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Modeling Using Variation
∗
OpenStax College
OpenStax College Precalculus
This work is produced by OpenStax-CNX and licensed under the
Creative Commons Attribution License 4.0
†
Abstract
In this section, you will:
•
•
•
Solve direct variation problems.
Solve inverse variation problems.
Solve problems involving joint variation.
A used-car company has just oered their best candidate, Nicole, a position in sales. The position oers
16% commission on her sales. Her earnings depend on the amount of her sales. For instance, if she sells
a vehicle for $4,600, she will earn $736. She wants to evaluate the oer, but she is not sure how. In this
section, we will look at relationships, such as this one, between earnings, sales, and commission rate.
1 Solving Direct Variation Problems
In the example above, Nicole's earnings can be found by multiplying her sales by her commission. The
formula e = 0.16s tells us her earnings, e, come from the product of 0.16, her commission, and the sale price
of the vehicle. If we create a table, we observe that as the sales price increases, the earnings increase as well,
which should be intuitive. See Table 1.
s,
sales prices
e = 0.16s
Interpretation
$4,600
e = 0.16 (4, 600) = 736
A sale of a $4,600 vehicle results in $736 earnings.
$9,200
e = 0.16 (9, 200) = 1, 472
A sale of a $9,200 vehicle results in $1472 earnings.
$18,400
e = 0.16 (18, 400) = 2, 944
A sale of a $18,400 vehicle results in $2944 earnings.
Table 1
Notice that earnings are a multiple of sales. As sales increase, earnings increase in a predictable way.
Double the sales of the vehicle from $4,600 to $9,200, and we double the earnings from $736 to $1,472. As
the input increases, the output increases as a multiple of the input. A relationship in which one quantity is a
constant multiplied by another quantity is called direct variation. Each variable in this type of relationship
varies directly with the other.
∗ Version
1.7: Feb 25, 2015 1:04 pm -0600
† http://creativecommons.org/licenses/by/4.0/
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Figure 1 represents the data for Nicole's potential earnings. We say that earnings vary directly with the
sales price of the car. The formula y = kxn is used for direct variation. The value k is a nonzero constant
greater than zero and is called the constant of variation. In this case, k = 0.16 and n = 1.
Figure 1
A General Note:
If x and y are related by an equation of the form
y = kxn
then we say that the relationship is
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direct variation and y varies directly with the nth power
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of x. In direct variation relationships, there is a nonzero constant ratio k = xyn , where k is called the
constant of variation, which help denes the relationship between the variables.
How To:
Given a description of a direct variation problem, solve for an unknown.
1.Identify the input, x,and the output, y.
2.Determine the constant of variation. You may need to divide y by the specied power of x to
determine the constant of variation.
3.Use the constant of variation to write an equation for the relationship.
4.Substitute known values into the equation to nd the unknown.
Example 1
Solving a Direct Variation Problem
The quantity y varies directly with the cube of x. If y = 25 when x = 2, nd y when x is 6.
Solution
The general formula for direct variation with a cube is y = kx3 . The constant can be found by
dividing y by the cube of x.
k=
=
=
y
x3
25
23
25
8
(2)
Now use the constant to write an equation that represents this relationship.
y=
25 3
x
8
(3)
Substitute x = 6 and solve for y.
y=
3
25
8 (6)
= 675
Analysis
The graph of this equation is a simple cubic, as shown in Figure 2.
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Figure 2
Q&A:
Do the graphs of all direct variation equations look like Example 1?
No. Direct variation equations are power functionsthey may be linear, quadratic, cubic, quartic,
radical, etc. But all of the graphs pass through (0, 0) .
Try It:
Exercise 2
(Solution on p. 22.)
The quantity y varies directly with the square of x. If y = 24 when x = 3, nd y when x is
4.
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2 Solving Inverse Variation Problems
Water temperature in an ocean varies inversely to the water's depth. Between the depths of 250 feet and 500
gives us the temperature in degrees Fahrenheit at a depth in feet below Earth's
feet, the formula T = 14,000
d
surface. Consider the Atlantic Ocean, which covers 22% of Earth's surface. At a certain location, at the
depth of 500 feet, the temperature may be 28 ◦ F.
