exploring quadratic graphs
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Exploring Quadratic Graphs. ALGEBRA 1 LESSON 10-1. (For help, go to Lessons 1-2 and 5-3.). Evaluate each expression for h = 3, k = 2, and j = –4. 1. hkj 2. kh 2 3. hk 2 4. kj 2 + h. Graph each equation. 5. y = 2 x – 1 6. y = | x | 7. y = x 2 + 2. 10-1. - PowerPoint PPT PresentationTRANSCRIPT
Exploring Quadratic GraphsALGEBRA 1 LESSON 10-1
Evaluate each expression for h = 3, k = 2, and j = –4.
1. hkj 2. kh2 3. hk2 4. kj 2 + h
(For help, go to Lessons 1-2 and 5-3.)
Graph each equation.
5. y = 2x – 1 6. y = | x | 7. y = x2 + 2
10-1
Exploring Quadratic GraphsALGEBRA 1 LESSON 10-1
1. hkj for h = 3, k = 2, j = –4: (3)(2)(–4) = 6(–4) = –24
2. kh2 for h = 3, k = 2: 2(32) = 2(9) = 18
3. hk2 for h = 3, k = 2: 3(22) = 3(4) = 12
4. kj 2 + h for h = 3, k = 2, j = –4: 2(–4)2 + 3 = 2(16) + 3 = 32 + 3 = 35
5. y = 2x – 1 6. y = | x |
7. y = x2 + 2
Solutions
10-1
Identify the vertex of each graph. Tell whether it is a minimum or a maximum.
Exploring Quadratic Graphs
a.
The vertex is (1, 2).
b.
The vertex is (2, –4).
It is a maximum. It is a minimum.
ALGEBRA 1 LESSON 10-1
10-1
Make a table of values and graph the quadratic
function y = x2.13
Exploring Quadratic GraphsALGEBRA 1 LESSON 10-1
x y = x2 (x, y)13
0 (0)2 = 0 (0, 0)13
2 (2)2 = 1 (2, 1 )13
13
13
3 (3)2 = 3 (3, 3)13
10-1
Use the graphs below. Order the quadratic functions
(x) = –x2, (x) = –3x2, and (x) = x2 from widest to narrowest graph.
So, the order from widest to narrowest is (x) = x2, (x) = –x2,(x) = –3x2.
12
12
(x) = –x2 (x) = x2 12
Of the three graphs, (x) = x2 is the widest and (x) = –3x2 is the narrowest.
12
(x) = –3x2
Exploring Quadratic GraphsALGEBRA 1 LESSON 10-1
10-1
Graph the quadratic functions y = 3x2 and y = 3x2 – 2. Compare the graphs.
The graph of y = 3x2 – 2 has the same shape as the graph of y = 3x2, but it is shifted down 2 units.
Exploring Quadratic GraphsALGEBRA 1 LESSON 10-1
x y = 3x2 y = 3x2 – 2
2 12 10
1 3 1
0 0 2
–1 3 1
2 12 10
10-1
Height h is dependent on time t.
Graph t on the x-axis and h on the y-axis.
Use positive values for t.
Exploring Quadratic GraphsALGEBRA 1 LESSON 10-1
A monkey drops an orange from a branch 26 ft above the ground. The force of gravity causes the orange to fall toward Earth. The function h = –16t2 + 26 gives the height of the orange, h, in feet after t seconds. Graph this quadratic function.
t h = –16t2 + 260 261 102 –38
10-1
9.
10. y = x2, y = 3x2, y = 4x2
11. ƒ(x) = x2, ƒ(x) = x2, ƒ(x) = 5x2
12. y = – x2, y = – x2, y = 5x2
13. ƒ(x) = – x2, ƒ(x) = –2x2,
ƒ(x) = –4x2
14.
Exploring Quadratic GraphsALGEBRA 1 LESSON 10-1
pages 513–516 Exercises
1. (2, 5); max.
2. (–3, –2); min.
3. (2, 1); min.
4.
5.
6.
7.
8.
12
13
14
12
23
10-1
Exploring Quadratic GraphsALGEBRA 1 LESSON 10-1
15.
16.
17.
18.
19.
20.
21. E
22. A
23. F
24. B
25. C
26. D
27. The graph of y = 2x2 is narrower.
28. The graph of y = –x2 opens downward.
29. The graph of y = 1.5x2 is narrower.
30. The graph of y = x2 is wider.
12
10-1
Exploring Quadratic GraphsALGEBRA 1 LESSON 10-1
31.
32.
33.
34.
35.
36.
37.
38. a.
b. 184 ftc. 56 ft
39. a. 0 < r < 6b. 0 < A < 36 113.1c.
10-1
Exploring Quadratic GraphsALGEBRA 1 LESSON 10-1
40. K, L
41. M
42. K
43. M
44. Answers may vary. Sample:a. y = 5x2 b. y = –5x2 c. y = 3x2
45. a.
b. 16 ftc. No; the apple falls 48 ft
from t = 1 to t = 2, because it is accelerating.
46. a. c 0 and a and c have opp. signs.b. c 0 and a and c have the same signs.
47. a.
b. 0 < x < 12; the side length of the square garden must be less than the width of the patio.
c. 96 < A < 240; as the side length of the garden increases from 0 to 12, the area of the patio decreases from 240 to 96.
d. about 6 ft
10-1
=/=/
Exploring Quadratic GraphsALGEBRA 1 LESSON 10-1
48. a. a > 0b. |a| > 1
49. a.
b.
50. B
51. G
52. D
53. [4] a.
b. 3.5 s[3] estimate incorrect or missing[2] error in table or graph[1] table OR reasonable graph only
t h0 2001 1842 1363 564 –56
10-1
Exploring Quadratic GraphsALGEBRA 1 LESSON 10-1
54. (x2 + 2)(x – 4)
55. (3a2 – 2)(5a – 6)
56. (7b2 + 1)(b + 2)
57. (y + 2)(y – 2)(y + 3)
58. 2n(n + 3)(n – 4)
59. 3m(2m + 3)(5m + 1)
60. 15x2 – 20x
61. 9n2 – 63n
62. –12t 3 + 22t 2
63. 12m6 – 4m5 + 20m2
64. –15y6 – 10y4 + 20y
65. –12c5 + 21c4 – 24c3
66. 15 balloons
10-1
1. a. Graph y = – x2 – 1.
b. Identify the vertex. Tell whether it is a
maximum or a minimum.
c. Compare this graph to the graph of y = x2.
2. a. Graph y = 4x2 + 3.
b. Identify the vertex. Tell whether it is a
maximum or a minimum.
c. Compare this graph to the graph of y = x2.
Exploring Quadratic GraphsALGEBRA 1 LESSON 10-1
12
(0, -1); maximum
This graph is wider, opens downward, and is shifted one unit down.
(0, 3); minimum
This graph is narrower and shifted 3 units up.
3. Order the quadratic functions y = –4x2, y = x2, and y = 2x2 from widest to narrowest graph.
14
y = x2, y = 2x2, y = –4x214
10-1
Quadratic Functions
(For help, go to Lessons 1-6 and 10-1.)ALGEBRA 1 LESSON 10-2
Evaluate the expression for the following values of a and b.
1. a = –6, b = 4 2. a = 15, b = 20
3. a = –8, b = –56 4. a = –9, b = 108
Graph each function.
5. y = x2 6. y = –x2 + 2 7. y = x2 – 1
–b2a
12
10-2
Quadratic FunctionsALGEBRA 1 LESSON 10-2
Solutions b2a for a = –6, b = 4:1.
b2a for a = 15, b = 20:2.
b2a for a = –8, b = –56:3.
b2a for a = –9, b = 108:4.
–42(–6) = –4
–12 = 13
–202(15) = –20
30 = 23–
–(–56)2(–8) = 56
–16 = 72– = –3 1
2
–
–
–
– –1082(–9) = –108
–18 = 6
5. y = x2 6. y = –x2 + 2 7. y = x2 – 112
10-2
Quadratic FunctionsALGEBRA 1 LESSON 10-2
Graph the function y = 2x2 + 4x – 3.
