
c.
PDF Pass
Chapter 4 A16 Glencoe Algebra 2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
4-5
Cha
pter
4
34
Gle
ncoe
Alg
ebra
2
Wor
d Pr
oble
m P
ract
ice
Co
mp
leti
ng
th
e S
qu
are
1. C
OM
PLET
ING
TH
E SQ
UA
RESa
man
tha
need
s to
sol
ve t
he e
quat
ion
x2-
12x
= 4
0.
Wha
t m
ust
she
do t
o ea
ch s
ide
of t
he
equa
tion
to
com
plet
e th
e sq
uare
?
2. A
RT T
he a
rea
in s
quar
e in
ches
of t
he
draw
ing
Folia
ge b
y Pa
ul C
ézan
ne is
ap
prox
imat
ed b
y th
e eq
uati
on
y=
x2– 4
0x+
396
. Com
plet
e th
e sq
uare
an
d fin
d th
e tw
o ro
ots,
whi
ch a
re e
qual
to
the
app
roxi
mat
e le
ngth
and
wid
th o
f th
e dr
awin
g.
3. C
OM
POU
ND
INTE
REST
Nik
ki in
vest
ed
$100
0 in
a s
avin
gs a
ccou
nt w
ith
inte
rest
co
mpo
unde
d an
nual
ly. A
fter
tw
o ye
ars
the
bala
nce
in t
he a
ccou
nt is
$12
10.
Use
the
com
poun
d in
tere
st fo
rmul
a A
=P(
1 +
r)t t
o fin
d th
e an
nual
in
tere
st r
ate.
4. R
EACT
ION
TIM
E La
uren
was
eat
ing
lunc
h w
hen
she
saw
her
frie
nd J
ason
ap
proa
ch. T
he r
oom
was
cro
wde
d an
d Ja
son
had
to li
ft h
is t
ray
to a
void
ob
stac
les.
Sud
denl
y, a
glas
s on
Jas
on’s
lunc
h tr
ay t
ippe
d an
d fe
ll of
f the
tra
y. La
uren
lung
ed fo
rwar
d an
d m
anag
ed t
o ca
tch
the
glas
s ju
st b
efor
e it
hit
the
gr
ound
. The
hei
ght
h, in
feet
, of t
he
glas
s t s
econ
ds a
fter
it w
as d
ropp
ed is
gi
ven
by h
=-
16t2
+ 4
.5. L
aure
n ca
ught
th
e gl
ass
whe
n it
was
six
inch
es o
ff th
e gr
ound
. How
long
was
the
gla
ss in
the
ai
r be
fore
Lau
ren
caug
ht it
?
5. P
ARA
BOLA
S A
par
abol
a is
mod
eled
by
y=
x2-
10x
+ 2
8. J
ane’
s ho
mew
ork
prob
lem
req
uire
s th
at s
he fi
nd t
he
vert
ex o
f the
par
abol
a. S
he u
ses
the
com
plet
ing
squa
re m
etho
d to
exp
ress
th
e fu
ncti
on in
the
form
y
= (x
-h)
2+
k, w
here
(h, k
) is
the
vert
ex o
f the
par
abol
a. W
rite
the
fu
ncti
on in
the
form
use
d by
Jan
e.
6. A
UD
ITO
RIU
M S
EATI
NG
The
sea
ts in
an
aud
itor
ium
are
arr
ange
d in
a s
quar
e gr
id p
atte
rn. T
here
are
45
row
s an
d 45
co
lum
ns o
f cha
irs.
For
a s
peci
al c
once
rt,
orga
nize
rs d
ecid
e to
incr
ease
sea
ting
by
addi
ng n
row
s an
d n
colu
mns
to
mak
e a
squa
re p
atte
rn o
f sea
ting
45
+ n
sea
ts o
n a
side
.
a. H
ow m
any
seat
s ar
e th
ere
afte
r th
e ex
pans
ion?
b. W
hat
is n
if o
rgan
izer
s w
ish
to a
dd
1000
sea
ts?
c. I
f org
aniz
ers
do a
dd 1
000
seat
s,
wha
t is
the
sea
ting
cap
acit
y of
th
e au
dito
rium
?
0
.5 s
econ
d
y
= (x
- 5
)2 + 3
n
2 + 9
0n +
202
5
3
025
A
dd 3
6.
1
8 in
ches
by
22 in
ches
1
0%
1
0
030_
041_
ALG
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M_C
04_C
R_6
6078
5.in
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/20/
10
9:06
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 4-5
4-5
Cha
pter
4
35
Gle
ncoe
Alg
ebra
2
Enri
chm
ent
Th
e G
old
en
Qu
ad
rati
c E
qu
ati
on
s
A g
olde
n re
ctan
gle
has
the
prop
erty
tha
t it
s le
ngth
a
a a
b b
a
can
be w
ritt
en a
s a
+ b
, whe
re a
is t
he w
idth
of t
he
rect
angl
e an
d a
+ b
−
a =
a −
b . A
ny g
olde
n re
ctan
gle
can
be
divi
ded
into
a s
quar
e an
d a
smal
ler
gold
en r
ecta
ngle
, as
sho
wn.
The
prop
orti
on u
sed
to d
efin
e go
lden
rec
tang
les
can
be
used
to
deri
ve t
wo
quad
rati
c eq
uati
ons.
The
se a
re
som
etim
es c
alle
d go
lden
qua
drat
ic e
quat
ions
.
Solv
e ea
ch p
robl
em.
1. I
n th
e pr
opor
tion
for
the
gold
en r
ecta
ngle
, let
a e
qual
1. W
rite
the
res
ulti
ng
quad
rati
c eq
uati
on a
nd s
olve
for
b.
2. I
n th
e pr
opor
tion
, let
b e
qual
1. W
rite
the
res
ulti
ng q
uadr
atic
equ
atio
n an
d so
lve
for
a.
3. D
escr
ibe
the
diffe
renc
e be
twee
n th
e tw
o go
lden
qua
drat
ic e
quat
ions
you
fo
und
in e
xerc
ises
1 a
nd 2
.
4. S
how
tha
t th
e po
siti
ve s
olut
ions
of t
he t
wo
equa
tion
s in
exe
rcis
es 1
and
2
are
reci
proc
als.
5. U
se t
he P
ytha
gore
an T
heor
em t
o fin
d a
radi
cal e
xpre
ssio
n fo
r th
e di
agon
al
of a
gol
den
rect
angl
e w
hen
a =
1.
6. F
ind
a ra
dica
l exp
ress
ion
for
the
diag
onal
of a
gol
den
rect
angl
e w
hen
b =
1.
b
2 + b
- 1
= 0
b =
-1
+ √
� 5
−
2
a
2 - a
- 1
= 0
a =
1 +
√ �
5 −
2
T
he s
igns
of t
he fi
rst-
degr
ee te
rms
are
oppo
site
.
(
-1
+ √
� 5
−
2 )
( 1
+ √
� 5
−
2 )
= -
( 1 2 ) +
( √
� 5 ) 2
−
4
= -
1 +
5
−
4 =
1
d
= √
��
��
�
10 -
2 √
� 5
−
2
d =
√ �
��
��
10 +
2 √
� 5
−
2
030_
041_
ALG
2_A
_CR
M_C
04_C
R_6
6078
5.in
dd
3512
/20/
10
9:06
PM
Answers (Lesson 4-5)
A14_A26_ALG2_A_CRM_C04_AN_660785.indd A16A14_A26_ALG2_A_CRM_C04_AN_660785.indd A16 12/20/10 10:56 PM12/20/10 10:56 PM
Copyri
ght
© G
lencoe/M
cG
raw
-Hill
, a
div
isio
n o
f T
he
McG
raw
-Hill
Co
mp
an
ies,
Inc.
PDF Pass
Chapter 4 A17 Glencoe Algebra 2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
4-6
Cha
pter
4
36
Gle
ncoe
Alg
ebra
2
Stud
y Gu
ide
and
Inte
rven
tion
Th
e Q
uad
rati
c F
orm
ula
an
d t
he D
iscri
min
an
t
Qua
drat
ic F
orm
ula
The
Qua
drat
ic F
orm
ula
can
be u
sed
to s
olve
any
qua
drat
ic
equa
tion
onc
e it
is w
ritt
en in
the
form
ax2 +
bx
+ c
= 0
.
Qua
drat
ic F
orm
ula
The
solu
tions
of a
x2 +
bx
+ c
= 0
, with
a ≠
0, a
re g
iven
by
x =
-
b ±
√
��
��
b2 -
4ac
−
2a
.
So
lve
x2 - 5
x =
14
by u
sing
the
Qua
drat
ic F
orm
ula.
Rew
rite
the
equ
atio
n as
x2 -
5x
- 1
4 =
0.
x =
-b
± √
��
��
b
2 - 4
ac
−
2a
Q
uadr
atic
For
mul
a
=
-
(-5)
± √
��
��
��
�
(-5)
2 - 4
(1)(-
14)
−−
2(1)
Rep
lace
a w
ith 1
, b w
ith -
5, a
nd c
with
-14
.
=
5
± √
��
81
−
2
Sim
plify
.
=
5
± 9
−
2
=
7
or
-2
The
solu
tion
s ar
e -
2 an
d 7.
Exer
cise
sSo
lve
each
equ
atio
n by
usi
ng t
he Q
uadr
atic
For
mul
a.
1. x
2 + 2
x -
35
= 0
2.
x2 +
10x
+ 2
4 =
0
3. x
2 - 1
1x +
24
= 0
4. 4
x2 + 1
9x -
5 =
0
5. 1
4x2 +
9x
+ 1
= 0
6.
2x2 -
x -
15
= 0
7. 3
x2 + 5
x =
2
8. 2
y2 + y
- 1
5 =
0
9. 3
x2 - 1
6x +
16
= 0
10. 8
x2 + 6
x -
9 =
0
11. r
2 - 3r
−
5 + 2 −
25 =
0
12. x
2 - 1
0x -
50
= 0
13. x
2 + 6
x -
23
= 0
14
. 4x2 -
12x
- 6
3 =
0
15. x
2 - 6
x +
21
= 0
Exam
ple
5
, -7
-4,
-6
3, 8
1
− 4 ,
-5
- 1 −
2 , -
1 −
7 3
, - 5 −
2
-
2, 1 −
3 5
− 2 ,
-3
4, 4
− 3
-
3 −
2 , 3 −
4 2
− 5 ,
1 −
5 5
± 5
√ �
3
-
3 ±
4 √
� 2
3 ±
6 √
� 2
−
2
3 ±
2i √
� 3
030_
041_
ALG
2_A
_CR
M_C
04_C
R_6
6078
5.in
dd
3612
/20/
10
9:06
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 4-6
4-6
Cha
pter
4
37
Gle
ncoe
Alg
ebra
2
Root
s an
d th
e D
iscr
imin
ant
Dis
crim
inan
tTh
e ex
pres
sion
und
er th
e ra
dica
l sig
n, b
2 - 4
ac, i
n th
e Q
uadr
atic
For
mul
a is
cal
led
the
disc
rimin
ant.
