1. Unit 4 Lecture 33 Review Test 4

2. Objectives
To get the most out of this presentation, complete the review test first.
Use this presentation to check your work.
If you get a problem wrong, go back to the appropriate section in the textbook.
Lecture 33

3. State the quadratic formula.
Lecture 33

4. State the quadratic formula.
The solutions of the quadratic equation,
0 = ax2 + bx + c, are
Lecture 33

5. State the formula for the x coordinate of the vertex.
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6. State the formula for the x coordinate of the vertex.
The x-coordinate of the vertex for the equation y = ax2 + bx + c, is,
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7. A farmer wants to enclose adjacent rectangular fields with 1000 feet of barbed wire fencing as indicated below. Find the equation for the area of the fields. W L
Lecture 33

8. 1000 - 3x = what is left for the length of two sides
A farmer wants to enclose adjacent rectangular fields with 1000 feet of barbed wire fencing as indicated below. Find the equation for the area of the fields. W L
Let x = width
x = what is left for the length of two sides
Lecture 33

9. 1000 - 3x = what is left for the length of two sides
A farmer wants to enclose adjacent rectangular fields with 1000 feet of barbed wire fencing as indicated below. Find the equation for the area of the fields. W L
Let x = width
x = what is left for the length of two sides
Lecture 33

10. 1000 - 3x = what is left for the length of two sides
A farmer wants to enclose adjacent rectangular fields with 1000 feet of barbed wire fencing as indicated below. Find the equation for the area of the fields. W L
Let x = width
x = what is left for the length of two sides
A= width *length
Lecture 33

11. 1000 - 3x = what is left for the length of two sides
A farmer wants to enclose adjacent rectangular fields with 1000 feet of barbed wire fencing as indicated below. Find the equation for the area of the fields. W L
Let x = width
x = what is left for the length of two sides
A= width *length
Lecture 33

12. 1000 - 3x = what is left for the length of two sides
A farmer wants to enclose adjacent rectangular fields with 1000 feet of barbed wire fencing as indicated below. Find the equation for the area of the fields. W L
Let x = width
x = what is left for the length of two sides
A= width *length
Lecture 33

13. 1000 - 3x = what is left for the length of two sides
A farmer wants to enclose adjacent rectangular fields with 1000 feet of barbed wire fencing as indicated below. Find the equation for the area of the fields. W L
Let x = width
x = what is left for the length of two sides
Lecture 33

