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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.
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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. hkj2. kh23. hk24. kj 2+ h Graph each equation. 5.y= 2x – 1 6.y = | x | 7.y = x2 + 2 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 Exploring Quadratic Graphs ALGEBRA 1 LESSON 10-1 Solutions 10-1
Exploring Quadratic Graphs ALGEBRA 1 LESSON 10-1 Identify the vertex of each graph. Tell whether it is a minimum or a maximum. a. b. The vertex is (1, 2). The vertex is (2, –4). It is a maximum. It is a minimum. 10-1
1 3 xy= x2 (x, y) 1 3 0 (0)2= 0(0, 0) 1 3 1 3 1 3 2 (2)2= 1(2, 1 ) 1 3 3 (3)2= 3(3, 3) Exploring Quadratic Graphs ALGEBRA 1 LESSON 10-1 Make a table of values and graph the quadratic function y= x2. 1 3 10-1
1 2 (x) = –x2 (x) = –3x2 (x) = x2 1 2 Of the three graphs, (x) = x2 is the widest and (x) = –3x2 is the narrowest. 1 2 So, the order from widest to narrowest is (x) = x2, (x) = –x2, (x) = –3x2. Exploring Quadratic Graphs ALGEBRA 1 LESSON 10-1 Use the graphs below. Order the quadratic functions (x) = –x2, (x) = –3x2,and (x) = x2 from widest to narrowest graph. 1 2 10-1
xy= 3x2 y= 3x2 – 2 2 12 10 1 3 1 0 0 2 –1 3 1 2 12 10 Exploring Quadratic Graphs ALGEBRA 1 LESSON 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. 10-1
Height h is dependent on time t. th = –16t2 + 26 0 26 1 10 2 –38 Graph t on the x-axis and h on the y-axis. Use positive values for t. Exploring Quadratic Graphs ALGEBRA 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. 10-1
Exploring Quadratic Graphs ALGEBRA 1 LESSON 10-1 pages 513–516 Exercises 1. (2, 5); max. 2. (–3, –2); min. 3. (2, 1); min. 4. 5. 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. 6. 7. 8. 1 2 1 3 1 4 1 2 2 3 10-1
Exploring Quadratic Graphs ALGEBRA 1 LESSON 10-1 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. 15. 16. 17. 18. 19. 20. 1 2 10-1
Exploring Quadratic Graphs ALGEBRA 1 LESSON 10-1 31. 32. 33. 34. 35. 36. 37. 38.a. b. 184 ft c. 56 ft 39.a. 0 < r < 6 b. 0 < A < 36 113.1 c. 10-1
= = / / Exploring Quadratic Graphs ALGEBRA 1 LESSON 10-1 40.K, L 41.M 42.K 43.M 44. Answers may vary. Sample: a.y = 5x2b.y = –5x2c.y = 3x2 45.a. b. 16 ft c. 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
t h 0 200 1 184 2 136 3 56 4 –56 Exploring Quadratic Graphs ALGEBRA 1 LESSON 10-1 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 48.a.a > 0 b. |a| > 1 49.a. b. 10-1
Exploring Quadratic Graphs ALGEBRA 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. –12t3 + 22t2 63. 12m6 – 4m5 + 20m2 64. –15y6 – 10y4 + 20y 65. –12c5 + 21c4 – 24c3 66. 15 balloons 10-1
1 4 3. Order the quadratic functions y = –4x2, y = x2, and y = 2x2 from widest to narrowest graph. 1 4 y = x2, y = 2x2, y = –4x2 Exploring Quadratic Graphs ALGEBRA 1 LESSON 10-1 1 2 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. (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. 10-1
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 = x26.y = –x2 + 2 7.y = x2 – 1 1 2 –b 2a Quadratic Functions ALGEBRA 1 LESSON 10-2 (For help, go to Lessons 1-6 and 10-1.) 10-2
1 3 for a = –9, b = 108: for a = –8, b = –56: for a = 15, b = 20: for a = –6, b = 4: = = –108 –18 56 –16 –4 –12 –20 30 –(–56) 2(–8) –20 2(15) –4 2(–6) –108 2(–9) 1 2 7. y = x2 – 1 5. y = x2 6. y = –x2 + 2 = 6 = b 2a b 2a b 2a b 2a Quadratic Functions ALGEBRA 1 LESSON 10-2 Solutions – 1. – 2 3 2. = – = – 7 2 1 2 3. = – = –3 = – 4. 10-2
Step 1: Find the equation of the axis of symmetry and the coordinates of the vertex. Find the equation of the axis of symmetry. x = = = – 1 The x-coordinate of the vertex is –1. y = 2x2 + 4x – 3 y = 2(–1)2 + 4(–1) – 3 = –5 b 2a – To find the y-coordinate of the vertex, substitute –1 for x. –4 2(2) The vertex is (–1, –5). Quadratic Functions ALGEBRA 1 LESSON 10-2 Graph the function y = 2x2 + 4x – 3. 10-2
