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F. Applications of Quadratic Functions and Inequalities

# F. Applications of Quadratic Functions and Inequalities

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Section 3.2 Solving Quadratic Equations and Inequalities

307

b. Since the height h(t) must be greater than or equal to 496 ft, we use the

function h1t2 ϭ Ϫ16t2 ϩ 160t ϩ 240 to write the inequality Ϫ16t2 ϩ 160t ϩ

240 Ն 496. In standard form, we obtain Ϫ16t2 ϩ 160t Ϫ 256 Ն 0 (subtract

496 from both sides). We begin by finding the zeroes of Ϫ16t2 ϩ 160t Ϫ 256 ϭ 0,

noting the related graph opens downward since the leading coefficient is

negative.

Ϫ16t2 ϩ 160t Ϫ 256 ϭ 0

t2 Ϫ 10t ϩ 16 ϭ 0

1t Ϫ 22 1t Ϫ 82 ϭ 0

t Ϫ 2 ϭ 0 or t Ϫ 8 ϭ 0

t ϭ 2 or t ϭ 8

related equation

divide by ؊16

factor

zero factor theorem

result

This shows the rocket is at exactly 496 ft after 2 sec (on its ascent) and after

8 sec (during its descent). We conclude the rocket’s height was greater than

496 ft for 8 Ϫ 2 ϭ 6 sec.

c. When the rocket hits the ground, its height is h ϭ 0. Substituting 0 for h1t2 and

solving gives

h1t2 ϭ Ϫ16t2 ϩ 160t ϩ 240

0 ϭ Ϫ16t2 ϩ 160t ϩ 240

0 ϭ t2 Ϫ 10t Ϫ 15

original function

substitute 0 for h(t )

divide by ؊16

The equation is nonfactorable, so we use the quadratic equation to solve, with

a ϭ 1, b ϭ Ϫ10, and c ϭ Ϫ15:

ϭ

Ϫb Ϯ 2b2 Ϫ 4ac

2a

Ϫ1Ϫ102 Ϯ 21Ϫ102 2 Ϫ 41121Ϫ152

2112

10 Ϯ 2160

2

10

4 210

ϭ

Ϯ

2

2

ϭ

substitute 1 for a, ؊10 for b, ؊15 for c

simplify

2160 ϭ 4 210

ϭ 5 Ϯ 2 210

simplify

Since we need the time t in seconds, we

use the approximate form of the answer,

obtaining t Ϸ Ϫ1.32 and t Ϸ 11.32. The

11 sec (since t represents time, the solution

t ϭ Ϫ1.32 does not apply). A calculator

check is shown in Figure 3.39 using the

Zeroes Method.

Figure 3.39

800

0

15

Ϫ200

Now try Exercises 157 through 166

EXAMPLE 12

Solving Applications of Inequalities Using the Quadratic Formula

For the years 1995 to 2006, the amount A of annual international telephone traffic

(in billions of minutes) can be modeled by A ϭ 0.17x2 ϩ 8.43x ϩ 64.58 where

x ϭ 0 represents the year 1995 [Source: Data from the 2009 Statistical Abstract of

the United States, Table 1344, page 846]. If this trend continues, in what year will

the annual number of minutes reach or surpass 275 billion?

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CHAPTER 3 Quadratic Functions and Operations on Functions

Analytical Solution

We are essentially asked to solve the inequality 0.17x2 ϩ 8.43x ϩ 64.58 Ն 275.

0.17x2 ϩ 8.43x ϩ 64.58 Ն 275

0.17x2 ϩ 8.43x Ϫ 210.42 Ն 0

given inequality

subtract 275

For a ϭ 0.17, b ϭ 8.43, and c ϭ Ϫ210.42, the quadratic formula gives

Ϫb Ϯ 2b2 Ϫ 4ac

2a

Ϫ8.43 Ϯ 218.432 2 Ϫ 410.1721Ϫ210.422

210.172

Ϫ8.43 Ϯ 2214.1505

0.34

x Ϸ 18.25 or x Ϸ Ϫ67.83

substitute known values

simplify

result

We disregard the negative solution (since x represents time), and find the annual

number of international telephone minutes will reach or surpass 275 billion about

18 years after 1995, or in the year 2013.

