Transcript Modeling Data With Quadratic Functions 5-1 1. 2.

Modeling Data With Quadratic Functions
ALGEBRA 2 LESSON 5-1
(For help, go to Lessons 1-2 and 2-2.)
Evaluate each function for x = –3, –1, 0, 1, and 3.
1. ƒ(x) = x
2. ƒ(x) = x2
3. ƒ(x) = –x
4. ƒ(x) = –x2
5. ƒ(x) =
1 2
x
3
6. ƒ(x) = – 1 x2
3
Write each equation in slope-intercept form.
7. 3x + 4y = 8
8. 2x – y = –7
9. 1 x + 3y = 9
2
5-1
Modeling Data With Quadratic Functions
ALGEBRA 2 LESSON 5-1
Solutions
1. ƒ(–3) = –3; ƒ(–1) = –1; ƒ(0) = 0; ƒ(1) = 1; ƒ(3) = 3
2. ƒ(–3) = (–3)2 = (–3)(–3) = 9; ƒ(–1) = (–1)2 = (–1)(–1) = 1;
ƒ(0) = 02 = 0 • 0 = 0;
ƒ(1) = 12 = 1 • 1 = 1; ƒ(3) = 32 = 3 • 3 = 9
3. ƒ(–3) = –(–3) = 3; ƒ(–1) = –(–1) = 1; ƒ(0) = –0 = 0; ƒ(1) = –1; ƒ(3) = –3
4. ƒ(–3) = –(–3)2 = –(–3)(–3) = –9;
ƒ(–1) = –(–1)2 = –(–1)(–1) = –1;
ƒ(0) = –02 = –(0 • 0) = –0 = 0;
ƒ(1) = –12 = –(1 • 1) = –1; ƒ(3) = –32 = –(3 • 3) = –9
5-1
Modeling Data With Quadratic Functions
ALGEBRA 2 LESSON 5-1
Solutions (continued)
1
1
5. ƒ(–3) = 3; ƒ(–1) = 3 ; ƒ(0) = 0; ƒ(1) = 3 ; ƒ(3) = 3
1
1
6. ƒ(–3) = –3; ƒ(–1) = – 3 ; ƒ(0) = 0; ƒ(1) = – 3 ; ƒ(3) = –3
7. 3x + 4y = 8
4y = –3x + 8
4y = –3x + 8
4
4
y = – 3x + 2
8. 2x – y = –7
–y = –2x – 7
y = 2x + 7
4
9. y = – 1x + 3
6
5-1
Modeling Data With Quadratic Functions
ALGEBRA 2 LESSON 5-1
Determine whether each function is linear or quadratic. Identify
the quadratic, linear, and constant terms.
a. ƒ(x) = (2x – 1)2
= (2x – 1)(2x – 1)
Multiply.
= 4x2 – 4x + 1
Write in standard form.
This is a quadratic function.
Quadratic term: 4x2
Linear term: –4x
Constant term: 1
5-1
Modeling Data With Quadratic Functions
ALGEBRA 2 LESSON 5-1
(continued)
b. ƒ(x) = x2 – (x + 1)(x – 1)
= x2 – (x2 – 1)
Multiply.
=1
Write in standard form.
This is a linear function.
Quadratic term: none
Linear term: 0x (or 0)
Constant term: 1
5-1
Modeling Data With Quadratic Functions
ALGEBRA 2 LESSON 5-1
Below is the graph of y = x2 – 6x + 11. Identify the vertex and
the axis of symmetry. Identify points corresponding to P and Q.
The vertex is (3, 2).
The axis of symmetry is x = 3.
P(1, 6) is two units to the left of the axis
of symmetry.
Corresponding point P (5, 6) is two units
to the right of the axis of symmetry.
Q(4, 3) is one unit to the right of the axis
of symmetry.
Corresponding point Q (2, 3) is one unit
to the left of the axis of symmetry.
5-1
Modeling Data With Quadratic Functions
ALGEBRA 2 LESSON 5-1
Find the quadratic function to model the values in the table.
x
y
–2
–17
1
10
5
–10
Substitute the values of x and y into y = ax2 + bx + c.
The result is a system of three linear equations.
y = ax2 + bx + c
–17 = a(–2)2 + b(–2) + c = 4a – 2b + c
Use (–2, –17).
10 = a(1)2 + b(1) + c = a + b + c
Use (1, 10).
–10 = a(5)2 + b(5) + c = 25a + 5b + c
Use (5, –10).
Using one of the methods of Chapter 3, solve the system
4a – 2b + c = –17
a – b + c = 10
25a + 5b + c = –10
The solution is a = –2, b = 7, c = 5. Substitute these values into
standard form. The quadratic function is y = –2x2 + 7x + 5.
{
5-1
Modeling Data With Quadratic Functions
ALGEBRA 2 LESSON 5-1
The table shows data about the wavelength x (in meters)
and the wave speed y (in meters per second) of deep water ocean
waves. Use the graphing calculator to model the data with a quadratic
function. Graph the data and the function. Use the model to estimate
the wave speed of a deep water wave that has a wavelength of
6 meters.
Wavelength Wave Speed
(m)
(m/s)
3
6
5
16
7
31
8
40
5-1
Modeling Data With Quadratic Functions
ALGEBRA 2 LESSON 5-1
(continued)
Wavelength
(m)
3
5
7
Wave Speed
(m/s)
6
16
31
8
40
Step 1: Enter the data.
Use QuadReg.
Step 2: Graph the data
and the function.
Step 3: Use the table
feature to find
ƒ(6).
An approximate model of the quadratic function is y = 0.59x2 + 0.34x – 0.33.
At a wavelength of 6 meters the wave speed is approximately 23m/s.
5-1
Modeling Data With Quadratic Functions
ALGEBRA 2 LESSON 5-1
Pages 237–239 Exercises
1. linear; none, x, 4
9. quadratic; –2x2, –8x,
none
2. quadratic; 2x2, –3x, 5
17. y = x2 – 5x + 2
18. y = 2x2 – x + 3
10. (0, –4), x = 0
3. quadratic; 3x2, –6x,
none
19. y = x2 + 2x
11. (–1, 0), x = –1
20. y = –3x2 + 20
4. quadratic; x2, none, –7
12. (–1, –4), x = –1
5. quadratic; x2, 3x, –10
13. P'(6, 9), Q'(2, 1)
6. linear; none, –7x, 28
14. P'(1, 5), Q'(–2, 8)
7. quadratic; 6x2, none, 6
15. P'(–1, –1), Q'(–4, –4)
8. linear; none, x, –1
16. y = –x2 + 3x – 4
5-1
21. a. y = –16x2 + 33x +
46, where x is the
number of
seconds after
release and y is
height in feet.
b. 28.5 ft
Modeling Data With Quadratic Functions
ALGEBRA 2 LESSON 5-1
22. a. y = 0.0236x2 +
0.907x – 2.09
29.
1
1
,0 , x =
2
2
30. a. x: 4, 5; y: 6, 10
b. 58.5%
23. y =
b. y = 1 x2 – 1 x
2
4x2
c. 45 segments
24. y = –2x2 + 3x + 5
25. no
5
2
7
31. a. y = –0.01467x2 +
1.312x + 9.795
b. 1992
26. y = 8 x2 – 4 x + 1
31. (continued)
c. Never; the
quadratic model is
useful over a
limited number of
years, but
because it
increases and
then decreases, it
does not model
the data after
2021.
27. – 1 , – 1 , x = – 1
32. 3
28. (–1, 4), x = –1
33. 8
2
2
2
34. 6
5-1
Modeling Data With Quadratic Functions
ALGEBRA 2 LESSON 5-1
35. – 11
37. –10
38. (continued)
c. Answers may vary. Sample: Quadratic; the
quadratic model comes closer to most data
points than the linear model because the data
follows a curve.
38. a. y = 3.157x – 52.34
39. Answers may vary. Sample: y = –
8
36. 25
4
1
b. y = 0.04243x2 –
0.04080x + 0.8890
2
1
6
1 2
x,
25
y = 25 x2 – x, y = x2 – x
5
5
5
40. Answers may vary. Sample: You need at least 3
points; you are going to substitute x- and y-values
into y = ax2 + bx + c to set up and solve a linear
system for finding values of a, b, and c.
5-1
Modeling Data With Quadratic Functions
ALGEBRA 2 LESSON 5-1
41. Answers may vary. Sample: They are similar in
that both are symmetric with respect in the y-axis,
have only non-negative y-values, lie in Quadrants I
and II, and have minimums at (0, 0); they are
different in that the graph of y = x2 rises more
steeply, while y = |x| rises at a steady rate as |x|
increases.
42. (3, 5)
43. a. You can find how high the arrow was when it
was released.
b. The negative intercept tells you how much
earlier you would have to shoot from a height of
zero the arrow for its height to be described by
the same function. The positive intercept tells
you how many seconds after the release the
arrow will take to hit the ground.
5-1
44. A
45. H
46. A
47. I
Modeling Data With Quadratic Functions
ALGEBRA 2 LESSON 5-1
48. [4] System:
a(1)2 + b(1) + c = 6, a(2)2 + b(2) + c = 11,
a(3)2 + b(3) + c = 20; a = 2, b = –1, c = 5;
y = 2x2 – x + 5
50. 3
1
2
1
–3
2
51. [0 –4
[2] incorrect system solved correctly OR correct
system solved incorrectly
52. 59
33
[1] correct function, without work shown
53. (2, 5)
–1
2
7 ; (3, 2)
10
46
22
54. (5, 8)
55. (–1, –1)
5-1
–3
–2 ;
11
(1, 0, 3)
[3] appropriate methods, but with one
computational error
49. 3
2
–2
–1
3
17 –4]
Modeling Data With Quadratic Functions
ALGEBRA 2 LESSON 5-1
56. 4
5
1
57. 2
58. –1
5-1
Modeling Data With Quadratic Functions
ALGEBRA 2 LESSON 5-1
1. Below is the graph of y = x2 + 2x – 2. Identify the vertex and axis of
symmetry. Identify points corresponding to P and Q.
(–1, –3), x = –1; P (–2, –2), Q (1, 1)
Find a quadratic function to model the values in each table.
2.
x
y
–2
y = –3x2 + x – 4
3.
x
y
–18
–2
24
0
–4
–1
13
2
–14
3
9
5-1
y = 2x2 – 5x + 6
Properties of Parabolas
ALGEBRA 2 LESSON 5-2
(For help, go to Lessons 2-2 and 2-5.)