If we create Table 2, we observe that, as the depth increases, the water temperature decreases.
d,
depth
500 ft
350 ft
250 ft
T =
14,000
14,000
500
14,000
350
14,000
250
d
Interpretation
= 28
At a depth of 500 ft, the water temperature is 28 ◦ F.
= 40
At a depth of 350 ft, the water temperature is 40 ◦ F.
= 56
At a depth of 250 ft, the water temperature is 56 ◦ F.
Table 2
We notice in the relationship between these variables that, as one quantity increases, the other decreases.
The two quantities are said to be inversely proportional and each term varies inversely with the other.
Inversely proportional relationships are also called inverse variations.
For our example, Figure 3 depicts the inverse variation. We say the water temperature varies inversely
with the depth of the water because, as the depth increases, the temperature decreases. The formula y = xk for
inverse variation in this case uses k = 14, 000.
Figure 3
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A General Note:
6
If x and y are related by an equation of the form
k
(5)
xn
where k is a nonzero constant, then we say that y varies inversely with the nth power of x. In
inversely proportional relationships, or inverse variations, there is a constant multiple k =
xn y.
y=
Example 2
Writing a Formula for an Inversely Proportional Relationship
A tourist plans to drive 100 miles. Find a formula for the time the trip will take as a function of
the speed the tourist drives.
Solution
Recall that multiplying speed by time gives distance. If we let t represent the drive time in hours,
and v represent the velocity (speed or rate) at which the tourist drives, then vt = distance. Because
the distance is xed at 100 miles, vt = 100. Solving this relationship for the time gives us our
function.
t (v) =
100
v
(6)
= 100v −1
We can see that the constant of variation is 100 and, although we can write the relationship using
the negative exponent, it is more common to see it written as a fraction.
How To:
Given a description of an indirect variation problem, solve for an unknown.
1.Identify the input, x, and the output, y.
2.Determine the constant of variation. You may need to multiply y by the specied power of x to
determine the constant of variation.
3.Use the constant of variation to write an equation for the relationship.
4.Substitute known values into the equation to nd the unknown.
Example 3
Solving an Inverse Variation Problem
A quantity y varies inversely with the cube of x. If y = 25 when x = 2, nd y when x is 6.
Solution
The general formula for inverse variation with a cube is y =
multiplying y by the cube of x.
k
x3 . The
constant can be found by
k = x3 y
= 23 · 25
= 200
Now we use the constant to write an equation that represents this relationship.
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y=
y=
k
x3 ,
200
x3
k = 200
y=
200
63
25
27
(8)
Substitute x = 6 and solve for y.
=
(9)
Analysis
The graph of this equation is a rational function, as shown in Figure 4.
Figure 4
Try It:
Exercise 5
(Solution on p. 22.)
A quantity y varies inversely with the square of x. If y = 8 when x = 3, nd y when x is 4.
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3 Solving Problems Involving Joint Variation
Many situations are more complicated than a basic direct variation or inverse variation model. One variable
often depends on multiple other variables. When a variable is dependent on the product or quotient of two
or more variables, this is called joint variation. For example, the cost of busing students for each school
trip varies with the number of students attending and the distance from the school. The variable c, cost,
varies jointly with the number of students, n, and the distance, d.
A General Note:
variables.
Joint variation occurs when a variable varies directly or inversely with multiple
For instance, if x varies directly with both y and z, we have x = kyz. If x varies directly with y and
inversely withz,we have x = ky
z . Notice that we only use one constant in a joint variation equation.
Example 4
Solving Problems Involving Joint Variation
A quantity x varies directly with the square of y and inversely with the cube root of z. If x =
6 when y = 2 and z = 8, nd x when y = 1 and z = 27.
Solution
Begin by writing an equation to show the relationship between the variables.
ky 2
x= √
3
z
(10)
Substitute x = 6, y = 2, and z = 8 to nd the value of the constant k.
2
k2
√
3
8
4k
2
6=
6=
(11)
3=k
Now we can substitute the value of the constant into the equation for the relationship.
3y 2
x= √
3
z
(12)
To nd x when y = 1 and z = 27, we will substitute values for y and z into our equation.
x=
3(1)2
√
3
27
(13)
=1
Try It:
Exercise 7
(Solution on p. 22.)
x varies directly with the square of y and inversely with z. If x = 40 when y = 4 and z =
2, nd x when y = 10 and z = 25.
