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3. Lesson 10-1 Graphing Quadratic Functions
Lesson 10-2 Solving Quadratic Equations by Graphing
Lesson 10-4 Solving Quadratic Equations by Using the Quadratic Formula
Lesson 10-5 Exponential Functions
Lesson 10-6 Growth and Decay
Lesson 10-7 Geometric Sequences
Contents

4. Example 1 Graph Opens Upward Example 2 Graph Opens Downward
Example 3 Vertex and Axis of Symmetry
Example 4 Match Equations and Graphs
Lesson 1 Contents

5. Use a table of values to graph
Graph these ordered pairs and connect them with a smooth curve. x y –2 –4 –1 1 2 3 4
Answer:
Example 1-1a

6. Use a table of values to graph x y–3 6 –2 3 –1 2 1
Answer:
Example 1-1b

7. Use a table of values to graph
Graph these ordered pairs and connect them with a smooth curve. x y –2 –8 –1 4 1 2 3
Answer:
Example 1-2a

8. Use a table of values to graph x y–3 –5 –2 –1 3 4 1 2
Answer:
Example 1-2b

9. Write the equation of the axis of symmetry.
Consider the graph of
Write the equation of the axis of symmetry. In
Equation for the axis of symmetry of a parabola
and
Answer: The equation of the axis of symmetry is
Example 1-3a

10. Find the coordinates of the vertex.
Consider the graph of
Find the coordinates of the vertex.
Since the equation of the axis of symmetry is x = –2 and the vertex lies on the axis, the x-coordinate for the vertex is –2.
Original equation
Simplify.
Add.
Answer: The vertex is at (–2, 6).
Example 1-3b

11. Identify the vertex as a maximum or minimum.
Answer: Since the coefficient of the x2 term is negative, the parabola opens downward and the vertex is a maximum point.
Example 1-3c

12. (–2, 6)
Graph the function.
You can use the symmetry of the parabola to help you draw its graph. On a coordinate plane, graph the vertex and the axis of symmetry.
Choose a value for x other than –2. For example, choose –1 and find the y-coordinate that satisfies the equation.
Original equation
Simplify.
Example 1-3d

13. Graph the function. Graph (–1, 4).
(–2, 6)
Graph the function.
Graph (–1, 4).
(–3, 4)
(–1, 4)
Since the graph is symmetrical about its axis of symmetry x = –2, you can find another point on the other side of the axis of symmetry. The point at (–1, 4) is 1 unit to the right of the axis. Go 1 unit to the left of the axis and plot the point (–3, 4).
Example 1-3e

14. Repeat this for several other points.
(–2, 6)
Graph the function.
Repeat this for several other points.
(–3, 4)
(–1, 4)
Then sketch the parabola.
(–4, –2)
(0, –2)
Example 1-3f

15. a. Write the equation of the axis of symmetry.
Consider the graph of
a. Write the equation of the axis of symmetry.
b. Find the coordinates of the vertex.
c. Identify the vertex as a maximum or minimum.
Answer:
Answer: (1, –2)
Answer: Since the coefficient of the x2 term is positive, the parabola opens upward and the vertex is a minimum point.
Example 1-3g

16. Consider the graph of
d. Graph the function.
Answer:
Example 1-3h

17. Multiple-Choice Test Item Which is the graph of A B
C D
Example 1-4a

18. Solve the Test Item Find the axis of symmetry of the graph
Read the Test Item You are given a quadratic function, and you are asked to choose the graph that corresponds to it.
Solve the Test Item Find the axis of symmetry of the graph
Equation for the axis of symmetry
and
Example 1-4b

19. The axis of symmetry is –1. Look at the graphs
The axis of symmetry is –1. Look at the graphs. Since only choices C and D have this as their axis of symmetry, you can eliminate choices A and B. Since the coefficient of the x2 term is negative, the graph opens downward. Eliminate choice C.
Answer: D
Example 1-4c