Step 1: Find the equation of the axis of symmetry and thecoordinates of the vertex.
Find the equation of the axis of symmetry.x = b2a– = –4
2(2) = – 1
The x-coordinate of the vertex is –1.
10-2
y = 2x2 + 4x – 3
y = 2(–1)2 + 4(–1) – 3
= –5
To find the y-coordinate of the vertex, substitute –1 for x.
The vertex is (–1, –5).
Quadratic FunctionsALGEBRA 1 LESSON 10-2
(continued)
Step 2: Find two other points.Use the y-intercept.For x = 0, y = –3, so one point is (0, –3).
10-2
Choose a value for x on the same side of the vertex.Let x = 1y = 2(1)2 + 4(1) – 3 = 3For x = 1, y = 3, so another point is (1, 3).
Find the y-coordinate for x = 1.
Quadratic FunctionsALGEBRA 1 LESSON 10-2
(continued)
10-2
Step 3: Reflect (0, –3) and (1, 3) across the axis of symmetry to get two more points.
Then draw the parabola.
Quadratic FunctionsALGEBRA 1 LESSON 10-2
Aerial fireworks carry “stars,” which are made of a sparkler-like material, upward, ignite them, and project them into the air in fireworks displays. Suppose a particular star is projected from an aerial firework at a starting height of 610 ft with an initial upward velocity of 88 ft/s. How long will it take for the star to reach its maximum height? How far above the ground will it be?
The equation h = –16t2 + 88t + 610 gives the height of the star h in feet at time t in seconds.
Since the coefficient of t 2 is negative, the curve opens downward, and the vertex is the maximum point.
10-2
Quadratic FunctionsALGEBRA 1 LESSON 10-2
(continued)
Step 2: Find the h-coordinate of the vertex.h = –16(2.75)2 + 88(2.75) + 610 Substitute 2.75 for t.h = 731 Simplify using a calculator.
The maximum height of the star will be about 731 ft.
Step 1: Find the x-coordinate of the vertex.
After 2.75 seconds, the star will be at its greatest height.
b2a– = –88
2(–16) = 2.75
10-2
Graph the boundary curve, y = –x2 + 6x – 5.
Use a dashed line because the solution of the inequality y > –x2 + 6x – 5 does not include the boundary.
Quadratic FunctionsALGEBRA 1 LESSON 10-2
Graph the quadratic inequality y > –x2 + 6x – 5.
Shade above the curve.
10-2
Quadratic FunctionsALGEBRA 1 LESSON 10-2
pages 520–523 Exercises
1. x = 0, (0, 4)
2. x = –1, (–1, –7)
3. x = 4, (4, –25)
4. x = 1.5, (1.5, –1.75)
5. B
6. E
7. C
8. F
9. A
10. D
11.
12.
13.
14.
15. a. 20 ftb. 400 ft2
16. a. 1.25 s b. 31 ft
17.
10-2
Quadratic FunctionsALGEBRA 1 LESSON 10-2
18.
19.
20.
21.
22.
23.
24.
25.
26.
10-2
Quadratic FunctionsALGEBRA 1 LESSON 10-2
27.
28.
29.
30.
31.
32–34. Answers may vary. Samples are given.
32. y = 2x2 – 8x + 1
33. y = –3x2
34. y = 2x2 + 4
35. a. 1.2 mb. 7.2 m
36. a. y < –0.1x2 + 12b.
c. Yes; when x = 6, y = 8.4, so the camper will fit.
37. a. $12.50b. $10,000
38. 26 units2
39. 26 units2
10-2
40. Answers may vary. Sample: If the coefficient of the squared term is pos., the vertex point is a min.; if it is neg., the vertex point is a max.
41. Answers may vary. Sample: a affects whether the parabola opens up or down, b affects the axis of symmetry, and c affects the y-intercept.
42. a. w = 13 – b. A = – 2 + 13c. (6.5, 42.25)d. 6.5 ft by 6.5 ft
Quadratic FunctionsALGEBRA 1 LESSON 10-2
43. a. 0.4 sb. No; after 0.6 s, the ball will
have a height of about 2.23 m but the net has a height of 2.43 m.
44.
45. a. 0.4 sb. No; it takes about 0.8 s to return to
h = 0.5 m, so it will take more time to reach the ground.
10-2
Quadratic FunctionsALGEBRA 1 LESSON 10-2
10-2
46. a. (0, 2)b. x = –2.5c. 5d. y = x2 + 5x + 2e. Answers may vary.
Sample: Test (–4, –2).–2 (–4)2 + 5(–4) + 2–2 16 – 20 + 2–2 = –2
f. No; you would not be able to determine the b value using the vertex formula.
47. A
48. I
49. B
50. [2] axis of symmetry:
x = = 16.7
maximum height:y –0.009(16.7)2 + 0.3(16.7) + 4.5y 7 ft
[1] appropriate methods, but with a minor computational error
51. C
52. A
53. F
54. D
55. B
56. E
–b2a
–0.3 2(–0.009)
Quadratic FunctionsALGEBRA 1 LESSON 10-2
57. c2 – 5c – 36
58. 2x2 + 7x – 30
59. 20t 2 + 17t + 3
60. 21n4 – 62n2 + 16
61. 2a3 + 9a2 – a + 20
62. 6r 3 + 9r 2 – 20r + 7
10-2
Quadratic FunctionsALGEBRA 1 LESSON 10-2
Graph each relation. Label the axis of symmetry and the vertex.
2. ƒ(x) = –x2 + 4x – 2
14
<3. y – x2 – 2x – 6
1. y = x2 – 8x + 15
x = –4
10-2
Finding and Estimating Square Roots
(For help, go to Lessons 1-2 and 8-5.)ALGEBRA 1 LESSON 10-3
Simplify each expression.
1. 112 2. (–12)2 3. –(12)2 4. 1.52
5. 0.62 6. 2 7. 2 8. 212
23– 4
5
10-3
1. 112 = 11 • 11 = 121 2. (–12)2 = (–12)(–12) = 144
3. (–12)2 = –(12)(12) = –144 4. 1.52 = (1.5)(1.5) = 2.25
5. 0.62 = (0.6)(0.6) = 0.36 6. 2 = • =
7. 2 = = 8.
2 = • =
Solutions
12
23– 4
5
12
12
14
23–2
3– 45
45
1625
Finding and Estimating Square RootsALGEBRA 1 LESSON 10-3
10-3
49
a. 25
Finding and Estimating Square RootsALGEBRA 1 LESSON 10-3
Simplify each expression.
c. – 64
d. –49
e. ± 0
b. ± 925
10-3
positive square root= 5
35
35
35The square roots are and – .= ±
negative square root= –8
For real numbers, the square root of a negative number is undefined.
is undefined
There is only one square root of 0. = 0
a. ± 144 = ± 12
Finding and Estimating Square RootsALGEBRA 1 LESSON 10-3
Tell whether each expression is rational or irrational.
c. – 6.25 = –2.5
e. 7 = 2.64575131 . . .
b. – = –0.44721359 . . .15
d. = 0.319
10-3
rational
irrational
rational
rational
irrational
Between what two consecutive integers is 28.34 ?
Finding and Estimating Square RootsALGEBRA 1 LESSON 10-3
28.34 is between 5 and 6.
28.34 is between the two consecutive perfect squares 25 and 36.25 < 28.34 < 36
The square roots of 25 and 36 are 5 and 6, respectively.28.345 < < 6
10-3
Find 28.34 to the nearest hundredth.
Finding and Estimating Square RootsALGEBRA 1 LESSON 10-3
28.34 5.323532662 Use a calculator.
5.32 Round to the nearest hundredth.
10-3
Suppose a rectangular field has a length three times its width x. The formula d = x2 + (3x)2 gives the distance of the diagonal of a rectangle. Find the distance of the diagonal across the field if x = 8 ft.
Finding and Estimating Square RootsALGEBRA 1 LESSON 10-3
d = x2 + (3x)2
d = 82 + (3 • 8)2 Substitute 8 for x.
