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2. SECTION 2.6 QUADRATIC FUNCTIONS 2

3. A baseball is “popped” straight up by a batter. The height of the ball above the ground is given by the function: y = f(t) = −16t 2 + 64t + 3, where t is time in seconds after the ball leaves the bat and y is in feet. Let's use our calculator: 3Page 88

4. Let's use our calculator: Y= → \Y 1 = −16x 2 +64x+3 4Page N/A

5. Let's use our calculator: Y= → \Y 1 = −16x 2 +64x+3 WindowValue Xmin Xmax5 Xscl1 Ymin-10 Ymax80 Yscl8 Window → Graph 5Page N/A

6. Although the path of the ball is straight up and down, the graph of its height as a function of time is concave down. 6Page 88

7. The ball goes up fast at first and then more slowly because of gravity. 7Page 88

8. The baseball height function is an example of a quadratic function, whose general form is y = ax 2 + bx + c. 8Page 89

9. Finding the Zeros of a Quadratic Function 9Page 89

10. Finding the Zeros of a Quadratic Function Back to our baseball example, precisely when does the ball hit the ground? 10Page 89

11. Finding the Zeros of a Quadratic Function Back to our baseball example, precisely when does the ball hit the ground? Or: For what value of t does f(t) = 0? 11Page 89

12. Finding the Zeros of a Quadratic Function Back to our baseball example, precisely when does the ball hit the ground? Or: For what value of t does f(t) = 0? Input values of t which make the output f(t) = 0 are called zeros of f. 12Page 89

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18. Let's use our calculator: Y= → \Y 1 = −16x 2 +64x+3 WindowValue Xmin Xmax5 Xscl1 Ymin-10 Ymax80 Yscl8 Window → Graph 18Page N/A

19. Now let's use the TI to find the zeros of this quadratic function: 19Page N/A

20. 2nd Trace 2: zero Left Bound ? Right Bound? Guess? 20Page N/A

21. zero X=4.0463382 Y=-1E-11 21Page N/A

22. Example #1: Find the zeros of f(x) = x 2 − x − 6. 22Page 89

23. Example #1: Find the zeros of f(x) = x 2 − x − 6. Set f(x) = 0 and solve by factoring: x 2 − x − 6 = 0 (x-3)(x+2) = 0 x = 3 & x = -2 23Page 89

24. Example #1: Find the zeros of f(x) = x 2 − x − 6. Let's use our calculator: 24Page 89 Example #1

25. Let's use our calculator: Y= → \Y 1 = x 2 -x-6 25Page N/A

26. Let's use our calculator: Y= → \Y 1 = x 2 -x-6 WindowValue Xmin-10 Xmax10 Xscl1 Ymin-10 Ymax10 Yscl1 Zoom 6 gives: Graph 26Page N/A

27. 27Page N/A

28. Now let's use the TI to find the zeros of this quadratic function: 28Page N/A

29. 2nd Trace 2: zero Left Bound ? Right Bound? Guess? 29Page N/A

30. zero x=-2 y=0 30Page N/A

31. 2nd Trace 2: zero Left Bound ? Right Bound? Guess? 31Page N/A

32. zero x=3 y=0 32Page N/A

33. Example #3 Figure 2.29 shows a graph of:2.29 What happens if we try to use algebra to find its zeros? 33Page 89 Example #3

34. Let's try to solve: 34Page 89

35. 35Page 90

36. Conclusion? 36Page 90

37. Conclusion? There are no real solutions, so h has no real zeros. Look at the graph again... 37Page 90

38. x y What conclusion can we draw about zeros and the graph below? 38Page 89

39. x y h has no real zeros. This corresponds to the fact that the graph of h does not cross the x-axis. 39Page 89

40. Let's use our calculator: Y= → \Y 1 = (-1/2) x 2 -2 40Page N/A

41. Let's use our calculator: Y= → \Y 1 = (-1/2) x 2 -2 WindowValue Xmin-4 Xmax4 Xscl1 Ymin-10 Ymax2 Yscl1 Window Graph 41Page N/A

42. x y 42Page N/A

43. 2nd Trace 2: zero Left Bound ? Right Bound? Guess? 43Page N/A

44. 2nd Trace 2: zero Left Bound ? Right Bound? Guess? ERR:NO SIGN CHNG 1:Quit 44Page N/A

45. Concavity and Quadratic Functions 45Page 90

46. Concavity and Quadratic Functions Unlike a linear function, whose graph is a straight line, a quadratic function has a graph which is either concave up or concave down. 46Page 90

47. Example #4 Let f(x) = x 2. Find the average rate of change of f over the intervals of length 2 between x = −4 and x = 4. What do these rates tell you about the concavity of the graph of f ? 47Page 90 Example #4

48. Let f(x) = x 2 Between x = -4 & x = -2: 48Page 90

49. Let f(x) = x 2 Between x = -2 & x = 0: 49Page 90

50. Let f(x) = x 2 Between x = 0 & x = 2: 50Page 90

51. Let f(x) = x 2 Between x = 2 & x = 4: 51Page 90

52. Let's recap: 52Page 90

53. What do these rates tell you about the concavity of the graph of f ? 53Page 90

54. What do these rates tell you about the concavity of the graph of f ? Since these rates are increasing, we expect the graph of f to be bending upward. Figure 2.30 confirms that the graph is concave up. 2.30 54Page 90