14. Find the equation for Profit
Lighten Up Company makes light bulbs. The cost of making x thousand light bulbs per week is
C = .5x2 –14 x The revenue from selling x thousand light bulbs per week is R = 12x - .5x2
Find the equation for Profit
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15. Profit = Revenue - Cost Revenue: R = - .5x2 +12x
Lighten Up Company makes light bulbs. The cost of making x thousand light bulbs per week is
C = .5x2 –14 x The revenue from selling x thousand light bulbs per week is R = 12x - .5x2
Find the equation for Profit
Profit = Revenue - Cost
Revenue: R = - .5x2 +12x
Cost: C = .5x2 –14 x + 120
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16. Revenue: R = - .5x2 +12x Cost: C = .5x2 –14 x + 120
The equation for profit is,
Profit = Revenue - Cost
Revenue: R = - .5x2 +12x
Cost:
C = .5x2 –14 x + 120
P = (- .5x2 +12x) – (.5x2 –14 x + 120)
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17. Revenue: R = - .5x2 +12x Cost: C = .5x2 –14 x + 120
The equation for profit is,
Profit = Revenue - Cost
Revenue: R = - .5x2 +12x
Cost:
C = .5x2 –14 x + 120
P = (- .5x2 +12x) – (.5x2 –14 x + 120)
P = - .5x2 +12x -.5x x - 120
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18. Revenue: R = - .5x2 +12x Cost: C = .5x2 –14 x + 120
The equation for profit is,
Profit = Revenue - Cost
Revenue: R = - .5x2 +12x
Cost:
C = .5x2 –14 x + 120
P = (- .5x2 +12x) – (.5x2 –14 x + 120)
P = - .5x2 +12x -.5x x - 120
P = – 1x2
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19. Revenue: R = - .5x2 +12x Cost: C = .5x2 –14 x + 120
The equation for profit is,
Profit = Revenue - Cost
Revenue: R = - .5x2 +12x
Cost:
C = .5x2 –14 x + 120
P = (- .5x2 +12x) – (.5x2 –14 x + 120)
P = - .5x2 +12x -.5x x - 120
P =– 1x x
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20. Revenue: R = - .5x2 +12x Cost: C = .5x2 –14 x + 120
The equation for profit is,
Profit = Revenue - Cost
Revenue: R = - .5x2 +12x
Cost:
C = .5x2 –14 x + 120
P = (- .5x2 +12x) – (.5x2 –14 x + 120)
P = - .5x2 +12x -.5x x - 120
P =– 1x x - 120
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21. Graph the Profit Equation
P = – 1x x - 120
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22. Graph the Profit Equation
P = – 1x x - 120
Vertex:
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23. Graph the Profit Equation
P = – 1x x - 120
Vertex:
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24. P = – 1x2 + 26 x - 120 Vertex: P = – 1(13)2 + 26 (13) - 120
Graph the Profit Equation
P = – 1x x - 120
Vertex:
P = – 1(13) (13) - 120
Lecture 33

25. P = – 1x2 + 26 x - 120 Vertex: P = – 1(13)2 + 26 (13) - 120
Graph the Profit Equation
P = – 1x x - 120
Vertex:
P = – 1(13) (13) - 120
P = – 1(169)
P = 49
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26. P = – 1x2 + 26 x - 120 Vertex: P = – 1(13)2 + 26 (13) - 120 P = 49
Graph the Profit Equation
P = – 1x x - 120
Vertex:
P = – 1(13) (13) - 120
P = 49
Vertex: (13,49)
x=13, P = $49
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27. X-Intercepts: when P = 0, what is x?
Graph the Profit Equation
P = – 1x x - 120
X-Intercepts: when P = 0, what is x?
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28. X-Intercepts: when P = 0, what is x? P = – 1x2 + 26 x - 120
Graph the Profit Equation
P = – 1x x - 120
X-Intercepts: when P = 0, what is x?
P = – 1x x - 120
0 = – 1x x - 120
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29. X-Intercepts: when P = 0, what is x? P = – 1x2 + 26 x - 120
Graph the Profit Equation
P = – 1x x - 120
X-Intercepts: when P = 0, what is x?
P = – 1x x - 120
0 = – 1x x - 120
0 = – 1(x )
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30. X-Intercepts: when P = 0, what is x? P = – 1x2 + 26 x - 120
Graph the Profit Equation
P = – 1x x - 120
X-Intercepts: when P = 0, what is x?
P = – 1x x - 120
0 = – 1x x - 120
0 = – 1(x )
0 = – 1(x - 20)(x – 6)
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31. X-Intercepts: when P = 0, what is x? P = – 1x2 + 26 x - 120
Graph the Profit Equation
P = – 1x x - 120
X-Intercepts: when P = 0, what is x?
P = – 1x x - 120
0 = – 1x x - 120
0 = – 1(x )
0 = – 1(x - 20)(x – 6)
0 = x or
0 = x – 6
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32. X-Intercepts: when P = 0, what is x? 0 = – 1x2 + 26 x - 120
Graph the Profit Equation
P = – 1x x - 120
X-Intercepts: when P = 0, what is x?
0 = – 1x x - 120
0 = – 1(x - 20)(x – 6)
0 = x or
0 = x – 6
x = 20 or
x = 6
Two solutions make P = 0:
x = 20 and x = 6 items
Lecture 33