The diagonal is about 25.3 ft long.
Use a calculator. Round to the nearest tenth.d 25.3
10-3
Simplify.d = 64 + 576
d = 640
Finding and Estimating Square RootsALGEBRA 1 LESSON 10-3
pages 526–528 Exercises
1. 13
2. 20
3.
4. 30
5. 0.5
6.
7. –1.1
8. 1.4
9. 0.6
10. –12
11.
13
67
54
12. ± 0.1
13. irrational
14. rational
15. irrational
16. rational
17. 5 and 6
18. 5 and 6
19. –12 and –11
20. 13 and 14
21. 3.46
22. –14.25
23. 107.47
24. –12.25
25. 0.93
26. ± 20
27. 0
28. ± 25
29. ±
30. ± 1.3
31. ±
32. ± 27
33. ± 1.5
34. ± 16
35. ± 0.1
37
19
10-3
=/
54. false; 1 = 1
55. true
56. true
57. False; answers may vary. Sample: 4 + 9 4 + 9.
58. False; answers may vary. Sample: 12 and 3 are irrational but 36 is rational.
59. a. 4 units2
b. unit2
c. 2 units2
d. 2 units
Finding and Estimating Square RootsALGEBRA 1 LESSON 10-3
36. ±
37. ± 202
38. 1
39. a. 8450 kmb. 7684 km
40. 21
41. –
42. 1.41
43. 1.26
44. –5.48
45. –33
46. –0.8
8 11
25
47. 6.40
48. 8.66
49. Answers may vary. Sample: The first expression means the neg. square root of 1 and the second expression means the pos. square root of 1.
50. Answers may vary. Sample: 3 and 4
51. 452. a. 5 s
b. 10 sc. No; the object takes
twice as long to fall.
53. False; zero has one square root.
12
10-3
Finding and Estimating Square RootsALGEBRA 1 LESSON 10-3
60. 1.3
61. 6
62.
63. 9
64. 11
65. 128
66.
29
67.
68.
69.
70.
71.
72. d 2 – 81
73. 9t 2 – 25
74. 8075
10-3
Finding and Estimating Square RootsALGEBRA 1 LESSON 10-3
75. x2 + 26x + 169
76. 16y2 – 56y + 49
77. 10,201
78. 813,604
79. 36k2 – 49
80. 144b2 – 168b + 49
81. –2
82.
83. –
84. –972
85.
86.
3 16
34
1 12814
10-3
1. Simplify each expression.
a. 196 b. ±
2. Tell whether each expression is rational, irrational, or undefined.
a. ± b. –25 c. – 2.25
3. Between what two consecutive integers is – 54?
4. The formula s = 13.5d estimates the speed s in miles per hour that a car was traveling, when it applied its brakes and left a skid mark d feet long on a wet road. Estimate the speed of a car that left a 120 foot long skid mark.
Finding and Estimating Square RootsALGEBRA 1 LESSON 10-3
425
35
14 ± 25
irrational undefined rational
–8 and –7
about 40.25 mph
10-3
Solving Quadratic EquationsALGEBRA 1 LESSON 10-4
(For help, go to Lesson10-3.)
Simplify each expression.
1. 36 2. – 81 3. ± 121
4. 1.44 5. 0.25 6. ± 1.21
7. 8. ± 9. ±14
19
49100
10-4
Solving Quadratic EquationsALGEBRA 1 LESSON 10-4
Solutions
13
14
19
49100
12
710
10-4
1. 36 = 6 2. – 81 = –9
3. ± 121 = ±11 4. 1.44 = 1.2
5. 0.25 = 0.5 6. ± 1.21 = ±1.1
7. = 8. ± = ±
9. ± =
Solving Quadratic EquationsALGEBRA 1 LESSON 10-4
Solve each equation by graphing the related function.
a. 2x2 = 0 b. 2x2 + 2 = 0 c. 2x2 – 2 = 0
There is one solution, x = 0.
There is no solution. There are two solutions, x = ±1.
10-4
Graph y = 2x2 Graph y = 2x2 + 2 Graph y = 2x2 – 2
x = ± 25 Find the square roots.
Solving Quadratic EquationsALGEBRA 1 LESSON 10-4
Solve 3x2 – 75 = 0.
3x2 – 75 + 75 = 0 + 75 Add 75 to each side.
3x2 = 75
x2 = 25 Divide each side by 3.
x = ± 5 Simplify.
10-4
Solving Quadratic EquationsALGEBRA 1 LESSON 10-4
S = 4 r 2
315 = 4 r 2 Substitute 315 for S.
Put in calculator ready form.315 (4) = r 2
= r 2315 (4)
Find the principle square root.
5.00668588 r Use a calculator.
The radius of the sphere is about 5 ft.
10-4
A museum is planning an exhibit that will contain a large globe. The surface area of the globe will be 315 ft2. Find the radius of the sphere producing this surface area. Use the equation S = 4 r2, where S is the surface area and r is the radius.
Solving Quadratic EquationsALGEBRA 1 LESSON 10-4
pages 531–534 Exercises
1.
±32.
no solution3.
0
4.
±25.
no solution6.
±3
7.
no solution
8.
0
9.
±2
10-4
Solving Quadratic EquationsALGEBRA 1 LESSON 10-4
10. ± 7
11. ± 21
12. ± 15
13. 0
14. no solution
15. ±
16. ±
17. ± 2
18. ± 27
19. x2 = 256; 16 m
20. x2 = 90; 9.5 ft
21. r 2 = 80; 5.0 cm
14
52
22. a. 6.0 in.b. The length of a radius
cannot be negative.
23. none
24. two
25. one
26. 10.4 in. by 10.4 in.
27. a. 11.3 ftb. 16.0 ftc. No; the radius increases
by about 1.4 times.
28. no solution
29. ± 37
30. ±
31. ± 2.8
32. ± 0.4
33. ± 3.5
34. 3.5 s
35. 121
36. a. n > 0b. n = 0c. n < 0
37. Answers may vary. Sample: Michael subtracted 25 from the left side of the equation but added 25 to the right side.
16
10-4
Solving Quadratic EquationsALGEBRA 1 LESSON 10-4
38. a. 2, –2; 2, –2
b. The first equation multiplied by 2 on both sides equals the second equation.
39. a. square: 4r 2, circle: r 2
b. 4r 2 – r 2 = 80c. 9.7 in., 19.3 in.
40. Answers may vary. Sample:a. 5x2 + 10 = 0, no solutionb. 2x2 + 0 = 0, x = 0c. –20x2 + 80 = 0, x = ± 2
41. 6.3 ft
42. 11.0 cm
43. a. 0.2 mb. 2.5 sc. 3.0 sd. Shorten; as decreases, t decreases.
44. a. –7b. (–7, 0)c. Answers may vary.
Sample: h = 5, –5, (–5, 0)d. (4, 0); the vertex is at (–h, 0).
45. 28 cm
46. B
47. I
48. B
10-4
Solving Quadratic EquationsALGEBRA 1 LESSON 10-4
49. [2]
x-intercepts 1.5, –1.5
[1] minor error in table OR incorrect graph
x y–2 5–1 –40 –71 –42 5
50. [4] a. 96 = 6s2
s2 = 16s = 4, so side is 4 ft.
b. 6(8)2 = 6 • 64 = 384, so surface area is 384 ft2. The surface area is quadrupled.
[3] appropriate methods, but with one computational error
[2] part (a) done correctly[1] no work shown
51. 3
52. –13
53. 40
54. 15
55. 0.2
10-4
Solving Quadratic EquationsALGEBRA 1 LESSON 10-4
56. –1.6
57.
58.