20. Multiple-Choice Test Item Which is the graph of A B
C D
Answer: B
Example 1-4d

21. End of Lesson 1

22. Example 4 Rational Roots
Example 1 Two Roots
Example 2 A Double Root
Example 3 No Real Roots
Example 4 Rational Roots
Example 5 Estimate Solutions to Solve a Problem
Lesson 2 Contents

23. Solve by graphing.
Graph the related function The equation of the axis of symmetry is or When f(x) equals or So the coordinates of the vertex are
Example 2-1a

24. Make a table of values to find other points to sketch the graph. x
f(x) –3 8 –1 –6
–10 1
–12 2 3 4 6
Example 2-1b

25. To solve you need to know where the value of f (x) is 0
To solve you need to know where the value of f (x) is 0. This occurs at the x-intercepts. The x-intercepts of the parabola appear to be –2 and 5.
Example 2-1c

26. Check Solve by factoring.
Original equation
Factor.
Zero Product Property or
Solve for x.
Answer: The solutions of the equation are –2 and 5.
Example 2-1d

27. Solve by graphing.
Answer: {–2, 4}
Example 2-1e

28. First rewrite the equation so one side is equal to zero.
Solve by graphing.
First rewrite the equation so one side is equal to zero.
Original equation
Add 9 to each side.
Simplify.
Example 2-2a

29. Graph the related function x f(x)1 4 2 3 5
Example 2-2b

30. Try solving the equation by factoring.
Notice that the vertex of the parabola is the x-intercept. Thus, one solution is 3. What is the other solution?
Try solving the equation by factoring.
Example 2-2c

31. Answer: The solution is 3.
Original equation
Factor. or
Zero Product Property
Solve for x.
There are two identical factors for the quadratic function, so there is only one root, called a double root.
Answer: The solution is 3.
Example 2-2d

32. Solve by graphing.
Answer: {–1}
Example 2-2e

33. Graph the related function x f(x)
Solve by graphing.
Graph the related function x
f(x) –3 6 –2 3 –1 2 1
Answer: The graph has no x-intercept. Thus, there are no real number solutions for the equation.
Example 2-3a

34. Solve by graphing.
Answer: 
Example 2-3b

35. Graph the related function x f(x)
Solve by graphing. If integral roots cannot be found, estimate the roots by stating the consecutive integers between which the roots lie.
Graph the related function x
f(x) 2 1 –1 –2 3 4
Notice that the value of the function changes from negative to positive between the x values of 0 and 1 and between 3 and 4.
Example 2-4a

36. The x-intercepts of the graph are between 0 and 1 and between 3 and 4.
f(x) 2 1 –1 –2 3 4
The x-intercepts of the graph are between 0 and 1 and between 3 and 4.
Answer: One root is between 0 and 1, and the other root is between 3 and 4.
Example 2-4b

37. Solve
Answer: One root is between –1 and –2, and the other root is between 3 and 4.
Example 2-4c

38. Model Rockets Shelly built a model rocket for her science project
Model Rockets Shelly built a model rocket for her science project. The equation models the flight of the rocket, launched from ground level at a velocity of 250 feet per second, where y is the height of the rocket in feet after t seconds. For how many seconds was Shelly’s rocket in the air?
You need to find the solution of the equation Use a graphing calculator to graph the related function The x-intercept is between 15 and 16 seconds.
Example 2-5a

39. Answer: between 15 and 16 seconds
Example 2-5b

40. Answer: between 7 and 8 seconds
Golf Martin hits a golf ball with an upward velocity of 120 feet per second. The function models the flight of the golf ball, hit at ground level, where y is the height of the ball in feet after t seconds. How long was the golf ball in the air?
Answer: between 7 and 8 seconds
Example 2-5c

41. End of Lesson 2

42. Example 1 Integral Roots Example 2 Irrational Roots
Example 3 Use the Quadratic Formula to Solve a Problem
Example 4 Use the Discriminant
Lesson 4 Contents

