1. Lecture 321 Unit 4 Lecture 32 Quadratics Applications & Their Graphs

2. Lecture 322 Objectives Formulate a quadratic equation from a problem situationFormulate a quadratic equation from a problem situation Solve a quadratic equation by using the quadratic formulaSolve a quadratic equation by using the quadratic formula Graph the quadratic equationGraph the quadratic equation Interpret the graph in terms of the problem situationInterpret the graph in terms of the problem situation

3. Lecture 323 Revenue: R = 78x - x 2 Cost: C = x 2 –38 x + 400 Find the equation for Profit Profit = Revenue - Cost Earl Black makes tea bags. The cost of making x million teas bags per month is C = x 2 –38 x+400. The revenue from selling x million tea bags per month is R = 78x - x 2

4. Lecture 324 The equation for profit is, Profit = Revenue - Cost Revenue: R = 78x - x 2 P = ( 78x – x 2 ) – (x 2 –38 x + 400) Cost: C = x 2 –38 x + 400

5. Lecture 325 The equation for profit is, Profit = Revenue - Cost Revenue: R = 78x - x 2 P = ( 78x – x 2 ) – (x 2 –38 x + 400) Cost: C = x 2 –38 x + 400 P = 78x – x 2 – x 2 +38 x - 400

6. Lecture 326 The equation for profit is, Profit = Revenue - Cost Revenue: R = 78x - x 2 P = ( 78x – x 2 ) – (x 2 –38 x + 400) Cost: C = x 2 –38 x + 400 P = 78x – x 2 – x 2 + 38 x - 400 P = – 2x 2

7. Lecture 327 The equation for profit is, Profit = Revenue - Cost Revenue: R = 78x - x 2 P = ( 78x – x 2 ) – (x 2 –38 x + 400) Cost: C = x 2 –38 x + 400 P = 78x – x 2 – x 2 + 38 x - 400 P = – 2x 2 + 116 x

8. Lecture 328 The equation for profit is, Profit = Revenue - Cost Revenue: R = 78x - x 2 P = ( 78x – x 2 ) – (x 2 –38 x + 400) Cost: C = x 2 –38 x + 400 P = 78x – x 2 – x 2 + 38 x - 400 P = – 2x 2 + 116 x - 400

9. Lecture 329 Graph the Profit Equation P = – 2x 2 + 116 x - 400

10. Lecture 3210 Graph the Profit Equation Vertex: P = – 2x 2 + 116 x - 400

11. Lecture 3211 Graph the Profit Equation Vertex: P = – 2x 2 + 116 x - 400

12. Lecture 3212 Graph the Profit Equation Vertex: P = – 2x 2 + 116 x - 400 P = – 2(29) 2 + 116 (29) – 400=1282

13. Lecture 3213 Graph the Profit Equation Vertex: P = – 2x 2 + 116 x - 400 P = – 2(29) 2 + 116 (29) – 400=1282 Vertex: (29,1282) x=29, P = $1282 x=29, P = $1282

14. Lecture 3214 Graph the Profit Equation X-Intercepts: when P = 0, what is x? P = – 2x 2 + 116 x - 400

15. Lecture 3215 Graph the Profit Equation X-Intercepts: when P = 0, what is x? P = – 2x 2 + 116 x - 400 0 = – 2x 2 + 116 x - 400

16. Lecture 3216 Graph the Profit Equation X-Intercepts: when P = 0, what is x? P = – 2x 2 + 116 x - 400 0 = – 2x 2 + 116 x - 400

17. Lecture 3217 Graph the Profit Equation X-Intercepts: when P = 0, what is x? P = – 2x 2 + 116 x - 400 0 = – 2x 2 + 116 x - 400

18. Lecture 3218 Graph the Profit Equation X-Intercepts: when P = 0, what is x? P = – 2x 2 + 116 x - 400

19. Lecture 3219 Graph the Profit Equation X-Intercepts: when P = 0, what is x? P = – 2x 2 + 116 x - 400

20. Lecture 3220 Graph the Profit Equation X-Intercepts: when P = 0, what is x? P = – 2x 2 + 116 x - 400

21. Lecture 3221 Use the Quadratic Formula to solve: 0 = -2x 2 + 116x - 400

22. Lecture 3222 Use the Quadratic Formula to solve: 0 = -2x 2 + 116x - 400

23. Lecture 3223 Use the Quadratic Formula to solve: 0 = -2x 2 + 116x - 400 Two solutions to make P = 0: x = 3.7 and x = 54.3 items x = 3.7 and x = 54.3 items