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Media:
Access these online resources for additional instruction and practice with direct and
inverse variation.
• Direct Variation1
• Inverse Variation2
• Direct and Inverse Variation3
Visit this website4 for additional practice questions from Learningpod.
4 Key Equations
Direct variation
Inverse variation
y = k xn , k is a nonzero constant.
y = k xn ,k is a nonzero constant.
Table 3
5 Key Concepts
• A relationship where one quantity is a constant multiplied by another quantity is called direct variation.
See Example 1.
• Two variables that are directly proportional to one another will have a constant ratio.
• A relationship where one quantity is a constant divided by another quantity is called inverse variation.
See Example 2.
• Two variables that are inversely proportional to one another will have a constant multiple. See Example 3.
• In many problems, a variable varies directly or inversely with multiple variables. We call this type of
relationship joint variation. See Example 4.
6 Section Exercises
6.1 Verbal
Exercise 8
(Solution on p. 22.)
What is true of the appearance of graphs that reect a direct variation between two variables?
Exercise 9
If two variables vary inversely, what will an equation representing their relationship look like?
Exercise 10
Is there a limit to the number of variables that can jointly vary? Explain.
1 http://openstaxcollege.org/l/directvariation
2 http://openstaxcollege.org/l/inversevariatio
3 http://openstaxcollege.org/l/directinverse
4 http://openstaxcollege.org/l/PreCalcLPC03
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(Solution on p. 22.)
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6.2 Algebraic
For the following exercises, write an equation describing the relationship of the given variables.
Exercise 11
y varies directly as x and when x = 6, y = 12.
Exercise 12
y varies directly as the square of x and when x = 4, y = 80.
(Solution on p. 22.)
Exercise 13
y varies directly as the square root of x and when x = 36, y = 24.
Exercise 14
y varies directly as the cube of x and when x = 36, y = 24.
(Solution on p. 22.)
Exercise 15
y varies directly as the cube root of x and when x = 27, y = 15.
Exercise 16
y varies directly as the fourth power of x and when x = 1, y = 6.
(Solution on p. 22.)
Exercise 17
y varies inversely as x and when x = 4, y = 2.
Exercise 18
y varies inversely as the square of x and when x = 3, y = 2.
(Solution on p. 22.)
Exercise 19
y varies inversely as the cube of x and when x = 2, y = 5.
Exercise 20
y varies inversely as the fourth power of x and when x = 3, y = 1.
(Solution on p. 22.)
Exercise 21
y varies inversely as the square root of x and when x = 25, y = 3.
Exercise 22
y varies inversely as the cube root of x and when x = 64, y = 5.
(Solution on p. 22.)
Exercise 23
y varies jointly with x and z and when x = 2 and z = 3, y = 36.
Exercise 24
y varies jointly as x, z, and w and when x = 1, z = 2, w = 5, then y = 100.
(Solution on p. 22.)
Exercise 25
y varies jointly as the square of x and the square of z and when x = 3 and z = 4, then y = 72.
Exercise 26
(Solution on p. 22.)
y varies jointly as x and the square root of z and when x = 2 and z = 25, then y = 100.
Exercise 27
y varies jointly as the square of xthe cube of z and the square root of W. When x = 1, z = 2, and w =
36, then y = 48.
Exercise 28
(Solution on p. 22.)
y varies jointly as x and z and inversely as w. When x = 3, z = 5, and w = 6, then y = 10.
Exercise 29
y varies jointly as the square of x and the square root of z and inversely as the cube of w. When x =
3, z = 4, and w = 3, then y = 6.
Exercise 30
(Solution on p. 22.)
y varies jointly as x and z and inversely as the square root of w and the square of t .When x = 3, z =
1, w = 25, and t = 2, then y = 6.
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6.3 Numeric
For the following exercises, use the given information to nd the unknown value.
Exercise 31
y varies directly as x. When x = 3, then y = 12. Find y when x = 20.
Exercise 32
(Solution on p. 22.)
y varies directly as the square of x. When x = 2, then y = 16. Find y when x = 8.
Exercise 33
y varies directly as the cube of x. When x = 3, then y = 5. Find y when x = 4.