55. 55Page 90

56. Let's use our calculator: Y= → \Y 1 = x 2 2nd Mode = Quit ( Vars → Enter Enter (-2) - Vars → Enter Enter (-4)) / (-2 - -4) Enter 56Page N/A

57. Let's use our calculator: Y= → \Y 1 = x 2 2nd Mode = Quit ( Vars → Enter Enter (-2) - Vars → Enter Enter (-4)) / (-2 - -4) Enter -6 57Page N/A

58. ( Vars → Enter Enter (-2) - Vars → Enter Enter (-4)) / (-2 - -4) Enter -6 ( Vars → Enter Enter (0) - Vars → Enter Enter (-2)) / (0 - -2) Enter -2 58Page N/A

59. ( Vars → Enter Enter (2) - Vars → Enter Enter (0)) / (2- 0) Enter 2 ( Vars → Enter Enter (4) - Vars → Enter Enter (2)) / (4 - 2) Enter 6 59Page N/A

60. Example #5 A high diver jumps off a 10-meter springboard. For h in meters and t in seconds after the diver leaves the board, her height above the water is in Figure 2.31 and given by: (a) Find and interpret the domain and range of the function and the intercepts of the graph. (b) Identify the concavity. 60Page 91 Example #5

61. Let's use our calculator: Y= → \Y 1 = −4.9x 2 +8x+10 61Page N/A

62. Let's use our calculator: Y= → \Y 1 = −4.9x 2 +8x+10 WindowValue Xmin-2 Xmax5 Xscl1 Ymin-10 Ymax15 Yscl1 Window Graph 62Page N/A

63. 63 Now let's use the TI to find the zeros of this quadratic function. 2nd MODE Page N/A

64. 2nd Trace 2: zero Left Bound ? Right Bound? Guess? 64Page N/A

65. 65 Now let's use the TI to find the zeros of this quadratic function. Zero X= -.8290322 Y= 0 Page N/A

66. 2nd Trace 2: zero Left Bound ? Right Bound? Guess? 66Page N/A

67. 67 Now let's use the TI to find the zeros of this quadratic function. Zero X= 2.4616853 Y= 0 Page N/A

68. 68 So, our zeros (solutions) are: X= -.8290322 Y= 0 X= 2.4616853 Y= 0 Which make sense? Page N/A

69. 69 Which make sense? Since t ≥ 0: X= -.8290322 Y= 0 X= 2.4616853 Y= 0 Page 91

70. 70 X= 2.4616853 Y= 0 Domain? Page 91

71. 71 X= 2.4616853 Y= 0 Domain? The interval of time the diver is in the air, namely 0 ≤ t ≤ 2.462. Page 91

72. 72 X= 2.4616853 Y= 0 Range? Page 91

73. 73 X= 2.4616853 Y= 0 Range? Given that the domain is 0 ≤ t ≤ 2.462, what can f(t) be? Page 91

74. 74 X= 2.4616853 Y= 0 Range? What you see in yellow. Page 91

75. 75 X= 2.4616853 Y= 0 Range? What you see in yellow. What is the maximum value of f(t)? Page 91

76. 2nd Trace 4: maximum Left Bound ? Right Bound? Guess? 76Page N/A

77. 77 X= 2.4616853 Y= 0 Range? What is the maximum value of f(t)? Maximum X=.81632636 Y= 13.265306 Page 91

78. 78 X= 2.4616853 Y= 0 Therefore, the range is: 0 ≤ f(t) ≤ 13.265306 Page 91

79. 79 What are the intercepts of the graph? Page 91

80. 80 What are the intercepts of the graph? How can we calculate? Page 91

81. 81 What are the intercepts of the graph? How can we calculate? We already did Page 91

82. 82 What are the intercepts of the graph? How can we calculate? We already did t= 2.4616853 f(t)= 0 horiz int. Page 91

83. 83 What are the intercepts of the graph? How can we calculate? What about? Page 91

84. 84 What are the intercepts of the graph? How can we calculate? Substitute 0 for t in the above equation... Page 91

85. 85 What are the intercepts of the graph? t= 0, f(t) = 10 vert int. Page 91

86. Finally, let's identify the concavity. 86Page 91

87. 87 What can we say about concavity? Page 91

88. 88 What can we say about concavity? Concave down. Let's confirm via a table... Page 91

89. 89 t (sec)h (meters)Rate of change Δh/Δt 010 0.512.775 1.013.100 1.510.975 2.06.400 Page 91

90. 90 t (sec)h (meters)Rate of change Δh/Δt 010 5.55 0.512.775 0.65 1.013.100 −4.25 1.510.975 −9.15 2.06.400 Page 91

91. 91 t (sec)h (meters)Rate of change Δh/Δt 010 5.55 0.512.775 0.65 1.013.100 −4.25 1.510.975 −9.15 2.06.400decreasing Δh/Δt Page 91

92. End of Section 2.6 92