43. Use two methods to solve
Method 1 Factoring
Original equation
Factor
Zero Product Property or
Solve for x.
Example 4-1a

44. Method 2 Quadratic Formula
For this equation,
Quadratic Formula
Multiply.
Example 4-1b

45. Answer: The solution set is {–5, 7}.
Add.
Simplify. or
Answer: The solution set is {–5, 7}.
Example 4-1c

46. Use two methods to solve
Answer: {–6, 5}
Example 4-1d

47. Step 1 Rewrite the equation in standard form.
Solve by using the Quadratic Formula. Round to the nearest tenth if necessary.
Step 1 Rewrite the equation in standard form.
Original equation
Subtract 4 from each side.
Simplify.
Example 4-2a

48. Step 2 Apply the Quadratic Formula.
and
Multiply.
Example 4-2b

49. Add. or
Example 4-2c

50. Answer: The approximate solution set is {–0.3, 0.8}.
Check the solutions by using the CALC menu on a graphing calculator to determine the zeros of the related quadratic function.
Answer: The approximate solution set is {–0.3, 0.8}.
Example 4-2d

51. Solve. by using the Quadratic Formula
Solve by using the Quadratic Formula. Round to the nearest tenth if necessary.
Answer: {–0.5, 0.7}
Example 4-2e

52. Space Travel Two possible future destinations of astronauts are the planet Mars and a moon of the planet Jupiter, Europa. The gravitational acceleration on Mars is about 3.7 meters per second squared. On Europa, it is only 1.3 meters per second squared. Using the information and equation from Example 3 on page 548 in your textbook, find how much longer baseballs thrown on Mars and on Europa will stay above the ground than a similarly thrown baseball on Earth.
In order to find when the ball hits the ground, you must find when H = 0. Write two equations to represent the situation on Mars and on Europa.
Example 4-3a

53. Baseball Thrown on Mars Baseball Thrown on Europa
These equations cannot be factored, and completing the square would involve a lot of computation.
Example 4-3b

54. To find accurate solutions, use the Quadratic Formula.
Example 4-3c

55. Since a negative number of seconds is not reasonable, use the positive solutions.
Answer: A ball thrown on Mars will stay aloft 5.6 – 2.2 or about 3.4 seconds longer than the ball thrown on Earth. The ball thrown on Europa will stay aloft 15.6 – 2.2 or about 13.4 seconds longer than the ball thrown on Earth.
Example 4-3d

56. Space Travel The gravitational acceleration on Venus is about 8
Space Travel The gravitational acceleration on Venus is about 8.9 meters per second squared, and on Callisto, one of Jupiter’s moons, it is 1.2 meters per second squared. Suppose a baseball is thrown on Callisto with an upward velocity of 10 meters per second from two meters above the ground. Find how much longer the ball will stay in air than a similarly-thrown ball on Venus. Use the equation where H is the height of an object t seconds after it is thrown upward, v is the initial velocity, g is the gravitational pull, and h is the initial height.
Example 4-3e

57. Answer: 16.9 – 2.4 or about 14.5 seconds
Example 4-3f

58. State the value of the discriminant for
State the value of the discriminant for . Then determine the number of real roots of the equation.
and
Simplify.
Answer: The discriminant is –220. Since the discriminant is negative, the equation has no real roots.
Example 4-4a

59. Step 1 Rewrite the equation in standard form.
State the value of the discriminant for . Then determine the number of real roots of the equation.
Step 1 Rewrite the equation in standard form.
Original equation
Add 144 to each side.
Simplify.
Example 4-4b

60. Step 2 Find the discriminant.
and
Simplify.
Answer: The discriminant is 0. Since the discriminant is 0, the equation has one real root.
Example 4-4c