Find the y-intercept of the graph of each function.
1. y = 3x + 3
2. y = –2x – 1
3. 4x – 3y = 12
Find the vertex of the graph of each function.
4. y = | –2x |
5. y = | – 2 x – 1|
3
6. y = | 3x + 7 |
Graph each equation.
7. y = –4x – 3
1
8. 2 x + y = –2
9. y = | 5x – 5 |
5-2
Properties of Parabolas
ALGEBRA 2 LESSON 5-2
Solutions
1. y = 3x + 3; y-intercept = 3
2.
y = –2x – 1; y-intercept = –1
3. 4x – 3y = 12
–3y = –4x + 12
–3y = –4x + 12
–3
–3
4
y = – x – 4; y-intercept = –4
4.
y = | –2x |; –
b
0
=
= 0,
m –2
vertex = (0, 0)
3
2
b
5. y = | – 3 x – 1|; – m = –
vertex = (– 3 , 0)
(–1)
–2
3
3
= –2
6.
y = | 3x + 7 |; –
7
b
= – 3,
m
vertex = (– 7 , 0)
2
3
5-2
Properties of Parabolas
ALGEBRA 2 LESSON 5-2
Solutions (continued)
8. 1 x + y = –2
7. y = –4x – 3
slope = –4
y-intercept = –3
2
y = 1x – 2
2
slope = – 1
2
y-intercept = –2
9. y = | 5x – 5 |
– b = – –5 = 1
m
5
vertex = (1, 0)
y-intercept point = (0, 5),
its reflection = (2, 5)
5-2
Properties of Parabolas
ALGEBRA 2 LESSON 5-2
Graph y = 1 x2 + 1.
3
Step 1: Graph the vertex, which is the y-intercept (0, 1).
Step 2: Make a table of values to find
some points on one side of the
axis of symmetry x = 0. Graph
the points.
x
y
1
1 13
2
21
3
3
4
4
61
3
Step 3: Graph corresponding points on
the other side of the axis of
symmetry.
Step 4: Sketch the curve.
5-2
Properties of Parabolas
ALGEBRA 2 LESSON 5-2
Graph y =
1 2
x + x + 3. Name the vertex and axis of symmetry.
2
Step 1: Find and graph the axis of symmetry.
x=–
b
1
= – 1 = –1
2a
2( )
2
Step 2: Find and graph the vertex. The xcoordinate of the vertex is –1.
The y-coordinate is y = 1 (–1)2 +
2
1
(–1) + 3 = 2 . So the vertex is (–
2
1, 2 ).1
2
Step 3: Find and graph the y-intercept and its reflection. Since c = 3,
the y-intercept is (0, 3) and its reflection is (–2, 3).
5-2
Properties of Parabolas
ALGEBRA 2 LESSON 5-2
(continued)
Step 4: Evaluate the function for another
value of x, such as
y = 1 (2)2 + (2) + 3 = 7. Graph
2
(2, 7) and its reflection (–4, 7).
Step 5: Sketch the curve.
5-2
Properties of Parabolas
ALGEBRA 2 LESSON 5-2
Graph y = – 1 x2 + 2x – 3. What is the maximum value
4
of the function?
Since a < 0, the graph of the function opens down, and the vertex
represents the maximum value. Find the coordinates of the vertex.
x = – b = – 21 = 4
2a
2(– )
4
1 2
y = – 4 (4) + 2(4) – 3 = 1
Find the x-coordinate of the vertex.
Find the y-coordinate of the vertex.
Graph the vertex and the axis of symmetry x = 4.
Graph two points on one side of the axis of
symmetry, such as (6, 0) and (8, –3).
Then graph corresponding points (2, 0)
and (0, –3).
The maximum value of the function is 1.
5-2
Properties of Parabolas
ALGEBRA 2 LESSON 5-2
The number of weekend get-away packages a hotel can sell
is modeled by –0.12p + 60, where p is the price of a get-away
package. What price will maximize revenue? What is the maximum
revenue?
Relate: revenue equals price times number of get-away packages sold
Define: Line R = revenue. Let p = price of a get-away package.
Let –0.12p + 60 = number of a get-away packages sold.
Write:
R = p ( –0.12p + 60 )
= –0.12p2 + 60p
Write in standard form.
5-2
Properties of Parabolas
ALGEBRA 2 LESSON 5-2
(continued)
Find the maximum value of the function. Since a < 0, the graph of the
function opens down, and the vertex represents a maximum value.
b
60
p = – 2a = – 2(–0.12) = 250
Find p at the vertex.
R = –0.12(250)2 + 60(250)
Evaluate R for p = 250
= 7500
Simplify.
A price of $250 will maximize revenue. The maximum revenue is $7500.
5-2
Properties of Parabolas
ALGEBRA 2 LESSON 5-2
Pages 244–247 Exercises
1.
3.
5.
2.
4.
6.
5-2
Properties of Parabolas
ALGEBRA 2 LESSON 5-2
7.
9.
12.
10.
8.
13.
11.
5-2
Properties of Parabolas
ALGEBRA 2 LESSON 5-2
14.
16.
18.
19.
17.
15.
20.
5-2
Properties of Parabolas
ALGEBRA 2 LESSON 5-2
21.
22.
23.
min, – 10
24.
max,
26.
3
max, 6
max, 6
25.
41
8
27.
min, 5
min, 2
28. $10; $13,500
29. 2 s; 64 ft
5-2
1
4
Properties of Parabolas
ALGEBRA 2 LESSON 5-2
30. 1000 tires; $20
33.
35. 13, 13; 169
36. –5, 5; –25
31.
37. B
38. C
34.
39. A
32.
40. Answers may vary.
Sample:
y = x2 + 20x + 96
41. 2.25 ft by 2.25 ft;
5.0625 ft2
5-2
Properties of Parabolas
ALGEBRA 2 LESSON 5-2
1
3 2 1
x –
4
2
42. y = – 10 x2 + 10
50. y =
43. length = 9 cm,
width = 9 cm
51. y = 10x2 – 1
52. y = –
44. 5
5 2
x
2
53. y = 6x2 + 8
45. –3
54. a. $20
b. $6050
c. Check students’
work.
46. –2
47. 2
1
48. y = 3 x2 + 2
49. y = –4x2 – 3
55. 25 ft by 50 ft, area =
1250 ft2
56. a. Check students’
work.
b. 60 bricks by 60
bricks
57. y = x2 + 1; up
58. y = 5x2 + 1; up
1 2
x + 1; down
2
1
60. y = – x2 + 1; down
2
1
61. y = x2 + 1; up
3
59. y = –
62. y = – 1 x2 + 1; down
5
5-2
Properties of Parabolas
ALGEBRA 2 LESSON 5-2
63. y = – 1 x2 + 1; down
67. –6, 24
4
1 2
64. y = –
x + 1; down
12
1
65. a. y = 14,400 x2
1
1
b. y = 14,400 x2 – 20 x
1
1
c. y = 14,400 x2 + 20 x
68. 1, 2
69. 3, –12
70. –
2
4
,–
9
3
71. 10 2 square units
3
2
72. 10 3 square units
66. a. Check students’ work.
b. Answers may vary. Sample: The widths of
y = ax2 + bx + c and y = –ax2 + bx + c are the
same. As |a| increases, the width of
y = ax2 + bx + c and y = –ax2 + bx + c
decrease.
5-2
73. 10
74. D
75. F
2
square units
3
Properties of Parabolas
ALGEBRA 2 LESSON 5-2
76. B
80. Answers may vary.
Sample:
y = –0.14857x2 +
5.1714x + 16.9714
77. H
78. B
79. [2]
b
The axis of symmetry is x = –
2a
or x = 3. The y-coordinate of
the vertex is 32 – 6 • 3 + 2 = –7.
So the vertex is (3, –7). The
y-intercept is where x = 0 or
02 – 6 • 0 + 2 = 2. So (0, 2) is
on the parabola. The point
corresponding to (0, 2) with respect to x = 3 is
(6, 2) so (6, 2) is also on the parabola.
[1] incorrect graph OR no explanation
5-2
81. 3
3
82.
4
–2
–4
0
3
0
Properties of Parabolas
ALGEBRA 2 LESSON 5-2
83. Let x = amount of storage space and
y = amount of space to be covered by the roof.
x >
– 40,000
y <
– 25,000
5-2
Properties of Parabolas
ALGEBRA 2 LESSON 5-2
1. Graph y = –2x2 – 4x + 2. Label the vertex and axis of symmetry.
2. What is the minimum value of the function y = 3x2 + 2x – 8?
–
25
3
3. The equation h = 40t – 16t2 describes the height h, in feet, of a ball that
is thrown straight up as a function of the time t, in seconds, that the ball
has been in the air. At what time does the ball reach its maximum
height? What is the maximum height?
1.25 s; 25 ft
5-2
Translating Parabolas
ALGEBRA 2 LESSON 5-3
(For help, go to Lesson2-6.)
Identify the parent function of each function. Then graph the function by translating
the parent function.
1. y = –x + 2
2. y = |3x| + 2
3. y = –|x + 1| – 1
Write an equation for each translation.
4. y = 2x, 2 units down
5. y = x, 4 units up, 1 unit right
5-3
Translating Parabolas
ALGEBRA 2 LESSON 5-3
Solutions
1. y = –x + 2
parent function: y = –x
translate 2 units up
2. y = |3x| + 2
parent function: y = |3x|
translate 2 units up
3. y = –|x + 1| – 1
parent function: y = –x
translate 2 unit left and
one unit down
4. y = 2x, 2 units down: y = 2x – 2
5. y = x, 4 units up and 1 unit right: y = (x – 1) + 4, or y = x + 3
5-3
Translating Parabolas
ALGEBRA 2 LESSON 5-3
Graph y =
2
(x + 1)2 – 2.
3
The graph of y = 2 (x + 1)2 – 2 is a translation of the graph of the
3
parent function y = 2 x2.
3
You can graph it by translating the parent function or by finding the
vertex and the axis of symmetry.
Step 1: Graph the vertex (–1, –2).
Draw the axis of symmetry x = –1.
Step 2: Find another point. When x = 2,
y = 2 (2 + 1)2 – 2 = 4. Graph (2, 4).
