Answers to Chapter 4 student companions

Quadratic Functions
and Transformations
4-1
Vocabulary
Review
1. Circle the vertex of each absolute value graph.
y
y
y
x
x
x
Vocabulary Builder
parabola
y
y
Related Words: vertex, axis of symmetry,
quadratic function
x
vertex
Definition: A parabola is the graph of a
quadratic function, a function of the form
y 5 ax 2 1 bx 1 c.
x
axis of
symmetry
Main Idea: A parabola is symmetrical around
its axis of symmetry, a line passing through
the vertex. A parabola can open upward or downward.
Use Your Vocabulary
2. Circle each function whose graph is a parabola.
y 5 26x 1 9
y 5 22x2 2 15x 2 18
y 5 x2 1 4x 1 4
y 5 16x 2 22
3. Cross out the function(s) whose graph is NOT a parabola.
y 5 5x 2 2 3x 1 6
Chapter 4
y5x23
y 5 2x 2 1 6x 2 7
82
y 5 0.2x 1 7
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parabola (noun) puh RAB uh luh
Problem 2 Graphing Translations of f(x) 5 x2
Got It? Graph g(x) 5 x 2 1 3. How is it a translation of f(x) 5 x 2 ?
Use the graphs of f(x) 5 x 2 and g(x) 5 x 2 1 3 at the right for
Exercises 4 and 5.
6
4
4. Circle the ordered pairs that are solutions of g(x) 5 x 2 1 3.
Underline the ordered pairs that are solutions of f(x) 5 x 2 .
(23, 0)
(23, 9)
(0, 0)
(3, 0)
(3, 9)
2
(0, 23)
(0, 3)
4
(0, 9)
y
2
O
x
2
4
2
(3, 12)
5. Underline the correct word to complete each sentence.
For each value of x, the value of g(x) 5 x 2 1 3 is 3 more / less than
the value of f(x) 5 x 2 .
The graph of g(x) 5 x 2 1 3 is a translation 3 units up / down of the
graph of f(x) 5 x 2 .
The graph shows f(x) 5 x 2 in red / blue and g(x) 5 x 2 1 3 in red / blue .
Problem 3 Interpreting Vertex Form
Got It? What are the vertex, axis of symmetry, minimum or maximum, and
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domain and range of the function y 5 22(x 1 1)2 1 4?
6. Compare y 5 2(x 2 1)2 1 4 with the vertex form y 5 a(x 2 h)2 1 k. Identify
a, h, and k.
h 5 21
a 5 22
k5
4
7. The vertex of the parabola is (h, k) 5 Q 21 , 4 R .
8. The axis of symmetry is the line x 5 21 .
9. Underline the correct word or symbol to complete each sentence.
Since a is , / . 0, the parabola opens upward / downward .
The parabola has a maximum / minimum value of 4 when x 5 21 .
10. Circle the domain.
all real numbers
x # 21
x#4
x$4
x # 21
x#4
x$4
11. Circle the range.
all real numbers
83
Lesson 4-1
Translation of the Parabola
12. Use one of the functions below to label each graph.
y 5 (x 1 3)2
y 5 x2 2 1
y 5 (x 2 2)2 1 3
y
y
y
x
y
x
y 5 (x 1 1)2 2 2
Problem 4
y 5 (x 1 1)2 2 2
x
y 5 x2 2 1
y 5 (x 1 3)2
x
y 5 (x 2 2)2 1 3
Using Vertex Form
Got It? What is the graph of f (x) 5 2(x 1 2)2 2 5?
13. Multiple Choice What steps transform the graph of y 5 x 2 to y 5 2(x 1 2)2 2 5?
Stretch by the
factor 2 and
translate 2 units
to the right and
5 units up.
Stretch by the
factor 2 and
translate 2 units
to the left and
5 units down.
Reflect across the
x-axis, stretch by
the factor 2, and
translate 2 units to
the left and
5 units down.
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Reflect across the
x-axis, stretch by the
factor 2, and translate
2 units to the left and
5 units up.
14. Circle the graph of f(x) 5 2(x 1 2)2 2 5.
y
Problem 5
y
x
y
x
y
Writing a Quadratic Function in Vertex Form
Got It? The graph shows the jump of a dolphin. The axis of symmetry
is x 5 2, and the height is 7. If the path of the jump passes through the
point (5, 5), what quadratic function models the path of the jump?
6
4
16. Substitute h and k in the vertex form f (x) 5 a(x 2 h)2 1 k.
2
Chapter 4
2
2 R 1
7
0
84
y
8
15. The vertex is Q 2 , 7 R .
y 5 aQx 2
x
x
x
0
2
4
6
8
17. Substitute (5, 5) for (x, y) and solve for a.
y 5 a(x 2 2)2 1 7
5 5 a(5 2 2)2 1 7
5 5 9a 1 7
22 5 9a
a 5 229
18. Write the quadratic function that models the path of the water.
f(x) 5 229(x 2 2)2 1 7
Lesson Check • Do you UNDERSTAND?
Vocabulary When does the graph of a quadratic function have a minimum value?
19. Circle the parabola that has a minimum value.
y
y
x
x
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20. The graph of y 5 x 2 is a parabola that opens upward / downward .
21. The graph of y 5 2x 2 is a parabola that opens upward / downward .
22. When does the graph of a quadratic function have a minimum value?
Answers may vary. Sample: The graph of a quadratic function has a
_______________________________________________________________________
minimum value when a S 0 and the parabola opens upward.
_______________________________________________________________________
Math Success
Check off the vocabulary words that you understand.
parabola
vertex form
quadratic function
axis of symmetry
Rate how well you can graph a quadratic function in vertex form.
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Lesson 4-1
Standard Form of a
Quadratic Function
4-2
Vocabulary
Review
1. Circle the functions in standard form.
y 5 2x2 2 4x 1 2
y 5 13 (x 2 4)
y 5 24x 1 1
5
y 5 223 x 2 3 x 1 4
Write each equation in standard form.
2. x 1 2y 5 17
3. 2x 5 5
y 5 212 x 1 17
2
4. 5 2 x 5 y 1 2
2x 2 5 5 0
y 5 2x 1 3
Vocabulary Builder
Standard Form of a
Quadratic Function
quadratic (adjective) kwah DRAT ik
y ax2 bx c,
Related Words: parabola, vertex, axis of symmetry
Examples: quadratic functions, y 5 x2 , y 5 23x2 1 7, f (x) 5 2x2 1 5x 2 4,
g(x) 5 12 (x 2 4)2 1 5
Nonexamples: not quadratic functions, y 5
2x2
x2 1 5x 1 10
1
,
3x
1 4x 1 5
Use Your Vocabulary
5. Circle the graphs of quadratic functions.
y
y
y
x
x
x
Chapter 4
y
86
x
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a0
Definition: A quadratic function is a function that can be
written in the form y 5 ax2 1 bx 1 c where a 2 0. The
graph of a quadratic function is a parabola.