24. Lecture 3224 Graph the Profit Equation P-Intercept: when x = 0, what is P? P = – 2x 2 + 116 x - 400

25. Lecture 3225 Graph the Profit Equation P-Intercept: when x = 0, what is P? P = – 2x 2 + 116 x - 400 P = – 2(0) 2 + 116 (0) – 400 = - 400

26. Lecture 3226 Graph the Profit Equation P-Intercept: when x = 0, what is P? P = – 2x 2 + 116 x - 400 P = – 2(0) 2 + 116 (0) – 400 = - 400 P = - 400

27. Lecture 3227 Graph the Profit Equation P = – 2x 2 + 116 x - 400 1.Because a = -2, parabola opens down 2.Vertex: ( 29,1282) 3.X-Intercepts: (3.7, 0) and (54.3,0) 4.P-Intercept: (0,-400)

28. Lecture 3228 102030405060 400 800 1200 -400 -800 (29,1282) Graph: P = – 2x 2 + 116 x - 400

29. Lecture 3229 102030405060 400 800 1200 -400 -800 (29,1282) (54.3,0) (3.7,0) Graph: P = – 2x 2 + 116 x - 400

30. Lecture 3230 102030405060 400 800 1200 -400 -800 (29,1282) (54.3,0) (3.7,0) (0,-400) Graph: P = – 2x 2 + 116 x - 400

31. Lecture 3231 102030405060 400 800 1200 -400 -800 (29,1282) (54.3,0) (3.7,0) (0,-400) Graph: P = – 2x 2 + 116 x - 400

32. Lecture 3232 How many items must be made and sold to generate $500 profit. P = – 2x 2 + 116 x - 400

33. Lecture 3233 How many items must be made and sold to generate $500 profit. P = – 2x 2 + 116 x - 400 500 = -2x 2 + 116x - 400

34. Lecture 3234 How many items must be made and sold to generate $500 profit. P = – 2x 2 + 116 x - 400 500 = -2x 2 + 116x - 400 Subtract 500 from both sides

35. Lecture 3235 How many items must be made and sold to generate $500 profit. P = – 2x 2 + 116 x - 400 500 = -2x 2 + 116x - 400 0 = -2x 2 + 116x - 900 Subtract 500 from both sides

36. Lecture 3236 Use the Quadratic Formula to solve: 0 = -2x 2 + 116x - 900 0 = ax 2 + bx + c

37. Lecture 3237 Use the Quadratic Formula to solve: 0 = -2x 2 + 116x - 900 a = -2 b = 116 c = -900 0 = ax 2 + bx + c

38. Lecture 3238 Use the Quadratic Formula to solve: 0 = -2x 2 + 116x - 900 a = -2 b = 116 c = -900 0 = ax 2 + bx + c

39. Lecture 3239 Use the Quadratic Formula to solve: 0 = -2x 2 + 116x - 900 a = -2 b = 116 c = -900 0 = ax 2 + bx + c

40. Lecture 3240 Use the Quadratic Formula to solve: 0 = -2x 2 + 116x - 900 0 = ax 2 + bx + c

41. Lecture 3241 Use the Quadratic Formula to solve: 0 = -2x 2 + 116x - 900 0 = ax 2 + bx + c

42. Lecture 3242 Use the Quadratic Formula to solve: 0 = -2x 2 + 116x - 900

43. Lecture 3243 Use the Quadratic Formula to solve: 0 = -2x 2 + 116x - 900

44. Lecture 3244 Use the Quadratic Formula to solve: 0 = -2x 2 + 116x - 900

45. Lecture 3245 Use the Quadratic Formula to solve: 0 = -2x 2 + 116x - 900 Two solutions to make P = $500: x = 9.2 and x = 48.8 items x = 9.2 and x = 48.8 items

46. Lecture 3246 102030405060 400 800 1200 -400 -800 Graph: P = – 2x 2 + 116 x - 400

47. Lecture 3247 102030405060 400 800 1200 -400 -800 Graph: P = – 2x 2 + 116 x - 400

48. Lecture 3248 102030405060 400 800 1200 -400 -800 Graph: P = – 2x 2 + 116 x - 400 (9.2,500) (48.8,500)

49. Lecture 3249 An angry student stands on the top of a 250 foot cliff and throws his book upward with a velocity of 46 feet per second. The height of the book from the ground is given by the equation: h = -16t 2 + 46t + 250 h is in feet and t is in seconds. Graph the equation.