Graphical Solution

F. You’ve just seen how

we can solve applications of

inequalities

350

After entering 0.17x2 ϩ 8.43x ϩ 64.58 as Y1,

our next task is to determine an appropriate

window size. With 1995 as year x ϭ 0 and the

data taken from year 2006 1x ϭ 112 it seems

25

Ϫ2

that x ϭ Ϫ2 to x ϭ 25 would be an

appropriate start for the domain (we can later

adjust the window if needed). As the primary

question is the year when telephone traffic

Ϫ100

surpasses 275 billion minutes, the range must

include this value and y ϭ Ϫ100 to y ϭ 350

would be a good start (as before, the negative values were used to create a frame

around the desired window). Using the Intersection-of-Graphs method, the resulting

graph is shown in the figure and indicates that telephone traffic surpassed 275

billion minutes in the 18th year (2013), in line with our analytical solution.

Now try Exercises 167 and 168

3.2 EXERCISES

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section, if necessary.

1. The solution x ϭ 2 ϩ 13 is called an

form of the solution. Using a calculator, we find

the

form is x Ϸ 3.732.

4. According to the summary on page 22, the

equation 4x2 Ϫ 5x might best be solved by

out the

.

2. To solve a quadratic equation by completing the

square, the coefficient of the

term must

be a

.

5. Discuss/Explain the relationship between solutions

to f 1x2 ϭ 0 and the x-intercepts of y ϭ f 1x2 . Be

sure to include an example.

3. The quantity b2 Ϫ 4ac is called the

of

the quadratic equation. If b2 Ϫ 4ac 7 0, there are

real roots.

6. Discuss/Explain why this version of the quadratic

formula is incorrect:

2b2 Ϫ 4ac

x ϭ Ϫb Ϯ

2a

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309

Six functions are shown below, along with their respective zeroes. Use these zeroes to solve the equations in Exercises

7 through 12.

f 1x2 ϭ x2 ϩ x Ϫ 12,

f 1Ϫ42 ϭ f 132 ϭ 0;

g1x2 ϭ x2 Ϫ 4x Ϫ 5,

g1Ϫ12 ϭ g152 ϭ 0;

h1x2 ϭ Ϫ2x2 Ϫ 4x ϩ 6,

h1Ϫ32 ϭ h112 ϭ 0;

j1x2 ϭ Ϫ4x2 ϩ 4x ϩ 8,

j1Ϫ12 ϭ j122 ϭ 0;

k1x2 ϭ 2x2 ϩ x Ϫ 3,

k1 Ϫ3

2 2 ϭ k112 ϭ 0;

l1x2 ϭ 3x2 ϩ 8x Ϫ 3,

l1Ϫ32 ϭ l1 13 2 ϭ 0

7. x2 ϩ x Ϫ 12 ϭ 0

8. x2 Ϫ 4x Ϫ 5 ϭ 0

10. 3x2 ϩ 8x ϭ 3

11. 4x2 ϭ 4x ϩ 8

Solve the following equations graphically by locating the

zeroes of a related quadratic function. Round to

hundredths as necessary.

9. 2x2 ϩ x ϭ 3

12. 2x2 ϩ 4x ϭ 6

both exact form and approximate form rounded to the

hundredths place. If there are no real solutions, so state.

13. x2 ϩ 3x Ϫ 5 ϭ 0

43. x2 ϩ 6x ϭ Ϫ5

44. m2 ϩ 8m ϭ Ϫ12

14. x2 Ϫ 7x Ϫ 2 ϭ 0

45. p2 Ϫ 6p ϩ 3 ϭ 0

46. n2 ϭ 4n ϩ 10

15. 0.4x2 Ϫ 0.6x Ϫ 2 ϭ 0

47. p2 ϩ 6p ϭ Ϫ4

48. x2 Ϫ 8x Ϫ 1 ϭ 0

16. 5x2 ϩ 0.2x Ϫ 1.1 ϭ 0

49. m2 ϩ 3m ϭ 1

50. n2 ϩ 5n Ϫ 2 ϭ 0

17. 2x2 Ϫ 3x ϭ 6

51. n2 ϭ 5n ϩ 5

52. w2 Ϫ 7w ϩ 3 ϭ 0

18. 3x2 ϩ x ϭ 3

Solve the following quadratic equations (a) algebraically

by completing the square and (b) graphically by using a

graphing calculator and the Zeroes Method. Round

answers to nearest hundredth when necessary.