Problem 1 Finding the Features of a Quadratic Function
Got It? What are the vertex, axis of symmetry, maximum and minimum values,
and range of y 5 23x2 2 4x 1 6?
6. Circle the graph of y 5 23x2 2 4x 1 6.
maximum,
+ 23 , 7 13 ,
axis of
symmetry,
x 2
3
7. Draw and label the axis of symmetry on the graph you circled in Exercise 6.
8. Circle and label the maximum or minimum value on the graph.
9. Circle the range of the function.
y $ 5.0
y # 6.0
all real numbers # 7.3
all real numbers # 9.2
Properties Quadratic Function in Standard Form
• The graph of f (x) 5 ax2 1 bx 1 c, a 2 0, is a parabola.
• If a . 0, the parabola opens upward. If a , 0, the parabola opens downward.
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b
• The axis of symmetry is the line x 5 22a
.
b
b
• The x-coordinate of the vertex is 22a
. The y-coordinate of the vertex is f Q 22a
R.
• The y-intercept is (0, c).
10. The y-intercept of the graph of f (x) 5 5x2 2 3x 2 4 is Q 0 , 24 R .
Problem 2 Graphing a Function of the Form y 5 ax2 1 bx 1 c
Got It? What is the graph of y 5 22x 2 1 2x 2 5?
b
11. The axis of symmetry is x 5 22a 5 2
2
5
2 ? 22
1
2
8
.
4
12. Substitute to find the y-coordinate of the vertex.
2
f Q 12 R 5 22 Q 12 R 1 2 Q 12 R 2 5 5 2412
1
4
1
y
2
O
13. The vertex is Q 2 , 242 R
4
14. The y-intercept is 25 . The reflection of the y-intercept across
8
x
2
4
the axis of symmetry is Q 1 , 25 R .
15. Plot the points from Exercises 13 and 14. Draw a smooth curve.
87
Lesson 4-2
Problem 3
Converting Standard Form to Vertex Form
Got It? What is the vertex form of y 5 2x2 1 4x 2 5?
16. Use the justifications at the right to find the vertex.
y 5 Q 21 R x2 1 Q 4 R x 1 Q 25 R
4
b
x 5 22a 5 2
2 ? 21
Write the function in the form y 5 ax2 1 bx 1 c.
5 2
Find the x-coordinate of the vertex.
y 5 21(4) 1 4(2) 2 5
Substitute the x-coordinate value into the equation and simplify.
y 5 21
17. The vertex is Q 2 , 21 R .
18. Use the general form of the equation, y 5 a(x 2 h)2 1 k. Substitute for a, h, and k.
2
y 5 21 B x 2 Q 2 R R 1 Q 21 R
y 5 2(x 2 2)2 2 1
19. The vertex form of the function is
Problem 4
.
Interpreting a Quadratic Graph
Got It? The Zhaozhou Bridge in China is the oldest known arch bridge, dating to 605 a.d.
You can model the support arch with the function f (x) 5 20.001075x2 1 0.131148x,
where x and y are measured in feet. How high is the arch above its supports?
Answers may vary. Sample: the middle, the vertex, the maximum
_______________________________________________________________________
21. Find the x-coordinate of the vertex.
0.131148
b
x 5 22a 5 2
2?
20.001075
5 61
22. The axis of symmetry of the parabola is x 5 61 .
23. The length of the bridge is
122
ft.
24. Use the x-coordinate of the vertex to find the y-coordinate.
y 5 20.001075(61)2 1 0.131148(61)
5 4.0000
25. The vertex is about Q 61 , 4 R , so the arch is 4 feet above its support.
Chapter 4
88
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20. What point on the parabola gives the height of the arch above its supports?
Lesson Check • Do you UNDERSTAND?
Error Analysis A student graphed the function
y 5 2x2 2 4x 2 3. Find and correct the error.
x = -4 = -1
b
b
26. The vertex of y 5 ax2 1 bx 1 c is Q22a , f Q22a RR .
Find the x- and y-coordinates of the vertex of
y5
2x2
2 4x 2 3.
x-coordinate:
2224
? 4
6
2(2)
y = 2(-1)2 - 4(-1) - 3
=2+4-3
=3
vertex (-1, 3)
(1, 3) 2
4 2
0
2
51
y-coordinate:
2(1)2
2 4(1) 2 3 5 25
27. Find the y-intercept of y 5 2x2 2 4x 2 3.
2(0)2 2 4(0) 2 3 5 23
y
28. Describe the student’s error and graph the function correctly.
Answers may vary. Sample: The student miscalculated the
________________________________________________________
4
x-coordinate of the vertex.
________________________________________________________
2
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2
O
x
2
4
2
4
Math Success
Check off the vocabulary words that you understand.
quadratic
standard form
vertex
axis of symmetry
y-intercept
Rate how well you can graph quadratic functions written in standard form.
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Lesson 4-2
Modeling With
Quadratic Functions
4-3
Vocabulary
Review
1. Cross out the graphs that are NOT parabolas.
y
y
y
x
y
x
x
x
Vocabulary Builder
model (verb)
MAH
dul
Definition: A function or equation models an action or relationship by describing its
behavior or the data associated with that relationship.
Example: The equation a 5 3g models the relationship between the number of
apples, a, and the number of oranges, g, when the number of apples is triple the
number of oranges.
Use Your Vocabulary
Draw a line from each description in Column A to the equation that models it in
Column B.
Column A
Column B
2. The string section of the orchestra has twice
as many violins as cellos.
y 5 2x 1 1
3. There are two eggs per person with one
extra for good measure.
y 5 100 2 2x
4. There were 100 shin guards in the closet, and
each player took two.
y 5 2x
Chapter 4
90
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Main Idea: Modeling is a way of using math to describe a real-world situation.
Problem 1 Writing an Equation of a Parabola
Got It? What is the equation of a parabola containing the points (0, 0), (1, 22),
and (21, 24)?
5. Substitute the three points one at a time into y 5 ax 2 1 bx 1 c to write a system
of equations.
2
0 5 aQ 0 R 1 b Q 0 R 1 c
Use (0, 0).
2
Use (1, 22).
22 5 a Q 1 R 1 b Q 1 R 1 c
Use (21, 24).