3
Step 3: Graph the point corresponding to
(2, 4). It is three units to the left of
the axis of symmetry at (–4, 4).
Step 4: Sketch the curve.
5-3
Translating Parabolas
ALGEBRA 2 LESSON 5-3
Write the equation of the parabola shown below.
y = a(x – h)2 + k
y = a(x – 2)2 – 5
–3 = a(0 – 2)2 – 5
2 = 4a
1
=a
2
Use the vertex form.
Substitute h = 2 and k = –5.
Substitute (0, –3).
Simplify.
Solve for a.
1
2
The equation of the parabola is y = (x – 2)2 – 5.
5-3
Translating Parabolas
ALGEBRA 2 LESSON 5-3
A long strip of colored paper is attached as a party
decoration at exactly opposite corners of the back wall of a
rectangular party room. The strip approximates a parabola with
equation y = 0.008(x – 25)2 + 10. The bottom left corner of the back
wall is the origin and x and y are measured in feet. How far apart are
the side walls? How high are they?
Start by drawing a diagram.
The function is in vertex form. Since
h = 25 and k = 10, the vertex is at
(25, 10).
The vertex is halfway between the two
corners of the wall, so the width of the
wall is 2(25 ft) = 50 ft.
5-3
Translating Parabolas
ALGEBRA 2 LESSON 5-3
(continued)
To find the wall’s height, find y for x = 0.
y = 0.008 (0 – 25)2 + 10
y = 0.008 (–25)2 + 10
= 15
The wall is 50 ft long and 15 ft high.
5-3
Translating Parabolas
ALGEBRA 2 LESSON 5-3
Write y = –7x2 – 70x – 169 in vertex form.
b
2a
= – (–70)
2(–7)
x=–
Find the x-coordinate of the vertex.
Substitute for a and b.
= –5
y = –7 (–5)2 – 70(–5) – 169
Find the y-coordinate of the vertex.
=6
The vertex is at (–5, 6).
y = a(x – h)2 + k
= –7(x – (–5))2 + 6
Write in vertex form.
Substitute for a, h and k.
= –7(x + 5)2 + 6
The vertex form of the function is y = –7(x + 5)2 + 6.
5-3
Translating Parabolas
ALGEBRA 2 LESSON 5-3
Pages 244–247 Exercises
1.
3.
5.
2.
6.
4.
5-3
Translating Parabolas
ALGEBRA 2 LESSON 5-3
7.
10.
13. y =
1 2
x
4
14. y = –x2 + 4
15. y = –(x – 2)2
8.
16. y = –(x + 2)2
11.
17. y = (x – 2)2
18. y = –2x2
9.
19. y = 6(x + 3)2 – 2
12.
20. y = –(x – 1)2 + 2
21. (–20, 0), –600
5-3
Translating Parabolas
ALGEBRA 2 LESSON 5-3
22. (3.2, 0), 1.024
31. y = 4 x + 7
2
23. (–5.5, 0), 726
32. y = 2 x + 1
2
33. y = 4 x – 5
2
8
16
4
24. (–1, –1), –0.9965
26. (125, 125), 15,750
27. y = (x – 2)2 + 2
34. y = –2(x –
9
– 1
+ 71
8
2)2
+ 11
2
–2
35. y = 4 x + 3
28. y = (x + 1)2 + 4
29. y = 6x2 – 10
30. y = –5x2 + 12
5-3
36.
8
4
25. (4, –25), –41
– 49
2
37.
Translating Parabolas
ALGEBRA 2 LESSON 5-3
38.
40.
42. a. All nonnegative
numbers; a price
cannot be
negative; it would
imply that the
bakery pays
people to take
bread.
41.
b. $277.50; $210.00
c. $0.55
d. $300.00
39.
43. y = –7(x – 1)2 + 2
4
9
44. y = – (x – 3)2 + 6
45. y = – 1 (x + 3)2 + 6
2
5-3
Translating Parabolas
ALGEBRA 2 LESSON 5-3
46. y = 3 (x + 2)2 + 6
55. y = –10x2 – 40x – 40
47. y = 7(x + 1)2 – 4
56. y = 16x2 – 8x + 2
48. y = –7x2 + 5
57. a. first: x = 4, second: x = 2.5
b. For the first spreadsheet the x1-values 3 and 5
are equidistant from 4 and their y1-values are
both –3. In the second spreadsheet, the x2values 2 and 3 are equidistant from 2.5 and
their y2-values are both 2.
2
c. y = –4(x – 4)2 + 1; y = 4 x – 5 + 1
2
49. y = –10 x –
1
10
50. y = 8 x – 1
2
4
2
–
– 3
2
51. y = 25x2 + 60x + 27
9
10
2
52. y = –9x2 + 24x – 10
53. y = 2x2 + 22x
1
2
54. y = x2 – 5x +
35
2
5-3
Translating Parabolas
ALGEBRA 2 LESSON 5-3
58. Answers may vary. Sample: The vertex is (3, 4),
so graph that point first. Then substitute 2 for x to
find (2, 2) is on the graph. Plot that point and the
symmetrically opposite point from the line of
symmetry, (4, 2), and sketch the parabola. Plot
more and more symmetric pairs if a more accurate
curve is desired.
64. yes
59. yes
68. Any real numbers a
and k such that a + k =
1 will work. However, if
a = 0 and k = 1, the
function will be linear
rather than quadratic.
60. yes
1
61. no; y = –3 x + 3
2
4
+ 3
65. no; y = –4 x –
2
+
21
4
66. yes
67. no; y = 100 x – 1
5
62. yes
69. a = 3, k = –1
63. no; y = (x + 1)2 + 7
70. a = –6, k = 35
5-3
3
4
2
+6
Translating Parabolas
ALGEBRA 2 LESSON 5-3
71. a = 1 , k = 1
5
72. a = –
11
47
,k=
3
3
73. a = 1, k = –650
77. Answers may vary.
Sample: The graph of
y = (x – 6)2 + 7 is the
graph of y = (x + 6)2
translated right 12
units and up 7 units.
74. y = –(x – 4)2 + 2
75. Check students’
work.
76. minimum; 150
81. y = – 1 (x – 3)2
4
1
82. y = – (x – 4)2
4
83. y = 2(x – 1)2
84. y = –4(x + 3)2
78. a. ah2 + k
b. h = 0 or a = 0
(Note, however,
that if a = 0, the
function will not be
quadratic.)
85. 4
86. 4
87. 7
7
1
79. y = 4 x2
88. 6
1
2
80. y = (x + 3)2
5-3
89. –3
Translating Parabolas
ALGEBRA 2 LESSON 5-3
90. 2
93.
91.
92.
94. 30
20
95. –4
0
96. max: 6, min: 0
5-3
Translating Parabolas
ALGEBRA 2 LESSON 5-3
1. Graph the function y = 4(x – 3)2.
2. Identify the vertex and the y-intercept of the graph of y = –2(x + 5)2 + 8.
(–5, 8), –42
3. Write the equation y = 3x2 + 12x – 1 in vertex form.
y = 3(x + 2)2 – 13
5-3
Factoring Quadratic Expressions
ALGEBRA 2 LESSON 5-4
(For help, go to Lessons 1-2 and 5-1.)
Simplify each expression.
1. x2 + x + 4x – 1
2. 6x2 – 4(3)x + 2x – 3
3. 4x2 – 2(5 – x) – 3x
Multiply.
4. 2x(5 – x)
5. (2x – 7)(2x – 7)
6. (4x + 3)(4x – 3)
5-4
Factoring Quadratic Expressions
ALGEBRA 2 LESSON 5-4
Solutions
1. x2 + x + 4x – 1 = x2 + (1 + 4)x – 1 = x2 + 5x – 1
2. 6x2 – 4(3)x + 2x – 3 = 6x2 – 12x + 2x – 3 = 6x2 + (–12 + 2)x – 3 =
6x2 – 10x – 3
3. 4x2 – 2(5 – x) – 3x = 4x2 – 10 + 2x – 3x = 4x2 + (2 – 3)x – 10 = 4x2 – x – 10
4. 2x(5 – x) = 2x(5) – 2x(x) = 10x – 2x2 = –2x2 + 10x
5. (2x – 7)(2x – 7) = (2x)2 – 2(2x)(7) + (7)2 = 4x2 – 28x + 49
6. (4x + 3)(4x – 3) = (4x)2 – (3)2 = 16x2 – 9
5-4
Factoring Quadratic Expressions
ALGEBRA 2 LESSON 5-4
Factor each expression.
a. 15x2 + 25x + 100
15x2 + 25x + 100 = 5(3x2) + 5(5x) + 5(20)
= 5(3x2 + 5x + 20)
Factor out the GCF, 5
Rewrite using the
Distributive Property.
b. 8m2 + 4m
8m2 + 4m = 4m(2m) + 4m(1)
Factor out the GCF, 4m
= 4m(2m + 1)
Rewrite using the
Distributive Property.
5-4
Factoring Quadratic Expressions
ALGEBRA 2 LESSON 5-4
Factor x2 + 10x + 24.
Step 1: Find factors with product ac and sum b.
Since ac = 24 and b = 10, find positive factors with product 24
and sum 11.
Factors of 24
Sum of factors
1, 24
25
2, 12
14
3, 8
11
6, 4
10
}
}
Step 2: Rewrite the term bx using the factors you found. Group the
remaining terms and find the common factors for each group.
x2 + 10x + 24
x2 + 4x + 6x + 24
Rewrite bx : 10x = 4x + 6x.
x(x + 4) + 6(x + 4)
Find common factors.
5-4
Factoring Quadratic Expressions
ALGEBRA 2 LESSON 5-4
(continued)
Step 3: Rewrite the expression as a product of two binominals.
x(x + 4) + 6(x + 4)
(x + 6)(x + 4)
Rewrite using the Distributive Property.
Check: (x + 6)(x + 4) = x2 + 4x + 6x + 24
= x2 + 10x + 24
5-4
Factoring Quadratic Expressions
ALGEBRA 2 LESSON 5-4
Factor x2 – 14x + 33.
Step 1: Find factors with product ac and sum b.
Since ac = 33 and b = –14, find negative factors with product 33
and sum b.
Factors of 33
Sum of factors
–1, –33
–34
–3, –11
–14
Step 2: Rewrite the term bx using the factors you found. Then find common
factors and rewrite the expression as a product of two binomials.
x2 + 14x + 33
x2 – 3x – 11x + 33Rewrite bx.