19. 2.9x2 ϭ 1.3x Ϫ 5.7

20. 8.1x2 ϭ 5.3x Ϫ 4.2

Solve the following equations using the square root

property of equality. Write answers in exact form

and approximate form rounded to hundredths. If

there are no real solutions, so state. Verify solutions

graphically.

21. m ϭ 16

22. p ϭ 49

23. y2 Ϫ 28 ϭ 0

24. m2 Ϫ 20 ϭ 0

25. p2 ϩ 36 ϭ 0

26. n2 ϩ 5 ϭ 0

27. x2 ϭ 21

16

28. y2 ϭ 13

9

2

2

29. 1n Ϫ 32 2 ϭ 36

31. 1w ϩ 52 ϭ 3

2

33. 1x Ϫ 32 ϩ 7 ϭ 2

2

35. 1m Ϫ 22 ϭ

2

18

49

30. 1p ϩ 52 2 ϭ 49

32. 1m Ϫ 42 ϭ 5

2

34. 1m ϩ 112 ϩ 5 ϭ 3

2

36. 1x Ϫ 52 ϭ

2

12

25

Fill in the blank so the result is a perfect square

trinomial, then factor into a binomial square.

37. x ϩ 6x ϩ _______

2

38. y ϩ 10y ϩ _______

2

39. n ϩ 3n ϩ _______ 40. x Ϫ 5x ϩ _______

2

2

2

3

41. p2 ϩ p ϩ _______ 42. x2 Ϫ x ϩ _______

3

2

53. 2x2 ϭ Ϫ7x ϩ 4

54. 3w2 Ϫ 8w ϩ 4 ϭ 0

55. 2n2 Ϫ 3n Ϫ 9 ϭ 0

56. 2p2 Ϫ 5p ϭ 1

57. 4p2 Ϫ 3p Ϫ 2 ϭ 0

58. 3x2 ϩ 5x Ϫ 6 ϭ 0

59. m2 ϭ 7m Ϫ 4

60. a2 Ϫ 15 ϭ 4a

Solve each equation using the most efficient method:

factoring, square root property of equality, or the

approximate form (rounded to hundredths). Check one

of the exact solutions in the original equation.

61. x2 Ϫ 3x ϭ 18

62. w2 ϩ 6w Ϫ 1 ϭ 0

63. 4m2 Ϫ 25 ϭ 0

64. 4a2 Ϫ 4a ϭ 1

65. 4n2 Ϫ 8n Ϫ 1 ϭ 0

66. 2x2 Ϫ 4x ϩ 5 ϭ 0

67. 6w2 Ϫ w ϭ 2

68. 3a2 Ϫ 5a ϩ 6 ϭ 0

69. 4m2 ϭ 12m Ϫ 15

70. 3p2 ϩ p ϭ 0

71. 4n2 Ϫ 9 ϭ 0

72. 4x2 Ϫ x ϭ 3

73. 5w2 ϭ 6w ϩ 8

74. 3m2 Ϫ 7m Ϫ 6 ϭ 0

75. 3a2 Ϫ a ϩ 2 ϭ 0

76. 3n2 Ϫ 2n Ϫ 3 ϭ 0

77. 5p2 ϭ 6p ϩ 3

78. 2x2 ϩ x ϩ 3 ϭ 0

79. 5w2 Ϫ w ϭ 1

80. 3m2 Ϫ 2 ϭ 5m

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81. 2a2 ϩ 5 ϭ 3a

82. n2 ϩ 4n Ϫ 8 ϭ 0

83. 2p2 Ϫ 4p ϩ 11 ϭ 0 84. 8x2 Ϫ 5x Ϫ 1 ϭ 0

5

2

1

8

1

85. w2 ϩ w ϭ

86. m2 Ϫ m ϩ ϭ 0

3

9

4

3

6

87. 0.2a2 ϩ 1.2a ϩ 0.9 ϭ 0

88. Ϫ5.4n2 ϩ 8.1n ϩ 9 ϭ 0

89.

