F. Applications of Quadratic Functions and Inequalities
Tải bản đầy đủ - 0trang
cob19545_ch03_293-313.qxd
8/10/10
6:08 PM
Page 307
College Algebra G&M—
3–27
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:
tϭ
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
quadratic formula
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
rocket will return to the ground in just over
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?
cob19545_ch03_293-313.qxd
11/25/10
3:29 PM
Page 308
College Algebra G&M—
308
3–28
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
xϭ
xϭ
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
xϭ
quadratic formula
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
quadratic functions and
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
cob19545_ch03_293-313.qxd
8/10/10
6:08 PM
Page 309
College Algebra G&M—
3–29
ᮣ
Section 3.2 Solving Quadratic Equations and Inequalities
309
DEVELOPING YOUR SKILLS
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
Solve by completing the square. Write your answers in
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
quadratic formula. Write your answer in both exact and
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
cob19545_ch03_293-313.qxd
11/25/10
3:30 PM
Page 310
College Algebra G&M—
310
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
method. Write your answer in interval notation when
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
cob19545_ch03_293-313.qxd
8/10/10
6:09 PM
Page 311
College Algebra G&M—
3–31
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 ؍2r2 ؉ 2rh
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 2r2 ϩ 2rh, 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.
cob19545_ch03_293-313.qxd
8/10/10
6:09 PM
Page 312
College Algebra G&M—
312
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
cob19545_ch03_293-313.qxd
8/10/10
6:10 PM
Page 313
College Algebra G&M—
3–33
ᮣ
Section 3.3 Quadratic Functions and Applications
313
MAINTAINING YOUR SKILLS
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
Quadratic Functions and Applications
LEARNING OBJECTIVES
In Section 3.3 you will see how we can:
A. Graph quadratic functions
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
B. Graph quadratic functions
using the vertex formula
C. Find the equation of a
quadratic function from
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.
cob19545_ch03_314-328.qxd
8/10/10
8:14 PM
Page 314
College Algebra G&M—
314
3–34
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
ᮣ