33. P-Intercept: when x = 0, what is P? P = – 1x2 + 26 x - 120
Graph the Profit Equation
P = – 1x x - 120
P-Intercept: when x = 0, what is P?
P = – 1x x - 120
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34. P-Intercept: when x = 0, what is P? P = – 1x2 + 26 x - 120
Graph the Profit Equation
P = – 1x x - 120
P-Intercept: when x = 0, what is P?
P = – 1x x - 120
P = – 1(0) (0) – 120 = - 120
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35. P-Intercept: when x = 0, what is P? P = – 1x2 + 26 x - 120
Graph the Profit Equation
P = – 1x x - 120
P-Intercept: when x = 0, what is P?
P = – 1x x - 120
P = – 1(0) (0) – 120 = - 120
P = - 120
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36. P = – 1x2 + 26 x - 120 Graph the Profit Equation
Because a = -1, parabola opens down
Vertex: (13, 49)
X-Intercepts: (6, 0) and (20, 0)
P-Intercept: (0, -120)
Lecture 33

37. P = – 1x2 + 26 x - 120 Graph: profit 100 75 50 25 -25 3 6 9 12 15 1821
items
-50
-75
-100
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38. P = – 1x2 + 26 x - 120 Graph: profit 100 75 (13,49) 50 25 -25 3 6 9 1215 18 21
items
-50
-75
-100
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39. P = – 1x2 + 26 x - 120 Graph: profit 100 75 (13,49) 50 25 (6,0) (20,0)
-25 3 6 9 12 15 18 21
items
-50
-75
-100
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40. P = – 1x2 + 26 x - 120 Graph: profit 100 75 (13,49) 50 25 (6,0) (20,0)
-25 3 6 9 12 15 18 21
items
-50
-75
-100
(0,-120)
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41. P = – 1x2 + 26 x - 120 Graph: profit 100 75 (13,49) 50 25 (6,0) (20,0)
-25 3 6 9 12 15 18 21
items
-50
-75
-100
(0,-120)
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42. P = – 1x2 + 26 x - 120 Explain the Profit Equation in business terms
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43. P = – 1x2 + 26 x - 120 Explain the Profit Equation in business terms
Vertex: (13, 49) maximum profit is $49 when 13 items are sold
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44. P = – 1x2 + 26 x - 120 Explain the Profit Equation in business terms
Vertex: (13, 49) maximum profit is $49 when 13 items are sold
X-Intercepts: (6, 0) and (20, 0) breakeven points, 0 profit is made
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45. P = – 1x2 + 26 x - 120 Explain the Profit Equation in business terms
Vertex: (13, 49) maximum profit is $49 when 13 items are sold
X-Intercepts: (6, 0) and (20, 0) breakeven points, 0 profit is made
P-Intercept: (0, -120) fixed costs or startup costs
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46. Herb’s Company has the following profit equation:
P = – 2x x - 50
profit 60
(7,48) 45 30 15
(2.1,0)
(11.9,0)
-15 2 4 6 8 10 12 14
items
-30
-45
(0,-50)
-60
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47. Herb’s Company need to make a profit of $30
Herb’s Company need to make a profit of $30. Graph and find where the lines intersect.
profit 60
(7,48) 45 30 15
(2.1,0)
(11.9,0)
-15 2 4 6 8 10 12 14
items
-30
-45
(0,-50)
-60
Lecture 33

48. P = – 2x2 + 28 x - 50 When P = 30, what is x? P = – 2x2 + 28 x - 50
Graph the Profit Equation
P = – 2x x - 50
When P = 30, what is x?
P = – 2x x - 50
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49. P = – 2x2 + 28 x - 50 When P = 30, what is x? P = – 2x2 + 28 x - 50
Graph the Profit Equation
P = – 2x x - 50
When P = 30, what is x?
P = – 2x x - 50
30 = – 2x x - 50
Lecture 33