}
}
x(x – 3) – 11(x – 3)
Find common factors.
(x – 11)(x – 3)
Rewrite using the Distributive
Property.
5-4
Factoring Quadratic Expressions
ALGEBRA 2 LESSON 5-4
Factor x2 + 3x –28.
Step 1: Find factors with product ac and sum b.
Since ac = –28 and b = 3, find factors 2 with product –28 and
sum 3.
Factors of –28
Sum of factors
1, –28
–27
–1, 28
27
2, –14
–12
–2, 14
12
4, –7
–3
–4, 7
3
Step 2: Since a = 1, you can write binomials using the factors you found.
x2 + 3x – 28
(x – 4)(x + 7)
Use the factors you found.
5-4
Factoring Quadratic Expressions
ALGEBRA 2 LESSON 5-4
Factor 6x2 – 31x + 35.
Step 1: Find factors with product ac and sum b.
Since ac = 210 and b = –31, find negative factors with product
210 and sum –31.
Factors of 210
Sum of factors
–1, –210 –2, –105 –3, –70 –5, –42 –10, –21
–211
–107
–73
–47
–31
Step 2: Rewrite the term bx using the factors you found. Then find common
factors and rewrite the expression as the product of two binomials.
6x2 – 31x + 35
2x(3x – 5) – 7(3x – 5)
Find common factors.
(2x – 7)(3x – 5)
Rewrite using the Distributive Property.
}
Rewrite bx.
}
6x2 – 10x – 21x + 35
5-4
Factoring Quadratic Expressions
ALGEBRA 2 LESSON 5-4
Factor 6x2 + 11x – 35.
Step 1: Find factors with product ac and sum b.
Since ac = 210 and b = 11, find factors with product –210 and
sum 11.
Factors of –210 –1, –210
Sum of factors
–209
Factors of –210
Sum of factors
–3, 70
67
–1, 210
209
5, –42
–37
2, –105 –2, 105
–103
103
–5, 42
37
10, –21
–11
3, –70
–67
–10, 21
11
Step 2: Rewrite the term bx using the factors you found. Then find common
factors and rewrite the expression as the product of two binomials.
6x2 + 11x + 35
6x2 – 10x + 21x – 35
Rewrite bx.
2x(3x – 5) + 7(3x – 5)
Find common factors.
(2x + 7)(3x – 5)
Rewrite using the Distributive Property.
5-4
Factoring Quadratic Expressions
ALGEBRA 2 LESSON 5-4
Factor 100x2 + 180x + 81.
100x2 + 180x + 81 = (10x)2 + 180 + (9)2
Rewrite the first and third
terms as squares.
= (10x)2 + 180 + (9)2
Rewrite the middle term to
verify the perfect square
trinomial pattern.
= (10x + 9)2
a2 + 2ab + b2 = (a + b)2
5-4
Factoring Quadratic Expressions
ALGEBRA 2 LESSON 5-4
A square photo is enclosed in a square frame, as shown in
the diagram. Express the area of the frame (the shaded area) in
completely factored form.
Relate: frame area equals the outer area minus the inner area
Define: Let x = length of side of frame.
Write:
area = x2 – (7)2
= (x + 7)(x – 7)
The area of the frame in factored form is (x + 7)(x – 7) in2.
5-4
Factoring Quadratic Expressions
ALGEBRA 2 LESSON 5-4
Pages 259–261 Exercises
1. 3; 3(a2 + 3)
9. (x + 2)(x + 5)
17. (d – 3)(d – 9)
2. 5; 5(5b2 – 7)
10. (x + 2)(x + 8)
18. (x – 4)(x – 9)
3. x; x(x – 2)
11. (y + 3)(y + 12)
19. (x – 7)(x + 2)
4. t; t(5t + 7)
12. (x + 2)(x + 20)
20. (x + 5)(x – 4)
5. 7y; 7y(2y + 1)
13. (x – 1)(x – 2)
21. (x – 8)(x + 5)
6. 9p; 9p(3p – 1)
14. (x – 12)(x – 1)
22. (c + 9)(c – 7)
7. (x + 1)(x + 2)
15. (r – 2)(r – 9)
23. (x + 15)(x – 5)
8. (x + 2)(x + 3)
16. (x – 4)(x – 6)
24. (t – 11)(t + 4)
5-4
Factoring Quadratic Expressions
ALGEBRA 2 LESSON 5-4
25. (3x + 4)(x + 9)
34. (z + 4)(2z – 7)
43. (x + 2)(x – 2)
26. (x – 8)(2x – 3)
35. (x + 4)(3x – 4)
44. (c + 8)(c – 8)
27. (r + 2)(5r + 13)
36. (4k + 3)(7k – 2)
45. (3x + 1)(3x – 1)
28. (m – 3)(2m – 5)
37. (x + 1)2
46. x2 – 16; (x + 4)(x – 4)
29. (t + 4)(5t + 8)
38. (t – 7)2
47. 5x – 1 by 5x – 1
30. (x – 12)(2x – 3)
39. (x – 9)2
48. (3x – 17) cm
31. (x + 4)(3x – 5)
40. (2n – 5)2
49. (x + y)2 – y2; x(x + 2y)
32. (y + 4)(5y – 8)
41. (3x + 8)2
50. (x – 7) ft
33. (x – 2)(7x + 6)
42. (9z + 2)2
51. 9(x + 2)(x – 2)
5-4
Factoring Quadratic Expressions
ALGEBRA 2 LESSON 5-4
52. 2(3z + 2)(3z – 2)
61. 3(y + 3)(y + 5)
53. 3(2y + 5)(2y – 5)
62. –(x – 1)(x – 4)
54. 16(2t + 1)(2t – 1)
63. 2(x – 5)(2x – 1)
55. 3(2x + 3)2
64. 1 (x + 1)(x – 1)
56. 4(2x – 5)2
65. –6(z2 + 100)
57. 2(a – 4)2
66.
58. 3(x – 9)(x + 1)
67. (x – 70) ft
2
h(R + r )(R – r )
59. 2(3b – 1)(3b + 5)
60. 4(n – 2)(n – 3)
5-4
68. Factor 3 from the terms
to get 3(x2 + 2x – 24).
Look for numbers
whose product is –24
and whose sum is 2.
The numbers –4 and 6
work. The complete
factorization is
3(x – 4)(x + 6).
69. Check students’ work.
70. The third line should be
x(2x – 5) – (2x – 5), and
the final line should be
(x – 1)(2x – 5).
Factoring Quadratic Expressions
ALGEBRA 2 LESSON 5-4
71. First factor out 4x2 to
get 4x2(x2 + 6x + 8).
To factor x2 + 6x + 8,
note that the numbers
2 and 4 have a
product of 8 and a
sum of 6. The
complete factorization
is 4x2(x + 2)(x + 4).
76. (x – 10)(x – 9)
72. (0.5t + 0.4)(0.5t - 0.4)
81. D
73. 100(9x – 10)(9x + 10)
82. H
77. (2x + 9)(3x + 14)
78. 2(a + 1)(6a – 7)
79. A
80. G
83. [2] First separate out
–25 so it reads
(a2 – 2ab + b2) – 25,
and then factor
a2 – 2ab + b2. This
leaves you with
(a – b)2 – 25. Factor
this as the difference
of 2 perfect squares
and you get
(a – b + 5)(a – b – 5).
[1] no work shown, only
answer, OR work
shown with minor
error
74. 100(6z – 7)(6z + 7)
75. (x + 12)(x – 3)
5-4
Factoring Quadratic Expressions
ALGEBRA 2 LESSON 5-4
84. [4] a. By entering two given lists into a graphing
calculator and then having it calculate the
quadratic regression, you get
h = –16t2 + 22t + 3 as the quadratic model
for the ball’s height as a function of time.
b. –16t2 + 22t + 3
1
Multiply a  c.
–16  3 = –48
2
Find factors of –48 that add to be 22.
–1, 48; –2, 24; –3, 16; –4, 12 . . .
–2 + 24 = 22
3
Rewrite as –16t2 – 2t + 24t + 3.
4
Find common factors. –2t(8t + 1) + 3(8t + 1)
5
Rewrite using Distributive Property.
(8t + 1)(–2t + 3)
5-4
[3] appropriate steps,
with one
computational error
[2] explanation in either
part (a) OR (b)
[1] correct solution,
without explanation
of steps
Factoring Quadratic Expressions
ALGEBRA 2 LESSON 5-4
85. y = (x – 1)2
86. y = –2 x –
1
2
2
+
11
2
87. y = 5x2 – 1
88. 1
89. –29
90. 2
91. penny: 2.5 g,
nickel: 5 g,
dime 2.3 g
5-4
Factoring Quadratic Expressions
ALGEBRA 2 LESSON 5-4
Factor each expression completely.
1. 12x2 + 6x + 18
2. m2 + 11m + 18
6(2x2 + x + 3)
(m + 2)(m+ 9)
3. x2 – 14x – 15
4. x2 – 13x + 42
(x – 15)(x + 1)
5. 64x2 + 144x + 81
(8x + 9)2
7. 5k2 – 125
5(k + 5)(k – 5)
(x – 6)(x – 7)
6. 3x2 + 5x – 50
(3x – 10)(x – 5)
8. 15n2 – 8n + 1
(5n – 1)(3n – 1)
5-4
Quadratic Equations
ALGEBRA 2 LESSON 5-5
(For help, go to Lessons 5-2 and 5-4.)
Factor each expression.
1. x2 + 5x – 14
2. 4x2 – 12
3. 9x2 – 16
5. y = x2 – 4x + 4
6. y = x2 – 4x
Graph each function.
4. y = x2 – 2x – 5
5-5
Quadratic Equations
ALGEBRA 2 LESSON 5-5
(For help, go to Lessons 8-3 and 7-4.)
Solutions
1. Factors of –14 with a sum of 5: 7 and –2
x2 + 5x – 14 = (x + 7)(x – 2)
2. 4x2 – 12x = 4x(x – 3)
3. 9x2 – 16 = (3x)2 – (4)2 = (3x – 4)(3x + 4)
4. y = x2 – 2x – 5
(–2)
axis of symmetry: x = – 2(1) = 1
vertex: (1, –6)
5-5
Quadratic Equations
ALGEBRA 2 LESSON 5-5
(For help, go to Lessons 8-3 and 7-4.)