24 5 a Q 21 R 1 b Q 21 R 1 c
2
6. Solve the system of equations.
Answers may vary. Sample:
c50
m
22 5 a 1 b
24 5 a 2 b
m
26 5 2a
23 5 a
22 5 23 1 b
15b
m
7. The equation of the parabola is y 5 23 x 2 1 1 x 1 0 .
Problem 2 Using a Quadratic Model
Got It? The parabolic path of a thrown ball can be modeled by the table. The top
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of a wall is at (5, 6). Will the ball go over the wall? If not, will it hit the wall on the way
up, or the way down?
8. Circle the system of equations you find by substituting the three given points that
are on the parabola.
1 5 9a 1 3b 1 c
2 5 25a 1 5b 1 c
3 5 36a 1 6b 1 c
35a1b1c
5 5 2a 1 2b 1 c
6 5 9a 2 3b 1 c
x
y
1
3
2
5
3
6
35a1b1c
5 5 4a 1 2b 1 c
6 5 9a 1 3b 1 c
9. Now, solve the system of equations.
Answers may vary.
c532a2b
a 5 212
Sample:
5 5 4a 1 2b 1 3 2 a 2 b
b 5 32 1 2 5 72
c 5 3 1 12 2 72 5 0
b 5 23a 1 2 m c 5 2a 1 1
6 5 9a 2 9a 1 6 1 2a 1 1
1
10. The solution of the system is a 5 22 , b 5
11. The quadratic model for the ball’s path is
7
2
,c 5
0 .
y 5 212x2 1 72x
.
12. How can you determine whether the ball goes over the wall? Place a ✓ if the
statement is correct. Place an ✗ if it is not.
✓
The value of the model at x 5 5 is at least 6.
✗
The value of the model at x 5 6 is at least 5.
91
Lesson 4-3
13. Will the ball go over the wall? Explain.
Answers may vary. Sample: No. The value of the model at x 5 5 is 5,
_______________________________________________________________________
which is less than the height of the wall.
_______________________________________________________________________
14. The value of the model at x 5 6 is less than / greater than value of the model at
x 5 5, therefore the ball was on its way down / up as it approached the wall.
Problem 3
Using Quadratic Regression
Got It? The table shows a meteorologist’s predicted temperatures
for a summer day in Denver, Colorado. What is a quadratic model
for the data? Predict the high temperature for the day. At what time
does the high temperature occur?
15. Using the LIST feature on a graphing calculator, identify the data that
you will enter.
L1 5
time
L2 5
predicted temperature
Time
Predicted
Temperature (°F)
6 A.M.
63
9 A.M.
76
12 P.M.
86
3 P.M.
89
6 P.M.
85
9 P.M.
76
6 a.m.:
6
3 a.m.:
15
9 a.m.:
9
6 p.m.:
18
12 p.m.:
12
9 p.m.:
21
17. Circle the calculator screen that shows the correct data entry.
L1
6
9
12
3
6
9
L2(6)76
L2
63
76
86
89
85
76
L3
L1
6
9
12
15
18
21
L2
63
76
86
89
85
76
L3
L2(6)63
L2(6)76
18. Enter the data from the table into your calculator. Use
the QuadReg function. Your screen should look like the
one at the right.
Write the quadratic model for temperature.
y 5 20.33x2 1 9.8x 1 15.6
Chapter 4
L1
6
9
12
15
18
21
92
QuadReg
y = ax 2bxc
a = –.329365
b = 9.797619
c = 15.571429
L2
76
86
89
85
76
63
L3
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16. Using a 24-hour clock, write the values for the L1 column.
19. Use your calculator to find the maximum value of the model. The vertex of the
parabola is Q 14.87 , 88.43 R .
20. The high temperature will be
88.43 °F.
21. At what time will the high temperature occur?
just before 3 P.M.
_______________________________________________________________________
Lesson Check • Do you UNDERSTAND?
y
Error Analysis Your classmate says he can write the equation
of a quadratic function that passes through the points (3, 4),
(5, 22), and (3, 0). Explain his error.
4
22. Graph the points (3, 4), (5, 22), and (3, 0).
2
23. Underline the correct words to complete the rule for finding
a quadratic model.
Two / Three noncollinear points, no two / three of which
are in line horizontally / vertically , are on the graph of exactly
2
O
x
2
4
6
2
one quadratic function.
24. What is your classmate’s error?
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The points (3, 4) and (3, 0) are on the same vertical line.
_______________________________________________________________________
_______________________________________________________________________
_______________________________________________________________________
Math Success
Check off the vocabulary words that you understand.
model
quadratic model
Rate how well you can use a quadratic model.
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Lesson 4-3
Factoring Quadratic
Expressions
4-4
Vocabulary
Review
1. Complete each factor tree.
54
24
2
9
12
3
4
2
3
6
3
3
2
2
Vocabulary Builder
factor (noun)
FAK
tur
expression
factors
Other Word Forms: factor (verb)
Main Idea: The factors of an expression are similar to the factors of a number.
Definition: The factors of a given expression are expressions whose product equals
the given expression. When you factor an expression, you break it into smaller
expressions whose product equals the given expression.
Example: The factors of the expression 2x 2 2 x 2 10 are 2x 2 5 and x 1 2.
Use Your Vocabulary
2. Circle the prime factors of 24xy.
24 ? x ? y
2?4?x?y
23 ? 3 ? x ? y
5 ? 4 ? a2 ? b
2 ? 33 ? a2 ? b
3. Circle the prime factors of 54a 2b.
54 ? a2 ? b
Chapter 4
94
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2x 2 x 10 (2x 5)(x 2)
Problem 1 Factoring ax 2 1 bx 1 c when a 5 t1
Got It? What is the expression x 2 1 14x 1 40 in factored form?
4. Complete the factor table. Then circle the pair of factors whose sum is 14.
Factors of 40
1, 40
2, 20
4, 10
5, 8
41
22
14
13
Sum of Factors
5. Circle the expression written as the product of two binomials.
(x 1 1)(x 1 40)
(x 1 2)(x 1 20)
(x 1 4)(x 1 10)
(x 1 5)(x 1 8)
Got It? What is the expression x 2 2 11x 1 30 in factored form?
6. Underline the correct word(s) to complete each sentence.
I need to find factors that multiply / sum to 30 and multiply / sum to 211.
At least one of the factors that sum to 211 must be positive / negative .
The two factors that multiply to 30 must both be positive / negative .
7. Circle the factors of 30 that sum to 211.
1 and 30
2 and 15
3 and 10
5 and 6
21 and 230
22 and 215
23 and 210
25 and 26
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8. Factor the expression.
x 2 2 11x 1 30 5 Q x 2
5 RQ x 2
6 R
Got It? What is the expression 2x 2 1 14x 1 32 in factored form?