59. (x + 4)(x + 1)
60. (y – 13)(y – 2)
61. (a + 5)(a – 2)
62. (z – 12)(z + 6)
63. (c – 12d)(c – 2d)
64. (t + 2u)(t – u)
65. 3.6135 106
66. 3.48 10–5
67. –8.12 100
68. 31,000
69. 701,000
70. 0.00062
5879
10-4
Solving Quadratic EquationsALGEBRA 1 LESSON 10-4
1. Solve each equation by graphing the related function. If the equation has no solution, write no solution.
a. 2x2 – 8 = 0
b. x2 + 2 = –2
2. Solve each equation by finding square roots.
a. m2 – 25 = 0
b. 49q2 = 9
3. Find the speed of a 4-kg bowling ball with a kinetic energy of 160 joules. Use the equation E = ms2, where m is the object’s mass in kg, E is its kinetic energy, and s is the speed in meters per second.
±2
no solution
±537±
12
about 8.94 m/s
10-4
Factoring to Solve Quadratic EquationsALGEBRA 1 LESSON 10-5
(For help, go to Lessons 2-2 and 9-6.)
Solve and check each equation.
1. 6 + 4n = 2 2. – 9 = 4 3. 7q + 16 = –3
Factor each expression.
4. 2c2 + 29c + 14 5. 3p2 + 32p + 20 6. 4x2 – 21x – 18
a8
10-5
Factoring to Solve Quadratic EquationsALGEBRA 1 LESSON 10-5
Solutions1. 6 + 4n = 2
4n = –4 n = –1Check: 6 + 4(–1) = 6 + (–4) = 2
2. – 9 = 4
= 13
a = 104
Check: – 9 = 13 – 9 = 4
3. 7q + 16 = –3 7q = –19
q = –2
Check: 7 (–2 ) + 16 = 7(– ) + 16 = –19 + 16 = –3
a8 a
8
1048
57
57
197
10-5
Factoring to Solve Quadratic EquationsALGEBRA 1 LESSON 10-5
Solutions (continued)
4. 2c2 + 29c + 14 = (2c + 1)(c + 14)Check: (2c + 1)(c + 14) = 2c2 + 28c + c + 14 = 2c2 + 29c + 14
5. 3p2 + 32p + 20 = (3p + 2)(p + 10)Check: (3p + 2)(p + 10) = 3p2 + 30p + 2p + 20 = 3p2 + 32p + 20
6. 4x2 – 21x – 18 = (4x + 3)(x – 6)Check: (4x + 3)(x – 6) = 4x2 – 24x + 3x – 18 = 4x2 – 21x – 18
10-5
Factoring to Solve Quadratic EquationsALGEBRA 1 LESSON 10-5
Solve (2x + 3)(x – 4) = 0 by using the Zero Product Property.
(2x + 3)(x – 4) = 0
2x + 3 = 0 or x – 4 = 0 Use the Zero-Product Property.
2x = –3 Solve for x.
x = – 32 or x = 4
Check: Substitute – for x.32 Substitute 4 for x.
(2x + 3)(x – 4) = 0 (2x + 3)(x – 4) = 0
[2(– ) + 3](– – 4) 032
32 [2(4) + 3](4 – 4) 0
(0)(– 5 ) = 012 (11)(0) = 0
10-5
Factoring to Solve Quadratic EquationsALGEBRA 1 LESSON 10-5
Solve x2 + x – 42 = 0 by factoring.
x2 + x – 42 = 0
(x + 7)(x – 6) = 0 Factor using x2 + x – 42
x + 7 = 0 or x – 6 = 0 Use the Zero-Product Property.
x = –7 or x = 6 Solve for x.
10-5
Factoring to Solve Quadratic EquationsALGEBRA 1 LESSON 10-5
Solve 3x2 – 2x = 21 by factoring.
3x2 – 2x = 21 Subtract 21 from each side.
(3x + 7)(x – 3) = 0 Factor 3x2 – 2x – 21.
3x + 7 = 0 or x – 3 = 0 Use the Zero-Product Property
3x = –7 Solve for x.
x = – or x = 373
10-5
Factoring to Solve Quadratic EquationsALGEBRA 1 LESSON 10-5
The diagram shows a pattern for an open-top box. The total area of the sheet of materials used to make the box is 130 in.2. The
height of the box is 1 in. Therefore, 1 in. 1 in. squares are cut from each corner. Find the dimensions of the box.
Define: Let x = width of a side of the box.Then the width of the material = x + 1 + 1 = x + 2The length of the material = x + 3 + 1 + 1 = x + 5
Relate: length width = area of the sheet
10-5
Factoring to Solve Quadratic EquationsALGEBRA 1 LESSON 10-5
(continued)
Write: (x + 2) (x + 5) = 130
x2 + 7x + 10 = 130 Find the product (x + 2) (x + 5).
x2 + 7x – 120 = 0 Subtract 130 from each side.
(x – 8) (x + 15) = 0 Factor x2 + 7x – 120.
x – 8 = 0 or x + 15 = 0 Use the Zero-Product Property.
x = 8 or x = –15 Solve for x.
The only reasonable solution is 8. So the dimensions of the box are 8 in. 11 in. 1 in.
10-5
12. 3, –4
13. –3, –5
14. –4, 7
15. 0, 6
16. 1, 2.5
17. –5, –
18. –2.5, 2.5
19. , –4
20. 2, 3
21. , –4
22. 5 cm
23. 5
25
32
Factoring to Solve Quadratic EquationsALGEBRA 1 LESSON 10-5
pages 538–540 Exercises
1. 3, 7
2. –4, 4.5
3. 0, –1
4. 0, 2.5
5. – , –
6. , –
7. 1, –4
8. –2, 7
9. 0, 8
10. 5, 11
11. –2, 5
27
45
74
83
24. 6 ft 15 ft
25. base: 10 ftheight: 22 ft
26. 2 and 3 or 7 and 8
27. 2q2 + 22q + 60 = 0; –6, –5
28. 6n2 – 5n – 4 = 0; , –
29. 4y2 + 12y + 9 = 0; –
30. a2 + 6a + 9 = 0; –3
31. 2t 2 + 11t + 12 = 0; –1.5, –4
32. x2 – 10x + 24 = 0; 4, 6
33. 2k2 + 11k – 63 = 0; , –9
34. 20y2 + 41y – 9 = 0; , –
7215
94
43
12
32
10-5
13
Factoring to Solve Quadratic EquationsALGEBRA 1 LESSON 10-5
35. 8 in. 10 in.
36. a. 2 sb. about 19 ft
37. Answers may vary. Sample: To solve a quadratic equation, write the equation in standard form, factor the quadratic expression, use the Zero-Product Property, and solve for the variable.
x2 + 8x = –15x2 + 8x + 15 = 0(x + 3)(x + 5) = 0x + 3 = 0 or x + 5 = 0x = –3 or x = –5
38. Answers may vary. Sample: x = 6, a = 2, b = 1; x = 3, a = 1, b = 11
39. Answers may vary. Sample:x2 – 2x – 8 = 0
(x – 4)(x + 2) = 0x – 4 = 0 or x + 2 = 0 x = 4 or x = –2
40. a. 0, 1; –1, 0b. 0
41. 0, 4, 6
42. 0, 1, 4
43. 0, 3
44. 0, 7, –10
45. 0, 1, 9
46. 0, 4, –5
47. 4
10-5
Factoring to Solve Quadratic EquationsALGEBRA 1 LESSON 10-5
48. Answers may vary. Samples:a. x2 – 3x – 40 = 0b. x2 – x – 6 = 0c. 2x2 + 19x – 10 = 0d. 21x2 + x – 10 = 0
49. –1, 1, –5
50. –2, 2, –1
51. D
52. H
53. A
54. C
55. D
56. [2] 3x2 + 20x + 1 = 83x2 + 20x – 7 = 0(3x – 1)(x + 7) = 0x = , x = –7
[1] appropriate methods, but with one computational error
57. x2 = 320; 17.9 ft
58. 38 = r 2; 3.5 ft
59. (2x + 3)(x + 5)
60. (3y – 1)(y – 3)
61. (4t – 3)(t + 2)
62. (3n – 1)(2n + 3)
63. 5a(3a + 2)(a – 4)
64. –2b(3b – 2)(3b – 5)
10-5
13
Factoring to Solve Quadratic EquationsALGEBRA 1 LESSON 10-5
1. Solve (2x – 3)(x + 2) = 0.
Solve by factoring.