Dis
crim
inan
tTy
pe a
nd N
umbe
r of
Roo
ts
b2 -
4ac
> 0
and
a p
erfe
ct s
quar
e 2
ratio
nal r
oots
b2 -
4ac
> 0
, but
not
a p
erfe
ct s
quar
e2
irrat
iona
l roo
ts
b2 -
4ac
= 0
1 ra
tiona
l roo
t
b2 -
4ac
< 0
2 co
mpl
ex ro
ots
F
ind
the
valu
e of
the
dis
crim
inan
t fo
r ea
ch e
quat
ion.
The
n de
scri
be
the
num
ber
and
type
of
root
s fo
r th
e eq
uati
on.
a. 2
x2 + 5
x +
3Th
e di
scri
min
ant
is
b2 - 4
ac =
52 -
4(2
) (3)
or
1.Th
e di
scri
min
ant
is a
per
fect
squ
are,
so
the
equa
tion
has
2 r
atio
nal r
oots
.
b. 3
x2 - 2
x +
5Th
e di
scri
min
ant
is
b2 - 4
ac =
(-2)
2 - 4
(3) (
5) o
r -
56.
The
disc
rim
inan
t is
neg
ativ
e, s
o th
e eq
uati
on h
as 2
com
plex
roo
ts.
Exer
cise
sC
ompl
ete
part
s a-
c fo
r ea
ch q
uadr
atic
equ
atio
n.a.
Fin
d th
e va
lue
of t
he d
iscr
imin
ant.
b. D
escr
ibe
the
num
ber
and
type
of
root
s.c.
Fin
d th
e ex
act
solu
tion
s by
usi
ng t
he Q
uadr
atic
For
mul
a.
1. p
2 + 1
2p =
-4
2.
9x2 -
6x
+ 1
= 0
3.
2x2 -
7x
- 4
= 0
4. x
2 + 4
x -
4 =
0
5. 5
x2 - 3
6x +
7 =
0
6. 4
x2 - 4
x +
11
= 0
7. x
2 - 7
x +
6 =
0
8. m
2 - 8
m =
-14
9.
25x
2 - 4
0x =
-16
10. 4
x2 + 2
0x +
29
= 0
11
. 6x2 +
26x
+ 8
= 0
12
. 4x2 -
4x
- 1
1 =
0
Exam
ple
Stud
y Gu
ide
and
Inte
rven
tion
(con
tinu
ed)
Th
e Q
uad
rati
c F
orm
ula
an
d t
he D
iscri
min
an
t
2
irra
tiona
l ro
ots;
1
rat
iona
l roo
t; 1 −
3 2
rat
iona
l roo
ts; -
1 −
2 , 4
-6
± 4
√ �
2
2
irra
tiona
l roo
ts;
2 r
atio
nal r
oots
;
-16
0; 2
com
plex
roo
ts;
-
2 ±
2 √
� 2
1 −
5 , 7
1 ±
i √
��
10
−
2
2
rat
iona
l roo
ts;
2 ir
ratio
nal r
oots
; 1
rat
iona
l roo
t; 4 −
5
1, 6
4
± √
� 2
2
com
plex
roo
ts;
2
rat
iona
l roo
ts;
2 ir
ratio
nal r
oots
;
- 5 −
2 ± i
-4,
- 1 −
3
1 −
2 ± √
� 3
030_
041_
ALG
2_A
_CR
M_C
04_C
R_6
6078
5.in
dd
3712
/20/
10
9:06
PM
Answers (Lesson 4-6)
A14_A26_ALG2_A_CRM_C04_AN_660785.indd A17A14_A26_ALG2_A_CRM_C04_AN_660785.indd A17 12/20/10 10:56 PM12/20/10 10:56 PM
Co
pyrig
ht ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
pa
nie
s, In
c.
PDF Pass
Chapter 4 A18 Glencoe Algebra 2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
4-6
Cha
pter
4
38
Gle
ncoe
Alg
ebra
2
Skill
s Pr
actic
eTh
e Q
uad
rati
c F
orm
ula
an
d t
he D
iscri
min
an
t
Com
plet
e pa
rts
a-c
for
each
qua
drat
ic e
quat
ion.
a. F
ind
the
valu
e of
the
dis
crim
inan
t.b.
Des
crib
e th
e nu
mbe
r an
d ty
pe o
f ro
ots.
c. F
ind
the
exac
t so
luti
ons
by u
sing
the
Qua
drat
ic F
orm
ula.
1. x
2 - 8
x +
16
= 0
2.
x2 -
11x
- 2
6 =
0
3. 3
x2 - 2
x =
0
4. 2
0x2 +
7x
- 3
= 0
5. 5
x2 - 6
= 0
6.
x2 -
6 =
0
7. x
2 + 8
x +
13
= 0
8.
5x2 -
x -
1 =
0
9. x
2 - 2
x -
17
= 0
10
. x2 +
49
= 0
11. x
2 - x
+ 1
= 0
12
. 2x2 -
3x
= -
2
Solv
e ea
ch e
quat
ion
by u
sing
the
Qua
drat
ic F
orm
ula.
13. x
2 = 6
4 14
. x2 -
30
= 0
15. x
2 - x
= 3
0
16. 1
6x2 -
24x
- 2
7 =
0
17. x
2 - 4
x -
11
= 0
18
. x2 -
8x
- 1
7 =
0
19. x
2 + 2
5 =
0
20. 3
x2 + 3
6 =
0
21. 2
x2 + 1
0x +
11
= 0
22
. 2x2 -
7x
+ 4
= 0
23. 8
x2 + 1
= 4
x
24. 2
x2 + 2
x +
3 =
0
25. P
ARA
CHU
TIN
G I
gnor
ing
win
d re
sist
ance
, the
dis
tanc
e d(
t) in
feet
tha
t a
para
chut
ist
falls
in t
seco
nds
can
be e
stim
ated
usi
ng t
he fo
rmul
a d(
t) =
16t
2 . If
a p
arac
huti
st ju
mps
fr
om a
n ai
rpla
ne a
nd fa
lls fo
r 11
00 fe
et b
efor
e op
enin
g he
r pa
rach
ute,
how
man
y se
cond
s pa
ss b
efor
e sh
e op
ens
the
para
chut
e?
0
; 1 r
atio
nal r
oot;
4
225;
2 r
atio
nal r
oots
; -2,
13
-
3; 2
com
plex
roo
ts;
1 ±
i √ �
3 −
2
-
7; 2
com
plex
roo
ts;
3 ±
i √ �
7 −
4
4
; 2 r
atio
nal r
oots
; 0, 2
− 3
28
9; 2
rat
iona
l roo
ts; -
3 −
5 , 1 −
4
1
20; 2
irra
tiona
l roo
ts; ±
√ �
�
30
−
5
24
; 2 ir
ratio
nal r
oots
; ±
√ �
6
1
2; 2
irra
tiona
l roo
ts; -
4 ±
√ �
3
21; 2
irra
tiona
l roo
ts;
1 ±
√ �
�
21
−
10
7
2; 2
irra
tiona
l roo
ts; 1
± 3
√ �
2 -
196;
2 c
ompl
ex r
oots
; ±7i
±8
-5,
6 2 ±
√ �
�
15
4 ±
√ �
�
33
±5i
± 2
i √ �
3
abou
t 8.3
s
-5
± √
� 3
−
2
1 ±
i −
4
-
1 ±
i √ �
5 −
2
7 ±
√ �
�
17
−
4
± √
��
30 9
− 4 , -
3 −
4
030_
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PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 4-6
4-6
Cha
pter
4
39
Gle
ncoe
Alg
ebra
2
Prac
tice
Th
e Q
uad
rati
c F
orm
ula
an
d t
he D
iscri
min
an
t
Solv
e ea
ch e
quat
ion
by u
sing
the
Qua
drat
ic F
orm
ula.
1. 7
x2 - 5
x =
0
2. 4
x2 - 9
= 0
3. 3
x2 + 8
x =
3
4. x
2 - 2
1 =
4x
5. 3
x2 - 1
3x +
4 =
0
6. 1
5x2 +
22x
= -
8
7. x
2 - 6
x +
3 =
0
8. x
2 - 1
4x +
53
= 0
9. 3
x2 = -
54
10. 2
5x2 -
20x
- 6
= 0
11. 4
x2 - 4
x +
17
= 0
12
. 8x
- 1
= 4
x2
13. x
2 = 4
x -
15
14
. 4x2 -
12x
+ 7
= 0
Com
plet
e pa
rts
a-c
for
each
qua
drat
ic e
quat
ion.
a. F
ind
the
valu
e of
the
dis
crim
inan
t.b.
Des
crib
e th
e nu
mbe
r an
d ty
pe o
f ro
ots.
c. F
ind
the
exac
t so
luti
ons
by u
sing
the
Qua
drat
ic F
orm
ula.
15. x
2 - 1
6x +
64
= 0
16
. x2 =
3x
17. 9
x2 - 2
4x +
16
= 0
18. x
2 - 3
x =
40
19. 3
x2 + 9
x -
2 =
0
20. 2
x2 + 7
x =
0
21. 5
x2 - 2
x +
4 =
0
22. 1
2x2 -
x -
6 =
0
23. 7
x2 + 6
x +
2 =
0
24. 1
2x2 +
2x
- 4
= 0
25
. 6x2 -
2x
- 1
= 0
26
. x2 +
3x
+ 6
= 0
27. 4
x2 - 3
x2 - 6
= 0
28
. 16x
2 - 8
x +
1 =
0
29. 2
x2 - 5
x -
6 =
0
30. G
RAVI
TATI
ON
The
hei
ght h
(t) in
feet
of a
n ob
ject
t se
cond
s af
ter
it is
pro
pelle
d st
raig
ht
up fr
om th
e gr
ound
wit
h an
init
ial v
eloc
ity
of 6
0 fe
et p
er s
econ
d is
mod
eled
by
the
equa
tion
h(
t) =
-16
t2 + 6
0t. A
t wha
t tim
es w
ill th
e ob
ject
be
at a
hei
ght o
f 56
feet
?
31. S
TOPP
ING
DIS
TAN
CE T
he fo
rmul
a d
= 0
.05s
2 + 1
.1s
esti
mat
es t
he m
inim
um s
topp
ing
dist
ance
d in
feet
for
a ca
r tr
avel
ing
s m
iles
per
hour
. If a
car
sto
ps in
200
feet
, wha
t is
the
fa
stes
t it c
ould
hav
e be
en tr
avel
ing
whe
n th
e dr
iver
app
lied
the
brak
es?
1.75
s, 2
s
abou
t 53.