9. Rewrite the expression to show a trinomial with a leading coefficient 1.
2x 2 1 14x 1 32 5
(21)(x 2 2 14x 2 32)
10. Reasoning You are looking for factors of 232 that sum to 214. Which of the factors
has the greater absolute value, the negative factor or the positive factor? How do
you know?
Answers may vary. Sample: Since the sum is negative, the negative
_______________________________________________________________________
factor must have the greater absolute value.
_______________________________________________________________________
11. Circle the factors of 232 that sum to 214.
21 and 32
22 and 16
24 and 8
1 and 232
2 and 216
4 and 28
12. Write the factored form of the expression.
2x 2 1 14x 1 32 5 (x 1 2)(x 2 16)
95
Lesson 4-4
Finding Common Factors
Problem 2
Got It? What is the expression 7n2 2 21 in factored form?
13. The GCF of 7n2 and 21 is 7 .
14. Use the Distributive Property to factor the expression.
7n2 1 21 5
7 Q n2 1 3 R
Factoring ax 2 1 bx 1 c when »a… u 1
Problem 3
Got It? What is the expression 4x 2 1 7x 1 3 in factored form? Check your answers.
15. Complete the diagram below.
4x 2 7x 3
4
12
3
r
16. Complete the factor pairs of ac. Then circle the pair that sums to 7.
Q 1, 12 R
Q 2, 6 R
Q 3, 4 R
17. Use your answer to Exercise 16 to complete the diagram below.
7x
4x
4x2 (4x)
3x
3x 3
3
4x + x 1 , 3 + x 1 ,
3
3
The expressions inside the parentheses must be equal.
Use the Distributive Property to factor out
the GCF, the part inside the parentheses.
+4x 3, + x
Problem 4
1 ,
Factoring a Perfect Square Trinomial
Got It? What is 64x 2 2 16x 1 1 in factored form?
18. Circle the form your answer will have.
(a 1 b)2
Chapter 4
(a 2 b)2
96
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4x 2
19. Use the justifications to complete each step.
64x 2 2 16x 1 1
Write the original expression.
2
Q 8 x R 2 16x 1 Q 1 R
2
Write the first and third terms as squares.
2
Q 8 x R 2 2 Q 8 RQ 1 R x 1 Q 1 R
2
Write the middle term as (2ac)x.
20. Write the expression as the square of a binomial.
(8x 2 1)2
Lesson Check • Do you UNDERSTAND?
Reasoning Explain how to rewrite the expression a2 2 2ab 1 b 2 2 25 as the
product of two trinomial factors. (Hint: Group the first three terms. What type of
expression are they?)
21. Complete: The first three terms of the expression are a 9.
perfect square trinomial
difference of two squares
22. Factor the first three terms of the expression.
(a 2 b)2
23. Rewrite the original expression using the factored form of the first three terms.
(a 2 b)2 2 25
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24. Complete: The expression you wrote in Exercise 23 is a 9.
perfect square trinomial
difference of two squares
25. Circle the expression written as the product of two trinomial factors.
a2 2 2ab 1 b2
(a 2 b)2 2 25
(a 2 b)(225)
(a 2 b 2 5)(a 2 b 1 5)
Math Success
Check off the vocabulary words that you understand.
factor of an expression
perfect square trinomial
difference of two squares
Rate how well you can factor quadratic expressions.
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Lesson 4-4
Quadratic Equations
4-5
Vocabulary
Review
1. Cross out the equation below that is not a function.
f (x) 5 2x 2 7
y 2 5 3x 2 2 4x
y 5 2x 2 1 14x 2 7
g(x) 5 u x 3 u
Vocabulary Builder
zero of a function (noun)
oh
ZEER
Main Idea: Wherever the graph of a function y 5 f (x) intersects the x axis, f (x) 5 0.
The value of x at any of these intersection points is called a zero of the function.
Definition: A value of x for which f (x) 5 0 is a zero of the function f (x) .
Use Your Vocabulary
Write the zero(s) of each function.
y
2.
y
3.
2
2
Zero(s):
4.
2
x
O
2
x
O
2
Zero(s):
Key Concept
2
2
22
Zero(s):
Zero-Product Property
If ab 5 0, then a 5 0 or b 5 0.
Example: If (x 1 7)(x 2 2) 5 0, then (x 1 7) 5 0 or (x 2 2) 5 0.
5. If either x 1 7 5 0 or x 2 2 5 0, circle all of the possible values of x.
27
Chapter 4
2
22
2
98
27
O
y
x
2
21, 1
Copyright © by Pearson Education, Inc. or its affiliates. All Rights Reserved.
Example: x 5 2 is a zero of f (x) 5 3x 2 6, because f (2) 5 3(2) 2 6 5 0.
Problem 1 Solving Quadratic Equations by Factoring
Got It? What are the solutions of the quadratic equation x 2 2 7x 5 212?
6. The equation is solved below. Write a justification for each step.
x 2 2 7x 5 212
Write the original equation.
x 2 2 7x 1 12 5 0
Write in standard form.
(x 2 3)(x 2 4) 5 0
x 2 3 5 0 or x 2 4 5 0
x 5 3 or x 5 4
Factor the expression.
Use the Zero-Product Property
Solve for x.
Problem 2 Solving Quadratic Equations With Tables
Got It? What are the solutions of the quadratic equation 4x 2 2 14x 1 7 5 4 2 x?
7. Write the equation in standard form.
4
x2 1
213 x 1
50
3
8. Enter the equation into your calculator. Use the results to complete the table below.
Copyright © by Pearson Education, Inc. or its affiliates. All Rights Reserved.
TABLE SETUP
TblStart = 0
$Tbl = 1
IndPnt: Auto Ask
DePend: Auto Ask
x
0
1
2
3
4
y1
3
6
7
0
15
9. Based on the table, one solution of the equation is x 5
3 .
10. Another solution occurs between 0 and 1 . Change the x-interval to 0.05.
Complete the table.
TABLE SETUP
TblStart = 0
$Tbl = .05
IndPnt: Auto Ask
DePend: Auto Ask
x
0.1
0.15
0.2
0.25
0.3
y1
1.74
1.14
0.56
0
0.54
11. Based on the table, the other solution to the equation is x 5
99
0.25 .
Lesson 4-5
Solving a Quadratic Equation by Graphing
Problem 3
Got It? What are the solutions of the quadratic equation x 2 1 2x 2 24 5 0?
12. The graph at the right shows the equation.
Circle the zeros of the function.
13. The solutions of the quadratic equation
are 26 and 4 .