50. Lecture 3250Graph:Vertex: h = -16t 2 + 46t + 250

51. Lecture 3251Graph:Vertex: h = -16t 2 + 46t + 250

52. Lecture 3252Graph:Vertex: h = -16t 2 + 46t + 250 h = – 16(1.4) 2 + 46 (1.4) + 250 = 283

53. Lecture 3253Graph:Vertex: h = -16t 2 + 46t + 250 h = – 16(1.4) 2 + 46 (1.4) + 250 = 283 Vertex: (1.4, 283) x=1.4 sec, h = 283 feet

54. Lecture 3254Graph: t-Intercepts: when h = 0, what is t? h = -16t 2 + 46t + 250

55. Lecture 3255Graph: t-Intercepts: when h = 0, what is t? h = – 16t 2 + 46 t + 250 h = -16t 2 + 46t + 250

56. Lecture 3256Graph: t-Intercepts: when h = 0, what is t? h = – 16t 2 + 46 t + 250 h = -16t 2 + 46t + 250 0 = – 16t 2 + 46 t + 250

57. Lecture 3257Graph: t-Intercepts: when h = 0, what is t? h = – 16t 2 + 46 t + 250 h = -16t 2 + 46t + 250 0 = – 16t 2 + 46 t + 250

58. Lecture 3258Graph: t-Intercepts: when h = 0, what is t? h = – 16t 2 + 46 t + 250 h = -16t 2 + 46t + 250 0 = – 16t 2 + 46 t + 250

59. Lecture 3259Graph: t-Intercepts: when P = 0, what is t? h = -16t 2 + 46t + 250

60. Lecture 3260Graph: t-Intercepts: when P = 0, what is t? h = -16t 2 + 46t + 250

61. Lecture 3261Graph: t-Intercepts: when P = 0, what is t? h = -16t 2 + 46t + 250

62. Lecture 3262Graph: t-Intercepts: when P = 0, what is t? h = -16t 2 + 46t + 250

63. Lecture 3263Graph: t-Intercepts: when P = 0, what is t? h = -16t 2 + 46t + 250

64. Lecture 3264 Use the Quadratic Formula to solve: 0 = -16t 2 + 46t + 250

65. Lecture 3265 Use the Quadratic Formula to solve: 0 = -16t 2 + 46t + 250

66. Lecture 3266 Use the Quadratic Formula to solve: Two solutions to make h = 0: t = -2.8 and t = 5.6 seconds t = -2.8 and t = 5.6 seconds t = -2.8 does not make sense t = -2.8 does not make sense 0 = -16t 2 + 46t + 250

67. Lecture 3267Graph: h-Intercept: when t = 0, what is h? h = -16t 2 + 46t + 250

68. Lecture 3268Graph: h-Intercept: when t = 0, what is h? h = -16t 2 + 46t + 250 h = – 16t 2 + 46 t + 250

69. Lecture 3269Graph: h-Intercept: when t = 0, what is h? h = – 16(0) 2 + 46 (0) + 250 h = -16t 2 + 46t + 250 h = – 16t 2 + 46 t + 250

70. Lecture 3270Graph: h-Intercept: when t = 0, what is h? h = – 16(0) 2 + 46 (0) + 250 h = -16t 2 + 46t + 250 h = – 16t 2 + 46 t + 250 h = 250 feet

71. Lecture 3271 Graph: 1.Because a = -16, parabola opens down 2.Vertex: ( 1.4, 283) 3.t-Intercept: (5.6, 0) 4.h-Intercept: (0, 250) h = -16t 2 + 46t + 250

72. Lecture 3272 123456 250 200 300 100 50 (1.4,283) Graph: h = -16t 2 + 46t + 250 150 Time (sec) Height (feet)

73. Lecture 3273 123456 250 200 300 100 50 (1.4,283) (5.6,0) Graph: h = -16t 2 + 46t + 250 150 Time (sec) Height (feet)

74. Lecture 3274 123456 250 200 300 100 50 (1.4,283) (5.6,0) (0,250) Graph: h = -16t 2 + 46t + 250 150 Time (sec) Height (feet)

75. Lecture 3275 123456 250 200 300 100 50 (1.4,283) (5.6,0) (0,250) Graph: h = -16t 2 + 46t + 250 150 Time (sec) Height (feet)

76. Lecture 3276 Ms. Piggie wants to enclose two adjacent chicken coops of equal size against the hen house wall. She has 66 feet of chicken-wire fencing and would like to make the chicken coup as large as possible. Find the formula for the area of the chicken coops. Wall L WWW