3–30

CHAPTER 3 Quadratic Functions and Operations on Functions

2 2

8

p Ϫ3ϭ p

7

21

90.

5 2 16

3

x Ϫ xϭ

9

15

2

Use the discriminant to determine whether the

given equation has irrational, rational, repeated,

or nonreal roots. Also state whether the original

equation is factorable using integers, but do not

solve for x.

91. Ϫ3x2 ϩ 2x ϩ 1 ϭ 0 92. 2x2 Ϫ 5x Ϫ 3 ϭ 0

93. Ϫ4x ϩ x2 ϩ 13 ϭ 0 94. Ϫ10x ϩ x2 ϩ 41 ϭ 0

95. 15x2 Ϫ x Ϫ 6 ϭ 0

96. 10x2 Ϫ 11x Ϫ 35 ϭ 0

97. Ϫ4x2 ϩ 6x Ϫ 5 ϭ 0 98. Ϫ5x2 Ϫ 3 ϭ 2x

99. 2x2 ϩ 8 ϭ Ϫ9x

100. x2 ϩ 4 ϭ Ϫ7x

101. 4x ϩ 12x ϭ Ϫ9

102. 9x ϩ 4 ϭ 12x

2

2

Solve the equations given. Simplify each result.

103. Ϫ6x ϩ 2x2 ϩ 5 ϭ 0 104. 17 ϩ 2x2 ϭ 10x

105. 5x2 ϩ 5 ϭ Ϫ5x

106. x2 ϭ Ϫ2x Ϫ 19

107. Ϫ2x ϭ Ϫ5x ϩ 11 108. 4x Ϫ 3 ϭ 5x

2

2

Solve each quadratic inequality by locating the

x-intercept(s) (if they exist), and noting the end-behavior

of the graph. Begin by writing the inequality in function

form as needed.

109. f 1x2 ϭ Ϫx2 ϩ 4x; f 1x2 7 0

110. g1x2 ϭ x Ϫ 5x; g1x2 6 0

125. g1x2 ϭ Ϫx2 ϩ 10x Ϫ 25; g1x2 6 0

126. h1x2 ϭ Ϫx2 ϩ 14x Ϫ 49; h1x2 6 0

127. Ϫx2 7 2

128. x2 6 Ϫ4

129. x2 Ϫ 2x 7 Ϫ5

130. Ϫx2 ϩ 3x 6 3

131. 2x2 Ն 6x Ϫ 9

132. 5x2 Ն 4x Ϫ 4

Solve each quadratic inequality using the interval test

possible.

133. 1x ϩ 321x Ϫ 12 6 0 134. 1x ϩ 221x Ϫ 52 6 0

135. 2x2 Ϫ x Ϫ 6 Ն 0

137. 1x ϩ 1.32 2 7 0

136. Ϫ3x2 ϩ x ϩ 4 Յ 0

138. Ϫ1x Ϫ 2.22 2 6 0

139. Ϫ1x Ϫ 2.92 2 Ն 0

140. 1x ϩ 3.22 2 Յ 0

3 2

141. ax Ϫ b 6 0

5

1 2

142. Ϫax ϩ b 7 0

8

143. x2 ϩ 14x ϩ 49 Ն 0 144. Ϫx2 ϩ 6x Ϫ 9 Յ 0

Recall that for a square root expression to represent a

real number, the radicand must be greater than or equal

to zero. Applying this idea results in an inequality that

can be solved using the skills from this section.

Determine the domain of the following radical

functions.