61. Step 1 Rewrite the equation in standard form.
State the value of the discriminant for . Then determine the number of real roots of the equation.
Step 1 Rewrite the equation in standard form.
Original equation
Subtract 12 from each side.
Simplify.
Example 4-4d

62. Step 2 Find the discriminant.
and
Simplify.
Answer: The discriminant is 244. Since the discriminant is positive, the equation has two real roots.
Example 4-4e

63. Answer: –4; no real roots
State the value of the discriminant for each equation. Then determine the number of real roots for the equation. a. b. c.
Answer: –4; no real roots
Answer: 0; 1 real root
Answer: 120; 2 real root
Example 4-4f

64. End of Lesson 4

65. Example 1 Graph an Exponential Function with a > 1
Example 2 Graph Exponential Functions with 0 < a < 1
Example 3 Use Exponential Functions to Solve Problems
Example 4 Identify Exponential Behavior
Lesson 5 Contents

66. Graph State the y-intercept.9 32 2 3 31 1 30
3–1 –1 y 3x x
Graph the ordered pairs and connect the points with a smooth curve.
Answer: The y-intercept is 1.
Example 5-1a

67. Use the graph to determine the approximate value of
The graph represents all real values of x and their corresponding values of y for
Answer: The value of y is about 5 when
Use a calculator to confirm this value.
Example 5-1b

68. a. Graph State the y-intercept.
b. Use the graph to determine the approximate value of
Answer: The y-intercept is 1.
Answer: about 1.5
Example 5-1c

69. Graph State the y-intercept.
Graph the ordered pairs and connect the points with a smooth curve. 1 4 –1 y x
Answer: The y-intercept is 1.
Example 5-2a

70. Use the graph to determine the approximate value of
Answer: The value of y is about 8 when
Use a calculator to confirm this value.
Example 5-2b

71. Answer: The y-intercept is 1.
a. Graph State the y-intercept. b. Use the graph to determine the approximate value of
Answer: The y-intercept is 1.
Answer: about 3
Example 5-2c

72. Depreciation People joke that the value of a new car decreases as soon as it is driven off the dealer’s lot. The function models the depreciation of the value of a new car that originally cost $25,000. V represents the value of the car and t represents the time in years from the time the car was purchased. Graph the function. What values of V and t are meaningful in the function.
Example 5-3a

73. Use a graphing calculator to graph the function.
Answer:
Only the values of V  25,000 and t  0 are meaningful in the context of the problem.
Example 5-3b

74. What is the value of the car after one year?
Original equation
Use a calculator.
Answer: After one year, the car’s value is about $20,500.
Example 5-3c

75. What is the value of the car after five years?
Original equation
Use a calculator.
Answer: After five years, the car’s value is about $9270.
Example 5-3d

76. Depreciation The function
Depreciation The function models the depreciation of the value of a new car that originally cost $22,000. V represents the value of the car and t represents the time in years from the time the car was purchased.
a. Graph the function. What values of V and t are meaningful in the function.
Answer: Only the values of V  22,000 and t  0 are meaningful in the context of the problem.
Example 5-3e

77. b. What is the value of the car after one year?
c. What is the value of the car after three years?
Answer: $18,040
Answer: $12,130
Example 5-3f

78. Determine whether the set of data displays exponential behavior.10 20 30 y 25
62.5
156.25
Method 1 Look for a Pattern
The domain values are at regular intervals of 10. Look for a common factor among the range values.
Example 5-4a

79. Answer: Since the domain values are at regular intervals and the range values have a common factor, the data are probably exponential. The equation for the data may involve (2.5)x.
Method 2 Graph the Data
Answer: The graph shows a rapidly increasing value of y as x increases. This is a characteristic of exponential behavior.
Example 5-4b

80. Determine whether the set of data displays exponential behavior.10 20 30 y 25 40 55
Method 1 Look for a Pattern
The domain values are at regular intervals of 10. The range values have a common difference of 15.
+15
+15
+15
Example 5-4c