77. Lecture 3277 Ms. Piggie wants to enclose two adjacent chicken coops of equal size against the hen house wall. She has 66 feet of chicken-wire fencing and would like to make the chicken coup as large as possible. Find the formula for the area of the chicken coops. Wall L WW WidthLengthArea 5 10 15 X W

78. Lecture 3278 Ms. Piggie wants to enclose two adjacent chicken coops of equal size against the hen house wall. She has 66 feet of chicken-wire fencing and would like to make the chicken coup as large as possible. Find the formula for the area of the chicken coops. Wall L WW WidthLengthArea 5 66-3(5)=56 5(51)=251 10 15 X W

79. Lecture 3279 Ms. Piggie wants to enclose two adjacent chicken coops of equal size against the hen house wall. She has 66 feet of chicken-wire fencing and would like to make the chicken coup as large as possible. Find the formula for the area of the chicken coops. Wall L WW WidthLengthArea 5 66-3(5)=56 5(51)=251 10 66-3(10)=36 10(36)=360 15 X W

80. Lecture 3280 Ms. Piggie wants to enclose two adjacent chicken coops of equal size against the hen house wall. She has 66 feet of chicken-wire fencing and would like to make the chicken coup as large as possible. Find the formula for the area of the chicken coops. Wall L WW WidthLengthArea 5 66-3(5)=56 5(51)=251 10 66-3(10)=36 10(36)=360 15 66-3(15)=21 15(21)=315 X W

81. Lecture 3281 Ms. Piggie wants to enclose two adjacent chicken coops of equal size against the hen house wall. She has 66 feet of chicken-wire fencing and would like to make the chicken coup as large as possible. Find the formula for the area of the chicken coops. Wall L WW WidthLengthArea 5 66-3(5)=56 5(51)=251 10 66-3(10)=36 10(36)=360 15 66-3(15)=21 15(21)=315 X 66-3(X) X(66-3X)= 66X-3X 2 W

82. Lecture 3282Graph:Vertex: A = - 3X 2 + 66X

83. Lecture 3283Graph:Vertex: A = - 3X 2 + 66X

84. Lecture 3284Graph:Vertex: A = - 3X 2 + 66X A = – 3(11) 2 + 66 (11) = 363

85. Lecture 3285Graph:Vertex: A = - 3X 2 + 66X A = – 3(11) 2 + 66 (11) = 363 Vertex: (11, 363) x=11 feet, A = 363 sq. feet

86. Lecture 3286Graph:Vertex: A = - 3X 2 + 66X A = – 3(11) 2 + 66 (11) = 363 Vertex: (11, 363) x=11 feet, A = 363 sq. feet

87. Lecture 3287Graph: x-Intercepts: when A = 0, what is x? A = - 3X 2 + 66X

88. Lecture 3288Graph: x-Intercepts: when A = 0, what is x? A = - 3X 2 + 66X

89. Lecture 3289Graph: x-Intercepts: when A = 0, what is x? A = - 3X 2 + 66X 0 = - 3X 2 + 66X

90. Lecture 3290Graph: x-Intercepts: when A = 0, what is x? A = - 3X 2 + 66X 0 = - 3X 2 + 66X 0 = - 3X(X - 22)

91. Lecture 3291Graph: x-Intercepts: when A = 0, what is x? A = - 3X 2 + 66X 0 = - 3X 2 + 66X 0 = - 3X(X - 22) 0 = - 3X or

92. Lecture 3292Graph: x-Intercepts: when A = 0, what is x? A = - 3X 2 + 66X 0 = - 3X 2 + 66X 0 = - 3X(X - 22) 0 = - 3X or 0 = (X - 22)

93. Lecture 3293Graph: x-Intercepts: when A = 0, what is x? A = - 3X 2 + 66X 0 = - 3X 2 + 66X 0 = - 3X(X - 22) 0 = - 3X or 0 = (X - 22) 0 = X or X = 22

94. Lecture 3294Graph: A-Intercept: when x = 0, what is A? A = - 3X 2 + 66X

95. Lecture 3295Graph: A-Intercept: when x = 0, what is A? A = - 3X 2 + 66X

96. Lecture 3296Graph: A-Intercept: when x = 0, what is A? A = – 3(0) 2 + 66 (0) A = - 3X 2 + 66X

97. Lecture 3297Graph: A-Intercept: when x = 0, what is A? A = – 3(0) 2 + 66 (0) A = 0 sq. feet A = - 3X 2 + 66X

98. Lecture 3298 Graph: 1.Because a = -3, parabola opens down 2.Vertex: ( 11, 363) 3.x-Intercepts: (0, 0) and (22, 0) 4.A-Intercept: (0, 0) A = - 3X 2 + 66X