145. h1x2 ϭ 2x2 Ϫ 25

146. p1x2 ϭ 225 Ϫ x2

147. q1x2 ϭ 2x2 Ϫ 5x

148. r1x2 ϭ 26x Ϫ x2

149. t1x2 ϭ 2Ϫx2 ϩ 3x Ϫ 4

150. Y1 ϭ 2x2 Ϫ 6x ϩ 9

2

111. h1x2 ϭ x2 ϩ 4x Ϫ 5; h1x2 Ն 0

112. p1x2 ϭ Ϫx2 ϩ 3x ϩ 10; p1x2 Յ 0

113. q1x2 ϭ 2x2 Ϫ 5x Ϫ 7; q1x2 6 0

114. r1x2 ϭ Ϫ2x2 Ϫ 3x ϩ 5; r1x2 7 0

115. 7 Ն x2

116. x2 Յ 13

117. x2 ϩ 3x Յ 6

118. x2 Ϫ 2 Յ 5x

119. 3x2 Ն Ϫ2x ϩ 5

120. 4x2 Ն 3x ϩ 7

121. s1x2 ϭ x2 Ϫ 8x ϩ 16; s1x2 Ն 0

122. t1x2 ϭ x2 Ϫ 6x ϩ 9; t1x2 Ն 0

123. r1x2 ϭ 4x ϩ 12x ϩ 9; r1x2 6 0

2

124. f 1x2 ϭ 9x2 Ϫ 6x ϩ 1; f 1x2 6 0

Match the correct solution with the inequality and

graph given.

151. f 1x2 7 0

a. x ʦ 1Ϫq, Ϫ22 ´ 11, q2

b. x ʦ 1Ϫq, Ϫ2 4 ´ 31, q 2

c. 1Ϫ2, 12

d. 3Ϫ2, 14

e. none of these

152. g1x2 Ն 0

a. x ʦ 1Ϫq, Ϫ42 ´ 11, q 2

b. x ʦ 1Ϫq, Ϫ4 4 ´ 31, q 2

c. 1Ϫ4, 12

d. 3Ϫ4, 1 4

e. none of these

y

5

f(x)

Ϫ5

5 x

Ϫ5

y

5

g(x)

Ϫ5

5 x

Ϫ5

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153. r1x2 Յ 0

a. x ʦ 1Ϫq, 32 ´ 13, q 2

b. x ʦ 1Ϫq, q2

c. {3}

d. { }

e. none of these

y

3

Ϫ2

8 x

r (x)

Ϫ7

154. s1x2 6 0

a. x ʦ 1Ϫq, 02 ´ 10, q 2

b. x ʦ 1Ϫq, q 2

c. {0}

d. { }

e. none of these

y

7

s(x)

Ϫ5

5 x

Ϫ3

WORKING WITH FORMULAS

155. Height of a projectile: h ‫ ؍‬؊16t 2 ؉ vt

If an object is projected vertically upward from

ground level with no continuing source of propulsion,

the height of the object (in feet) is modeled by the

equation shown, where v is the initial velocity, and t is

the time in seconds. Use the quadratic formula to

solve for t in terms of v and h. (Hint: Set the equation

equal to zero and identify the coefficients as before.)

311

Section 3.2 Solving Quadratic Equations and Inequalities

156. Surface area of a cylinder: A ‫ ؍‬2␲r2 ؉ 2␲rh

The surface area of a cylinder is given by the formula

shown, where h is the height and r is the radius of the

base. The equation can be considered a quadratic in

the variable r. Use the quadratic formula to solve for r

in terms of h and A. (Hint: Rewrite the equation in

standard form and identify the coefficients as before.)

APPLICATIONS

157. Height of a projectile: The height of an object

thrown upward from the roof of a building 408 ft

tall, with an initial velocity of 96 ft/sec, is given by

the equation h ϭ Ϫ16t 2 ϩ 96t ϩ 408, where h

represents the height of the object after t seconds.

How long will it take the object to hit the ground?

Answer in exact form and decimal form rounded to

the nearest hundredth.

158. Height of a projectile: The height of an object

thrown upward from the floor of a canyon 106 ft

deep, with an initial velocity of 120 ft/sec, is given

by the equation h ϭ Ϫ16t 2 ϩ 120t Ϫ 106, where

h represents the height of the object after t seconds.

How long will it take the object to rise to the height

of the canyon wall 1h ϭ 02 ? Answer in exact form

and decimal form rounded to hundredths.