50. P = – 2x2 + 28 x - 50 When P = 30, what is x? P = – 2x2 + 28 x - 50
Graph the Profit Equation
P = – 2x x - 50
When P = 30, what is x?
P = – 2x x - 50
Subtract 30 from both sides
30 = – 2x x - 50
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51. P = – 2x2 + 28 x - 50 When P = 30, what is x? P = – 2x2 + 28 x - 50
Graph the Profit Equation
P = – 2x x - 50
When P = 30, what is x?
P = – 2x x - 50
Subtract 30 from both sides
30 = – 2x x - 50
0 = – 2x x - 80
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52. Use the Quadratic Formula to solve:
0 = – 2x x - 80
0 = ax2 + bx + c
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53. Use the Quadratic Formula to solve:
0 = – 2x x - 80
0 = ax2 + bx + c
a = -2
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54. Use the Quadratic Formula to solve:
0 = – 2x x - 80
0 = ax2 + bx + c
a = -2
b = 28
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55. 0 = – 2x2 + 28 x - 80 0 = ax2 + bx + c a = -2 b = 28 c = -80
Use the Quadratic Formula to solve:
0 = – 2x x - 80
0 = ax2 + bx + c
a = -2
b = 28
c = -80
Lecture 33

56. 0 = – 2x2 + 28 x - 80 0 = ax2 + bx + c a = -2 b = 28 c = -80
Use the Quadratic Formula to solve:
0 = – 2x x - 80
0 = ax2 + bx + c
a = -2
b = 28
c = -80
Lecture 33

57. 0 = – 2x2 + 28 x - 80 0 = ax2 + bx + c a = -2 b = 28 c = -80
Use the Quadratic Formula to solve:
0 = – 2x x - 80
0 = ax2 + bx + c
a = -2
b = 28
c = -80
Lecture 33

58. 0 = – 2x2 + 28 x - 80 0 = ax2 + bx + c a = -2 b = 28 c = -80
Use the Quadratic Formula to solve:
0 = – 2x x - 80
0 = ax2 + bx + c
a = -2
b = 28
c = -80
Lecture 33

59. Use the Quadratic Formula to solve:
0 = – 2x x - 80
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60. Use the Quadratic Formula to solve:
0 = – 2x x - 80
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61. Use the Quadratic Formula to solve:
0 = – 2x x - 80
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62. Two solutions to make P = $30: x = 4 and x = 10 items
Use the Quadratic Formula to solve:
0 = – 2x x - 80
Two solutions to make P = $30:
x = 4 and x = 10 items
Lecture 33

63. Herb’s Company need to make a profit of $30
Herb’s Company need to make a profit of $30. Graph and find where the lines intersect.
profit 60 45 30 15
-15 2 4 6 8 10 12 14
items
-30
-45
-60
Lecture 33

64. Herb’s Company need to make a profit of $30
Herb’s Company need to make a profit of $30. Graph and find where the lines intersect.
profit 60 45
(4,30)
(10,30) 30 15
-15 2 4 6 8 10 12 14
items
-30
-45
-60
Lecture 33

65. Herb’s Company need to make a profit of $30
Herb’s Company need to make a profit of $30. Graph and find where the lines intersect.
profit 60 45
(4,30)
(10,30) 30 15
-15 2 4 6 8 10 12 14
items
-30
-45
-60
Lecture 33

66. Simplify:
5(2x2 – x + 1) – 3(6x2 – 7x + 2)
Lecture 33

67. 5(2x2 – x + 1) – 3(6x2 – 7x + 2) 10x2 – 5x + 5 – 18x2 + 21x – 6
Simplify:
5(2x2 – x + 1) – 3(6x2 – 7x + 2)
10x2 – 5x + 5 – 18x2 + 21x – 6
Lecture 33