2. 6 = a2 – 5a 3. 12x + 4 = –9x2 4. 4y2 = 25
–2, 32
–1, 6 – 23 ± 5
2
10-5
Completing the SquareALGEBRA 1 LESSON 10-6
(For help, go to Lessons 9-4 and 9-7.)
Find each square.
1. (d – 4)2 2. (x + 11)2 3. (k – 8)2
Factor.
4. b2 + 10b + 25 5. t2 + 14t + 49 6. n2 – 18n + 81
10-6
Completing the SquareALGEBRA 1 LESSON 10-6
1. (d – 4)2 = d2 – 2d(4) + 42 = d2 – 8d + 16
2. (x + 11)2 = x2 + 2x(11) + 112 = x2 + 22x + 121
3. (k – 8)2 = k2 – 2k(8) + 82 = k2 – 16k + 64
4. b2 + 10b + 25 = b2 + 2b(5) + 52 = (b + 5)2
5. t2 + 14t + 49 = t2 + 2t(7) + 72 = (t + 7)2
6. n2 – 18n + 81 = n2 – 2n(9) + 92 = (n – 9)2
Solutions
10-6
Completing the SquareALGEBRA 1 LESSON 10-6
Find the value of c to complete the square for x2 – 16x + c.
The value of b in the expression x2 – 16x + c is –16.
The term to add to x2 – 16x is or 64. –162
2
10-6
Completing the SquareALGEBRA 1 LESSON 10-6
Solve the equation x2 + 5 = 50 by completing the square.
Step 1: Write the left side of x2 + 5 = 50 as a perfect square.
x2 + 5 = 50
x2 + 5 + = 50 +52
2 52
2 Add , or , to each side of the
equation.
52
2 254
52
2x + 200
4 + 254= Write x2 + 5x + as a square. 5
22
Rewrite 50 as a fraction with denominator 4.
52
2x + = 225
4
10-6
Completing the SquareALGEBRA 1 LESSON 10-6
(continued)
Step 2: Solve the equation.
Find the square root of each side.52
2x + = 225
4±
52
x + = 152
± Simplify.
52
x + = 152 or
52x + = 15
2– Write as two equations.
x = 5 or x = –10 Solve for x.
10-6
Completing the SquareALGEBRA 1 LESSON 10-6
Solve x2 + 10x – 16 = 0 by completing the square. Round to the nearest hundredth.
Step 1: Rewrite the equation in the form x2 + bx = c and complete the square.
x2 + 10x – 16 = 0
x2 + 10x = 16 Add 16 to each side of the equation.
(x + 5)2 = 41 Write x2 + 10x +25 as a square.
x2 + 10x + 25 = 16 + 25 Add , or 25, to each side of the equation.102
2
10-6
Completing the SquareALGEBRA 1 LESSON 10-6
(continued)
Step 2: Solve the equation.
x + 5 = ± 41 Find the square root of each side.
Use a calculator to find 41x + 5 ± 6.40
x + 5 6.40 or x + 5 –6.40 Write as two equations.
Subtract 5 from each side.x 6.40 – 5 or x –6.40 – 5
x 1.40 or x –11.40 Simplify
10-6
Completing the SquareALGEBRA 1 LESSON 10-6
Suppose you wish to section off a soccer field as shown in the diagram to run a variety of practice drills. If the area of the field is 6000 yd2, what is the value of x?
Define: width = x + 10 + 10 = x + 20length = x + x + 10 + 10 = 2x + 20
Relate: length width = area
Write: (2x + 20)(x + 20) = 60002x2 + 60x + 400 = 6000
Step 1: Rewrite the equation in the form x2 + bx = c.
2x2 + 60x + 400 = 60002x2 + 60x = 5600 Subtract 400 from each side.x2 + 30x = 2800 Divide each side by 2.
10-6
Completing the SquareALGEBRA 1 LESSON 10-6
(continued)
Step 2: Complete the square.
x2 + 30x + 255 = 2800 + 225 Add , or 225, to each side.302
2
(x + 15)2 = 3025 Write x2 + 30x + 255 as a square.
Step 3: Solve each equation.
(x + 15) = ± 3025 Take the square root of each side.
x + 15 = ± 55 Use a calculator.
x + 15 = 55 or x + 15 = –55x = 40 or x = –70 Use the positive answer for this problem.
The value of x is 40 yd.
10-6
Completing the SquareALGEBRA 1 LESSON 10-6
pages 544–546 Exercises
1. 49
2. 16
3. 400
4. 9
5. 144
6. 324
7. 4, –12
8. 13.06, –3.06
9. –5, –17
10. 1.24, –7.24
11. 9, –29
12. 19, –17
13. 7, –5
14. –2.17, –7.83
15. 11, 1
16. 1.19, –4.19
17. 4.82, –5.82
18. 22, –31
19. 1
20. 4
21.
22. 2.16, –4.16
23. 5, –1
24. 7, –2
25. a. (2x + 1)(x + 1)b. 2x2 + 3x + 1 = 28c. 3
26. –0.27, –3.73
27. –3, –4
28. 4, –10
29. 6, 2
30. 8.32, 1.68
31. no solution
32. 9.37, –1.87
33. 8.12, –0.12
34. –4, –5
10-6
81 100
Completing the SquareALGEBRA 1 LESSON 10-6
35. a. = 50 – 2wb. w(50 – 2w) = 150; 21.5, 3.5c. 7 ft 21.5 ft or 43 ft 3.5 ftd. No; the answers in part
(b) were rounded.
36. The student did not divide each side of the equation by 4.
37. Answers may vary. Sample: Add 1 to each side of the equation, and then complete the square by adding 225 to each side of the equation. Write x2 + 30x + 225 as the square (x + 15)2 and add 1 and 225 to get 226. Then take square roots and solve the resulting equations.
38. Answers may vary. Sample:x2 + 10x – 50 = 0
x2 + 10x = 50x2 + 10x + 25 = 50 + 25
(x + 5)2 = ± 75x + 5 = ± 8.7x + 5 ± 8.7
x + 5 8.7 or x + 5 –8.7 x 3.7 or x –13.7
39. 5.16, –1.16
40. 6.83, 1.17
41. 5.6 ft by 14.2 ft
42. a. 6x2 + 28xb. 6x2 + 28x = 384c. 13 in. 6 in. 6 in.
10-6
Completing the SquareALGEBRA 1 LESSON 10-6
43. a. A = x2 + 5x + 1
b. 6.86c. 207.5 ft2
44. a. 3 ± 5 b. (3, –5)c. Answers may vary.
Sample: p is the x-coordinate of the vertex.
45. B
46. I
47. D
48. [2] (x)(x + x + 4) = 200
(x)(2x + 4) = 200
(x)(x + 2) = 200
x2 + 2x = 200
x2 + 2x + 1 = 201
(x + 1)2 = 201
x + 1 ±14.18
x 13.18 x –15.18
The value of x is about 13.18 cm.
[1] appropriate methods, but with one computational error
72
12
12
10-6
Completing the SquareALGEBRA 1 LESSON 10-6
49. [4] a. (8 + x)(12 + x) = 2 • (8 • 12)x2 + 20x – 96 = 0
b. x2 + 20x = 96x2 + 20x + 100 = 196
(x + 10)2 = 196x + 10 = ±14
x = 4c. 12 ft by 16 ft
[3] appropriate methods, but with one computational error
[2] part (c) not done[1] no work shown
50. –3, 7
51. –6, –5
52. 0, 5
53. – ,
54. –
55. – ,
56. (x + 2)2
57. (t – 11)2
58. (b + 5)(b – 5)
59. (4c + 3)2
60. (7s + 13)(7s – 13)
61. 2m(2m + 3)(2m – 3)
62. (5m + 12)2
63. (20k + 3)(20k – 3)
64. (16g – 11)(16g + 11)
83
83
3216
52
65. r 12
66. p13
67. –y
68.