Using a Quadratic Equation
Problem 4
Got It? The function y 5 20.03x 2 1 1.60x models the path of a kicked soccer ball.
The height is y, the distance is x, and the units are meters. How far does the soccer
ball travel? How high does the soccer ball go? Describe a reasonable domain and
range for the function.
14. The graph below shows the function. Circle the point on the graph where the soccer
ball is at its highest point and the point where the soccer ball lands. Label each
point with its coordinates.
20
2
+ 26 3 ,
y
1
21 3 ,
10
8
12
16
20
24
28
32
36
40
44
48
52
1
+ 53 3 ,
Reasoning Circle the phrase that completes each sentence.
15. The distance the soccer ball travels is the 9.
x-coordinate of
the vertex
y-intercept
x-coordinate of the
positive zero
y-coordinate of
the vertex
16. The maximum height of the soccer ball is the 9.
x-coordinate of
the vertex
y-intercept
x-coordinate of the
positive zero
y-coordinate of
the vertex
17. Underline the correct word to complete each sentence.
The domain should include positive / negative numbers only.
The range should include positive / negative numbers only.
18. Complete.
Domain:
Chapter 4
0
#x#
5313
Range:
100
0
#y#
2113
0
,
Copyright © by Pearson Education, Inc. or its affiliates. All Rights Reserved.
x
4
Lesson Check • Do you know HOW?
Solve the equation x 2 2 9 5 0 by factoring.
19. Circle the phrase that best describes the expression on the left side of the equals sign.
binomial
difference of two squares
parabola
quadratic expression
20. Factor the expression on the left side of the equal sign.
(x 2 3)(x 1 3)
21. The solutions of the equation are 3 and 23 .
Lesson Check • Do you UNDERSTAND?
Vocabulary If 5 is a zero of the function y 5 x 2 1 bx 2 20, what is the value
of b? Explain.
22. If 5 is a zero of the function then whenever x
x 2 1 bx 2 20
5 5,
5 0.
23. Substitute 5 for x and solve for b .
x 2 1 bx 2 20 5 0
1 b ? 5 2 20 5 0
25 1 5b 2 20 5 0
5 1 5b 5 0
5b 5 25
b 5 21
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52
24. The coefficient b 5 21 .
Math Success
Check off the vocabulary words that you understand.
zero of a function
Zero-Product property
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101
Lesson 4-5
4-6
Completing the Square
Vocabulary
Review
Draw a line from each expression to its square root.
1. 25x 2
x12
2. x 2 1 4x 1 4
6x 2 3
3. 36x 2 2 36x 1 9
2x 2 5y
4. 4x 2 2 20xy 1 25y 2
45x
Vocabulary Builder
trinomial (noun) try NOH mee ul
Related Words: perfect square
Main Idea: You can use perfect square trinomials to solve quadratic equations.
Examples: 4x 2 2 7x 1 5, ax 2 1 bx 1 c, and 2x 2 5y 1 4z are all trinomials.
x 2 1 4x 1 4 is a perfect square trinomial because it is the square of the binomial
x 1 2.
Use Your Vocabulary
5. Write the number of terms in each expression.
2
x11
3
t 2 2 2t 2 6
1
y3
3
p 2 2 6p 1 9
6. Put a T next to each expression that is a trinomial. Put an N next to each expression
that is not a trinomial.
N
x2
T
g3 1 g 2 4
T
N
x 2 2 2x 1 5
x 2 2 4x
7. Cross out the expression that is NOT a perfect square trinomial.
x 2 1 2x 1 1
Chapter 4
9x 2 2 6x 1 1
4x 2 2 4x 2 4
102
25x 2 2 30x 1 9
Copyright © by Pearson Education, Inc. or its affiliates. All Rights Reserved.
Definition: A trinomial is an expression consisting of three terms.
Problem 2 Determining Dimensions
Got It? The lengths of the sides of a rectangular window have the ratio 1.6 to 1.
The area of the window is 2822.4 in 2 . What are the window dimensions?
8. Circle the equation that represents this situation.
x(1.6x) 5 2822.4
1.6x 2 1 x 5 2822.42
(1 1 1.6)x 2 5 2822.4
9. The equation is solved below. Write a justification from the box for each step.
Divide each side by 1.6.
Simplify.
Simplify the left side.
Take the square root of each side.
x(1.6x) 5 2822.4
Write the original equation.
1.6x 2 5 2822.4
Simplify the left side.
1.6x 2
2822.4
1.6 5 1.6
Divide each side by 1.6.
Simplify.
x 2 5 1764
Take the square root of each side.
x 5 442
10. One side of the window measures
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1.6 Q
42
in. The other side measures
R , or 67.2 in.
42
Problem 3 Solving a Perfect Square Trinomial Equation
Got It? What is the solution of x 2 2 14x 1 49 5 25?
11. Use the justifications at the right to solve the equation.
x 2 2 14x 1 49 5 25
Write the original equation.
2
Q x 2 7 R 5 25
Factor the perfect square trinomial.
Qx 2 7 R 5 4 5
x 2
7
5
5
or x 2
x 5 12 or x 5
Key Concept
Take the square root of each side.
7 5 25
Write as two equations.
2
Solve for x.
Completing the Square
2
You can turn the expression x 2 1 bx into a perfect square trinomial by adding Q b2 R .
b 2
b 2
x 2 1 bx 1 Q 2 R 5 Q x 1 2 R
103
Lesson 4-6
12. Circle the value that completes the square for x 2 1 16x.
24
4
216
64
Completing the Square
Problem 4
Got It? What value completes the square for x 2 1 6x?
13. In the expression, the value b 5
6 .
14. Circle the expression for the value that completes the square.
6
22
62
62
2
6 2
Q2 R
15. Complete the square and write the expression as a perfect square.
Answers may vary. Sample:
6 2
Q 2 R 5 32 5 9
x 2 1 6x 1 9
x 2 1 6x 1 9 5 (x 1 3)2
Solving by Completing the Square
Problem 5
16. Use the justifications at the right to solve the equation.
2x 2 2 x 1 3 5 x 1 7
2 x 2 1 22 x 5
4
22
4
x2 1 °
2
¢x5
x 2 1 21 x 5
Write the original equation.
Rewrite so that all terms with x on
one side of the equation.
2
Divide each side by a so that
the coefficient of x 2 is 1.
2
Simplify.
1
4
2
Find Q b2 R .
2
b 2
Q2 R 5
Q 21
R
2
x 2 1 21 x 1
5
9
1
4
2
Add Q b2 R to each side.
5
4
2
Q x 1 212 R 5
9
Factor the trinomial.