2 m
i/h
0
; 1 r
atio
nal;
8 9
; 2 r
atio
nal;
0, 3
0
; 1 r
atio
nal;
4 −
3
1
69; 2
rat
iona
l; -
5, 8
2
irra
tiona
l; -
9 ±
√
��
10
5 −
6
4
9; 2
rat
iona
l; 0,
- 7 −
2
2
com
plex
; 1
± i
√
�
19
−
5
2
rat
iona
l; 3 −
4 , -
2 −
3 2
com
plex
; -
3 ±
i √
�
5 −
7
2
rat
iona
l; 1 −
2 , -
2 −
3 2
irra
tiona
l; 1
± √
�
7 −
6
2
com
plex
; -
3 ±
i √
�
15
−
2
2
irra
tiona
l; 3
± √
��
10
5 −
8
0
; 1 r
atio
nal;
1 −
4 2
irra
tiona
l; 5
± √
��
73
−
4
± 3 −
2
1 −
3 , 4
7 ±
2i
2 ±
√
�
10
−
5
2 ±
√
�
3 −
2
3 ±
√
�
2 −
2
- 2 −
3 , - 4 −
5
0, 5 −
7
1 −
3 , -3
3 ±
√
�
6
± 3
i √
�
2
2 ±
i √
�
11
-3,
7
1 ±
4i
−
2
030_
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Answers (Lesson 4-6)
A14_A26_ALG2_A_CRM_C04_AN_660785.indd A18A14_A26_ALG2_A_CRM_C04_AN_660785.indd A18 12/20/10 10:56 PM12/20/10 10:56 PM
Copyri
ght
© G
lencoe/M
cG
raw
-Hill
, a
div
isio
n o
f T
he
McG
raw
-Hill
Co
mp
an
ies,
Inc.
PDF Pass
Chapter 4 A19 Glencoe Algebra 2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
4-6
Cha
pter
4
40
Gle
ncoe
Alg
ebra
2
Wor
d Pr
oble
m P
ract
ice
Th
e Q
uad
rati
c F
orm
ula
an
d t
he D
iscri
min
an
t
1. P
ARA
BOLA
S Th
e gr
aph
of a
qua
drat
ic
equa
tion
of t
he fo
rm y
=ax
2+
bx+
c is
sh
own
belo
w.
y
xO5
-5
Is t
he d
iscr
imin
ant
b2-
4ac
pos
itiv
e,
nega
tive
, or
zero
? E
xpla
in.
2. T
AN
GEN
T K
athl
een
is t
ryin
g to
find
bso
tha
t th
e x-
axis
is t
ange
nt t
o th
e pa
rabo
la y
=x2
+bx
+ 4
. She
find
s on
e va
lue
that
wor
ks, b
= 4
. Is
this
the
onl
y va
lue
that
wor
ks?
Exp
lain
.
3. S
PORT
S In
199
0, A
mer
ican
Ran
dy
Bar
nes
set
the
wor
ld r
ecor
d fo
r th
e sh
ot
put.
His
thr
ow c
an b
e de
scri
bed
by t
he
equa
tion
y=
–16
x2+
368
x. U
se t
he
Qua
drat
ic F
orm
ula
to fi
nd h
ow fa
r hi
s th
row
was
to
the
near
est
foot
.
4. E
XA
MPL
ES G
ive
an e
xam
ple
of a
qu
adra
tic
func
tion
f(x
) tha
t ha
s th
e fo
llow
ing
prop
erti
es.
I. Th
e di
scri
min
ant
of f
is z
ero.
II. T
here
is n
o re
al s
olut
ion
of t
he
equa
tion
f(x
) = 1
0.
Sket
ch t
he g
raph
of x
=f(
x).
yx
O
-2
-2
-4
-6
-8
-4
24
5. T
AN
GEN
TS T
he g
raph
of y
= x
2 is
a pa
rabo
la t
hat
pass
es t
hrou
gh t
he p
oint
at
(1, 1
). Th
e lin
e y
= m
x -
m +
1,
whe
re m
is a
con
stan
t, al
so p
asse
s th
roug
h th
e po
int
at (1
, 1).
a. T
o fin
d th
e po
ints
of i
nter
sect
ion
betw
een
the
line
y =
mx
- m
+ 1
an
d th
e pa
rabo
la y
= x
2 , se
t x2 =
m
x -
m +
1 a
nd t
hen
solv
e fo
r x.
R
earr
angi
ng t
erm
s, t
his
equa
tion
be
com
es x
2 - m
x +
m -
1 =
0. W
hat
is t
he d
iscr
imin
ant
of t
his
equa
tion
?
b. F
or w
hat
valu
e of
m is
the
re o
nly
one
poin
t of
inte
rsec
tion
? E
xpla
in t
he
mea
ning
of t
his
in t
erm
s of
the
co
rres
pond
ing
line
and
the
para
bola
.
N
egat
ive;
the
equa
tion
has
no
real
sol
utio
ns s
o th
e di
scrim
inan
t is
nega
tive.
N
o, b
= -
4 al
so w
orks
; the
x-
axis
will
be
tang
ent w
hen
the
disc
rimin
ant b
2 - 1
6 is
zer
o.
This
hap
pens
whe
n b
= 4
or
-4.
2
3 ft
S
ampl
e an
swer
: f(x
) = -
x2
x
2 - 4
m +
4
m
= 2
; the
par
abol
a y
= x
2 and
th
e lin
e y
= 2
x -
1 h
ave
exac
tly
one
poin
t of i
nter
sect
ion
at
(1, 1
). In
oth
er w
ords
, thi
s lin
e is
tang
ent t
o th
e pa
rabo
la a
t (1
, 1).
030_
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M_C
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R_6
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10
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PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 4-6
4-6
Cha
pter
4
41
Gle
ncoe
Alg
ebra
2
Enri
chm
ent
Su
m a
nd
Pro
du
ct
of
Ro
ots
Som
etim
es y
ou m
ay k
now
the
roo
ts o
f a q
uadr
atic
equ
atio
n w
itho
ut k
now
ing
the
equa
tion
it
self.
Usi
ng y
our
know
ledg
e of
fact
orin
g to
sol
ve a
n eq
uati
on, y
ou c
an w
ork
back
war
d to
fin
d th
e qu
adra
tic
equa
tion
. The
rul
e fo
r fin
ding
the
sum
and
pro
duct
of r
oots
is a
s fo
llow
s:
Sum
and
Pro
duct
of R
oots
If th
e ro
ots
of a
x2 + b
x +
c =
0, w
ith a
≠ 0
, are
s1 a
nd
s 2, th
en s
1 + s
2 = -
a −
b and
s1 �
s2 =
c −
a .
W
rite
a q
uadr
atic
equ
atio
n th
at h
as t
he r
oots
3 a
nd -
8.
The
root
s ar
e x
= 3
and
x =
-8.
x
y
O
(–5 – 2, –30
–1 4)
10 –10
–20
–30
24
–2–4
–6–8
3 +
(-8)
= -
5 A
dd th
e ro
ots.
3(
-8)
= -
24
Mul
tiply
the
root
s.
Equ
atio
n: x
2 + 5
x -
24
= 0
Exer
cise
sW
rite
a q
uadr
atic
equ
atio
n th
at h
as t
he g
iven
roo
ts.
1. 6
, -9
2. 5
, -1
3. 6
, 6
4. 4
± √
�
3 5.
- 2 −
5 , 2 −
7 6.
-
2 ±
3 √
�
5 −
7
Fin
d k
such
tha
t th
e nu
mbe
r gi
ven
is a
roo
t of
the
equ
atio
n.
7. 7
; 2x2 +
kx
- 2
1 =
0
8. -
2; x
2 - 1
3x +
k =
0
Exam
ple
x
2 + 3
x -
54
= 0
x
2 - 4
x -
5 =
0
x2 -
12x
+ 3
6 =
0
x
2 - 8
x +
13
= 0
3
5x2 +
4x
- 4
= 0
4
9x2 -
42x
+ 2
05 =
0
-
11
-
30
030_
041_
ALG
2_A
_CR
M_C
04_C
R_6
6078
5.in
dd
4112
/20/
10
9:06
PM
Answers (Lesson 4-6)
A14_A26_ALG2_A_CRM_C04_AN_660785.indd A19A14_A26_ALG2_A_CRM_C04_AN_660785.indd A19 12/20/10 10:56 PM12/20/10 10:56 PM
Co
pyrig
ht ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
pa
nie
s, In
c.
PDF Pass
Chapter 4 A20 Glencoe Algebra 2
Answers (Lesson 4-6 and Lesson 4-7)
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
4-6
Cha
pter
4
42
Gle
ncoe
Alg
ebra
2
Spre
adsh
eet A
ctiv
ityA
pp
roxi
mati
ng
th
e R
eal
Zero
s o
f P
oly
no
mia
ls
You
have
lear
ned
the
Loca
tion
Pri
ncip
le, w
hich
can
be
used
to
appr
oxim
ate
the
real
zer
os o
f a p
olyn
omia
l.
A1 2 3 4 5
C x f(x)
B 5
G 0 –2
D –5 23
J 5 23
E
–3.3
3333
339.
1111
111
H
1.66
6666
70.
7777
778
I
3.33
3333
39.
1111
111
F
–1.6
6666
670.
7777
778
–5
She
et 1
She
et 2
She
et 3
In t
he s
prea
dshe
et a
bove
, the
pos
itiv
e re
al z
ero
of ƒ
(x) =
x2 -
2 c
an b
e ap
prox
imat
ed in
the
fo
llow
ing
way
. Set
the
spr
eads
heet
pre
fere
nce
to m
anua
l cal
cula
tion
. The
val
ues
in A
2 an
d B
2 ar
e th
e en
dpoi
nts
of a
ran
ge o
f val
ues.
The
val
ues
in D
2 th
roug
h J2
are
val
ues
equa
lly
in t
he in
terv
al fr
om A
2 to
B2.
The
form
ulas
for
thes
e va
lues
are
A2,
A2
+ (B
2 -
A2)
/6, A
2 +
2*
(B2
- A
2)/6
, A2
+ 3
*(B
2 -
A2)
/6, A
2 +
4*(
B2
- A
2)/6
, A2
+ 5
*(B
2 -
A2)
/6, a
nd B
2,
resp
ecti
vely
.
Row
3 g
ives
the
func
tion
val
ues
at t
hese
poi
nts.
The
func
tion
ƒ(x
) = x
2 - 2
is e
nter
ed in
to
the
spre
adsh
eet
in C
ell D
3 as
D2^
2 -
2. T
his
func
tion
is t
hen
copi
ed t
o th
e re
mai
ning
cel
ls
in t
he r
ow.
You
can
use
this
spr
eads
heet
to
stud
y th
e fu
ncti
on v
alue
s at
the
poi
nts
in c
ells
D2
thro
ugh
J2. T
he v
alue
in c
ell F
3 is
pos
itiv
e an
d th
e va
lue
in c
ell G
3 is
neg
ativ
e, s
o th
ere
mus
t be
a
zero
bet
wee
n -
1.66
67 a
nd 0
. Ent
er t
hese
val
ues
in c
ells
A2
and
B2,
res
pect
ivel
y, an
d re
calc
ulat
e th
e sp
read
shee
t. (Y
ou w
ill h
ave
to r
ecal
cula
te a
num
ber
of t
imes
.) Th
e re
sult
is
a ne
w t
able
from
whi
ch y
ou c
an s
ee t
hat
ther
e is
a z
ero
betw
een
1.41
414
and
1.41
4306
. B
ecau
se t
hese
val
ues
agre
e to
thr
ee d
ecim
al p
lace
s, t
he z
ero
is a
bout
1.4
14. T
his
can
be
veri
fied
by u
sing
alg
ebra
.B
y so
lvin
g x2 -
2 =
0, w
e ob
tain
x =
± √
�
2 . T
he p
osit
ive
root
is
x=
± √
�
2 =
1.4
1421
3. .