81. Answer: The graph is a line, not an exponential function.
Answer: Since the domain values are at regular intervals and there is a common difference of 15, the data display linear behavior.
Method 2 Graph the Data
Answer: The graph is a line, not an exponential function.
Example 5-4d

82. Determine whether the set of data displays exponential behavior. a.
12.5 25 50
100 y 30 20 10 x
Answer: The domain values are at regular intervals and the range values have a common factor of so it is probably exponential. Also the graph shows rapidly decreasing values of y as x increases.
Example 5-4e

83. Determine whether the set of data displays exponential behavior. b.10 5 –5 y 30 20 x
Answer: The domain values are at regular intervals and the range values have a common difference of 5. The data display linear behavior.
Example 5-4f

84. End of Lesson 5

85. Example 1 Exponential Growth Example 2 Compound Interest
Example 3 Exponential Decay
Example 4 Depreciation
Lesson 6 Contents

86. General equation for exponential growth
Population In 2000, the United States had a population of about 280 million, and a growth rate of about 0.85% per year.
Write an equation to represent the population of the United States since the year 2000.
General equation for exponential growth
and or
Example 6-1a

87. Simplify.
Answer: where y represents the population and t represents the number of years since 2000
Example 6-1b

88. Equation for the population of the U.S.
According to the equation, what will be the population of the United States in the year 2010?
Equation for the population of the U.S.
or 10
Answer: In 2010, the population will be about 304,731,295.
Example 6-1v

89. Answer: about 4540 students
Population In 2000, Scioto School District had a student population of about 4500 students, and a growth rate of about 0.15% per year.
a. Write an equation to represent the student population of the Scioto School District since the year 2000.
b. According to the equation, what will be the student population of the Scioto School District in the year 2006?
Answer:
Answer: about 4540 students
Example 6-1f

90. Compound interest equation
Compound Interest When Jing May was born, her grandparents invested $1000 in a fixed rate savings account at a rate of 7% compounded annually. The money will go to Jing May when she turns 18 to help with her college expenses. What amount of money will Jing May receive from the investment?
Compound interest equation
Example 6-2a

91. Compound interest equation
Simplify.
Answer: She will receive about $3380.
Example 6-2n

92. Compound Interest When Lucy was 10 years old, her father invested $2500 in a fixed rate savings account at a rate of 8% compounded semiannually. When Lucy turns 18, the money will help to buy her a car. What amount of money will Lucy receive from the investment?
Answer: about $4682
Example 6-2v

93. General equation for exponential decay
Charity During an economic recession, a charitable organization found that its donations dropped by 1.1% per year. Before the recession, its donations were $390,000.
Write an equation to represent the charity’s donations since the beginning of the recession.
General equation for exponential decay
Answer:
Simplify.
Example 6-3a

94. Equation for the amount of donations
Charity During an economic recession, a charitable organization found that its donations dropped by 1.1% per year. Before the recession, its donations were $390,000.
Estimate the amount of the donations 5 years after the start of the recession.
Equation for the amount of donations
Answer: The amount of donations should be about $369,017.
Example 6-3n

95. Charity A charitable organization found that the value of its clothing donations dropped by 2.5% per year. Before this downturn in donations, the organization received clothing valued at $24,000.
a. Write an equation to represent the value of the charity’s clothing donations since the beginning of the downturn.
b. Estimate the value of the clothing donations 3 years after the start of the downturn.
Answer:
Answer: about $22,245
Example 6-3v

96. General equation for exponential decay
Depreciation Jackson and Elizabeth bought a house when they first married ten years ago. Since that time the value of the real estate in their neighborhood has declined 3% per year. If they initially paid $179,000 for their house, what is its value today?
General equation for exponential decay
Simplify.
Example 6-4a

97. Answer: Their house will be worth about $131,999.
Use a calculator.
Answer: Their house will be worth about $131,999.
Example 6-4n