Choose a value for x on the same side of the vertex. Let x = 1 y = 2(1)2 + 4(1) – 3 = 3 For x = 1, y = 3, so another point is (1, 3). Find the y-coordinate for x = 1. Quadratic Functions ALGEBRA 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
Then draw the parabola. Quadratic Functions ALGEBRA 1 LESSON 10-2 (continued) Step 3:Reflect (0, –3) and (1, 3) across the axis of symmetry to get two more points. 10-2
Quadratic Functions ALGEBRA 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 t2 is negative, the curve opens downward, and the vertex is the maximum point. 10-2
b 2a – Step 1: Find the x-coordinate of the vertex. After 2.75 seconds, the star will be at its greatest height. –88 2(–16) = = 2.75 Quadratic Functions ALGEBRA 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. 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. Shade above the curve. Quadratic Functions ALGEBRA 1 LESSON 10-2 Graph the quadratic inequality y > –x2 + 6x – 5. 10-2
Quadratic Functions ALGEBRA 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 ft b. 400 ft2 16.a. 1.25 s b. 31 ft 17. 10-2
Quadratic Functions ALGEBRA 1 LESSON 10-2 18. 19. 20. 21. 22. 23. 24. 25. 26. 10-2
Quadratic Functions ALGEBRA 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 m b. 7.2 m 36.a.y < –0.1x2 + 12 b. c. Yes; when x = 6, y = 8.4, so the camper will fit. 37.a. $12.50 b. $10,000 38. 26 units2 39. 26 units2 10-2
Quadratic Functions ALGEBRA 1 LESSON 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 + 13 c. (6.5, 42.25) d. 6.5 ft by 6.5 ft 43.a. 0.4 s b. 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 s b. 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
50.[2] axis of symmetry: x = = 16.7 maximum height: y –0.009(16.7)2 + 0.3(16.7) + 4.5 y 7 ft [1] appropriate methods, but with a minor computational error 51. C 52. A 53. F 54. D 55. B 56. E 46.a. (0, 2) b.x = –2.5 c. 5 d.y = x2 + 5x + 2 e. 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 –b 2a –0.3 2(–0.009) Quadratic Functions ALGEBRA 1 LESSON 10-2 10-2
Quadratic Functions ALGEBRA 1 LESSON 10-2 57.c2 – 5c – 36 58. 2x2 + 7x – 30 59. 20t2 + 17t + 3 60. 21n4 – 62n2 + 16 61. 2a3 + 9a2 – a + 20 62. 6r3 + 9r2 – 20r + 7 10-2
< 1 4 3.y – x2 – 2x – 6 x = –4 Quadratic Functions ALGEBRA 1 LESSON 10-2 Graph each relation. Label the axis of symmetry and the vertex. 1.y = x2 – 8x + 15 2. ƒ(x) = –x2 + 4x – 2 10-2
Simplify each expression. 1. 1122. (–12)23. –(12)24. 1.52 5. 0.626.27.28.2 2 3 – 1 2 4 5 Finding and Estimating Square Roots ALGEBRA 1 LESSON 10-3 (For help, go to Lessons 1-2 and 8-5.) 10-3
2 3 1 2 4 5 – 2 3 2 3 – – Finding and Estimating Square Roots ALGEBRA 1 LESSON 10-3 Solutions 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 = • = 1 4 1 2 1 2 4 5 4 5 16 25 4 9 10-3
9 25 positive square root a. 25 = 5 3 5 3 5 The square roots are and – . b. ± = ± 3 5 negative square root c. – 64 = –8 For real numbers, the square root of a negative number is undefined. d. –49 is undefined There is only one square root of 0. e. ± 0 = 0 Finding and Estimating Square Roots ALGEBRA 1 LESSON 10-3 Simplify each expression. 10-3
a. ± 144 = ± 12 b. – = –0.44721359 . . . c. – 6.25 = –2.5 1 5 1 9 d. = 0.3 e. 7 = 2.64575131 . . . Finding and Estimating Square Roots ALGEBRA 1 LESSON 10-3 Tell whether each expression is rational or irrational. rational irrational rational rational irrational 10-3
28.34 is between the two consecutive perfect squares 25 and 36. 25 < 28.34 < 36 5 < < 6 28.34 The square roots of 25 and 36 are 5 and 6, respectively. 28.34 is between 5 and 6. Finding and Estimating Square Roots ALGEBRA 1 LESSON 10-3 Between what two consecutive integers is 28.34 ? 10-3