69. –
70. t 29
1 m40
1w
10-6
Completing the SquareALGEBRA 1 LESSON 10-6
1. x2 + 14x = –43
2. 3x2 + 6x – 24 = 0
3. 4x2 + 16x + 8 = 40
Solve each equation by completing the square. If necessary, round to the nearest hundredth.
–9.45, –4.55
–4, 2
–5.46, 1.46
10-6
Using the Quadratic Formula
(For help, go to Lesson 10-6.)ALGEBRA 1 LESSON 10-7
Find the value of c to complete the square for each expression.
1. x2 + 6x + c 2. x2 + 7x + c 3. x2 – 9x + c
Solve each equation by completing the square.
4. x2 – 10x + 24 = 0 5. x2 + 16x – 36 = 0
6. 3x2 + 12x – 15 = 0 7. 2x2 – 2x – 112 = 0
10-7
Using the Quadratic FormulaALGEBRA 1 LESSON 10-7
1. x2 + 6x + c; c = 2 = 32 = 9 2. x2 + 7x + c; c = 2 =
3. x2 – 9x + c; c = 2 =
4. x2 – 10x + 24 = 0;5. x2 + 16x – 36 = 0;
c = 2 = (–5)2 = 25 c = 2 = 82 = 64
x2 – 10x = –24 x2 + 16x = 36x2 – 10x + 25 = –24 + 25 x2 + 16x + 64 = 36 + 64 (x – 5)2 = 1 (x + 8)2 = 100 (x – 5) = ±1 (x + 8) = ±10x – 5 = 1 or x – 5 = –1 x + 8 = 10 or x + 8 = –10 x = 6 or x = 4 x = 2 or x = –18
62
72
494
–92
814
–102
162
Solutions
10-7
Using the Quadratic FormulaALGEBRA 1 LESSON 10-7
Solutions (continued)
6. 3x2 + 12x – 15 = 0 7. 2x2 – 2x – 112 = 0
3(x2 + 4x – 5) = 0 2(x2 – x – 56) = 0
x2 + 4x – 5 = 0; x2 – x – 56 = 0;
c = 2 = 22 = 4 c = 2 =
x2 + 4x = 5 x2 – x = 56
x2 + 4x + 4 = 5 + 4 x2 – x + = 56 +
(x + 2)2 = 9 (x – )2 =
(x + 2) = ±3 x – = ±
x + 2 = 3 or x + 2 = –3 x – = or x – =
x = 1 or x = –5 x = 8 or x = –7
42
–12
14
14
14
12
2254
12
152
12
152
12
–152
10-7
Using the Quadratic FormulaALGEBRA 1 LESSON 10-7
Solve x2 + 2 = –3x using the quadratic formula.
x2 + 3x + 2 = 0 Add 3x to each side and write in standard form.
x = –b ± b2 – 4ac2a Use the quadratic formula.
x = –3 ± (–3)2 – 4(1)(2)2(1) Substitute 1 for a, 3 for b, and 2 for c.
x = –3 ± 12 Simplify.
x = –3 + 12 x = –3 – 1
2or Write two solutions.
x = –1 or x = –2 Simplify.
10-7
Using the Quadratic FormulaALGEBRA 1 LESSON 10-7
(continued)
Check: for x = –1 for x = –2
(–1)2 + 3(–1) + 2 0 (–2)2 + 3(–2) + 2 0
1 – 3 + 2 0 4 – 6 + 2 0
0 = 0 0 = 0
10-7
Using the Quadratic FormulaALGEBRA 1 LESSON 10-7
Solve 3x2 + 4x – 8 = 0. Round the solutions to the nearest hundredth.
x = –b ± b2 – 4ac2a Use the quadratic formula.
x = –4 ± 42 – 4(3)(–8)2(3) Substitute 3 for a, 4 for b, and –8 for c.
–4 ± 1126x =
Use a calculator.x –4 + 10.5830052446 x –4 – 10.583005244
6or
x 1.10 or x –2.43 Round to the nearest hundredth.
10-7
x = or x = Write two solutions.–4 + 1126
–4 – 1126
Using the Quadratic FormulaALGEBRA 1 LESSON 10-7
A child throws a ball upward with an initial upward velocity of 15 ft/s from a height of 2 ft. If no one catches the ball, after how many seconds will it land? Use the vertical motion formula h = –16t2 + vt + c, where h = 0, v = velocity, c = starting height, and t = time to land. Round to the nearest hundredth of a second.
Step 1: Use the vertical motion formula.h = –16t2 + vt + c0 = –16t2 + 15t + 2 Substitute 0 for h, 15 for v, and 2 for c.
Step 2: Use the quadratic formula.
x = –b ± b2 – 4ac2a
10-7
Using the Quadratic FormulaALGEBRA 1 LESSON 10-7
(continued)
t = –15 ± 152 – 4(–16)(2)2(–16) Substitute –16 for a, 15 for b, 2 for c, and t for x.
t = –15 + 18.79–32 or t = –15 – 18.79
–32 Write two solutions.
t –0.12 or t 1.06 Simplify. Use the positive answer because it is the only reasonable answer in this situation.
The ball will land in about 1.06 seconds.
–15 ± 225 + 128–32t = Simplify.
–15 ± 353–32t =
10-7
Using the Quadratic FormulaALGEBRA 1 LESSON 10-7
Which method(s) would you choose to solve each equation? Justify your reasoning.
10-7
a. 5x2 + 8x – 14 = 0 Quadratic formula; the equation cannot be factored easily.
b. 25x2 – 169 = 0 Square roots; there is no x term.
c. x2 – 2x – 3 = 0 Factoring; the equation is easily factorable.
d. x2 – 5x + 3 = 0 Quadratic formula, completing the square, or graphing; the x2 term is 1, but the equation is not factorable.
e. 16x2 – 96x + 135 = 0 Quadratic formula; the equation cannot be factored easily and the numbers are large.
Using the Quadratic FormulaALGEBRA 1 LESSON 10-7
pages 550–552 Exercises
1. –1, –1.5
2. 2.8, –6
3. 1.5
4. –0.67, –15
5. 6.67, –0.25
6. –4, –9
7. 2.67, –16
8. 13, –8.5
9. 16, –2.4
10. 0.07, –2.67
11. 10.42, 1.58
13. 1.14, –0.7714. 2.20, –3.0315. 3.84, –0.1716. a. 0 = –16t 2 + 10t + 3
b. t 0.8; 0.8 s17. a. 0 = –16t 2 + 50t + 3.5
b. t 3.2; 3.2 s18. Completing the square or graphing;
the x2 term is 1 but the equation is not factorable.19. Factoring or square roots; the equation
is easily factorable and there is no x term.20. Quadratic formula; the equation
cannot be factored.21. Quadratic formula; the equation
cannot be factored.22. Factoring; the equation is easily factorable.
10-7
Using the Quadratic FormulaALGEBRA 1 LESSON 10-7
23. Quadratic formula; the equation cannot be factored.
24. 6, –6
25. 0.87, –1.54
26. 1.41, –1.41
27. 1.28, –2.61
28. 2
29. 3, –3
30. 1.72, –0.39
31. 1.4, –1
32. 2.23, –1.43
33. a. 7 ft 8 ftb. x(x + 1) = 60, 7.26 ft 8.26 ft
34. About 2.1s35. Answers may vary. Sample: You solve
the linear equation using transformations and you solve the quadratic equation using the quadratic formula.
36. 7.40 ft and 5.40 ft
37. 13.44 cm and 7.44 cm
38. Answers may vary. Sample: A rectangle has length x. Its width is 5 feet longer than three times the length. Find the dimensions if its area is 182 ft2.