4
x 1 212 5 w32
x5
Chapter 4
1
2
Take the square root of each side.
w 32
Solve for x.
104
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Got It? What is the solution of 2x 2 2 x 1 3 5 x 1 7?
Problem 6 Writing in Vertex Form
Got It? What is y 5 x 2 1 3x 2 6 in vertex form? Name the vertex and y-intercept.
17. Circle the equation that you can use to complete the square.
3 2
3 2
y 5 x 2 1 3x 1 Q 2 R 2 6 2 Q 2 R
3
3
y 5 x 2 1 3x 2 2 2 6 1 2
3 2
y 5 x 2 1 3x 2 Q 2 R
3
y 5 x 2 1 3x 2 2 2 6
18. Simplify the equation.
y 5 Qx 1
3
2
2
2
3
R 2 6 1 Q2 R 5 Qx 1
3
2
2
15
R 1 24
15
3
19. The vertex is Q 22 , 2 4 R .
20. The y-intercept is 26 .
Lesson Check • Do you UNDERSTAND?
How can you rewrite the equation x 2 1 12x 1 5 5 3 so that the left side of the
equation is in the form (x 1 a)2 ?
21. Use the justifications at the right to rewrite the equation.
x 2 1 12x 1 5 5 3
Write the original equation.
Copyright © by Pearson Education, Inc. or its affiliates. All Rights Reserved.
x 2 1 12x 5 22
x 2 1 12x 1
Q
12
2
Rewrite the equation as x 2 1 bx 5 c.
2
R
5 22 1
Q
12
2
2
R
Complete the square.
x 2 1 12x 1 36 5 22 1 36
Simplify powers.
x 2 1 12x 1 36 5 34
Add.
2
Q x 1 6 R 5 34
Write as (x 1 a)2 5 c.
Math Success
Check off the vocabulary words that you understand.
trinomial
perfect square trinomial
completing the square
Rate how well you can simplify quadratic expressions by completing the square.
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105
Lesson 4-6
The Quadratic Formula
4-7
Vocabulary
Review
Draw a line from each formula to its description.
1. a 5 s 2
area of a circle
2. c 5 2pr
circumference of a circle
3. p 5 2(l 1 w)
area of a square
4. a 5 pr 2
perimeter of a rectangle
Vocabulary Builder
discriminant
discriminant (noun) dih SKRIM uh nunt
Definition: The discriminant of a quadratic
equation in the form ax 2 1 bx 1 c 5 0
is the value of the expression b2 2 4ac .
Main Idea: The discriminant helps you determine how many real solutions
a quadratic function has.
Use Your Vocabulary
Circle the discriminant of each equation.
5. 2x2 1 (27x) 2 4 5 0
72 2 4(2)24
7 2 4(24)
(27)2 2 4(2)(24)
6. 3x 2 1 4x 1 2 5 0
4 2 4(3)(2)
12 1 4(3)(2)
42 2 4(3)(2)
2 2 4(1)(1)
1(2) 2 (1)(1)
7. x 2 1 x 2 1 5 0
12 2 4(1)(21)
8. 4x2 1 (212x) 1 9 5 0
12 2 4(4)(9)
Chapter 4
(212)2 2 4(4)(9)
106
(212)2 1 4(4)(9)
Copyright © by Pearson Education, Inc. or its affiliates. All Rights Reserved.
b2 4ac 0 means 2 real solutions.
b2 4ac 0 means 1 real solution.
b2 4ac 0 means 0 real solutions.
Key Concept
The Quadratic Formula
To solve the quadratic equation ax 2 1 bx 1 c 5 0, use the Quadratic Formula.
x5
2b 4 "b2 2 4ac
2a
9. Cross out the value of a that does NOT give a solution to the quadratic formula.
a54
a 5 21
a51
a50
Problem 1 Using the Quadratic Formula
Got It? What are the solutions to x 2 1 4x 5 24? Use the Quadratic Formula.
10. Circle the standard form of the equation.
x 2 1 4x 5 24
x 2 1 4x 2 4 5 0
x 2 1 4x 1 4 5 0
11. Identify the values of a, b, and c.
a5
b5
1
c5
4
4
12. Substitute the values of a, b, and c into the Quadratic Formula. Use the justifications
to solve the equation.
Copyright © by Pearson Education, Inc. or its affiliates. All Rights Reserved.
x5
5
5
2b 4
2
"Q
b R 2 4Q a R c
Write the Quadratic Formula.
2Q a R
24 4
2
"Q
4 R 2 4 Q 1 RQ 4 R
2Q 1 R
24 4
"
Substitute for a, b, and c.
0
Simplify under the radical.
2
5 22
Simplify.
13. Substitute the value you found in Exercise 12 into the original equation to check
your solution.
x 2 1 4x 1 4 5 0
Q 22 R 2 1 4 Q 22 R 1 4 0 0
0 50 ✓
Write the original equation.
Substitute for x.
Check for equality.
107
Lesson 4-7
Problem 2
Applying the Quadratic Formula
Got It? Fundraising Your School’s jazz band is selling CDs as a fundraiser. The
total profit p depends on the amount x that your band charges for each CD. The
equation p 5 2x 2 1 48x 2 300 models the profit of the fundraiser. What’s the least
amount, in dollars, you can charge for a CD to make a profit of $100?
14. Circle the equation that represents the situation.
0 5 2x 2 1 48x 2 200
0 5 x 2 1 48x 1 500
0 5 2x 2 1 48x 2 400
15. Cross out the value that will NOT be substituted into the Quadratic Formula to
solve the problem.
21
1
2400
48
16. Substitute values for a, b, and c into the Quadratic Formula and simplify.
x5
x5
248
4
" Q 48 R
2
2 4 Q 21 R Q 2400 R
2 Q 21 R
248
4
"
704
22
< 37.27 or x < 10.73
17. The smallest amount you can charge is 10.73 for each CD to make a profit
of $100.
Using the Discriminant
Got It? What is the number of real solutions of 2x 2 2 3x 1 7 5 0?
18. Complete the reasoning model below.
Think
Write
Find the values of a, b, and c.
a 2 , b 3 , c 7
Evaluate b2 4ac.
b2 4ac + 3 , 2 4 + 2 , + 7 , 47
Interpret the discriminant.
Problem 4
The discriminant is positive / negative / zero .
The equation has 2 / 1 / 0 real solution(s).
Using the Discriminant to Solve a Problem
Got It? Reasoning You hit a golf ball into the air from a height of 1 in. above the
1
ground with an initial vertical velocity of 85 ft/s. The function h 5 216t 2 1 85t 1 12
models the height, in feet, of the ball at time t in seconds. Will the ball reach a height
of 120 ft? Explain.