159. Cost, revenue, and

profit: The cost of

raw materials to

produce plastic

toys is given by the

cost equation

C ϭ 2x ϩ 35,

where x is the

number of toys in

hundreds. The total

income (revenue)

from the sale of

these toys is given

by R ϭ Ϫx 2 ϩ 122x Ϫ 1965. (a) Determine the

profit equation 1profit ϭ revenue Ϫ cost2. (b) During

the Christmas season, the owners of the company

decide to manufacture and donate as many toys as

they can, without taking a loss (i.e., they break

even: profit or P ϭ 02. How many toys will they

produce for charity?

160. Cost, revenue, and profit: The cost to produce

bottled spring water is given by the cost equation

C ϭ 16x ϩ 63, where x is the number of bottles in

thousands. The total revenue from the sale of these

bottles is given by the equation

R ϭ Ϫx2 ϩ 326x Ϫ 18,463. (a) Determine the

profit equation 1profit ϭ revenue Ϫ cost2. (b) After

a bad flood contaminates the drinking water of a

nearby community, the owners decide to bottle and

donate as many bottles of water as they can,

without taking a loss (i.e., they break even: profit

or P ϭ 0). Approximately how many bottles will

they produce for the flood victims?

161. Height of an arrow: If an object is projected

vertically upward from ground level with no

continuing source of propulsion, its height (in feet)

is modeled by the equation h ϭ Ϫ16t2 ϩ vt, where

v is the initial velocity and t is the time in seconds.

Use the quadratic formula to solve for t, given an

arrow is shot into the air with v ϭ 144 ft/sec and

h ϭ 260 ft. See Exercise 155.

162. Surface area of a cylinder: The surface area of a

cylinder is given by A ϭ 2␲r2 ϩ 2␲rh, where h is

the height and r is the radius of the base. The

equation can be considered a quadratic in the

variable r. Use the quadratic formula to solve for

r, given A ϭ 4710 cm2 and h ϭ 35 cm. See

Exercise 156.

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163. Tennis court dimensions: A

regulation tennis court for a

doubles match is laid out so that

its length is 6 ft more than two

times its width. The area of the

doubles court is 2808 ft2. What

is the length and width of the

doubles court?

Exercises 163

and 164

164. Tennis court dimensions: A

regulation tennis court for a

Singles

singles match is laid out so that

its length is 3 ft less than three

Doubles

times its width. The area of the

2

singles court is 2106 ft . What is the length and

width of the singles court?

165. Cost, revenue, and profit: The revenue for a

manufacturer of microwave ovens is given by the

equation R ϭ x140 Ϫ 13x2, where revenue is in

thousands of dollars and x thousand ovens are

manufactured and sold. What is the minimum

number of microwave ovens that must be sold to

bring in a revenue of at least \$900,000?

166. Cost, revenue, and profit: The revenue for a

manufacturer of computer printers is given by the

3–32

CHAPTER 3 Quadratic Functions and Operations on Functions

equation R ϭ x130 Ϫ 0.4x2 , where revenue is in

thousands of dollars and x thousand printers are

manufactured and sold. What is the minimum

number of printers that must be sold to bring in a

revenue of at least \$440,000?

167. Cell phone subscribers: For the years 1995 to

2002, the number N of cellular phone subscribers

(in millions) can be modeled by the equation

N ϭ 17.4x2 ϩ 36.1x ϩ 83.3, where x ϭ 0

represents the year 1995 [Source: Data from the

2005 Statistical Abstract of the United States, Table

1372, page 870]. If this trend continued, in what

year did the number of subscribers reach or surpass

3750 million?

168. U.S. international trade balance: For the years

1995 to 2003, the international trade balance B (in

millions of dollars) can be approximated

by the equation B ϭ Ϫ3.1x2 ϩ 4.5x Ϫ 19.9,

where x ϭ 0 represents the year 1995 [Source:

Data from the 2005 Statistical Abstract of the

United States, Table 1278, page 799]. If this trend

continues, in what year will the trade balance reach

a deficit of \$750 million dollars or more?

EXTENDING THE CONCEPT

169. Using the discriminant: Each of the following

equations can easily be solved by factoring. Using

the discriminant, we can create factorable equations

with identical values for b and c, but where a 1.