Solutions (continued)
5. y = x2 – 4x + 4
(–4)
axis of symmetry: x = – 2(1) = 2
vertex: (2, 0)
6. y = x2 – 4x
axis of symmetry: x = 2
vertex: (2, –4)
5-5
Quadratic Equations
ALGEBRA 2 LESSON 5-5
3x2 – 20x – 7 = 0
3x2 – 20x – 7 = 0
Write in standard form.
3x2 – 21x + x – 7 = 0
Rewrite the bx term.
3x(x – 7) + (x – 7) = 0
Find common factors.
(3x + 1)(x – 7) = 0
3x + 1 = 0 or x – 7 = 0
x=–
1
or x = 7
3
The solutions are –
Factor using the Distributive Property.
Use the Zero-Product Property.
Solve for x.
1
and 7.
3
5-5
Quadratic Equations
ALGEBRA 2 LESSON 5-5
(continued)
3x2 – 20x = 7
Check:
3–1
3
2
– 20 – 1
3
1 + 20
3
3
3x2 – 20x = 7
7
3(7)2 – 20(7)
7
7
147 – 140
7
7=7
7= 7
5-5
Quadratic Equations
ALGEBRA 2 LESSON 5-5
Solve 6x2 – 486 = 0.
6x2 – 486 = 0
6x2 486
=
6
6
Rewrite in the form ax2 = c.
Isolate x2.
x2 = 81
Simplify.
x = ±9
Take the square root of each side.
5-5
Quadratic Equations
ALGEBRA 2 LESSON 5-5
The function y = –16x2 + 270 models the height y in feet of a
heavy object x seconds after it is dropped from the top of a building
that is 270 feet tall. How long does it take the object to hit the ground?
y = –16x2 + 270
0 = –16x2 + 270
–270 = –16x2
Substitute 0 for y.
Isolate x2.
16.875 = x2
±4.1
x
Take the square root of each side.
The object takes about 4.1 seconds to hit the ground.
Check: Is the answer reasonable? The negative number –4.1 is also a
solution to the equation. However, since a negative value for time
has no meaning in this case, only the positive solution is
reasonable.
5-5
Quadratic Equations
ALGEBRA 2 LESSON 5-5
Use a graphing calculator to solve 2x2 + 7x – 1 = 0. Round the
solution to the nearest hundredth.
The solutions are x
–3.64 and x
5-5
0.14.
Quadratic Equations
ALGEBRA 2 LESSON 5-5
A carpenter wants to cut a piece of plywood in the shape of a
right triangle. He wants the hypotenuse of the triangle to be 6 feet long,
as shown in the diagram. About how long should the perpendicular
sides be?
Relate: From the Pythagorean Theorem we know for a
right triangle that the hypotenuse squared equals
the sum of the squares of the other two sides.
Define: Let x = the shorter leg.
Then x + 1 = the longer leg.
Write:
62 = x2 + ( x + 1 )2
36 = x2 + x2 + 2x + 1
0 = 2x2 + 2x – 35
5-5
Multiply.
Write in standard form.
Quadratic Equations
ALGEBRA 2 LESSON 5-5
(continued)
Graph the related function y = 2x2 + 2x – 35. Use the CALC feature to
find the positive solution.
The sides of the triangle are 3.7 ft and 4.7 ft.
5-5
Quadratic Equations
ALGEBRA 2 LESSON 5-5
Pages 266-268 Exercises
1. –4, –2
9. –4, 4
2. 3, 6
10. – ,
3. –1,
3
2
4. 5
5. –2, –1
6.
– 2, 6
3
7. –4, 4
8. –2, 2
3
17. –1, 2
5 5
3 3
11. –
5,
12. –2
18. –2
5
2, 2
2
5, 2
5
19. a. about 6.61 s
b. about 6.89 s
20. –4.30, –0.70
13. 0, 4
21. –1.32, 8.32
2
14. – 3 , 0
22. –0.78, 1.28
7 7
15. – ,
2 2
23. –1.67, –1.5
16. –4, 4
24. –0.59, 2.26
5-5
Quadratic Equations
ALGEBRA 2 LESSON 5-5
25. –0.94, 2.34
26. –5.53, 0.36
w
27. –1, 0.25
28. –3.12, 5.12
29. –1.46, 5.46
30. –5.16, 1.16
31. –1.16, 2.16
38. –3, 3
32. a. Answers may
vary. Sample:
7.2 cm
4.4 cm
1.6
b. the tree trunk
33. a. 5 10 or about 1.76 s
9
b. 10 10 or about 10.54 s
3
34. Check students’ work.
1
2
39. – , 3
40. –8.69, 0.69
41. – 3 , – 2
2
42. –4,
3
5
2
43. –5.89, 5.89
44. –3.25, 0.92
35. 3 ft
45. –3.58, 3.58
36. –10, 4
46. –4, 0
37. 3, 8
5-5
Quadratic Equations
ALGEBRA 2 LESSON 5-5
47. 1, 7
56. – 5 , – 4 , (2, 2)
48. –5, 3
57. To find the x-coordinates, set the right side of the
first equation equal to the right side of the second
equation and solve for x. Then find the
corresponding y-values by substituting each
solution into the simpler of the two original
equations.
49. –10, –1
50. –1.5, 0.5
3
9
51. –6, 0
58. Answers may vary. Sample: x2 – 8x + 15 = 0
52. –1, 4
59. Answers may vary. Sample: x2 + x – 6 = 0
53. –4, 3.5
60. Answers may vary. Sample: x2 + 7x + 6 = 0
54. (–1, 1), (2, 4)
61. Answers may vary. Sample: 6x2 – 7x + 2 = 0
55. (0, –2), (2, 2)
5-5
Quadratic Equations
ALGEBRA 2 LESSON 5-5
62. x = 4, y = 1 or x = –4, y = 9
66. a. Answers may vary. Sample: If
x =/ h, then x – h will be nonzero,
63. a. y = x2 + 6x + 14
(x – h)2 will be positive, and
a(x – h)2 will be positive. Adding a
b. about –4.732, about –1.268
positive to k will always result in a
number greater than k. So the
64. Solve (x – 4)(x – 6) = 0 to find that
point (h, k) is the lowest point
the zeros of y = x2 – 10x + 24 are 4
when x = h on the graph of
and 6. Average 4 and 6 to get 5. This
y = (x – h)2 + k.
is the x-coordinate of the vertex.
Substitute 5 for x in x2 – 10x + 24 to
b. No; (x – h)3 can be negative.
find that –1 is the y-coordinate of the
vertex. The vertex is (5, –1).
67. B
65. a. 100 ft
68. G
b. 5 s
69. D
5-5
Quadratic Equations
ALGEBRA 2 LESSON 5-5
75. (5z + 3)(5z – 3)
70. B
71. A
76. 3s(2s + 3)
72. B
77. 3  3; 5
73. [2] 6x2 – 15x – 9 = 0
3(2x2 – 5x – 3) = 0
3(2x + 1)(x – 3) = 0
2x + 1 = 0
x–3=0
2x = –1
x=3
78. 2  3; 2
79. 3  3; 0
x=–1
2
80. Distributive Property
x = – 1, 3
2
81. Comm. Prop. of Add.
[1] solution only, no work shown
82. Assoc. Prop. of Mult.
74. (3x – 1)(x – 1)
83. Additive Inverse Prop.
5-5
Quadratic Equations
ALGEBRA 2 LESSON 5-5
Solve each equation by factoring.
1. 4x2 – 17x – 15 = 0
2. 10x2 + 19x + 6 = 0
3
4
3
2
– ,5
– ,–
2
5
3. Solve 3x2 = 4800 by using square roots.
±40
4. Use a graphing calculator to solve 2x2 + 5x – 9 = 0. Round the
solutions to the nearest hundredth.
–3.71, 1.21
5-5
Complex Numbers
ALGEBRA 2 LESSON 5-6
(For help, go to Skills Handbook pg 855.)
Factor each expression.
1.
32 + 42
2.
(–2)2 + 82
3.
52 + (–12)2
4.
62 + 102
5.
x2 + x2
6.
(3x)2 + (4x)2
5-6
Complex Numbers
ALGEBRA 2 LESSON 5-6
Solutions
1.
32 + 42 =
2.
(–2)2 + 82 =
3.
52 + (–12)2 =
4.
62 + 102 =
5.
x2 + x2 =
6.
(3x)2 + (4x)2 =
9 + 16 =
25 = 5
4 + 64 =
68 =
25 + 144 =
36 + 100 =
2x2 =
2•
4 • 17 = 2
17
169 = 13
136 =
x2 =
9x2 + 16x2 =
4 • 34 = 2
2•x=x
25x2 = 5x
5-6
2
34
Complex Numbers
ALGEBRA 2 LESSON 5-6
Simplify
–54 =
=
–1 • 54
–1 •
=i•
54
54
=i•3
= 3i
–54 by using the imaginary number i.
6
6
5-6
Complex Numbers
ALGEBRA 2 LESSON 5-6
Write
–121 – 7 in a + bi form.
–121 – 7 = 11i – 7
= –7 + 11i
Simplify the radical expression.
Write in the form a + bi.
5-6
Complex Numbers
ALGEBRA 2 LESSON 5-6
Find each absolute value.
a.
|–7i|
–7i is seven units from the origin on the imaginary axis.
So |–7i| = 7
b.
|10 + 24i|
|10 + 24i| =
=
102 + 242
100 + 576 = 26
5-6
Complex Numbers
ALGEBRA 2 LESSON 5-6
Find the additive inverse of –7 – 9i.
–7 – 9i
–(–7 – 9i)
Find the opposite.
7 + 9i
Simplify.
5-6
Complex Numbers
ALGEBRA 2 LESSON 5-6
Simplify (3 + 6i) – (4 – 8i).
(3 + 6i) – (4 – 8i) = 3 + (–4) + 6i + 8i
= –1 + 14i
Use commutative and
associative properties.
Simplify.
5-6
Complex Numbers
ALGEBRA 2 LESSON 5-6
Find each product.
a. (3i)(8i)
(3i)(8i) = 24i 2
Multiply the real numbers.
= 24(–1)
Substitute –1 for i 2.
= –24
Multiply.
b. (3 – 7i )(2 – 4i )
(3 – 7i )(2 – 4i ) = 6 – 14i – 12i + 28i 2
Multiply the binomials.