. , w
hich
ver
ifies
the
res
ult.
Exer
cise
s 1
. Use
a s
prea
dshe
et li
ke t
he o
ne a
bove
to
appr
oxim
ate
the
zero
of ƒ
(x) =
3x
- 2
to
thre
e de
cim
al p
lace
s. T
hen
veri
fy y
our
answ
er b
y us
ing
alge
bra
to fi
nd t
he e
xact
val
ue o
f the
ro
ot.
2. U
se a
spr
eads
heet
like
the
one
abo
ve t
o ap
prox
imat
e th
e re
al z
eros
of f
(x) =
x2 +
2x
+ 0
.5.
Rou
nd y
our
answ
er t
o fo
ur d
ecim
al p
lace
s. T
hen,
ver
ify y
our
answ
er b
y us
ing
the
quad
rati
c fo
rmul
a.
3. U
se a
spr
eads
heet
like
the
one
abo
ve t
o ap
prox
imat
e th
e re
al z
ero
of
ƒ(x)
= x
3 - 3 −
2 x2 -
6x
- 2
bet
wee
n -
0.4
and
-0.
3.
T
he s
prea
dshe
et g
ives
x =
0.6
67. B
y so
lvin
g fo
r x
alge
brai
cally
,
x =
2 −
3 . So,
the
appr
oxim
atio
n is
cor
rect
.
The
proc
ess
give
s -
1.70
71 a
nd -
0.29
29 to
the
near
est
t
en-t
hous
andt
h. T
he q
uadr
atic
form
ula
give
s x
= -
1 ±
√ �
2 −
2 .
-1
- √
� 2
−
2 ≈
-
1.70
71 a
nd -
1 +
√ �
2 −
2
≈ -
0.29
29.
-0.
3781
to th
e ne
ares
t ten
-thou
sand
th
042_
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Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 4-7
4-7
Cha
pter
4
43
Gle
ncoe
Alg
ebra
2
Stud
y Gu
ide
and
Inte
rven
tion
Tran
sfo
rmati
on
s o
f Q
uad
rati
c G
rap
hs
Wri
te Q
uadr
atic
Equ
atio
ns in
Ver
tex
Form
A q
uadr
atic
func
tion
is e
asie
r to
gr
aph
whe
n it
is in
ver
tex
form
. You
can
wri
te a
qua
drat
ic fu
ncti
on o
f the
form
y
= a
x2 + b
x +
c in
ver
tex
from
by
com
plet
ing
the
squa
re.
W
rite
y =
2x2 -
12x
+ 2
5 in
ver
tex
form
. The
n gr
aph
the
func
tion
.
y =
2x2 -
12x
+ 2
5
x
y
O2
2468
46
y =
2(x
2 - 6
x) +
25
y =
2(x
2 - 6
x +
9) +
25
- 1
8y
= 2
(x -
3)2 +
7
The
vert
ex fo
rm o
f the
equ
atio
n is
y =
2(x
- 3
)2 + 7
.
Exer
cise
s
Wri
te e
ach
equa
tion
in v
erte
x fo
rm. T
hen
grap
h th
e fu
ncti
on.
1. y
= x
2 - 1
0x +
32
2. y
= x
2 + 6
x 3.
y =
x2 -
8x
+ 6
x
y
O2468
24
6
x
y
O
-2
-4
-6
-8
-2
-4
-6
x
y
O4
-4
8
8 4
-4
-8
-12
4. y
= -
4x2 +
16x
- 1
1 5.
y =
3x2 -
12x
+ 5
6.
y =
5x2 -
10x
+ 9
x
y
O-
2
-2
-4
2
246
4
x
y
O-
2
-2
-4
-6
2
2
4
x
y
O-
22
246810
4
Exam
ple
y
= -
4(x
- 2
)2 + 5
y
= 3
(x -
2)2 -
7
y=
5(x
- 1
)2 + 4
y
= (x
- 5
)2 + 7
y
= (x
+ 3
)2 - 9
y
= (x
- 4
)2 - 1
0
042_
056_
ALG
2_A
_CR
M_C
04_C
R_6
6078
5.in
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4312
/20/
10
9:06
PM
A14_A26_ALG2_A_CRM_C04_AN_660785.indd A20A14_A26_ALG2_A_CRM_C04_AN_660785.indd A20 12/20/10 10:56 PM12/20/10 10:56 PM
Copyri
ght
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lencoe/M
cG
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-Hill
, a
div
isio
n o
f T
he
McG
raw
-Hill
Co
mp
an
ies,
Inc.
PDF Pass
Chapter 4 A21 Glencoe Algebra 2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
4-7
Cha
pter
4
44
Gle
ncoe
Alg
ebra
2
Stud
y Gu
ide
and
Inte
rven
tion
(con
tinu
ed)
Tran
sfo
rmati
on
s o
f Q
uad
rati
c G
rap
hs
Tran
sfor
mat
ions
of
Qua
drat
ic G
raph
s Pa
rabo
las
can
be t
rans
form
ed b
y ch
angi
ng t
he v
alue
s of
the
con
stan
ts a
, h, a
nd k
in t
he v
erte
x fo
rm o
f a q
uadr
atic
equ
atio
n:
y =
a(x
– h
) 2 + k
.
•
The
sign
of a
det
erm
ines
whe
ther
the
gra
ph o
pens
upw
ard
(a >
0) o
r do
wnw
ard
(a <
0).
•
The
abso
lute
val
ue o
f a a
lso
caus
es a
dila
tion
(enl
arge
men
t or
red
ucti
on) o
f the
pa
rabo
la. T
he p
arab
ola
beco
mes
nar
row
er if
⎪a⎥
>1
and
wid
er if
⎪a⎥
< 1
.
•
The
valu
e of
h t
rans
late
s th
e pa
rabo
la h
oriz
onta
lly. P
osit
ive
valu
es o
f h s
lide
the
grap
h to
the
rig
ht a
nd n
egat
ive
valu
es s
lide
the
grap
h to
the
left
.
•
The
valu
e of
k t
rans
late
s th
e gr
aph
vert
ical
ly. P
osit
ive
valu
es o
f k s
lide
the
grap
h up
war
d an
d ne
gati
ve v
alue
s sl
ide
the
grap
h do
wnw
ard.
G
raph
y =
(x
+ 7
)2 + 3
.
•
Rew
rite
the
equ
atio
n as
y =
[x –
(–7)
]2 + 3
.
•
Bec
ause
h =
–7
and
k =
3, t
he v
erte
x is
at
(–7,
3).
The
axis
of
sym
met
ry is
x =
–7.
Bec
ause
a =
1, w
e kn
ow t
hat
the
grap
h op
ens
up, a
nd t
he g
raph
is t
he s
ame
wid
th a
s th
e gr
aph
of y
= x
2 .
•
Tran
slat
e th
e gr
aph
of y
= x
2 sev
en u
nits
to
the
left
and
th
ree
unit
s up
.
Exer
cise
s
Gra
ph e
ach
func
tion
.
1.
y =
–2x
2 +
2
2. y
= –
3(x
– 1)
2 3.
y =
2(x
+ 2
)2 + 3
Exam
ple
x
15 5
-15
-15-5
515
y
-5
x
68 4 2
-6-
4-
8
-6
-4
-2
-8
24
68
y
-2
x
68 4 2
-6-
4-
8
-6
-4
-2
-8
24
68
y
-2
x
68 4 2
-6-
4-
8
-6
-4
-2
-8
24
68
y
-2
042_
056_
ALG
2_A
_CR
M_C
04_C
R_6
6078
5.in
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10
9:06
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
4-7
Cha
pter
4
45
Gle
ncoe
Alg
ebra
2
Skill
s Pr
actic
eTr
an
sfo
rmati
on
s o
f Q
uad
rati
c G
rap
hs
Wri
te e
ach
quad
rati
c fu
ncti
on in
ver
tex
form
. The
n id
enti
fy t
he v
erte
x, a
xis
of
sym
met
ry, a
nd d
irec
tion
of
open
ing.
1. y
= (x
- 2
)2 2.
y =
-x2 +
4
3. y
= x
2 - 6
4. y
= -
3(x
+ 5
)2 5.
y =
-5x
2 + 9
6.
y =
(x -
2)2 -
18
7. y
= x
2 - 2
x -
5
8. y
= x
2 + 6
x +
2
9. y
= -
3x2 +
24x
Gra
ph e
ach
func
tion
.
10. y
= (x
- 3
)2 - 1
11
. y =
(x +
1)2 +
2
12. y
= -
(x -
4)2 -
4
x
y
O6 4 2
26
4
x
y
O4
-4
6 4 2
-2
2
x
y
O2
–2
–4
–6
46
13. y
= -
1 −
2 (x
+ 2
)2 14
. y =
-3x
2 + 4
15
. y =
x2 +
6x
+ 4
x
y
O
-2
-4
-6
-2
2-
6-
4
x
y
O4 2
24
–2
–4
-4
-2
x
y
O2
2
–2
–4
-4
-2
-6
y
= -
3(x
+ 5
)2 + 0
; y
= -
5(x
- 0
)2 + 9
; y
= (x
- 2
)2 - 1
8;
(
-5,
0);
x =
-5;
dow
n (
0, 9
); x
= 0
; dow
n (
2, -
18);
x =
2; u
p
y
= (x
- 2
)2 + 0
; y
= -
(x -
0)2 +
4;
y =
(x -
0)2 -
6;
(
2, 0
); x
= 2
; up
(0,
4);
x =
0; d
own
(0,
-6)
; x =
0; u
p
y
= (x
- 1
)2 - 6
; y
= (x
+ 3
)2 - 7
; y
= -
3(x
- 4
)2 + 4
8;
(
1, -
6); x
= 1
; up
(-
3, -
7); x
= -
3; u
p (
4, 4
8); x
= 4
; dow
n
042_
056_
ALG
2_A
_CR
M_C
04_C
R_6
6078
5.in
dd
4512
/20/
10
9:06
PM
Answers (Lesson 4-7)
A14_A26_ALG2_A_CRM_C04_AN_660785.indd A21A14_A26_ALG2_A_CRM_C04_AN_660785.indd A21 12/20/10 10:56 PM12/20/10 10:56 PM
Co
pyrig
ht ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
pa
nie
s, In
c.