7 ft 26 ft
10-7
Using the Quadratic FormulaALGEBRA 1 LESSON 10-7
39. if the expression b2 – 4ac equals zero
40. about 1.9 s
41. a. Check students’ work.b. 356.9 millionc. 2007
42. a. s = –b. 6.5
ab
46. [2]
–1.8, 3.7[1] correct substitution into
quadratic formula, with one calculation error
47. 1.54, 8.46
48. –1, –2
49. 0.1, –6.1
50. (2c + 5)(c + 3)
51. (3z – 2)(z + 4)
52. (5n + 2)(n – 7)
11 ± (112) – 4(6)(–40)2(6)
53. (6v – 5)(2v + 7)
54. (2x – 1)(3x – 5)
55. (5t + 3)(3t + 2)
x =
10-7
43. a. 7.62 107 lbb. 3.3 104 tonsc. 6.27 s
44. D
45. F
Using the Quadratic FormulaALGEBRA 1 LESSON 10-7
1. Solve 2x2 – 11x + 12 = 0 by using the quadratic formula.
2. Solve 4x2 – 12x = 64. Round the solutions to the nearest hundredth.
3. Suppose a model rocket is launched from a platform 2 ft above the ground with an initial upward velocity of 100 ft/s. After how many seconds will the rocket hit the ground? Round the solution to the nearest hundredth.
1.5, 4
–2.77, 5.77
6.27 seconds
10-7
Using the Discriminant
(For help, go to Lessons 1-6 and 10-7.)ALGEBRA 1 LESSON 10-8
Evaluate b2 – 4ac for the given values of a, b, and c.
1. a = 3, b = 4, c = 8 2. a = –2, b = 0, c = 9
3. a = 11, b = –5, c = 7
Solve using the quadratic formula. If necessary, round to the nearest hundredth.
4. 3x2 – 7x + 1 = 0 5. 4x2 + x – 1 = 0 6. x2 – 12x + 35
= 0
10-8
Using the DiscriminantALGEBRA 1 LESSON 10-8
Solutions
1. b2 – 4ac for a = 3, b = 4, c = 8: 42 – 4(3)(8) = 16 – 96 = –80
2. b2 – 4ac for a = –2, b = 0, c = 9: 02 – 4(–2)(9) = 0 + 72 = 72
3. b2 – 4ac for a = 11, b = –5, c = 7: (–5)2 – 4(11)(7) = 25 – 308 = –283
4. 3x2 – 7x + 1 = 0; x = for a = 3, b = –7, c = 1:
x = = =
x = 2.18 or x = 0.15
–b ± b2 – 4ac2a
–(–7) ± (–7)2 – 4(3)(1)2(3)
7 ± 49 – 126
7 ± 376
7 + 376
7 – 376
10-8
Using the DiscriminantALGEBRA 1 LESSON 10-8
Solutions (continued)
5. 4x2 + x – 1 = 0; x = for a = 4, b = 1, c = –1:
x = = =
x = 0.39 or x = –0.64
–b ± b2 – 4ac2a
–1 ± 12 – 4(4)(–1)2(4)
–1 ± 1 + 168
–1 ± 178
6. x2 – 12x + 35 = 0; x = for a = 1, b = –12, c = 35:
x = =
x = =
x = = = 7 or x = = = 5
–1 – 178
–1 + 178
–b ± b2 – 4ac2a
–(–12) ± (–12)2 – 4(1)(35)2(1)
12 ± 144 – 1402
12 ± 42
12 ± 22
12 + 22
142
12 – 22
102
10-8
Using the DiscriminantALGEBRA 1 LESSON 10-8
Find the number of solutions of x2 = –3x – 7 using the discriminant.
x2 + 3x + 7 = 0 Write in standard form.
b2 – 4ac = 32 – 4(1)(7) Evaluate the discriminant. Substitute for a, b, and c.
= 9 – 28 Use the order of operations.
= –19 Simplify.
Since –19 < 0, the equation has no solution.
10-8
Using the DiscriminantALGEBRA 1 LESSON 10-8
A football is kicked from a starting height of 3 ft with an initial upward velocity of 40 ft/s. Will the football ever reach a height of 30 ft?
h = –16t2 + vt + c Use the vertical motion formula.
30 = –16t2 + 40t + 3 Substitute 30 for h, 40 for v, and 3 for c.
0 = –16t2 + 40t – 27 Write in standard form.
b2 – 4ac = (40)2 – 4 (–16)(–27) Evaluate the discriminant.
= 1600 – 1728 Use the order of operations.
= –128 Simplify.
The discriminant is negative. The football will never reach a height of 30 ft.
10-8
Using the DiscriminantALGEBRA 1 LESSON 10-8
pages 556–558 Exercises
1. A
2. C
3. B
4. 0
5. 1
6. 2
7. 2
8. 2
9. 2
10. 0
11. 2
12. 2
13. 1
14. 2
15. 0
16. none
17. No; the discriminant is negative.
18. a. yesb. noc. nod. no
19. 0
20. 0
21. 2
22. 2
23. 0
24. 2
25. a. S = –0.75p2 + 54pb. noc. $36d. If a product is too expensive,
fewer people will buy it.
26. a. k > 4b. k = 4c. k < 4
27. a. A2 ^ 2 – 4;A2 ^ 2 – 8
b. |b| < 2
10-8
Using the DiscriminantALGEBRA 1 LESSON 10-8
30. a. 16; 5, 1b. 81; 4, –5c. 73; 3.89, –0.39d. Rational; the square root
of a discriminant that is a perfect square is a pos. integer.
31. no
32. no
33. yes; 1, –1.25
34. yes; –1,
35. no
36. yes; 2.5, –1
37. Answers may vary. Sample: Use values for a, b, and c such that the discriminant is positive.
38. never
39. sometimes
40. always
41. 2; since the parabola crosses the x-axis once, it must cross again.
42. y = 2x2 + 8x + 10 has a vertex closer to the x-axis; its discriminant is closer to zero.
43. C
44. G
45. B
46. D
47. A
10-8
23
Using the DiscriminantALGEBRA 1 LESSON 10-8
48. [2] x(25 – x) = 136x2 – 25x + 136 = 0
x =
= = 17 or 8
yes, 17 cm by 8 cm[1] no work shown OR appropriate
methods with one computational error
49. 0.5, –1.5
50. 1.83, –3.83
51. 0.61, –0.27
52. 1.79, –2.79
53. 2.54, 0.13
54. 2, 1.5
–(–25) ± (–25)2 – 4(1)(136)2(1)
25 ± 92
55. $1093.81
56. $312.88
57. $6104.48
58. arithmetic
59. arithmetic
60. geometric
61. arithmetic
10-8
Using the DiscriminantALGEBRA 1 LESSON 10-8
Find the number of solutions for each equation.
1. 3x2 – 4x = 7
2. 4x2 = 4x – 1
3. –3x2 + 2x – 12 = 0
4. A ball is thrown from a starting height of 4 ft with an initial upward velocity of 30 ft/s. Is it possible for the ball to reach a height of 18 ft?
two
one
none
yes
10-8
Choosing a ModelALGEBRA 1 LESSON 10-9
(For help, go to Lessons 6-2, 8-7 and 10-1.)
Graph each function.
1. y = 3x – 1 2. y = x + 2
3. y = 2x 4. y = x
5. y = x2 + 5 6. y = 2x2 – 1
14
13
10-9
Choosing a ModelALGEBRA 1 LESSON 10-9
1. y = 3x – 1 2. y = x + 2
3. y = 2x 4. y = x
5. y = x2 + 5 6. y = 2x2 – 1
14
13
Solutions
10-9
Choosing a ModelALGEBRA 1 LESSON 10-9
Graph each set of points. Which model is most appropriate for each set?
a. (–2, 2.25), (0, 3),(1, 4) (2, 6)
b. (–2, –2), (0, 2),(1, 4), (2, 6)
c. (–1, 5), (2, 11), (0, 3),(1, 5), (–2, 11)
exponential model linear modelquadratic model
10-9
Choosing a ModelALGEBRA 1 LESSON 10-9
a. Which kind of function best models the data below? Write an equation to model the data.
x y0 01 1.42 5.63 12.64 22.4
Step 1: Graph the data
There is a common second difference, 2.8.