= 6 – 26i + 28(–1)
Substitute –1 for i 2.
= –22 – 26i
Simplify.
5-6
Complex Numbers
ALGEBRA 2 LESSON 5-6
Solve 9x2 + 54 = 0.
9x2 + 54 = 0
9x2 = –54
Isolate x2.
x2 = –6
x = ±i
6
Find the square root of each side.
Check: 9x2 + 54 = 0
9(i
6)2 + 54
0
9(6)i 2 + 54
0
54(–1)
9x2 + 54 = 0
9(i(–
–54
6))2 + 54
0
9(6i 2) + 54
0
54(–1)
–54
–54 = –54
54 = 54
5-6
Complex Numbers
ALGEBRA 2 LESSON 5-6
Find the first three output values for f(z) = z2 – 4i. Use z = 0 as
the first input value.
Use z = 0 as the first input value.
f(0) = 02 – 4i
= –4i
f(–4i )= (–4i )2 – 4i
First output becomes second input.
Evaluate for z = –4i.
= –16 – 4i
f(–16 – 4i )= (–16 – 4i )2 – 4i
Second output becomes third input.
Evaluate for z = –16 – 4i.
= [(–16)2 + (–16)(–4i ) + (–16)(–4i) + (–4i )2] – 4i
= (256 + 128i – 16) – 4i
= 240 + 124i
The first three output values are –4i, –16 – 4i, 240 + 124i.
5-6
Complex Numbers
ALGEBRA 2 LESSON 5-6
Pages 274–276 Exercises
9. –10i
1. 2i
2. i
7
10. 6i
3. i
15
11. 2 + i
4. 9i
5. 5i
2
6. 4i
7. 4i
8. 9i
2
17. –2 – 5i
2
18. 4 + 6i
3
19. 2
12. 8 + 2i
2
20. 13
13. 6 – 2i
7
21. 2
2
14. 3 + 2i
22.
15. 7 – 5i
23. 3
16. 2 + i
24. –4i
5-6
17
5
2
2
Complex Numbers
ALGEBRA 2 LESSON 5-6
25. –5 + 3i
34. –7 – 10i
43. ± 8i
26. –9 – i
35. 10
44. ± i
27. 3 + 2i
36. 26 – 7i
45. ±6i
28. 4 – 7i
37. 9 + 58i
46. ±
29. 6 + 3i
38. 9 – 23i
47. –i, –1 – i, i
30. 1 – 7i
39. –36
48. –2i, –4 – 2i, 12 + 14i
31. 7 + 4i
40. 65 + 72i
49. 1 – i, 1 – 3i, –7 – 7i
32. –2 – 3i
41. ± 5i
50. ± i
33. 10 + 6i
42. ± i
2
i
51. ±7i
2
5-6
3
3
7
15
3
65
Complex Numbers
ALGEBRA 2 LESSON 5-6
52. ± i
53. No; the test scores
were real numbers.
He added the scores
and divided by the
number of scores.
The set of real
numbers is closed
with respect to
addition and division
so he should have
gotten a real
number.
54. a. A: –5, B: 3 + 2i, C: 2 – i, D: 3i, E: –6 – 4i, F: –1 + 5i
b. 5, –3 – 2i, –2 + i, –3i, 6 + 4i, 1 – 5i
55. a. Check students’ work.
b. a circle with radius 10 and center at the origin
56. –5, 5
57. 288i
58. –1 + 5i
59. 10 – 4i
60. 8 – 2i
61. 11 – 5i
5-6
Complex Numbers
ALGEBRA 2 LESSON 5-6
62. 6 + 10i
63. 7 – i
64. 10 + 11i
65. –27 + 8i
66. –13 + I
67. (continued)
b. Answers may vary. Sample: The sum of a
complex number a + bi and its conjugate is
2a. The product of a + bi and its conjugate
is the square of the absolute value of
a + bi. The absolute values of a complex
number and its conjugate are equal.
c.
They are symmetric
images of each other
with respect to the real
axis.
67. a. row 2: 2, 5, 5, 5
row 3: 6, 10, 10, 10
row 4: –12, 100, 10, 10
5-6
Complex Numbers
ALGEBRA 2 LESSON 5-6
67. (continued)
d. True; the additive
inverse of a + bi is
–a – bi, and the
conjugate of –a – bi
is –a + bi. The
conjugate of a + bi
is a – bi, and the
additive inverse of
a – bi is –a + bi.
68. x = –7, y = 3
69. x = 16 , y = – 19
3
8
71. (a + bi )(a – bi ) = a2 + b2; since a and b are real, so
is a2 + b2.
72. 0.383 + 0.11i, 0.517589 + 0.19426i
73. nonzero real numbers x and y
74. (3 + 4i )1 = 3 + 4i and 32 + 42 = 25;
(3 + 4i )2 = –7 + 24i and
(–7)2 + (24)2 = 625 = 252; (3 + 4i )3 = –117 + 44i and
(–117)2 + (44)2 = 15,625 = 253;
(3 + 4i)4 = –527 – 336i and
(–527)2 + (–336)2 = 390,625 = 254;
(3 + 4i )5 = –237 – 3116i and
(–237)2 + (–3116)2 = 9,765,625 = 255
70. x = –7, y = –3
5-6
Complex Numbers
ALGEBRA 2 LESSON 5-6
75.
78. B
79. H
If the points for the
origin, a + bi, c + di,
and the sum are not
collinear, then they
form the vertices of a
parallelogram.
80. [4] x4 – 16 = (x2 – 4)(x2 + 4) = 0
(x – 2)(x + 2)(x2 + 4) = 0
x–2=0
x+2=0
x2 = –4
x=2
x = –2
x = –4
x = ±2i
[3] appropriate methods, with one computational
error
76. D
[2] initial factoring correct but solvings are
incorrect
77. G
[1] answer only, without work shown
5-6
Complex Numbers
ALGEBRA 2 LESSON 5-6
81. –2.351, 0.851
86.
88.
82. –0.640, 0.390
83. –5.562, –1.438
84.
87.
89.
85.
5-6
Complex Numbers
ALGEBRA 2 LESSON 5-6
1. Simplify –
–169.
–13i
2. Find |–1 + i |
2
Simplify each expression.
3. (–7i )(–3i )
4. (9 + 10i) + (–7 + 4i)
–21
2 + 14i
5. (–3 + 4i ) – (–3 – 8i )
6. (–5 + 2i)(7 – 4i)
–27 + 34i
12i
7. Solve 1 x2 + 2 = 0.
2
±2i
5-6
Completing the Square
ALGEBRA 2 LESSON 5-7
(For help, go to Lessons 5-1 and page 262.)
Simplify each expression.
1. (x – 3)(x – 3)
2. (2x – 1)(2x – 1)
3. (x + 4)(x + 4) – 3
4. ±
25
5. ±
48
6. ±
–4
7. ±
9
16
5-7
Completing the Square
ALGEBRA 2 LESSON 5-7
Solutions
1. (x – 3)(x – 3) = x2 – 2(3)x + 32 = x2 – 6x + 9
2. (2x – 1)(2x – 1) = (2x)2 – 2(1)(2x) + 12 = 4x2 – 4x + 1
3. (x + 4)(x + 4) – 3 = x2 + 2(4)x + 42 – 3 = x2 + 8x + 16 – 3 = x2 + 8x + 13
4. ±
25 = ±5
5. ±
48 = ±
16 • 3 = ±4
6. ±
–4 = ±
–1 • 4 = ±2i
7. ±
9
=±
16
9
3
3
=± 4
16
5-7
Completing the Square
ALGEBRA 2 LESSON 5-7
Solve x2 – 12x + 36 = 9.
x2 – 12x + 36 = 9
(x – 6)2 = 9
x – 6 = ±3
x – 6 = 3 or x – 6 = –3
x = 9 or
Factor the trinomial.
Find the square root of each side.
Solve for x.
x=3
5-7
Completing the Square
ALGEBRA 2 LESSON 5-7
Find the missing value to complete the square: x2 + 20x +
b
2
2
= 20
2
2
= 100
x2 + 20x + 100
2
Find b . Substitute 20 for b.
2
Complete the square.
5-7
.
Completing the Square
ALGEBRA 2 LESSON 5-7
Solve x2 + 4x + 1 = 0.
4
2
2
=4
Find
x2 + 4x = –1
Rewrite so all terms containing x are on one side.
x2 + 4x + 4 = –1 + 4
Complete the square by adding 4 to each side.
(x + 2)2 = 3
x+2=±
b 2
.
2
Factor the perfect square trinomal.
3
x = –2 ±
Find the square root of each side.
3
Solve for x.
5-7
Completing the Square
ALGEBRA 2 LESSON 5-7
(continued)
x2 + 4x + 1
0
(–2 + 3)2 + 4(–2 +
3) + 1
(–2)2 + 2(–2 3) + ( 3)2 + (–8) + 4 3 + 1
0
0
Check:
4–4
3+3–8+4
(4 + 3 – 8 + 1) + (–4
3+1
3+4
3) 0
0 = 0
x2 + 4x + 1 0
(–2 – 3)2 + 4(–2 –
3) + 1
(–2)2 – 2(–2 3) + ( 3)2 – 8 – 4 3 + 1
4+4
3+3–8–4
(4 + 3 – 8 + 1) + (4
5-7
0
3+1
3–4
0
0
0
3) 0
0 = 0
Completing the Square
ALGEBRA 2 LESSON 5-7
Solve x2 + 6x + 12 = 0.
6
2
2
=9
Find
b 2
.
2
x2 + 6x = –12
Rewrite so all terms containing x are
on one side.
x2 + 6x + 9 = –12 + 9
Complete the square by adding 9 to
each side.
(x + 3)2 = –3
Factor the perfect square trinomial.
x + 3 = ± –3
x = –3 ±
= –3 ± i
Find the square root of each side.
–3
–3
Solve for x.
Simplify.
5-7
Completing the Square
ALGEBRA 2 LESSON 5-7
Solve 2x2 + 7x – 1 = 0.
7
x–1 =0
2
2
7
1
x2 + x =
2
2
x2 +
Divide each side by 2.
Rewrite so all terms containing x are on
one side.
7 2
49
2
=
2
Find b .