Chapter 4
108
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Problem 3
19. Circle the correct strategy to solve the problem.
Evaluate the discriminant
using the values a 5 216,
1
b 5 85, c 5 12
.
Substitute 120 for h in the equation
and evaluate the discriminant to
check for real solutions.
21. Evaluate the discriminant.
20. Write the equation in standard form.
216 t 2 1
85
Substitute 120 for t in
the equation and solve
for h.
t 1 211911
12 5 0
Answers may vary. Sample:
2
(85)2 2 4(216) Q 211911
12 R 5 7225 2 76743
5 244923
22. The discriminant is positive / negative / zero , so the equation has 2 / 1 / 0
real solutions.
23. The golf ball will / will not reach a height of 120 feet.
Lesson Check • Do you UNDERSTAND?
Reasoning For what values of k does the equation x 2 1 kx 1 9 5 0 have one real
solution? two real solutions?
2
24. If 9 completes the square, then Q k2 R 5
9 , so k2 5 3 and k 5
6 .
Copyright © by Pearson Education, Inc. or its affiliates. All Rights Reserved.
25. Place a ✓ if you can use the equation or inequality to solve this problem. Place
an ✗ if you cannot.
✓
✗
k 2 2 36 5 0
✓
k 2 2 36 , 0
k 2 2 36 . 0
✗
k 2 5 362
26. Now answer the question.
Answers may vary. Sample: For k 5 6 and k 5 26, the equation has one
_______________________________________________________________________
real solution. For k S 6 or k R 26, the equation has two real solutions.
_______________________________________________________________________
Math Success
Check off the vocabulary words that you understand.
quadratic formula
discriminant
real solutions
Rate how well you can use the quadratic formula to solve problems.
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Lesson 4-7
Complex Numbers
4-8
Vocabulary
Review
1. Circle the square root that is not a real number.
!64
!6 2 (2)(4)
!4 2 (2)(26)
"(25)2
Vocabulary Builder
conjugate (adjective)
KAHN
juh gut
Related Words: complex numbers, pairs, roots, imaginary solutions
Math Usage: The conjugate of the complex number a 1 bi is a 2 bi .
Use Your Vocabulary
Write C if the number pairs are complex conjugate or N if they are not.
C
2. 4 1 3i, 4 2 3i
N
3. 5 1 !2, 5 2 !2
C
4. !5 2 !3i, !5 1 !3i
N
5. 3 1 !5i, 3 1 !25i
Key Concept
Square Root of a Negative Real Number
The imaginary unit i is the complex number whose square is 21. So, i 2 5 21,
and i 5 !21.
For any positive real number a, !2a 5 !21 ? a 5 !21 ? !a 5 i!a.
Note that A !25
Chapter 4
B2
5 A i!5
B2
5 i 2 A !5
B2
5 21 ? 5 5 25 (not 5).
110
Example: !25 5 i!5
Copyright © by Pearson Education, Inc. or its affiliates. All Rights Reserved.
Main Idea: Complex solutions occur in conjugate pairs of the form a 1 bi and
a 2 bi . The product of complex conjugates is always a real number. You can use
complex conjugates to simplify division of complex numbers.
6. Use !21 5 i to complete each equation.
!22 5
i !2
!23 5 i Å
3
!26 5
i!6
28
Å
5 i!8
Problem 1 Simplifying a Number Using i
Got It? How do you write the number !212 using the imaginary unit i ?
7. Circle the expression that is equivalent to !212.
!21 ? 4(23)
4!21 ? !3
!21 ? 4 ? 3
22!3
8. Simplify the expression you circled in Exercise 7.
Answers may vary. Sample:
!21 ? 4 ? 3 5 2 !21 ? !3 5 2 ? i ? !3 5 2 !3i
9. Using the imaginary unit i, !212 5 j.
2!3i
4!i
6i
4!3i
Problem 2 Graphing in the Complex Number Plane
Copyright © by Pearson Education, Inc. or its affiliates. All Rights Reserved.
Got It? What are the graph and absolute value of 5 2 i?
11. Graph the point.
10. Underline the correct words
to complete the sentence.
imaginary axis
The graph of 5 2 i is 5 units
4i
left / right and 1 unit up / down .
2i
12. Find the absolute value.
2
u5 2 iu 5 Ä
5 Ä
5 Ä
Q 5 R 1Q 1 R
25 1
26
1
2
Use the
Distance Formula.
4
Simplify powers.
2
2
real axis
4
2i
4i
Add.
Problem 3 Adding and Subtracting Complex Numbers
Got It? What is the sum (7 2 2i) 1 (23 1 i)?
111
Lesson 4-8
13. The sum is found below. Write the justification for each step.
(7 2 2i) 1 (23 1 i)
Write the original expression.
7 1 (22i 2 3) 1 i
Associative
Property
7 1 (23 2 2i) 1 i
Commutative
Property
(7 2 3) 1 (22i 1 i)
Associative
Property
42i
Simplify.
Problem 4
Multiplying Complex Numbers
Got It? What is the product (7i)(3i)?
14. Complete the solution. Justifications are given.
(7i)(3i)
Write the original expression.
21 i 2
Multiply.
Q 21 RQ 21 R
Substitute 21 for i 2 .
221
Problem 5
Dividing Complex Numbers
2 2i
Got It? What is the quotient 35 1
4i ?
15. Circle the first step in simplifying the fraction.
Find the complex conjugate of 5 2 2i.
Find the complex conjugate of 3 1 4i.
Find the absolute value of 5 2 2i.
Find the absolute value of 3 1 4i.
16. Cross out the expression that is NOT equivalent to the quotient.
15 2 20i 2 6i 1 8i 2
9 2 12i 1 12i 2 16i 2
15 2 26i 1 8i 2
25
25 1 10i 2 10i 2 16i 2
9 2 12i 1 12i 2 16i 2
17. Simplify.
Answers may vary. Sample:
15 2 26i 1 8i 2
25
8 2 26i
7
5 15 2 25
5 25
2 26
25 i
26
5 2 2i
7
18. 3 1 4i 5 25 2 25 i
Chapter 4
112
Copyright © by Pearson Education, Inc. or its affiliates. All Rights Reserved.
Simplify.
Problem 6 Factoring using Complex Conjugates
Got It? What are the factors of each expression?
19. 5x 2 1 20
5
What is the GCF of 5 and 20?
Rewrite x 2 1 4 as a2 1 b2.
Write as a product using the GCF.
a5
x
b5
5(x2 1 4)
x 2 1 22
2
Use a2 1 b2 5 (a 1 bi)(a 2 bi).