99. Lecture 3299 369121518 250 200 300 100 50 (11,363) Graph: 150 Width (feet) Area (sq. feet) A = - 3X 2 + 66X 21 350

100. Lecture 32100 369121518 250 200 300 100 50 (11,363) (22,0)(0,0) Graph: 150 Width (feet) Area (sq. feet) A = - 3X 2 + 66X 21 350

101. Lecture 32101 369121518 250 200 300 100 50 (11,363) (22,0)(0,0) Graph: 150 Width (feet) Area (sq. feet) A = - 3X 2 + 66X 21 350

102. Lecture 32102 369121518 250 200 300 100 50 (11,363) (22,0)(0,0) Graph: 150 Width (feet) Area (sq. feet) A = - 3X 2 + 66X 21 350

103. Lecture 32103 Find the dimensions of the coop if the area can only be 360 sq feet. When A = 360, what is x?

104. Lecture 32104 Find the dimensions of the coop if the area can only be 360 sq feet. When A = 360, what is x? A = - 3X 2 + 66X

105. Lecture 32105 Find the dimensions of the coop if the area can only be 360 sq feet. When A = 360, what is x? A = - 3X 2 + 66X 360 = - 3X 2 + 66X

106. Lecture 32106 Find the dimensions of the coop if the area can only be 360 sq feet. When A = 360, what is x? A = - 3X 2 + 66X 360 = - 3X 2 + 66X Subtract 360 from both sides

107. Lecture 32107 Find the dimensions of the coop if the area can only be 360 sq feet. When A = 360, what is x? A = - 3X 2 + 66X 360 = - 3X 2 + 66X Subtract 360 from both sides 0 = - 3X 2 + 66X - 360

108. Lecture 32108 Find the dimensions of the coop if the area can only be 360 sq feet. When A = 360, what is x? A = - 3X 2 + 66X 360 = - 3X 2 + 66X Subtract 360 from both sides 0 = - 3X 2 + 66X - 360 0 = - 3(X 2 – 22X + 120)

109. Lecture 32109 Find the dimensions of the coop if the area can only be 360 sq feet. When A = 360, what is x? A = - 3X 2 + 66X 360 = - 3X 2 + 66X Subtract 360 from both sides 0 = - 3X 2 + 66X - 360 0 = - 3(X 2 – 22X + 120) 0 = - 3(X – 10)(X – 12)

110. Lecture 32110 Find the dimensions of the coop if the area can only be 360 sq feet. When A = 360, what is x? A = - 3X 2 + 66X 360 = - 3X 2 + 66X 0 = (X - 10) or Subtract 360 from both sides 0 = - 3X 2 + 66X - 360 0 = - 3(X 2 – 22X + 120) 0 = - 3(X – 10)(X – 12)

111. Lecture 32111 Find the dimensions of the coop if the area can only be 360 sq feet. When A = 360, what is x? A = - 3X 2 + 66X 360 = - 3X 2 + 66X 0 = (X - 10) or 0 = (X - 12) Subtract 360 from both sides 0 = - 3X 2 + 66X - 360 0 = - 3(X 2 – 22X + 120) 0 = - 3(X – 10)(X – 12)

112. Lecture 32112 Find the dimensions of the coop if the area can only be 360 sq feet. When A = 360, what is x? A = - 3X 2 + 66X 360 = - 3X 2 + 66X 0 = (X - 10) or 0 = (X - 12) X =10 or X = 12 Subtract 360 from both sides 0 = - 3X 2 + 66X - 360 0 = - 3(X 2 – 22X + 120) 0 = - 3(X – 10)(X – 12)

113. Lecture 32113 369121518 250 200 300 100 50 (22,0)(0,0) Graph: 150 Width (feet) Area (sq. feet) A = - 3X 2 + 66X 21 350 (11,363)

114. Lecture 32114 369121518 250 200 300 100 50 (12,360) (22,0)(0,0) Graph: 150 Width (feet) Area (sq. feet) A = - 3X 2 + 66X 21 350 (11,363) (10,360)

115. Lecture 32115 369121518 250 200 300 100 50 (12,360) (22,0)(0,0) Graph: 150 Width (feet) Area (sq. feet) A = - 3X 2 + 66X 21 350 (11,363) (10,360)

116. Lecture 32116

117. Lecture 32117

118. Lecture 32118

119. Lecture 32119

120. Lecture 32120

121. Lecture 32121