For instance, x2 Ϫ 3x Ϫ 10 ϭ 0 and

4x2 Ϫ 3x Ϫ 10 ϭ 0 can both be solved by

factoring. Find similar equations 1a 12 for the

quadratics given here. (Hint: The discriminant

b2 Ϫ 4ac must be a perfect square.)

a. x2 ϩ 6x Ϫ 16 ϭ 0

b. x2 ϩ 5x Ϫ 14 ϭ 0

c. x2 Ϫ x Ϫ 6 ϭ 0

170. Using the discriminant: For what values of c will

the equation 9x2 Ϫ 12x ϩ c ϭ 0 have

a. no real roots

b. one rational root

c. two real roots

d. two integer roots

Complex polynomials: Many techniques applied to solve

polynomial equations with real coefficients can be

applied to solve polynomial equations with complex

coefficients. Here we apply the idea to carefully chosen

quadratic equations, as a more general application must

wait until a future course, when the square root of a

complex number is fully developed. Solve each equation

1

using the quadratic formula, noting that ‫ ؍‬؊i.

i

171. z2 Ϫ 3iz ϭ Ϫ10

172. z2 Ϫ 9iz ϭ Ϫ22

173. 4iz2 ϩ 5z ϩ 6i ϭ 0

174. 2iz2 Ϫ 9z ϩ 26i ϭ 0

175. 0.5z2 ϩ 17 ϩ i2z ϩ 16 ϩ 7i2 ϭ 0

176. 0.5z2 ϩ 14 Ϫ 3i2z ϩ 1Ϫ9 Ϫ 12i2 ϭ 0

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Section 3.3 Quadratic Functions and Applications

313

177. (R.3) State the formula for the perimeter and area

of each figure illustrated.

a.

b.

L

r

W

c.

d.

b1

a

c

h

h

c

178. (R.4) Factor and solve the following equations:

a. x2 Ϫ 5x Ϫ 36 ϭ 0

b. 4x2 Ϫ 25 ϭ 0

c. x3 ϩ 6x2 Ϫ 4x Ϫ 24 ϭ 0

179. (1.4) A total of 900 tickets were sold for a recent

concert and \$25,000 was collected. If good seats

were \$30 and cheap seats were \$20, how many of

each type were sold?

180. (1.5) Solve for C: P ϭ C ϩ Ct.

b

b2

3.3

LEARNING OBJECTIVES

In Section 3.3 you will see how we can:

As our knowledge of functions grows, our ability to apply mathematics in new ways likewise grows. In this section, we’ll build on the foundation laid in Section 3.2 and previous

chapters, as we introduce additional tools used to apply quadratic functions effectively.

by completing the square

using the vertex formula

C. Find the equation of a

its graph

D. Solve applications

involving extreme values

Figure 3.40 f(x) ϭ ax2 ϩ bx ϩ c

Endbehavior

y

Axis of

symmetry

xϭh

y-intercept

(0, c)

(x1, 0)

(x2, 0)

x

x-intercepts

Vertex

(h, k)

A. Graphing Quadratic Functions by Completing the Square

Our earlier work suggests the graph of any quadratic function will be a parabola.

Figure 3.40 provides a summary of the graph’s characteristic features. As pictured,

1.

2.

3.

4.

5.

The parabola opens upward 1y ϭ ax2 ϩ bx ϩ c, a 7 02 .

The vertex is at (h, k) and y = k is a global minimum.

The vertex is below the x-axis, so there are two x-intercepts.

The axis of symmetry contains the vertex, with equation x ϭ h.

The y-intercept is (0, c), since f 102 ϭ c.

In Section 2.2, we graphed transformations of f 1x2 ϭ x2, using y ϭ a1x Ϯ h2 2 Ϯ k.

Here, we’ll show that by completing the square, we can graph any quadratic function as

a transformation of this basic graph.

When completing the square on a quadratic equation (Section 3.2), we applied the

standard properties of equality to both sides of the equation. When completing the

square on a quadratic function, the process is altered slightly, so that we operate on

only one side.

The basic ideas are summarized here.

Graphing f1x2 ‫ ؍‬ax2 ؉ bx ؉ c by Completing the Square

1. Group the variable terms apart from the constant c.

2. Factor out the leading coefficient a from this group.

3. Compute 3 12 1 ba 2 4 2 and add the result to the variable terms,

then subtract a # 3 12 1 ba 2 4 2 from c to maintain an equivalent expression.