(x 1 2i )(x 2 2i)
What are the factors of 5x 2 1 20?
5, (x 1 2i ), and (x 2 2i)
2
2
Rewrite x 2 1 81 in terms of a2 1 b2 . x 1 9
20. x 2 1 81
(x 1 9i )(x 2 9i)
Use a2 1 b2 5 (a 1 bi)(a 2 bi).
Lesson Check • Do you UNDERSTAND?
Error Analysis Describe and correct the error made in
simplifying the expression (4 2 7i)(4 1 7i).
Copyright © by Pearson Education, Inc. or its affiliates. All Rights Reserved.
21. Simplify the expression.
(4 - 7i) (4 + 7i) = 16 + 28i - 28i + 49i2
= 16 - 49
= -33
Student work may vary. Example:
(4 2 7i )(4 1 7i ) 5 16 1 28i 2 28i 2 49i 2 5 16 2 49(21) 5 16 1 49 5 65
22. Explain the error shown above.
Answers may vary. Example: Multiplying 27i by 7i, the answer is 249i 2.
_______________________________________________________________________
The answer shown above uses 49i 2.
_______________________________________________________________________
Math Success
Check off the vocabulary words that you understand.
imaginary number
complex number
complex conjugates
Rate how well you can find complex-number solutions to quadratic equations.
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Lesson 4-8
4-9
Quadratic Systems
Vocabulary
Review
Write T for true or F for false.
T
1. The solution of system y 5 3x 1 2 and y 5 5x is the point where the two
lines intersect.
F
2. The solution of a system of 2 linear equations has at most 2 points of
intersection.
F
3. The solution of a system of inequalities is the point where the lines intersect
with the y-axis.
T
4. The solution of a system of inequalities is the region where the graphs of the
inequalities overlap.
Quadratic-Linear System (noun) kwah DRAT ik LIN ee ur SIS tum
Related Words: System of equations, system of inequalities.
Main Idea: A system of equations can include an equation with a graph that is
not a line. Such a system can have more than one solution.
Definition: A quadratic-linear system is a system of one quadratic equation
and one linear equation. The system can have two, one, or no solutions (points
of intersection).
Use Your Vocabulary
5. Cross out the graph that does NOT illustrate a quadratic-linear system.
Chapter 4
114
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Vocabulary Builder
Problem 1 Solving a Linear-Quadratic System by Graphing
Got It? What is the solution of the system?
e
y 5 x 2 1 6x 1 9
y5x13
6. Complete the table of values for
both equations.
7. Use the points from the table to
graph the two equations.
x
x 2 6x 9
x3
6
4
1
1
4
3
0
0
2
2
1
1
1
4
2
0
9
3
1
16
4
4
y
O
2
x
2
4
2
8. The solutions are Q 23 , 0 R and Q 22 , 1 R .
Copyright © by Pearson Education, Inc. or its affiliates. All Rights Reserved.
9. Substitute into both equations to check the solutions.
0 5 (23)2 1 6(23) 1 9
0 5 9 2 18 1 9
050
1 5 (22)2 1 6(22) 1 9
1 5 4 2 12 1 9
151
0 5 23 1 3
050
1 5 22 1 3
151
Problem 2 Solving Using Substitution
Got It? What is the solution of the system?
e
y 5 2x 2 2 3x 1 10
y5x15
10. Use the justifications at the right to solve the system.
x1 5
5 21 x 2 1 23 x 1 10
Substitute x 1 5 for y in the
quadratic equation.
1 x 2 1 4 x 1 25 5 0
Write in standard form.
Q x 1 5 RQ x 1 21 R 5 0
Factor.
x 5 25 or x 5 1
Solve for x.
x 5 25 S y 5 25 1 5 5 0
Substitute for x in y 5 x 1 5.
x5 1 S y5 1
155 6
115
Lesson 4-9
11. Check the solutions.
Solution 1 Q 25 , 0 R
Solution 2 Q 1 , 6 R
y 5 2x 2 2 3x 1 10
y 5 2x 2 2 3x 1 10
0
2
5 2Q 25 R 2 3 Q 25 R 1 10
0 5
0
Problem 3
2
6
5 2Q 1 R 2 3 Q 1 R 1 10
6
5 6
Solving a Quadratic System of Equations
Got It? What is the solution of the system?
e
y 5 x 2 2 4x 1 5
y 5 2x 2 1 5
12. Circle the graph of the system. Each graph shows the standard viewing window.
13. Use the graph you circled. Circle the solution of the system.
Problem 4
(21, 4)
(0, 5)
(1, 2)
(2, 1)
(3, 24)
Solving a Quadratic System of Inequalities
Got It? What is the solution of this system of inequalities?
e
y K 2x 2 2 4x 1 3
y S x2 1 3
The graph at the right shows the boundaries of the inequalities.
8
14. Shade the region that represents y # 2x 2 2 4x 1 3.
6
15. Shade the region that represents y . x 2 1 3 in
another color.
16. Outline the region that represents the solution of the
system of inequalities.
2
4
Lesson Check • Do you know HOW?
Solve the system by substitution.
Chapter 4
e
y
y 5 x 2 2 2x 1 3
y5x 11
116
2
O
x
2
Copyright © by Pearson Education, Inc. or its affiliates. All Rights Reserved.
(22, 1)
17. Complete the solution. Justifications are given.
x 1 1 5 x 2 2 2x 1 3
0 5 x2 2 3 x 1
Substitute x 1 1 for y.
2
Addition property of equality.
0 5 Q x 2 1 RQ x 2 2 R
Factor.
x5 1
Solve for x.
y51115
2
or
x5
2
or
y5
2 115
Substitute for x in y 5 x 1 1 and solve for y.
3
The solutions are Q 1, 2 R and Q 2 , 3 R .
Lesson Check • Do you UNDERSTAND?
Reasoning How many points of intersection can you have between a linear function
and a quadratic function? Draw graphs to justify your answers.
18. If possible, draw linear function and a quadratic function with the number of
intersections specified. Answers may vary. Samples are shown.
0 points of intersection
1 point of intersection
y
y
2
Copyright © by Pearson Education, Inc. or its affiliates. All Rights Reserved.
y
2
x
O
2
2 points of intersection
2
2
2
O
2
x
2
O
2
2
x
2
2
19. Circle the number(s) of points of intersection you can have between a linear
function and a quadratic function.
0
1
2
3
4
Math Success
Check off the vocabulary words that you understand.
quadratic-linear system
system of equations
system of inequalities
Rate how well you can solve and graph systems of equations and inequalities.
Need to
review
0
2
4
6
8
Now I
get it!
10
117
Lesson 4-9

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