98. Depreciation A business owner bought a computer system for the office staff for $22,000. If the computer system depreciates 8% per year, find the value of the computer system in 4 years.
Answer: about $15,761
Example 6-4v

99. End of Lesson 6

100. Example 1 Recognize Geometric Sequences
Example 2 Continue Geometric Sequences
Example 3 Use Geometric Sequences to Solve a Problem
Example 4 nth Term of a Geometric Sequence
Example 5 Find Geometric Means
Lesson 7 Contents

101. Determine whether the sequence is geometric.
1, 4, 16, 64, 256, …
Determine the pattern.
In this sequence, each term in found by multiplying the previous term by 4.
Answer: This sequence is geometric.
Example 7-1a

102. Determine whether the sequence is geometric.
1, 3, 5, 7, 9, 11, …
Determine the pattern.
In this sequence, each term is found by adding 2 to the previous term.
Answer: This sequence is arithmetic, not geometric.
Example 7-1b

103. Determine whether each sequence is geometric. a.
b. 0, 11, 22, 33, 44, …
Answer: yes
Answer: no
Example 7-1c

104. Find the next three terms in the geometric sequence.
20, –28, 39.2, …
Divide the second term by the first.
The common factor is –1.4. Use this information to find the next three terms.
20, –28, 39.2
–54.88
76.832 –
Answer: The next three terms are –54.88, , and –
Example 7-2a

105. Find the next three terms in the geometric sequence.
64, 48, 36, …
Divide the second term by the first.
The common factor is 0.8. Use this information to find the next three terms.
64, 48, 36 27
20.25
Answer: The next three terms are 27, 20.25, and
Example 7-2b

106. Find the next three terms in each geometric sequence.
b. 100, –250, 625, …
Answer: 9.072, ,
Answer: –1562.5, , –
Example 7-2c

107. Geography The population of the African country of Liberia was about 2,900,000 in If the population grows at a rate of about 5% per year, what will the population be in the years 2003, 2004, and 2005?
The population is a geometric sequence in which the first term is 2,900,000 and the common ratio is 1.05.
Example 7-3a

108. Year
Population
1999
2,900,000
2000
2,900,000(1.05) or 3,045,000
2001
3,045,000(1.05) or 3,197,250
2002
3,197,250(1.05) or 3,357,112.5
2003
3,357,112.5(1.05) or 3,524,968.1
2004
3,524,968.1(1.05) or 3,701,216.5
2005
3,701,216.5(1.05) or 3,886,277.3
Answer: The population of Liberia in the years 2003, 2004, and 2005 will be about 3,524,968, 3,701,217, and 3,886,277, respectively.
Example 7-3b

109. Answer: about 2,760,202 in 2003 and about 2,815,406 in 2004
Geography The population of the Baltic State of Latvia was about 2,500,000 in If the population grows at a rate of about 2% per year, what will the population be in the years 2003 and 2004?
Answer: about 2,760,202 in 2003 and about 2,815,406 in 2004
Example 7-3c

110. Find the eighth term of a geometric sequence in which
Formula for the nth term of a geometric sequence
Example 7-4a

111. Answer: The eighth term in the sequence is 15,309.
Example 7-4b

112. Find the ninth term of a geometric sequence in which
Answer: 786,432
Example 7-4c

113. Find the geometric mean in the sequence 7, ___, 112.
In the sequence, and To find you must first find r.
Formula for the nth term of a geometric sequence
Divide each side by 7.
Example 7-5a

114. Take the square root of each side.
Simplify.
Take the square root of each side.
If the geometric mean is 7(4) or 28. If the geometric mean is 7(–4) or –28.
Answer: The geometric mean is 28 or –28.
Example 7-5b

115. Find the geometric mean in the sequence 9, ___, 576.
Answer: 72 or –72
Example 7-5c

116. End of Lesson 7

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