2
16
2
7
49
x2 + 2 x + 16 =
7 2
x+
=
4
7
x+4=
49
Complete the square by adding
to
16
each side.
1 49
2 + 16
57
16
± 57
4
x=–
7
±
4
Factor the perfect square trinomial.
Find the square root of each side.
57
4
Solve for x.
5-7
Completing the Square
ALGEBRA 2 LESSON 5-7
Write y = x2 + 5x + 2 in vertex form.
y = x2 + 5x + 2
=
x2
2
+ 5x + 5
2
+2–
5
2
Complete the square. Add and subtract
2
5 on the right side.
2
2
= x+ 5
2
2
= x+ 5
2
2
+ 2 – 25
Factor the perfect square trinomial.
– 17
Simplify.
4
4
The vertex form is y = x + 5
2
2
– 17.
4
5-7
Completing the Square
ALGEBRA 2 LESSON 5-7
The monthly profit P from the sales of rugs woven by a family
rug-making business depends on the price r that they charge for a rug.
The profit is model by P = –r 2 + 500r – 59,500. Write the function in
vertex form. Use the vertex form to find the price that yields the
maximum monthly profit and the amount of the maximum profit.
P = –r 2 + 500r – 59500
= –(r 2 – 500r) – 59500
Factor –1 from the first two
terms.
= –[r 2 – 500r + (–250)2] – 59500 + (–250)2 Add and subtract (–250)2 on
the right side.
= –(r – 250)2 – 59500 + 62500
Factor the perfect square
trinomial.
= –(r – 250)2 + 3000
Simplify in vertex form.
The vertex is (250, 3000). A price of $250 per rug gives a maximum
monthly profit of $3000.
5-7
Completing the Square
ALGEBRA 2 LESSON 5-7
Pages 281–283 Exercises
1. –4, –2
9. 144
17. –1, 9
2. –8, 12
10. 100
18. –3 ± i
3. –1, 3
11. 4
19. ±2i
12. 4
20. –1 ± 2i
5. –4, 3
13. –4, 7
21. 1 ±
6. 1, 11
14. –1, 4
22. 3 ± i 31
7. 81
15. –3 ± 4i
8. 1
16. 1 ± i
4. –
4
16
8
,–
3
3
9
2
2
5-7
13
5
2
23. 2 ± 15
3
24. – 5 ± i 3
4
4
Completing the Square
ALGEBRA 2 LESSON 5-7
2
25. –4, 7
34. y = –4 x + 5
26. 2 ± 1 i
1
1
35. y = (x – 5)2 – ; 5, – 1
2
8
2
3
3
27. – 3 , 1
2 2
16
2
8 16
2
1
5
36. y = – (x – 2)2 + 3; (2, 3)
28. y = (x + 2)2 – 11
37. a. (60, 5000)
b. $5000
29. y = –(x – 2)2 + 3
30. y = –2 x –
+ 73 ; – 5 , 73
3
2
2
+
31. y = (x + 2)2 – 3
32. y = 2(x – 2)2 – 7
33. y = –(x + 1)2 + 4
c. $60
11
2
38. Add –11 to each side of the given equation to
obtain x2 + 8x = –11. Then add the square of half
the coefficient of x to each side to obtain
x2 + 8x + 42 = –11 + 42, or x2 + 8x + 16 = 5.
Rewrite the left side of the last equation as
(x + 4)2 to obtain (x + 4)2 = 5.
5-7
Completing the Square
ALGEBRA 2 LESSON 5-7
39. a.
b.
c.
d.
e.
(59, 36.81)
36.81 ft
7.65 ft
about 120 ft
Answers may vary. Sample: The path is
parabolic. Also, the linear model does not
predict that the ball will eventually hit the
ground.
40. –10, 10
41. –20, 20
42. –22, 22
43. –16, 16
5-7
44. –18, 18
45. –10, 10
46. –1, 1
47. –12, 12
48. –12, 12
Completing the Square
ALGEBRA 2 LESSON 5-7
49. row 2: 47, 46, 45, 44, 43, 42, 41, 40
row 3: 96, 141, 184, 225, 264, 301, 336, 369, 400
a.
49. (continued)
d. 625 units2; 25
units by 25 units
A = –w2 + 50w
e. A = w(50 – w);
yes; both
equations are
quadratic and
model the same
situation.
b. Check students’ work.
c. The numbers w such that 0 < w < 50; since the
perimeter is 100 the width would have to be
less than 50, and since length can’t be negative
it would have to be greater than 0.
5-7
Completing the Square
ALGEBRA 2 LESSON 5-7
50. a.
64. A
56.
65. G
57.
(315, 630)
b. 630 ft
c. 630 ft
5
8
8
– 3 ± 41
8
8
3
– a,
2
2a
–a ± a 13
6
3 1
, , a =/ 0
a a
55. 1 ±
58.
59.
3
51. –12 ± 3
17
52. –4 ± 2
5
1
53. 4 ±
54. –
57
12
2 1
,
3 3
66. C
1
60. –
,–
, a =/
2a
2a
0
3a
61. 3, –
, a =/ –
a+3
3
2
5
62. – 3a , 2a , a =/ 0
63. –5 + 3
2, 5 + 3
5-7
2
Completing the Square
ALGEBRA 2 LESSON 5-7
67. [2] x2 + 36
= 14x
–36 – 14x
–14x – 36
x2 – 14x
= –36
x2 – 14x + (–7)2 = –36 + (–7)2
x2 – 14x + 49 = 13
(x – 7)2 = 13
x – 7 = 13
x = 7 ± 13
[1] one computational error, or correct answer
without work shown
5-7
Completing the Square
ALGEBRA 2 LESSON 5-7
68. [4] 3x2 + 7x = 6 Move variables to the left side and constant to the right by
using the Addition Property of Equality.
7
+ 3 x = 2 Divide each side by 3.
7
49 121
1 7
x2 + x +
=
Add the square of • to each side.
3
36
36
2 3
2
x + 7 = 121 Factor left side.
6
36
7
x + = ± 11 Take the square root of each side.
6
6
x = – 7 ± 11 Add – 7 to each side.
6
6
6
x = –3, 2 Simplify – 7 – 11 and – 7 + 11 .
3
6
6
6
6
x2
[3] appropriate methods, but with one computational error
[2] correct listing of steps with incorrect explanations
5-7
Completing the Square
ALGEBRA 2 LESSON 5-7
68. (continued)
[1] correct solution,
without work
shown
75. (2, 0)
76. (3,
1)
77. (3, 1)
69. –2 +
2i
70. 5 +
2i
71. 84 + 5i
1
7
72. y = 2 x2 + 2 x +
9
73. y = – 1 x2 + x + 2
2
74. y = 3x2 – 5x + 2
5-7
Completing the Square
ALGEBRA 2 LESSON 5-7
Complete the square.
1. x2 + 60x +
2. x2 – 7x +
49
9
900
Simplify each expression.
3. x2 – 6x – 16 = 0
–2, 8
5. 3x2 + 5x – 28 = 0
–4, 7
3
4. x2 – 14x + 74 = 0
7 ± 5i
6. 4x2 – 6x + 3 = 0
3 ±i
4
5-7
3
4
The Quadratic Formula
ALGEBRA 2 LESSON 5-8
(For help, go to Lessons 1-2 and 5-1.)
Write each quadratic equation in standard form.
1.
y = 8 – 10x2
2.
y = (x + 2)2 – 1
3.
y = –2x(x – 1) + (x + 1)2
4.
y = (3x)2 – (x – 1)2
Evaluate the expression b2 – 4ac for the given values of a, b, and c.
5.
a = 1, b = 6, c = 3
6.
a = –5, b = 2, c = 4
7.
a = 3, b = – 6, c = 7
8.
a = 2, b = 3, c = –10
5-8
The Quadratic Formula
ALGEBRA 2 LESSON 5-8
Solutions
1. y = 8 – 10x2; y = –10x2 + 8
2. y = (x + 2)2 – 1 = x2 + 2(2)x + 22 – 1 = x2 + 4x + 4 – 1 = x2 + 4x + 3
3. y = –2x(x – 1) + (x + 1)2 = –2x2 + 2x + x2 + 2x + 1 = –x2 + 4x + 1
4. y = (3x)2 – (x – 1)2 = 9x2 – (x2 – 2x + 1) = 8x2 + 2x – 1
5. b2 – 4ac for a = 1, b = 6, c = 3: 62 – 4(1)(3) = 36 – 12 = 24
6. b2 – 4ac for a = –5, b = 2, c = 4: 22 – 4(–5)(4) = 4 – (–80) = 84
7. b2 – 4ac for a = 3, b = –6, c = 7: (–6)2 – 4(3)(7) = 36 – 84 = –48
8. b2 – 4ac for a = 2, b = 3, c = –10: 32 – 4(2)(–10) = 9 – (–80) = 89
5-8
The Quadratic Formula
ALGEBRA 2 LESSON 5-8
Use the Quadratic Formula to solve 3x2 + 23x + 40 = 0.
3x2 + 23x + 40 = 0
a = 3, b = 23, c = 40
x=
=
=
–b ±
–23 ±
–23 ±
b2 – 4ac
2a
232 – 4(3)(40)
2(3)
529 – 480
6
Write in standard form.
Find values of a, b, and c.
Write the Quadratic Formula.
Substitute.
Simplify.
–23 ± 49
6
–23 ± 7
=
6
30
16
=–
or –
6
6
8
= –5 or –
3
=
5-8
The Quadratic Formula
ALGEBRA 2 LESSON 5-8
(continued)
Check: 3x2 + 23x + 40 = 0
3(–5)2 + 23(–5) + 40
0
75 – 115 + 40
0
3x2 + 23x + 40 = 0
3 –
0=0
8 2
8
+ 23 – + 40
3
3
0
64 184
–
+ 40
3
3
0
0=0
5-8
The Quadratic Formula
ALGEBRA 2 LESSON 5-8
Solve 3x2 + 2x = –4.
3x2 + 2x + 4 = 0
Write in standard form.
a = 3, b = 2, c = 4
Find the values of a, b, and c.
y=
=
=
=
–(2) ±
–2 ±
(2)2 – 4(3)(4)
2(3)
4 – 48
6
–2 ±
Substitute.
Simplify.