PDF Pass
Chapter 4 A22 Glencoe Algebra 2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
4-7
Cha
pter
4
46
Gle
ncoe
Alg
ebra
2
Prac
tice
Tran
sfo
rmati
on
s o
f Q
uad
rati
c G
rap
hs
Wri
te e
ach
equa
tion
in v
erte
x fo
rm. T
hen
iden
tify
the
ver
tex,
axi
s of
sym
met
ry,
and
dire
ctio
n of
ope
ning
. 1
. y =
-6x
2 -
24
x -
25
2. y
= 2
x2 + 2
3.
y =
-4x
2 + 8
x
4. y
= x
2 + 1
0x +
20
5. y
= 2
x2 + 1
2x +
18
6. y
= 3
x2 - 6
x +
5
7. y
= -
2x2 -
16x
- 3
2 8.
y =
-3x
2 + 1
8x -
21
9. y
= 2
x2 + 1
6x +
29
Gra
ph e
ach
func
tion
.
10. y
= (x
+ 3
)2 - 1
11
. y
= -
x2 + 6
x -
5
12. y
= 2
x2 - 2
x +
1
x
y
O-
2-
4-
6
246
x
y
O-
2
-2
-424
24
6
x
y O-
2-
42
246
4
13. W
rite
an
equa
tion
for
a pa
rabo
la w
ith
vert
ex a
t (1
, 3) t
hat
pass
es t
hrou
gh (-
2, -
15).
14. W
rite
an
equa
tion
for
a pa
rabo
la w
ith
vert
ex a
t (-
3, 0
) tha
t pa
sses
thr
ough
(3, 1
8).
15. B
ASE
BALL
The
hei
ght
h of
a b
aseb
all t
sec
onds
aft
er b
eing
hit
is g
iven
by
h(t)
= -
16t2 +
80t
+ 3
. Wha
t is
the
max
imum
hei
ght
that
the
bas
ebal
l rea
ches
, and
w
hen
does
thi
s oc
cur?
16. S
CULP
TURE
A m
oder
n sc
ulpt
ure
in a
par
k co
ntai
ns a
par
abol
ic a
rc t
hat
st
arts
at
the
grou
nd a
nd r
each
es a
max
imum
hei
ght
of 1
0 fe
et a
fter
a
hori
zont
al d
ista
nce
of 4
feet
. Wri
te a
qua
drat
ic fu
ncti
on in
ver
tex
form
th
at d
escr
ibes
the
sha
pe o
f the
out
side
of t
he a
rc, w
here
y is
the
hei
ght
of a
poi
nt o
n th
e ar
c an
d x
is it
s ho
rizo
ntal
dis
tanc
e fr
om t
he le
ft-h
and
star
ting
poi
nt o
f the
arc
. 10
ft
4 ft
y
= -
6(x
+ 2
)2 - 1
;
y =
2(x
+ 0
)2 + 2
; y
= -
4(x
- 1
)2 + 4
;
(-
2, -
1); x
= -
2; d
own
(0,
2);
x =
0; u
p (
1, 4
); x
= 1
; dow
n
y
= (x
+ 5
)2 - 5
; y
= 2
(x +
3)2 ;
(-3,
0);
y =
3(x
- 1
)2 + 2
;
(-
5, -
5); x
= -
5; u
p x
= -
3; u
p (
1, 2
); x
= 1
; up
y
= -
2(x
+ 4
)2 ; y
= -
3(x
- 3
)2 + 6
; y
= 2
(x +
4)2 -
3;
(
-4,
0);
x =
-4;
dow
n (
3, 6
); x
= 3
; dow
n (
-4,
-3)
; x =
-4;
up
y
= -
2(x
- 1
)2 + 3
y
= 1 −
2 (x +
3)2
103
ft; 2
.5 s
y =
- 5 −
8 (x -
4)2 +
10
042_
056_
ALG
2_A
_CR
M_C
04_C
R_6
6078
5.in
dd
4612
/20/
10
9:06
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
4-7
Cha
pter
4
47
Gle
ncoe
Alg
ebra
2
Wor
d Pr
oble
m P
ract
ice
Tran
sfo
rmati
on
s o
f Q
uad
rati
c G
rap
hs
1.A
RCH
ES A
par
abol
ic a
rch
is u
sed
as a
br
idge
sup
port
. The
gra
ph o
f the
arc
h is
sh
own
belo
w.
y
xO
5
5
-5
If t
he e
quat
ion
that
cor
resp
onds
to
this
gra
ph is
wri
tten
in t
he fo
rm
y+
a(x
-h)
2+
k, w
hat
are
h an
d k?
2. T
RAN
SLAT
ION
S Fo
r a
com
pute
r an
imat
ion,
Bar
bara
use
s th
e qu
adra
tic
func
tion
f(x)
= -
42(x
- 2
0)2 +
168
00 t
o he
lp h
er s
imul
ate
an o
bjec
t to
ssed
on
anot
her
plan
et. F
or o
ne s
kit,
she
had
to
use
the
func
tion
f(x
+ 5
) - 8
000
inst
ead
of f
(x).
Whe
re is
the
ver
tex
of t
he g
raph
of
y =
f(x
+ 5
) - 8
000?
3. B
RID
GES
The
sha
pe fo
rmed
by
the
mai
n ca
bles
of t
he G
olde
n G
ate
Bri
dge
appr
oxim
atel
y fo
llow
s th
e eq
uati
on
y =
0.0
002x
2 - 0
.23x
+ 2
27. G
raph
the
pa
rabo
la fo
rmed
by
one
of t
he c
able
s.
4.W
ATER
JET
S Th
e gr
aph
show
s th
e pa
th
of a
jet
of w
ater
.
y
xO
5
The
equa
tion
cor
resp
ondi
ng t
o th
is
grap
h is
y=
a(x
-h)
2+
k. W
hat
are
a,
h, a
nd k
?
5. P
ROFI
T A
the
ater
ope
rato
r pr
edic
ts t
hat
the
thea
ter
can
mak
e -
4x2 +
160
x do
llars
per
sho
w if
tic
kets
are
pri
ced
at x
do
llars
.
a. R
ewri
te t
he e
quat
ion
y =
-4x
2 + 1
60x
in t
he fo
rm y
= a
(x -
h) 2 +
k.
b. W
hat
is t
he v
erte
x of
the
par
abol
a an
d w
hat
is it
s ax
is o
f sym
met
ry?
c.G
raph
the
par
abol
a.
y
xO
800
1600
2040
h =
-1
and
k =
5
(15,
880
0)
a =
-2,
h =
2, k
= 6
y =
-4(
x -
20)
2 + 1
600
vert
ex a
t (20
, 160
0); a
xis
is x
= 2
0
Wid
th
Height
x
150
500
1000
y
300
042_
056_
ALG
2_A
_CR
M_C
04_C
R_6
6078
5.in
dd
4712
/20/
10
9:06
PM
Answers (Lesson 4-7)
A14_A26_ALG2_A_CRM_C04_AN_660785.indd A22A14_A26_ALG2_A_CRM_C04_AN_660785.indd A22 12/20/10 10:56 PM12/20/10 10:56 PM
Copyri
ght
© G
lencoe/M
cG
raw
-Hill
, a
div
isio
n o
f T
he
McG
raw
-Hill
Co
mp
an
ies,
Inc.
PDF Pass
Chapter 4 A23 Glencoe Algebra 2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
4-7
Cha
pter
4
48
Gle
ncoe
Alg
ebra
2
Enri
chm
ent
A S
ho
rtcu
t to
Co
mp
lex
Ro
ots
Whe
n gr
aphi
ng a
qua
drat
ic fu
ncti
on, t
he r
eal r
oots
are
sho
wn
in t
he g
raph
. Yo
u ha
ve le
arne
d th
at q
uadr
atic
func
tion
s ca
n al
so h
ave
imag
inar
y ro
ots
that
can
not
be s
een
on t
he g
raph
of t
he fu
ncti
on. H
owev
er, t
here
is a
way
to
grap
hica
lly r
epre
sent
the
com
plex
roo
ts o
f a q
uadr
atic
func
tion
.
F
ind
the
com
plex
roo
ts o
f th
e qu
adra
tic
func
tion
y =
x2 -
4x
+ 5
.
Step
1
Gra
ph t
he fu
ncti
on.
y
xO6
5
Step
2
Ref
lect
the
gra
ph o
ver
the
hori
zont
al li
ne c
onta
inin
g th
e ve
rtex
. In
this
exa
mpl
e,
the
vert
ex is
(2, 1
).
y
xO
Step
3
The
real
par
t of
the
com
plex
roo
t is
the
poi
nt h
alfw
ay b
etw
een
the
x-in
terc
epts
of t
he r
efle
cted
gra
ph a
nd t
he im
agin
ary
part
of t
he
com
plex
roo
ts a
re +
and
- h
alf t
he d
ista
nce
betw
een
the
x-in
terc
epts
of
the
ref
lect
ed g
raph
. So,
in t
his
exam
ple,
the
com
plex
roo
ts a
re
2 +
1i
and
2 -
1i.
Exer
cise
sU
sing
thi
s m
etho
d, f
ind
the
com
plex
roo
ts o
f th
e fo
llow
ing
quad
rati
c fu
ncti
ons.
1. y
= x
2 + 2
x +
5
2. y
= x
2 + 4
x +
8
3. y
= x
2 + 6
x +
13
4. y
= x
2 + 2
x +
17
Exam
ple
-
1 +
2i,
-1
- 2
i -
2 +
2i,
-2
- 2
i
-
3 +
2i,
-3
- 2
i
-1
+ 4
i, -
1 -
4i
042_
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PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 4-8
4-8
Cha
pter
4
49
Gle
ncoe
Alg
ebra
2
Stud
y Gu
ide
and
Inte
rven
tion
Qu
ad
rati
c I
neq
ualiti
es
Gra
ph Q
uadr
atic
Ineq
ualit
ies
To g
raph
a q
uadr
atic
ineq
ualit
y in
tw
o va
riab
les,
use
th
e fo
llow
ing
step
s:
1. G
raph
the
rel
ated
qua
drat
ic e
quat
ion,
y =
ax2 +
bx
+ c
. U
se a
das
hed
line
for
< o
r >
; use
a s
olid
line
for
≤ o
r ≥
.
2. T
est
a po
int
insi
de t
he p
arab
ola.
If i
t sa
tisf
ies
the
ineq
ualit
y, sh
ade
the
regi
on in
side
the
par
abol
a;
othe
rwis
e, s
hade
the
reg
ion
outs
ide
the
para
bola
.
G
raph
the
ineq
uali
ty y
> x
2 + 6
x +
7.
Firs
t gr
aph
the
equa
tion
y =
x2 +
6x
+ 7
. By
com
plet
ing
the
x
y
O2
-224
-4
-2
squa
re, y
ou g
et t
he v
erte
x fo
rm o
f the
equ
atio
n y
= (x
+ 3
)2 - 2
, so
the
ver
tex
is (-
3, -
2). M
ake
a ta
ble
of v
alue
s ar
ound
x =
-3,
an
d gr
aph.
Sin
ce t
he in
equa
lity
incl
udes
>, u
se a
das
hed
line.