Use a calculator. 5.323532662 28.34 5.32 Round to the nearest hundredth. Finding and Estimating Square Roots ALGEBRA 1 LESSON 10-3 Find 28.34 to the nearest hundredth. 10-3
Substitute 8 for x. d = x2 + (3x)2 Simplify. d = 82 + (3 • 8)2 d = 64 + 576 d = 640 d 25.3 Use a calculator. Round to the nearest tenth. Finding and Estimating Square Roots ALGEBRA 1 LESSON 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. The diagonal is about 25.3 ft long. 10-3
Finding and Estimating Square Roots ALGEBRA 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. 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 1 3 3 7 6 7 1 9 5 4 10-3
= / Finding and Estimating Square Roots ALGEBRA 1 LESSON 10-3 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. 4 52.a. 5 s b. 10 s c. No; the object takes twice as long to fall. 53. False; zero has one square root. 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 36. ± 37. ± 202 38. 1 39.a. 8450 km b. 7684 km 40. 21 41. – 42. 1.41 43. 1.26 44. –5.48 45. –33 46. –0.8 8 11 2 5 1 2 10-3
Finding and Estimating Square Roots ALGEBRA 1 LESSON 10-3 60. 1.3 61. 6 62. 63. 9 64. 11 65. 128 66. 67. 68. 69. 70. 71. 72.d2 – 81 73. 9t2 – 25 74. 8075 2 9 10-3
Finding and Estimating Square Roots ALGEBRA 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 3 4 1 12 81 4 10-3
2 5 ± 4 25 3 5 Finding and Estimating Square Roots ALGEBRA 1 LESSON 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. 14 irrational undefined rational –8 and –7 about 40.25 mph 10-3
Simplify each expression. 1. 36 2. – 81 3. ± 121 4. 1.44 5. 0.25 6. ± 1.21 7.8. ± 9. ± 1 4 1 9 49 100 Solving Quadratic Equations ALGEBRA 1 LESSON 10-4 (For help, go to Lesson10-3.) 10-4
1 4 1 9 49 100 7 10 Solving Quadratic Equations ALGEBRA 1 LESSON 10-4 Solutions 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. ± = 1 3 1 2 10-4
Graph y = 2x2 Graph y = 2x2 + 2 Graph y = 2x2 – 2 Solving Quadratic Equations ALGEBRA 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
3x2 – 75 + 75 = 0 + 75Add 75 to each side. 3x2 = 75 x = ± 25 Find the square roots. Solving Quadratic Equations ALGEBRA 1 LESSON 10-4 Solve 3x2 – 75 = 0. x2 = 25 Divide each side by 3. x = ± 5 Simplify. 10-4
S = 4 r2 315 = 4 r2 Substitute 315 for S. = r2 Put in calculator ready form. 315 (4) 315 (4) Find the principle square root. = r2 Use a calculator. 5.00668588 r Solving Quadratic Equations ALGEBRA 1 LESSON 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. The radius of the sphere is about 5 ft. 10-4
Solving Quadratic Equations ALGEBRA 1 LESSON 10-4 pages 531–534 Exercises 1. ±3 2. no solution 3. 0 7. no solution 8. 0 9. ±2 4. ±2 5. no solution 6. ±3 10-4
Solving Quadratic Equations ALGEBRA 1 LESSON 10-4 30. ± 31. ± 2.8 32. ± 0.4 33. ± 3.5 34. 3.5 s 35. 121 36. a.n > 0 b.n = 0 c.n < 0 37. Answers may vary. Sample: Michael subtracted 25 from the left side of the equation but added 25 to the right side. 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. r2 = 80; 5.0 cm 1 6 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 ft b. 16.0 ft c. No; the radius increases by about 1.4 times. 28. no solution 29. ± 5 2 1 4 3 7 10-4
Solving Quadratic Equations ALGEBRA 1 LESSON 10-4 41. 6.3 ft 42. 11.0 cm 43.a. 0.2 m b. 2.5 s c. 3.0 s d. Shorten; as decreases, t decreases. 44.a. –7 b. (–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 38.a. 2, –2; 2, –2 b. The first equation multiplied by 2 on both sides equals the second equation. 39.a. square: 4r2, circle: r2 b. 4r2 – r2 = 80 c. 9.7 in., 19.3 in. 40. Answers may vary. Sample: a. 5x2 + 10 = 0, no solution b. 2x2 + 0 = 0, x = 0 c. –20x2 + 80 = 0, x = ± 2 10-4
xy –2 5 –1 –4 0 –7 1 –4 2 5 Solving Quadratic Equations ALGEBRA 1 LESSON 10-4 49.[2] x-intercepts 1.5, –1.5 [1] minor error in table OR incorrect graph 50.[4] a. 96 = 6s2 s2 = 16 s = 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