4. Factor the grouped terms as a binomial square and combine constant terms.

5. Graph using transformations of f 1x2 ϭ x2.

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CHAPTER 3 Quadratic Functions and Operations on Functions

EXAMPLE 1

Graphing a Quadratic Function by Completing the Square

Given g1x2 ϭ x2 Ϫ 6x ϩ 5, complete the square to

rewrite g as a transformation of f 1x2 ϭ x2, then graph the function.

Solution

To begin we note the leading coefficient is a ϭ 1.

g1x2 ϭ x2 Ϫ 6x ϩ 5

ϭ 11x Ϫ 6x ϩ ___ 2 ϩ 5

2

ϭ 11x2 Ϫ 6x ϩ 92 Ϫ 9 ϩ 5

adds 1 # 9 ϭ 9

subtract 9

ϭ 1x Ϫ 32 2 Ϫ 4

given function

y

x ϭ3

group variable terms

2

1

c a b1Ϫ62 d ϭ 9

2

5

(6, 5)

(0, 5)

factor and simplify

The graph of g is the graph of f shifted 3 units right,

and 4 units down. The graph opens upward (a 7 0)

with the vertex at (3, Ϫ4), and axis of symmetry

x ϭ 3. From the original equation we find g102 ϭ 5,

giving a y-intercept of (0, 5). The point (6, 5) was

obtained using the symmetry of the graph. The graph

is shown in the figure.

g(x)

x

Ϫ3

(3, Ϫ4)

Ϫ5

Now try Exercises 7 through 10

Note that by adding 9 and simultaneously subtracting 9 (essentially adding “0”),

we changed only the form of the function, not its value. In other words, the resulting

expression is equivalent to the original. If the leading coefficient is not 1, we factor it

out from the variable terms, but take it into account when we add the constant needed

to maintain an equivalent expression (steps 2 and 3).

EXAMPLE 2

Graphing a Quadratic Function by Completing the Square

Given p1x2 ϭ Ϫ2x2 Ϫ 8x Ϫ 3, complete the square to rewrite p as a transformation

of f 1x2 ϭ x2, then graph the function.

Solution

p1x2 ϭ Ϫ2x2 Ϫ 8x Ϫ 3

ϭ 1Ϫ2x2 Ϫ 8x ϩ ___ 2Ϫ3

ϭ Ϫ21x2 ϩ 4x ϩ ___ 2Ϫ3

ϭ Ϫ21x2 ϩ 4x ϩ 42 Ϫ 1Ϫ82 Ϫ 3

adds Ϫ2 # 4 ϭ Ϫ8

WORTHY OF NOTE

In cases like f 1x2 ϭ 3x2 Ϫ 10x ϩ 5,

where the linear coefficient has no

integer factors of a, we factor out 3

and simultaneously divide the linear

coefficient by 3. This yields

10

h1x2 ϭ 3 ax2 Ϫ

x ϩ ____ b ϩ 5,

3

and the process continues as

5 2

2

before: 3 1 12 21 10

3 2 4 ϭ 13 2 ϭ

25

9,

and

so on. For more on this idea, see

Exercises 15 through 20.

subtract Ϫ8

ϭ Ϫ21x ϩ 22 2 ϩ 8 Ϫ 3

ϭ Ϫ21x ϩ 22 2 ϩ 5

given function

group variable terms

factor out a ϭ Ϫ2 (notice sign change)

2

1

c a b142 d ϭ 4

2

x ϭ Ϫ2

factor trinomial,

simplify

result

The graph of p is a parabola, shifted 2 units left,

stretched by a factor of 2, reflected across the x-axis

(opens downward), and shifted up 5 units. The

vertex is (Ϫ2, 5), and the axis of symmetry is

x ϭ Ϫ2. From the original function, the y-intercept

is (0, Ϫ3). The point (Ϫ4, Ϫ3) was obtained using

the symmetry of the graph. The graph is shown in

the figure.

(Ϫ2, 5)

y

5

p(x)

Ϫ5

3

(Ϫ4, Ϫ3)

x

(0, Ϫ3)

Ϫ5

Now try Exercises 11 through 20

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