Test
the
poi
nt (-
3, 0
), w
hich
is in
side
the
par
abol
a. S
ince
(-
3)2 +
6(-
3) +
7 =
-2,
and
0 >
-2,
(-3,
0) s
atis
fies
the
ineq
ualit
y. Th
eref
ore,
sha
de t
he r
egio
n in
side
the
par
abol
a.
Exer
cise
sG
raph
eac
h in
equa
lity
.
1. y
> x
2 - 8
x +
17
2. y
≤ x
2 + 6
x +
4
3. y
≥ x
2 + 2
x +
2
x
y
O2
46
-2
-4
-6
x
y
O2
-4
-2
-2
-4
-6
2
x
y
O2
246
-2
-4
4. y
< -
x2 + 4
x -
6
5. y
≥ 2
x2 + 4
x 6.
y >
-2x
2 - 4
x +
2
x
y
O2
-2
-4
-6
4-
2
x
y
O2
246
-2
-4
-2
x
y
O2
24
4-
2
-2
-4
-4
Exam
ple
042_
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Answers (Lesson 4-7 and Lesson 4-8)
A14_A26_ALG2_A_CRM_C04_AN_660785.indd A23A14_A26_ALG2_A_CRM_C04_AN_660785.indd A23 12/20/10 10:56 PM12/20/10 10:56 PM
Co
pyrig
ht ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
pa
nie
s, In
c.
PDF Pass
Chapter 4 A24 Glencoe Algebra 2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
4-8
Cha
pter
4
50
Gle
ncoe
Alg
ebra
2
Stud
y Gu
ide
and
Inte
rven
tion
(con
tinu
ed)
Qu
ad
rati
c I
neq
ualiti
es
Solv
e Q
uadr
atic
Ineq
ualit
ies
Qua
drat
ic in
equa
litie
s in
one
var
iabl
e ca
n be
sol
ved
grap
hica
lly o
r al
gebr
aica
lly.
Gra
phic
al M
etho
d
To s
olve
ax2 +
bx
+ c
< 0
: Fi
rst g
raph
y =
ax2 +
bx
+ c
. The
sol
utio
n co
nsis
ts o
f the
x-v
alue
s
for w
hich
the
grap
h is
bel
ow th
e x-
axis
.
To s
olve
ax2 +
bx
+ c
> 0
: Fi
rst g
raph
y =
ax2 +
bx
+ c
. The
sol
utio
n co
nsis
ts o
f the
x-v
alue
s fo
r whi
ch th
e gr
aph
is a
bove
the
x-ax
is.
Alg
ebra
ic M
etho
d
Find
the
root
s of
the
rela
ted
quad
ratic
equ
atio
n by
fact
orin
g,
com
plet
ing
the
squa
re; o
r usi
ng th
e Q
uadr
atic
For
mul
a.
2 ro
ots
divi
de th
e nu
mbe
r lin
e in
to 3
inte
rval
s.Te
st a
val
ue in
eac
h in
terv
al to
see
whi
ch in
terv
als
are
solu
tions
.
If t
he in
equa
lity
invo
lves
≤ o
r ≥
, the
roo
ts o
f the
rel
ated
equ
atio
n ar
e in
clud
ed in
the
so
luti
on s
et.
So
lve
the
ineq
uali
ty x
2 - x
- 6
≤ 0
.
Firs
t fin
d th
e ro
ots
of t
he r
elat
ed e
quat
ion
x2 - x
- 6
= 0
. The
x
y
O
equa
tion
fact
ors
as (x
- 3
)(x +
2) =
0, s
o th
e ro
ots
are
3 an
d -
2.
The
grap
h op
ens
up w
ith
x-in
terc
epts
3 a
nd -
2, s
o it
mus
t be
on
or b
elow
the
x-a
xis
for
-2
≤ x
≤ 3
. The
refo
re t
he s
olut
ion
set
is
{x|-
2 ≤
x ≤
3}.
Exer
cise
sSo
lve
each
ineq
uali
ty.
1. x
2 + 2
x <
0
2. x
2 - 1
6 <
0
3. 0
< 6
x -
x2 -
5
4. c
2 ≤ 4
5.
2m
2 - m
< 1
6.
y2 <
-8
7. x
2 - 4
x -
12
< 0
8.
x2 +
9x
+ 1
4 >
0
9. -
x2 + 7
x -
10
≥ 0
10. 2
x2 + 5
x- 3
≤ 0
11
. 4x2 -
23x
+ 1
5 >
0
12. -
6x2 -
11x
+ 2
< 0
13. 2
x2 - 1
1x +
12
≥ 0
14
. x2 -
4x
+ 5
< 0
15
. 3x2 -
16x
+ 5
< 0
Exam
ple
{
x ⎥
-2
< x
< 0
} {
x ⎥
-4
< x
< 4
} {
x ⎥
1 <
x <
5}
{
c ⎥
-2
≤ c
≤ 2
} {
m ⎥
- 1 −
2 < m
< 1
} �
{
x ⎥
-2
< x
< 6
} {
x ⎥
x <
-7
or x
> -
2}
{x
⎥ 2
≤ x
≤ 5
}
{
x⎥
-3
≤ x
≤ 1 −
2 }
{x
⎥ x
< 3 −
4 or
x >
5}
{
x ⎥
x <
-2
or x
> 1 −
6 }
{
x⎥
x <
3 −
2 or
x >
4}
�
{
x⎥
1 −
3 < x
< 5
}
042_
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PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 4-8
4-8
Cha
pter
4
51
Gle
ncoe
Alg
ebra
2
Skill
s Pr
actic
eQ
uad
rati
c I
neq
ualiti
es
Gra
ph e
ach
ineq
uali
ty.
1. y
≥ x
2 - 4
x +
4
2. y
≤ x
2 - 4
3.
y >
x2 +
2x
- 5
x
y
O4
6 4 2
-2
2
x
y
O4 2
–2
–4
-2
-4
24
x
y
O2
-4
4-
2
-2
-4
-6
Solv
e ea
ch in
equa
lity
by
grap
hing
.
4. x
2 - 6
x +
9 ≤
0
5. -
x2 - 4
x +
32
≥ 0
6.
x2 +
x -
10
> 1
0
xO8 6 4 2
2–2
46
y
x
y
O6
-6
15 5
–5
-2
2
–15
Solv
e ea
ch in
equa
lity
alg
ebra
ical
ly.
7. x
2 - 3
x -
10
< 0
8.
x2 +
2x
- 3
5 ≥
0
9. x
2 - 1
8x +
81
≤ 0
10
. x2 ≤
36
11. x
2 - 7
x >
0
12. x
2 + 7
x +
6 <
0
13. x
2 + x
- 1
2 >
0
14. x
2 + 9
x +
18
≤ 0
15. x
2 - 1
0x +
25
≥ 0
16
. -x2 -
2x
+ 1
5 ≥
0
17. x
2 + 3
x >
0
18. 2
x2 + 2
x >
4
19. -
x2 - 6
4 ≤
-16
x 20
. 9x2 +
12x
+ 9
< 0
x
y O2
6
-6
6-
2-
4-
6
183012
{
x ⎥ -
2 <
x <
5}
{x
⎥ x
≤ -
7 or
x ≥
5}
{
x ⎥
x <
-3
or x
> 0
}
{x⎥
x <
-2
or x
> 1
}
{
x ⎥
x <
0 o
r x
> 7
}
{x ⎥
-6
< x
< -
1}
{
all r
eal n
umbe
rs}
{x
⎥ -
5 ≤
x ≤
3}
{
x ⎥
x =
9}
{x
⎥ -
6 <
x <
6}
{
x ⎥
x <
-4
or x
> 3
}
{x ⎥
-6
≤ x
≤ -
3}
{
all r
eal n
umbe
rs}
�
{
x ⎥
x =
3}
{x
⎥ -
8 ≤
x ≤
4}
{x
⎥ x
< -
5 or
x >
4}
042_
056_
ALG
2_A
_CR
M_C
04_C
R_6
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5.in
dd
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10
9:06
PM
Answers (Lesson 4-8)
A14_A26_ALG2_A_CRM_C04_AN_660785.indd A24A14_A26_ALG2_A_CRM_C04_AN_660785.indd A24 12/20/10 10:56 PM12/20/10 10:56 PM
Copyri
ght
© G
lencoe/M
cG
raw
-Hill
, a
div
isio
n o
f T
he
McG
raw
-Hill
Co
mp
an
ies,
Inc.
PDF Pass
Chapter 4 A25 Glencoe Algebra 2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
4-8
Cha
pter
4
52
Gle
ncoe
Alg
ebra
2
Prac
tice
Qu
ad
rati
c I
neq
ualiti
es
Gra
ph e
ach
ineq
uali
ty.
1. y
≤ x
2 + 4
2.
y >
x2 +
6x
+ 6
3.
y <
2x2 -
4x
- 2
x
y
O2
268 4
4-
2-
4
x
y
O24
-2
-4
-6
-2
-4
x
y O2
2
4-
2
-2
-4
-4
Solv
e ea
ch in
equa
lity
.
4. x
2 + 2
x +
1 >
0
5. x
2 - 3
x +
2 ≤
0
6. x
2 + 1
0x +
7 ≥
0
7. x
2 - x
- 2
0 >
0
8. x
2 - 1
0x +
16
< 0
9.
x2 +
4x
+ 5
≤ 0
10. x
2 + 1
4x +
49
≥ 0
11
. x2 -
5x
> 1
4 12
. -x2 -
15
≤ 8
x
13. -
x2 + 5
x -
7 ≤
0
14. 9
x2 + 3
6x +
36
≤ 0
15
. 9x
≤ 1
2x2
16. 4
x2 + 4
x +
1 >
0
17. 5
x2 + 1
0 ≥
27x
18
. 9x2 +
31x
+ 1
2 ≤
0
19. F
ENCI
NG
Van
essa
has
180
feet
of f
enci
ng t
hat
she
inte
nds
to u
se t
o bu
ild a
rec
tang
ular
pl
ay a
rea
for
her
dog.
She
wan
ts t
he p
lay
area
to
encl
ose
at le
ast
1800
squ
are
feet
. Wha
t ar
e th
e po
ssib
le w
idth
s of
the
pla
y ar
ea?