68. 5(2x2 – x + 1) – 3(6x2 – 7x + 2) 10x2 – 5x + 5 – 18x2 + 21x – 6
Simplify:
5(2x2 – x + 1) – 3(6x2 – 7x + 2)
10x2 – 5x + 5 – 18x2 + 21x – 6
10x2 – 5x + 5 – 18x2 + 21x – 6
– 8x2
Lecture 33

69. 5(2x2 – x + 1) – 3(6x2 – 7x + 2) 10x2 – 5x + 5 – 18x2 + 21x – 6
Simplify:
5(2x2 – x + 1) – 3(6x2 – 7x + 2)
10x2 – 5x + 5 – 18x2 + 21x – 6
10x2 – 5x + 5 – 18x2 + 21x – 6
– 8x2 + 16x
Lecture 33

70. 5(2x2 – x + 1) – 3(6x2 – 7x + 2) 10x2 – 5x + 5 – 18x2 + 21x – 6
Simplify:
5(2x2 – x + 1) – 3(6x2 – 7x + 2)
10x2 – 5x + 5 – 18x2 + 21x – 6
10x2 – 5x + 5 – 18x2 + 21x – 6
– 8x2 + 16x – 1
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71. Multiply:
(2x-1)(x+5)
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72. Multiply:
F O I L
(2x-1)(x+5)
2x2
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73. Multiply:
F O I L
(2x-1)(x+5)
2x2 + 10x
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74. Multiply:
F O I L
(2x-1)(x+5)
2x2 + 10x - x
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75. Multiply:
F O I L
(2x-1)(x+5)
2x2 + 10x - x - 5
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76. F O I L (2x-1)(x+5) 2x2 + 10x - x - 5 2x2 + 9x - 5 Multiply:
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77. Multiply:
(x-3)2
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78. Multiply:
(x-3)2 = (x-3)(x-3)
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79. Multiply:
(x-3)2 = (x-3)(x-3)
F O I L
(x-3)(x-3) x2
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80. Multiply:
(x-3)2 = (x-3)(x-3)
F O I L
(x-3)(x-3)
x2 - 3x
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81. (x-3)2 = (x-3)(x-3) F O I L (x-3)(x-3) x2 - 3x - 3x Multiply:
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82. (x-3)2 = (x-3)(x-3) F O I L (x-3)(x-3) x2 - 3x - 3x + 9 Multiply:
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83. (x-3)2 = (x-3)(x-3) F O I L (x-3)(x-3) x2 - 3x - 3x + 9 x2 - 6x + 9
Multiply:
(x-3)2 = (x-3)(x-3)
F O I L
(x-3)(x-3)
x2 - 3x - 3x + 9
x2 - 6x + 9
Lecture 33

84. Factor: 9x2 + 6x
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85. Factor: 9x2 + 6x
9x2 + 6x = 3x( )
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86. Factor: 9x2 + 6x
9x2 + 6x = 3x( )
9x2 + 6x = 3x(3x + 2)
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87. Factor: x2 – 2x – 15
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88. Factor: x2 – 2x – 15
(x – )(x + )
x2 – 2x – 15 =
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89. Factor: x2 – 2x – 15
(x – 5)(x + 3)
x2 – 2x – 15 =
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90. Solve: x2 + 5x + 6=0
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91. Solve: x2 + 5x + 6=0
(x + 3)(x + 2)=0
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92. Solve: x2 + 5x + 6=0
(x + 3)(x + 2)=0
x + 3 =0 or x + 2 = 0
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93. (x + 3)(x + 2)=0 x + 3 =0 or x + 2 = 0 or x = -3 Solve: x2 + 5x + 6=0
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94. (x + 3)(x + 2)=0 x + 3 =0 or x + 2 = 0 or x = -2 x = -3
Solve: x2 + 5x + 6=0
(x + 3)(x + 2)=0
x + 3 =0 or x + 2 = 0 or
x = -2
x = -3
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95. Lecture 33

96. Lecture 33

97. Lecture 33