Step 2: The data appear to be quadratic. Test for a common second difference.
x y0 01 1.42 5.63 12.64 22.4
+1
+1
+1+1
+ 1.4
+ 4.2+ 7.0+ 9.8
+ 2.8
+ 2.8
+ 2.8
10-9
Choosing a ModelALGEBRA 1 LESSON 10-9
a. (continued)
Step 3: Write a quadratic model. y = ax2
5.6 = a(2)2 Use a point other than(0, 0) to find a.
5.6 = 4a Simplify.1.4 = a Divide each side
by 4. y = 1.4x2 Write a quadratic
function.
Step 4: Test two points other than (2, 5.6) and (0, 0)
y = 1.4(1)2 y = 1.4(3)2
y = 1.4 • 1 y = 1.4 • 9y = 1.4 y = 12.6(1, 1.4) and (3, 12.6) are both data points.
The equation y = 1.4x2 models the data.
10-9
Choosing a ModelALGEBRA 1 LESSON 10-9
b. Which kind of function best models the data below? Write an equation to model the data.
x y–1 4 0 2 1 1 2 0.5 3 0.25
Step 1: Graph the data Step 2: The data appear to suggest an exponential model. Test for a common ratio.
x y–1 4 0 2 1 1 2 0.5 3 0.25
+1
+1
+1+1
2 4 = 0.51 2 = 0.50.5 1 = 0.50.25 0.5 = 0.5
10-9
Choosing a ModelALGEBRA 1 LESSON 10-9
b. (continued)
Step 4: Test two points other than (0, 2).
y = 2 • 0.51 y = 2 • 0.52
y = 2 • 0.5 y = 2 • 0.25
y = 1.0 y = 0.5
(1, 1) and (2, 0.5) are both data points.
Step 3: Write an exponential model.
Relate: y = a • bx
Define: Let a = the initial value, 2.
Let b = the decay factor, 0.5.
Write: y = 2 • 0.5x
The equation y = 2 • 0.5x models the data.
10-9
Choosing a ModelALGEBRA 1 LESSON 10-9
Suppose you are studying deer that live in an area. The data in the table was collected by a local conservation organization. It indicates the number of deer estimated to be living in the area over a five-year period. Determine which kind of function best models the data. Write an equation to model the data.
Year Estimated Population
0 90 1 69 2 52 3 40 4 31
Step 1: Graph the data to decide which model is most appropriate.
The graph curves, and it does not look quadratic. It may be exponential.
10-9
Choosing a ModelALGEBRA 1 LESSON 10-9
(continued)
Step 2: Test for a common ratio.
Year Estimated Population
0 90 1 69 2 52 3 40 4 31
+1
+1
+1+1
69 90 0.76652 69 0.75340 52 0.76931 40 0.775
The common ratio is roughly 0.77.
The population of deer is roughly 0.77 times its value the previous year.
10-9
Choosing a ModelALGEBRA 1 LESSON 10-9
(continued)
Step 3: Write an exponential model.
Relate: y = a • bx
Define: Let a = the initial value, 90.
Let b = the decay factor, 0.77.
Write: y = 90 • 0.77x
Step 4: Test two points other than (0, 90).
y = 90 • 0.771 y = 90 • 0.772
y 69 y 53
The predicted value (1, 69) matches the corresponding data point. The point (2, 53) is close to the data point (2, 52).
The equation y = 90 • 0.77x models the data.
10-9
5.
exponential
6.
linear
7. quadratic; y = 1.5x2
8. linear; y = 2x – 5
9. quadratic; y = 2.8x2
Choosing a ModelALGEBRA 1 LESSON 10-9
pages 563–566 Exercises
1.
quadratic
2.
linear
3.
exponential
4.
quadratic
10-9
Choosing a ModelALGEBRA 1 LESSON 10-9
10. exponential; y = 1 • 1.2x
11. exponential; y = 5 • 0.4x
12. linear; y = – x + 2
13. a.
linearb. 65, 64, 64; yesc. 64d. y = 64x – 5
12
14. a. exponentialb. y = 16,500 • 0.88x
15. a. 41, 123, 206b. 82, 83c. d = 41t 2
d. 256.25 cm16. a.
linearb. 5 yearsc. 600, 600, 600; 120, 120, 120d. p = 120t + 5100
10-9
Choosing a ModelALGEBRA 1 LESSON 10-9
17. a. 5b. 398, 429, 407, 389;
79.6, 85.8, 81.4, 77.8c. 81.2d. p = 81.2t + 4457e. 6893 million,
or about 6.9 billion
18. Answers may vary. Sample: Linear data have a common first difference, quadratic data have a common second difference, and exponential data have a common ratio.
19. y = 0.875x2 – 0.435x + 1.515
20. y = 1.987 • 0.770x
21. y = 2.125x2 – 4.145x + 2.955
22. y = –0.336x2 – 0.219x + 4.666
23. y = –1.1x + 3.5
24. y = 0.102 • 2.582x
25. a. i.
ii.
x y1 –22 13 64 135 22
) 3 ) 2) 5 ) 2) 7 ) 2) 9
x y1 32 123 274 485 75
) 9 ) 6) 15 ) 6) 21 ) 6) 27
10-9
iii.
b. The second common difference is twice the coefficient of x2.
c. When second differences are the same, the data are quadratic. You can determine the coefficient of x2 by dividing the second difference by 2.
Choosing a ModelALGEBRA 1 LESSON 10-9
x y1 –12 63 214 445 75
) 7 ) 8) 15 ) 8) 23 ) 8) 31
26. Answers may vary. Sample:
27. a. quadratic b. d = 13.6t 2
c. 54.5 ft
28. Check students’ work.
x y0 52 134 296 53
10-9
Choosing a ModelALGEBRA 1 LESSON 10-9
29. a. 1.85, 1.28, 1.45, 1.43b. 139, 85, 174, 240c. –54, 89, 66d. 1.85; the ratio is much
larger than the other ratios.e. 85; the difference is much
smaller than the other differences.f.
30. B
31. H
32. [2] p = 33,500(1.014)n, 33,500(1.014)10 38,497
[1] correct formula, inaccurate evaluation
33. [4] a. linearb. d = –2.5n + 43.5c. 18
[3] appropriate methods, but with one computational error
[2] part (c) not answered[1] no work shown
34. 1
35. 0
36. 2
10-9
Choosing a ModelALGEBRA 1 LESSON 10-9
37.
38.
39.
40.
41.
42.
43. 0.125
44. –32
45.
46. 250
47. 16
48. 30.375
10-9
227
Choosing a ModelALGEBRA 1 LESSON 10-9
Which kind of function best models the data in each table? Write an equation to model the data.
x y–1 15 0 3 1 0.6 2 0.12 3 0.024
1. x y–1 –5 0 –3 1 –1 2 1 3 3
2. x y–1 2.2 0 0 1 2.2 2 8.8 3 19.8
3.
exponential;y = 3 • 0.2x
linear;y = 2x – 3
quadratic;y = 2.2x2
10-9
Quadratic Equations and FunctionsALGEBRA 1 CHAPTER 10
1. D2. C3. A4. B5. x = 0, (0, –7); min.6. x = 1.5, (1.5, –0.25); min.7. x = 2.5, (2.5, 11.5); max.8. x = –6, (–6, –18); min.9.
10.
11.
12.
13.
14.
15. Answers may vary. Sample: You can tell how wide it is and whether it opens upward or downward.
10-A
Quadratic Equations and FunctionsALGEBRA 1 CHAPTER 10
16. 117. 018. 219. 220.
21.
22. 1.2
23. 40
24.
25. 0.2
26. 5, 6
27. 11, 12
28. 18, 19
29. –3, –2
30. 1
31. no solution
32. 2
33. 2
23
34. 135. 5, –536. 3.33, –137. 2, –838. 1, –239. 0.24, –4.2440. 1, –2.3341. Answers may vary.
Sample: y = –x2 + 4;
42. 140 = 10 r 2, 2.1 ft
43. 800 = 2w 2, w = 20 ft,
= 40 ft
44. quadratic; y = x2
45. exponential; y = (2x)
46. linear; y = x + 2
47. exponential; y = 40(0.5x)
10-A
12
12