20. B
USI
NES
S A
bic
ycle
mak
er s
old
300
bicy
cles
last
yea
r at
a p
rofit
of $
300
each
. The
m
aker
wan
ts t
o in
crea
se t
he p
rofit
mar
gin
this
yea
r, bu
t pr
edic
ts t
hat
each
$20
incr
ease
in
pro
fit w
ill r
educ
e th
e nu
mbe
r of
bic
ycle
s so
ld b
y 10
. How
man
y $2
0 in
crea
ses
in p
rofit
ca
n th
e m
aker
add
in a
nd e
xpec
t to
mak
e a
tota
l pro
fit o
f at
leas
t $1
00,0
00?
a
ll re
als
{x
⎥ x
= -
2}
{x
⎥ x
≤ 0
or
x ≥
3 −
4 }
30 ft
to 6
0 ft
al
l rea
ls
{x ⎥
-1
≤ x
≤ 3
} {x
⎥ x
≤ 2
or
x ≥
5}
{
x ⎥
x <
-4
or x
> 5
} {
x ⎥
2 <
x <
8}
�
a
ll re
als
{x
⎥ x
< -
2 or
x >
7}
{x
⎥ -
5 ≤
x ≤
-3}
{
x⎥
x ≠
- 1 −
2 }
{x
⎥ x
≤ 2 −
5 or
x ≥
5}
{
x⎥
-3
≤ x
≤ -
4 −
9 }
from
5 to
10
042_
056_
ALG
2_A
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M_C
04_C
R_6
6078
5.in
dd
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10
9:06
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 4-8
4-8
Cha
pter
4
53
Gle
ncoe
Alg
ebra
2
Wor
d Pr
oble
m P
ract
ice
Qu
ad
rati
c I
neq
ualiti
es
1.H
UTS
The
spa
ce in
side
a h
ut is
sha
ded
in t
he g
raph
. The
par
abol
a is
des
crib
ed
by t
he e
quat
ion
y=
-
4−
5(x
- 1
)2+
4.
y
xO
Wri
te a
n in
equa
lity
that
des
crib
es t
he
shad
ed r
egio
n.
2. D
ISCR
IMIN
AN
TS C
onsi
der
the
equa
tion
ax2 +
bx
+ c
= 0
. Ass
ume
that
th
e di
scri
min
ant
is z
ero
and
that
a is
po
siti
ve. W
hat
are
the
solu
tion
s of
the
in
equa
lity
ax2 +
bx
+ c
≤ 0
?
3. K
IOSK
S C
aleb
is d
esig
ning
a k
iosk
by
wra
ppin
g a
piec
e of
she
et m
etal
wit
h di
men
sion
s x
+ 5
inch
es b
y 4x
+ 8
in
ches
into
a c
ylin
dric
al s
hape
. Ign
orin
g co
st, C
aleb
wou
ld li
ke a
kio
sk t
hat
has
a su
rfac
e ar
ea o
f at
leas
t 44
80 s
quar
e in
ches
. Wha
t va
lues
of x
sat
isfy
th
is c
ondi
tion
?
4. D
AM
S Th
e H
oove
r D
am is
a c
oncr
ete
arch
dam
des
igne
d to
hol
d th
e w
ater
of
Lake
Mea
d. A
t it
s ce
nter
, the
dam
’s he
ight
is a
ppro
xim
atel
y 72
5 fe
et, a
nd
the
dam
var
ies
from
45
to 6
60 fe
et t
hick
. Th
e da
rk li
ne o
n th
is s
ketc
h of
the
cro
ss-
sect
ion
of t
he d
am is
a p
arab
ola.
y
xTh
ickn
ess
Height
Cro
ss-S
ectio
n of
Hoov
er D
am
Dam
Lake
Mea
d
a. W
rite
an
equa
tion
for
the
Hoo
ver
Dam
par
abol
a. L
et t
he h
eigh
t be
the
y-
valu
e of
the
par
abol
a an
d th
e th
ickn
ess
be t
he x
-val
ue o
f the
pa
rabo
la. (
Hin
t: th
e eq
uati
on w
ill b
e in
the
form
: y =
k(x
– m
axim
um
thic
knes
s) +
max
imum
hei
ght.)
b. U
sing
you
r eq
uati
on, g
raph
the
pa
rabo
la o
f the
Hoo
ver
Dam
for
45 ≤
x ≤
660
. Thic
knes
s
Height
x
250
220
440
660
y
500
750
c. E
stim
ate
to t
he n
eare
st fo
ot t
he
thic
knes
s of
the
dam
whe
n th
e he
ight
is
200
feet
.
y =
–0.
0019
2 (x
– 6
60)2 +
725
0 ≤
y ≤
- 4 −
5 (x
- 1
)2 + 4
x =
- b
−
2a
x >
30
(Not
e th
at th
e va
lues
of
x <
-37
res
ult i
n a
high
er p
rodu
ct,
but n
egat
ive
leng
ths
do n
ot m
ake
sens
e.)
353
feet
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Answers (Lesson 4-8)
A14_A26_ALG2_A_CRM_C04_AN_660785.indd A25A14_A26_ALG2_A_CRM_C04_AN_660785.indd A25 12/20/10 10:56 PM12/20/10 10:56 PM
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-Hill C
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PDF Pass
Chapter 4 A26 Glencoe Algebra 2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
4-8
Ch
ap
ter
4
54
Gle
ncoe
Alg
ebra
2
Enri
chm
ent
Gra
ph
ing
Ab
solu
te V
alu
e I
neq
ualiti
es
You
can
solv
e ab
solu
te v
alue
ineq
ualit
ies
by g
raph
ing
in m
uch
the
sam
e m
anne
r yo
u gr
aphe
d qu
adra
tic
ineq
ualit
ies.
Gra
ph t
he r
elat
ed a
bsol
ute
func
tion
fo
r ea
ch in
equa
lity
by u
sing
a g
raph
ing
calc
ulat
or. F
or >
and
≥, i
dent
ify t
he
x-va
lues
, if a
ny, f
or w
hich
the
gra
ph li
es b
elow
the
x-a
xis.
For
< a
nd ≤
, ide
ntify
th
e x
valu
es, i
f any
, for
whi
ch t
he g
raph
lies
abo
ve t
he x
-axi
s.
For
each
ineq
uali
ty, m
ake
a sk
etch
of
the
rela
ted
grap
h an
d fi
nd t
he
solu
tion
s ro
unde
d to
the
nea
rest
hun
dred
th.
1. ⎪
x -
3⎥ >
0
2. ⎪
x⎥ -
6 <
0
3. -
⎪ x +
4⎥ +
8 <
0
4. 2
⎪ x +
6⎥ -
2 ≥
0
5. ⎪
3x -
3⎥ ≥
0
6. ⎪
x -
7⎥ <
5
7.⎪ 7
x -
1⎥>
13
8.⎪ x
- 3
.6⎥≤
4.2
9.⎪ 2
x+
5⎥≤
7
x
> 3
or
x <
3
-6 <
x <
6
-12
< x
< 4
x
≤ -
7 or
x ≥
-5
all
real
num
bers
2
< x
< 1
2
x
< -
1.71
or
x >
2
-0.
6 ≤
x ≤
7.8
-
6 ≤
x ≤
1
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PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DAT
E
P
ER
IOD
Lesson 4-8
4-8
Ch
ap
ter
4
55
Gle
ncoe
Alg
ebra
2
Grap
hing
Cal
cula
tor
Activ
ityQ
uad
rati
c I
neq
ualiti
es
an
d t
he T
est
Men
u
The
ineq
ualit
y sy
mbo
ls, c
alle
d re
lati
onal
ope
rato
rs, i
n th
e T
EST
men
u ca
n be
use
d to
dis
play
th
e so
luti
on o
f a q
uadr
atic
ineq
ualit
y. A
noth
er m
etho
d th
at c
an b
e us
ed t
o fin
d th
e so
luti
on
set
of a
qua
drat
ic in
equa
lity
is t
o gr
aph
each
sid
e of
an
ineq
ualit
y se
para
tely
. Exa
min
e th
e gr
aphs
and
use
the
inte
rsec
t fu
ncti
on t
o de
term
ine
the
rang
e of
val
ues
for
whi
ch t
he
ineq
ualit
y is
tru
e.
Exer
cise
sSo
lve
each
ineq
uali
ty.
1. -
x2 - 1
0x -
21 <
0
2. x
2 - 9
< 0
3.
x2 +
10x
+ 2
5 ≤
0
4. x
2 + 3
x ≤
28
5. 2
x2 + x
≥ 3
6.
4x2 +
12x
+ 9
> 0
7. 2
3 >
-x2 +
10x
8.
x2 -
4x -
13 ≤
0
9. (
x +
1)(x
-3)
> 0
So
lve
x2 +
x ≥
6.
Plac
e th
e ca
lcul
ator
in D
ot m
ode.
Ent
er t
he in
equa
lity
into
Y1.
Th
en t
race
the
gra
ph a
nd d
escr
ibe
the
solu
tion
as
an in
equa
lity.
Key
stro
kes:
Y=
x2
+
2
nd
[TE
ST] 4
6
ZO
OM
4.
Use
TR
AC
E t
o de
term
ine
the
endp
oint
s of
the
seg
men
ts.
Thes
es v
alue
s ar
e us
ed t
o ex
pres
s th
e so
luti
on o
f the
ineq
ualit
y,
{ x |
x ≥
- 3
or
x ≥
2 }.
So
lve
2x2 +
4x -
5 ≤
3.
Plac
e th
e le
ft s
ide
of t
he in
equa
lity
in Y
1 an
d th
e ri
ght
side
in Y
2.
Det
erm
ine
the
poin
ts o
f int
erse
ctio
n. U
se t
he in
ters
ecti
on p
oint
s to
exp
ress
the
sol
utio
n se
t of
the
ineq
ualit
y. B
e su
re t
o se
t th
e ca
lcul
ator
to
Con
nect
ed m
ode.
Key
stro
kes:
Y
= 2
x2
+ 4
—
5 E
NT
ER
3 E
NT
ER
Z
OO
M 6
.
Pres
s 2
nd
[CA
LC] 5
and
use
the
k
ey t
o m
ove
the
curs
or
to t
he le
ft o
f the
firs
t in
ters
ecti
on p
oint
. Pre
ss E
NT
ER
. The
n m
ove
the
curs
or t
o th
e ri
ght
of t
he in
ters
ecti
on p
oint
and
pre
ss E
NT
ER
EN
TE
R. O
ne o
f the
val
ues
used
in t
he s
olut
ion
set
is d
ispl
ayed
. R
epea
t th
e pr
oced
ure
on t
he o
ther
inte
rsec
tion
poi
nt.
The
solu
tion
is {
x | -
3.24
≤ x
≤ 1
.24}
.
Exam
ple
1
Exam
ple
2
[-4.
7, 4
.7] s
cl:1
by
[-3.
1, 3
.1] s
cl:1
[-10
, 10]
scl
:1 b
y [-
10, 1
0] s
cl:1
[-10
, 10]
scl
:1 b
y [-
10, 1
0] s
cl:1
{
x | x
< -
7 or
x >
-3}
{x | -
3 <
x <
3}
{
x | x
= -
5}
{
x | -
7 ≤
x ≤
4 }
{x | x
≤ -
1.5
or x
≥ 1
}
{x |
x ≠
-1.
5}
{
x | x
≤ 3
.59
or x
≥ 6
.41}
{x | -
2.12
≤ x
≤ 6
.12}
{x
| x <
-1
or x
> 3
}
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Answers (Lesson 4-8)
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Chapter 4 Assessment Answer Key
An
swer
s
Copyri
ght
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lencoe/M
cG
raw
-Hill
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-Hill
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Chapter 4 A27 Glencoe Algebra 2
Quiz 1 (Lessons 4-1 and 4-2) Quiz 3 (Lessons 4-5 and 4-6) Mid-Chapter TestPage 59 Page 60 Page 61
1.