Exercise 34
(Solution on p. 22.)
y varies directly as the square root of x. When x = 16, then y = 4. Find y when x = 36.
Exercise 35
y varies directly as the cube root of x. When x = 125, then y = 15. Find y when x = 1, 000.
Exercise 36
y varies inversely with x. When x = 3, then y = 2. Find y when x = 1.
(Solution on p. 22.)
Exercise 37
y varies inversely with the square of x. When x = 4, then y = 3. Find y when x = 2.
Exercise 38
(Solution on p. 22.)
y varies inversely with the cube of x. When x = 3, then y = 1. Find y when x = 1.
Exercise 39
y varies inversely with the square root of x. When x = 64, then y = 12. Find y when x = 36.
Exercise 40
(Solution on p. 22.)
y varies inversely with the cube root of x. When x = 27, then y = 5. Find y when x = 125.
Exercise 41
y varies jointly as x and z.When x = 4 and z = 2, then y = 16. Find y when x = 3 and z = 3.
Exercise 42
(Solution on p. 22.)
y varies jointly as x, z, and w.When x = 2, z = 1, and w = 12, then y = 72. Find y when x = 1, z =
2, and w = 3.
Exercise 43
y varies jointly as x and the square of z. When x = 2 and z = 4, then y = 144. Find y when x =
4 and z = 5.
Exercise 44
(Solution on p. 22.)
y varies jointly as the square of x and the square root of z. When x = 2 and z = 9, then y =
24. Find y when x = 3 and z = 25.
Exercise 45
y varies jointly as x and z and inversely as w. When x = 5, z = 2, and w = 20, theny = 4. Find y when x =
3 and z = 8, and w = 48.
Exercise 46
(Solution on p. 22.)
y varies jointly as the square of x and the cube of z and inversely as the square root of w. When x =
2, z = 2, and w = 64, then y = 12. Find y when x = 1, z = 3, and w = 4.
Exercise 47
y varies jointly as the square of x and of z and inversely as the square root of w and of t .When x =
2, z = 3, w = 16, and t = 3, then y = 1. Find y when x = 3, z = 2, w = 36, and t = 5.
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6.4 Technology
For the following exercises, use a calculator to graph the equation implied by the given variation.
Exercise 48
y varies directly with the square of x and when x = 2, y = 3.
(Solution on p. 22.)
Exercise 49
y varies directly as the cube of x and when x = 2, y = 4.
Exercise 50
y varies directly as the square root of x and when x = 36, y = 2.
(Solution on p. 23.)
Exercise 51
y varies inversely with x and when x = 6, y = 2.
Exercise 52
y varies inversely as the square of x and when x = 1, y = 4.
(Solution on p. 23.)
6.5 Extensions
For the following exercises, use Kepler's Law, which states that the square of the time, T, required for a
planet to orbit the Sun varies directly with the cube of the mean distance, a, that the planet is from the Sun.
Exercise 53
Using the Earth's time of 1 year and mean distance of 93 million miles, nd the equation relating
T and a.
Exercise 54
(Solution on p. 24.)
Use the result from the previous exercise to determine the time required for Mars to orbit the Sun
if its mean distance is 142 million miles.
Exercise 55
Using Earth's distance of 150 million kilometers, nd the equation relating T and a.
Exercise 56
(Solution on p. 24.)
Use the result from the previous exercise to determine the time required for Venus to orbit the
Sun if its mean distance is 108 million kilometers.
Exercise 57
Using Earth's distance of 1 astronomical unit (A.U.), determine the time for Saturn to orbit the
Sun if its mean distance is 9.54 A.U.
6.6 Real-World Applications
For the following exercises, use the given information to answer the questions.
Exercise 58
(Solution on p. 24.)
The distance s that an object falls varies directly with the square of the time, t, of the fall. If an
object falls 16 feet in one second, how long for it to fall 144 feet?
Exercise 59
The velocity v of a falling object varies directly to the time, t, of the fall. If after 2 seconds, the
velocity of the object is 64 feet per second, what is the velocity after 5 seconds?
Exercise 60
(Solution on p. 24.)
The rate of vibration of a string under constant tension varies inversely with the length of the
string. If a string is 24 inches long and vibrates 128 times per second, what is the length of a string
that vibrates 64 times per second?