–44
6
–2 ± 2i
6
=–
11
1 i 11
±
3
3
5-8
The Quadratic Formula
ALGEBRA 2 LESSON 5-8
The longer leg of a right triangle is 1 unit longer than the
shorter leg. The hypotenuse is 3 units long. What is the length of the
shorter leg?
x2 + (x + 1)2 = 32
x2 + x2 + 2x + 1 = 9
2x2 + 2x – 8 = 0
Write in standard form.
a = 2, b = 2, c = –8
Find the values of a, b, and c.
x=
=
=
–b ±
–(2) ±
b 2 – 4ac
2a
(2)2 – 4(2)(–8)
2(2)
–1 ± 17
2
The length of the shorter leg is
Use the Quadratic Formula.
Substitute for a, b, and c.
Simplify.
17 – 1
units.
2
5-8
The Quadratic Formula
ALGEBRA 2 LESSON 5-8
(continued)
–1–
17
Is the answer reasonable? Since
is a negative number, and a
2
length cannot be negative, that answer is not reasonable.
Since
17 – 1
2
1.56, that answer is reasonable.
5-8
The Quadratic Formula
ALGEBRA 2 LESSON 5-8
Determine the type and number of solutions of
x2 + 5x + 10 = 0.
a = 1, b = 5, c = 10
Find the values of a, b, and c.
b2 – 4ac = (5)2 – 4(1)(10)
Evaluate the discriminant.
= 25 – 40
Simplify.
= –15
Since the discriminant is negative, x2 + 5x + 10 = 0 has two imaginary
solutions.
5-8
The Quadratic Formula
ALGEBRA 2 LESSON 5-8
A player throws a ball up and toward a wall that is 17 ft high.
The height h in feet of the ball t seconds after it leaves the player’s
hand is modeled by h = –16t 2 + 25t + 6. If the ball makes it to where
the wall is, will it go over the wall or hit the wall?
h = –16t 2 + 25t + 6
17 = –16t 2 + 25t + 6
0 = –16t 2 + 25t – 11
Substitute 17 for h.
Write the equation in standard form.
a = –16, b = 25, c = –11
Find the values of a, b, and c.
b2 – 4ac = (25)2 – 4(–16)(–11)
Evaluate the discriminant.
= 625 – 704
Simplify.
= –79
Since the discriminant is negative, the equation 17 = –16t2 + 25t + 6 has no
real solution. The ball will hit the wall.
5-8
The Quadratic Formula
ALGEBRA 2 LESSON 5-8
Pages 289–291 Exercises
10
17. 1 ± i
2
10. – 2 , 4
3
18. – 2 –
i
3. – 2 , 1
11. 1, 4
19.
4. –1, 1
12. – 5 , 1
5. –5
13. 3 ± i
6. – 5 , 1
14. 1 ± 2i
7. 3w 5
15. –
1. 1, 3
9.
2. –6, –2
3
1
7
3
3
2
2
8. –3 ±
2±
14
3
i 3
5
±
2
2
i 15
7
20. 4 ± 4
3
2
3
i 11
±
2
2
16. –2 ± i
26
3
2
5-8
1
15
1
22. – ,
2
5
23. ±
3
± i 14
21. –
15
3
24. – 2 ±
3
10
; 0.61, 2.72
3
13
; –1.87, 0.54
3
The Quadratic Formula
ALGEBRA 2 LESSON 5-8
1
6
3 1
,
2 2
25. – , 1
34. –223; two, imaginary
43. –
26. 1 ±
35. 169; two, real
44. –3.45, 1.45
36. –116; two, imaginary
45. 1 ± i
37. 1; two, real
46. –1.70, 4.70
38. 0; one, real
47. –7, 7
39. 0; one, real
48. –8.47, 0.47
31. –4; two, imaginary
40. no
49. 3 ± i
32. 36; two, real
41. 1, 10
50. – ,
33. 0; one, real
42. 0, 42
51. –1, 6
337; –1.24, 1.38
14
14
71
4
27. – ±
; –2.49, 0.89
5
5
23
1
28. – 2 ± 2 ; –2.90, 1.90
5
29. 4 ±
30. – 1 ±
4
33
; –0.19, 2.69
4
5
; –0.81, 0.31
4
1
2
5-8
2
3
2
The Quadratic Formula
ALGEBRA 2 LESSON 5-8
52. –5.41, 2.41
53. a. w(18 – w) = 36
b. 2.29 in. by 15.71 in.
11
54. 3 or – 3
55. Answers may vary. Sample:
Assume the coefficients are
real numbers. If the
discriminant is negative,
then there are 2 imaginary
solutions. If the discriminant
is 0, then there is 1 real
solution. If the discriminant
is positive, then there are 2
real solutions.
56. a. Answers may vary. Sample:
Graph y = 0.0721x2 – 2.8867x + 117.061
and y = 100. Where they intersect is the
year when 100 million tons were released
in the air. Wherever y = 0.0721x2 –
2.8867x + 117.061 is below y = 100 is
where less than 100 million tons were
released.
b. Answers may vary. Sample: Where
y = 0.0721x2 – 2.8867x + 117.061 < 100
is the solution. Subtract 100 from both
sides and you get y = 0.0721x2 – 2.8867x
+ 17.061 < 0. You then use the quadratic
formula to solve.
c. Check students’ work.
5-8
The Quadratic Formula
ALGEBRA 2 LESSON 5-8
57. two
58. one
59. none
60. two
61. two
62. two
63. a. 12 or –12
b. k such that |k| < 12
c. k such that |k| > 12
64. a. k such that |k| < 6
b. k such that |k| > 6
c. 6, –6
65. Imaginary solutions always come in pairs
because they are the positive and
negative solution of the square root of
a negative number.
66. a. II
b. III
c. I
67. a. x2 = 100
b. 17.72 cm
5-8
The Quadratic Formula
ALGEBRA 2 LESSON 5-8
68. a. yes
b.
71. Answers may vary.
Sample:
x2 – 5x + 7 = 0
72.
3±i
2a
73. 5 ± 85
5a
c. 0 < t < 5
69. Answers may vary.
Sample:
x2 – 3x + 1 = 0
70. Answers may vary.
Sample:
x2 + 5x + 3 = 0
74. –a ± a
26
–b – b2 – 4ac
–2b
b
b2 – 4ac
+
=
=
–
2a
2a
a
2a
75. a. –b +
b. –b +
b2 – 4ac
2a
2a
–b
2a
2
–
5-8
–b
–
2a

b2 – 4ac
2a
2
=
4ac
4a2
b2 – 4ac
2a
c
= a
=
The Quadratic Formula
ALGEBRA 2 LESSON 5-8
76. C
80. [4] Rewrite the equation as
3x2 – x – 4 = 0. Then use
a = 3, b = –1 and c = –4 in
the quadratic formula. You get
77. F
78. A
–(–1) ±
79. [2] For an equation of the form
ax2 + bx + c = 0, where a =/ 0,
the discriminant is the number
b2 – 4ac. If the discriminant is
negative, then there are 2
imaginary solutions; if it is zero,
then there is 1 real solution;
if it is positive, then there are
2 real solutions.
(–1)2 – 4(3)(–4)
=
2(3)
1 – (–48)
1 ± 49
=
=
6
6
4
1+7 1–7
1±7
or
,–1
=
3
6 , 6
6
1±
[3] appropriate methods, with one
computational error
[2] insufficient explanation and
computational error
[1] only states that the discriminant
is b2 – 4ac
5-8
[1] correct solution, without work shown
The Quadratic Formula
ALGEBRA 2 LESSON 5-8
81. –2, 10
86.
82. 2 ± 2
27
–21
–3
1
–16
–7
52
24
2
87.
83. 3 ± 41
2
84.
5
–7
85. 3
0
6 2
–1 –2
0
–3
4
2
–2
–4
5-8
The Quadratic Formula
ALGEBRA 2 LESSON 5-8
Use the Quadratic Formula to solve each equation.
1. 6x2 + 19x + 8 = 0
–
8
1
,–
3
2
2. x2 – 2x – 11 = 0
1±2
3
3. x2 – 2x – 15 = 0
4. 3x2 – 7x + 5 = 0
7 ± i 11
–3, 5
6
Find the value of the discriminant for each quadratic equation. Tell how many
different solutions each equation has and whether the solutions are real or
imaginary.
5. 2x2 – 5x + 7 = 0
–31; two, imaginary
6. –3x2 – 14x – 8 = 0
100; two, real
7. 4x2 – 5x + 10 = 7x + 1
0; one, real
5-8
Quadratic Equations and Functions
ALGEBRA 2 CHAPTER 5
Page 296
1.
2
4
y = x2 – x;
9
3
4.
6.
(9, 6), (6, 0), (3, –2)
2.
x = 2.5, (2.5, 3.25)
x = 0, (0, –7)
7.
5.
3.
x = 0, (0, –8)
x = –1, (–1, 5)
5-A
Quadratic Equations and Functions
ALGEBRA 2 CHAPTER 5
8. 4i
16. 2 – i
9. –2 + 12i
17.
19.
10. 9 – 10i
4
53
11. 31 + 14i
20.
18.
12. –9 + i
13. 62 – 54i
14. h = 144.5 m;
t = 4.25 s
5
8
15. –3 + 7i
5-A
5
Quadratic Equations and Functions
ALGEBRA 2 CHAPTER 5
21.
2
13
22.
13
23. Answers may vary.
Sample: In the
coordinate plane you
graph ordered pairs
(a, b); in the complex
plane you graph
complex numbers
a + bi. For both, you
find a on the
horizontal axis and
you find b on the
vertical axis.
24. –5, 5
26. –9, 1
7–3
2
3
28. 0, 2
27.
29. – 1 – i
2
30. –
–
7+3
2
5
3
,–1+ i
2
3
2
1
i 39
+
10
10
3
5-A
2
,
1
– i 39 ,
10
10
31. – 1 –
25. –8, 3
5
22
,–1 –
2
3
22
2
Quadratic Equations and Functions
ALGEBRA 2 CHAPTER 5
32. – 3 + i
2
15 , – 3 – i 15
2
2
2
35. y = 2 x – 3
4
2
– 2 1;
8
37. 64; 2 real solutions
38. –35; 2 imaginary
solutions
33. y = (x – 3)2 – 4;
39. 24; 2 real solutions
40. 0; 1 real solution
36. y = – 1 (x – 4)2 – 1;
2
34. y = –(x – 4)2 + 6;
5-A
41. Check students’
work.