2.
3.
4.
5.
xO
f(x)
(0, -3)(-1, -4)
f(x) = x2 + 2x - 3
x = -1
4
2
2 4
–2
–4
-4 -2
-3; x = -1; -1
between 1 and 2;between -6 and -5
A
3, -1
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
{-5, 2 _ 3 }
{-9, 5}
C
x 2 + 4x - 12 = 0
3 x 2 + 10x - 8 = 0
4i √ � 5
-6 √ � 2
68 + 4i
-11 + 3i
1 _ 2 + 1
_ 2 i
1.
2.
3.
4.
5.
{-10, 2}
1 ± 3 √ � 5
B
2 ± √ � 5
-96; 2 complex roots
xO
y
(2, -1)
x = 24
2
2
–2
–4
-4 -2
y
xO
4
2
2 4
–2
-4 -2
y = 2 (x + 5) 2
1.
2.
3.
4.
5. A
{x ⎪1 < x < 5}
1.
2.
3.
4.
5.
B
H
A
F
D
y
xO
4
2
2 4-4 -2
8.
6.
7.
8.
9.
10.
1, 3
minimum, -9 1 _ 2
{-2, 9}
{0, 1 _ 4 }
- 25
_ 34
+ 15i
_ 34
Quiz 2 (Lessons 4-3 and 4-4)
Page 59
Quiz 4 (Lessons 4-7 and 4-8)
Page 60
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e M
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-Hill C
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Chapter 4 A28 Glencoe Algebra 2
Vocabulary Test Form 1Page 62 Page 63 Page 64
false; complex conjugates
false; constant termfalse; quadratic
inequality
false; roots
true
false; minimum value
false; quadratic term
true
true
false; discriminant
Sample answer: The value that determines the roots of a quadratic equation.
Sample answer: a quadratic equation written as ax2 + bx + c = 0
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
B
G
A
H
B
H
B
H
A
F
1.
2.
3.
4.
5.
6.
7.
8.
9.
10. 1 and 7; 14
B
J
D
B
G
F
C
J
B
G
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
B:
A27_A36_ALG2_A_CRM_C04_AN_660785.indd A28A27_A36_ALG2_A_CRM_C04_AN_660785.indd A28 12/20/10 10:56 PM12/20/10 10:56 PM
Chapter 4 Assessment Answer Key
An
swer
s
Copyri
ght
© G
lencoe/M
cG
raw
-Hill
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McG
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-Hill
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Chapter 4 A29 Glencoe Algebra 2
Form 2A Form 2BPage 65 Page 66 Page 67 Page 68
C
G
A
H
D
G
C
F
D
F
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
B
G
C
J
D
B
F
C
J
F
Sample answer:16x2 + 3 = 0
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
B:
B
H
B
J
A
H
A
J
A
H
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
B:
C
G
D
F
B
G
D
H
A
F
Sample answer:
9x 2 + 2 = 0
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Chapter 4 A30 Glencoe Algebra 2
Form 2CPage 69 Page 70
xO
f (x)
(3, 0)
(1, 5)
f(x) = -5x2 + 10x
x = 1
xO
4
2
2 4
–2
–4
-4 -2
y
maximum; 4
2, 4
{-3, 2 _ 5 }
9 in. by 16 in.
6 + 12j ohms
37
_ 17
- 22
_ 17
j amps
4x2 + 21x - 18 = 0
{-8, 2}
1.
2.
3.
4.
5.
6.
7.
8.
9.
y
xO 2 4-2-4
4
2
–2
–4
{-2 ± √ �� 13 }
{-2, 1 _ 2 }
3 ± i √ �� 31
_
10
0; 1 real, rational root
33; 2 real, irrational roots
(-5, -7); x = -5; down
y = 3
_ 2 (x - 2)2 - 1
h(t ) = -16(t - 1.5)2
+ 51; 51ft
y = (x - 3)2 - 1
{x ⎪x ≤ - 1 _ 2 or x ≥ 3}
9x 2 - 7 = 0
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
B:
A27_A36_ALG2_A_CRM_C04_AN_660785.indd A30A27_A36_ALG2_A_CRM_C04_AN_660785.indd A30 12/20/10 10:56 PM12/20/10 10:56 PM
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swer
s
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ght
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lencoe/M
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Chapter 4 A31 Glencoe Algebra 2
Form 2DPage 71 Page 72
y
xO
4
2
2 4
–2
–4
-4 -2
xO
f(x)
(0, 3)
(2, -1)
x = 2
4
2
2 4
–4
-4 -2
f(x) = x2 - 4x
+ 3
minimum; -17
1, -3
{-1, 4 _ 3 }
8 in. by 18 in.
9 - 6j ohms
23
_ 17
- 27
_ 17
j ohms
2x 2 + 5x - 12 = 0
1.
2.
3.
4.
5.
6.
7.
8.
y
xO
4
2
2 4
–2
-4 -2
{4 ± √ � 2 }
{-1, 2 _ 3 }
{ 9 ± √ �� 41
_
4 }
0; 1 real, rational root
-8; 2 complex roots
(6,-5); x = 6; down
y = - 1 _ 4 (x + 4)2 + 2
y = (x + 2)2 + 4
h(t ) = -16(t - 2)2 + 76; 76 ft
{x ⎪- 3 _
2 < x < 5}
16x 2 - 5 = 0
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
B:
{ -2 ± √ � 6
_ 3 }
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Chapter 4 A32 Glencoe Algebra 2
Form 3Page 73 Page 74
y
x
6
2
2 6-6 -2O
y
x
6
2
2 6-6 -2O
xO
f (x)
(0, 3)(- ),1
383
f(x) = 3x2 + 2x + 3
-13x =
4
2
2 4
–2
-4 -2
minimum; 22
_ 25
$8.00; $6400
3, 6
between -3 and -2;
between 4 and 5
{ 1 _ 2 ,
5
_ 3 }
12x2 - 13x - 14 = 0
4 - 6i
- 1 _ 9 -
4 √ � 5
_ 9 i
1.
2.
3.
4.
5.
6.
7.
8.
9.
y
xO 2 4-4
6
4
2
-2
-2 < k < 2
{-0.35, 0.85}
{ 5 ± i √ �� 39
_
8 }
{-3.5, 1}
6 ± 4 √ � 2
1.2; two real, irrational roots
h(t ) = -9.1(t - 32.5)2
+ 30,000; 30,000 ft
y = - 29 _
200 (x + 9)2 +
29
_ 2
{x ⎪x ≤ - 7 _ 2 or x = 1}
16x 2 + 24x + 29 = 0
y = - 3
_ 5 (x + 7
_ 2 )
2
- 1 _ 2 ;
(- 7 _ 2 , - 1
_ 2 ) ; x = -
7
_ 2 ; down
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
B:
A27_A36_ALG2_A_CRM_C04_AN_660785.indd A32A27_A36_ALG2_A_CRM_C04_AN_660785.indd A32 12/20/10 10:56 PM12/20/10 10:56 PM
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swer
s
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ght
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lencoe/M
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-Hill
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-Hill
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Chapter 4 A33 Glencoe Algebra 2
Score General Description Specifi c Criteria
4 SuperiorA correct solution that is supported by well-developed, accurate explanations
• Shows thorough understanding of the concepts of graphing, analyzing, and finding the maximum and minimum values of quadratic functions; solving quadratic equations; computing with complex numbers; and solving inequalities.
• Uses appropriate strategies to solve problems. • Computations are correct. • Written explanations are exemplary. • Goes beyond requirements of some of or all problems.
3 SatisfactoryA generally correct solution, but may contain minor flaws in reasoning or computation
• Shows an understanding of the concepts of graphing, analyzing, and finding the maximum and minimum values of quadratic functions; solving quadratic equations; computing with complex numbers; and solving inequalities.
• Uses appropriate strategies to solve problems. • Computations are mostly correct. • Written explanations are effective. • Satisfies all requirements of problems.
2 Nearly SatisfactoryA partially correct interpretation and/or solution to the problem
• Shows an understanding of most of the concepts of graphing, analyzing, and finding the maximum and minimum values of quadratic functions; solving quadratic equations; computing with complex numbers; and solving inequalities.
• May not use appropriate strategies to solve problems. • Computations are mostly correct. • Written explanations are satisfactory. • Satisfies the requirements of most of the problems.
1 Nearly Unsatisfactory A correct solution with no supporting evidence or explanation
• Final computation is correct. • No written explanations or work is shown to substantiate
the final computation. • Satisfies minimal requirements of some of the problems.
0 UnsatisfactoryAn incorrect solution indicating no mathematical understanding of the concept or task, or no solution is given
• Shows little or no understanding of most of the concepts of graphing, analyzing, and finding the maximum and minimum values of quadratic functions; solving quadratic equations; computing with complex numbers; and solving inequalities.
• Does not use appropriate strategies to solve problems. • Computations are incorrect. • Written explanations are unsatisfactory. • Does not satisfy the requirements of problems. • No answer may be given.
Page 75, Extended-Response Test Scoring Rubric
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Chapter 4 Assessment Answer KeyC
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Chapter 4 A34 Glencoe Algebra 2
1. Student responses should indicate that using the Square Root Property, as Mi-Ling’s group did, would take less time than the other method since the equation is already set up as a perfect square set equal to a constant. To solve using the other method, the binomial would need to be expanded and the constant on the right brought to the left side of the equal sign.
2a. Jocelyn had trouble because the problem is impossible. No such parabola exists.
2b. Student responses will vary. One of the three conditions must be omitted or modified. Sample answer: “...that passes through (-1, -12).”
2c. Answers will vary and depend on the answer for part b. For example, for the sample answer in part b above, a possible equation is: y = -2(x + 3)2 - 4.
3a. Answer must be of the form y = a(x - h)2 + 8 where h is any real number and a < 0.
3b. Answers must be of the form y = a[x - (h + n)]2 + 8 where h and a represent the same values as in part a. The student choice is for the value of n. The student should indicate that the graph will shift to the left n units if his or her value of n is negative, but will shift the graph to the right n units if the chosen value of n is positive.
4. Students should indicate that Joseph’s answer is not correct. In Step 2, when he completed the square by adding 9 inside the parentheses, he actually added 2(9) = 18 to the right side of the equation, so he must subtract 18 from the constant on the same side, rather than add 9, to keep the statements equivalent. The correct solution is f(x) = 2(x + 3)2 - 23.
5a. > The graph is strictly above the x-axis for all values of x other than 2.
5b. <; The graph is never below the x-axis.
5c. ≥; The graph is always on or above the x-axis.
In addition to the scoring rubric found on page A33, the following sample answers may be used as guidance in evaluating open-ended assessment items.
Page 75, Extended-Response Test Sample Answers
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Chapter 4 Assessment Answer Key
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Chapter 4 A35 Glencoe Algebra 2
Standardized Test PracticePage 76 Page 77
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6. F G H J
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Chapter 4 Assessment Answer KeyC
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Chapter 4 A36 Glencoe Algebra 2
Standardized Test PracticePage 78
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