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Exercise 61
The volume of a gas held at constant temperature varies indirectly as the pressure of the gas. If
the volume of a gas is 1200 cubic centimeters when the pressure is 200 millimeters of mercury, what
is the volume when the pressure is 300 millimeters of mercury?
Exercise 62
(Solution on p. 24.)
The weight of an object above the surface of the Earth varies inversely with the square of the
distance from the center of the Earth. If a body weighs 50 pounds when it is 3960 miles from
Earth's center, what would it weigh it were 3970 miles from Earth's center?
Exercise 63
The intensity of light measured in foot-candles varies inversely with the square of the distance
from the light source. Suppose the intensity of a light bulb is 0.08 foot-candles at a distance of 3
meters. Find the intensity level at 8 meters.
Exercise 64
(Solution on p. 24.)
The current in a circuit varies inversely with its resistance measured in ohms. When the current
in a circuit is 40 amperes, the resistance is 10 ohms. Find the current if the resistance is 12 ohms.
Exercise 65
The force exerted by the wind on a plane surface varies jointly with the square of the velocity of
the wind and with the area of the plane surface. If the area of the surface is 40 square feet surface
and the wind velocity is 20 miles per hour, the resulting force is 15 pounds. Find the force on a
surface of 65 square feet with a velocity of 30 miles per hour.
Exercise 66
(Solution on p. 24.)
The horsepower (hp) that a shaft can safely transmit varies jointly with its speed (in revolutions
per minute (rpm) and the cube of the diameter. If the shaft of a certain material 3 inches in
diameter can transmit 45 hp at 100 rpm, what must the diameter be in order to transmit 60 hp at
150 rpm?
Exercise 67
The kinetic energy K of a moving object varies jointly with its mass m and the square of its
velocity v. If an object weighing 40 kilograms with a velocity of 15 meters per second has a kinetic energy of 1000 joules, nd the kinetic energy if the velocity is increased to 20 meters per
second.
7 Chapter Review Exercises
You have reached the end of Chapter 3: Polynomial and Rational Functions. Let's review some of the Key
Terms, Concepts and Equations you have learned.
7.1 Complex Numbers
Perform the indicated operation with complex numbers.
Exercise 68
(Solution on p. 24.)
(4 + 3i) + (−2 − 5i)
Exercise 69
(6 − 5i) − (10 + 3i)
Exercise 70
(2 − 3i) (3 + 6i)
Exercise 71
2−i
2+i
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(Solution on p. 24.)
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Solve the following equations over the complex number system.
Exercise 72
(Solution on p. 24.)
x2 − 4x + 5 = 0
Exercise 73
x2 + 2x + 10 = 0
7.2 Quadratic Functions
For the following exercises, write the quadratic function in standard form. Then, give the vertex and axes
intercepts. Finally, graph the function.
Exercise 74
(Solution on p. 24.)
f (x) = x2 − 4x − 5
Exercise 75
f (x) = −2x2 − 4x
For the following problems, nd the equation of the quadratic function using the given information.
Exercise 76
The vertex is(− − 2, 3)and a point on the graph is (3, 6) .
(Solution on p. 25.)
Exercise 77
The vertex is (− − 3, 6.5) and a point on the graph is (2, 6) .
Answer the following questions.
Exercise 78
(Solution on p. 25.)
A rectangular plot of land is to be enclosed by fencing. One side is along a river and so needs no
fence. If the total fencing available is 600 meters, nd the dimensions of the plot to have maximum
area.
Exercise 79
An object projected from the ground at a 45 degree angle with initial velocity of 120 feet per
−32
2
second has height, h, in terms of horizontal distance traveled, x, given by h (x) = (120)
2 x + x. Find
the maximum height the object attains.
7.3 Power Functions and Polynomial Functions
For the following exercises, determine if the function is a polynomial function and, if so, give the degree and
leading coecient.
Exercise 80
(Solution on p. 25.)
f (x) = 4x5 − 3x3 + 2x − 1
Exercise 81
f (x) = 5x+1 − x2
Exercise 82
f (x) = x2 3 − 6x + x2
(Solution on p. 25.)
For the following exercises, determine end behavior of the polynomial function.
Exercise 83
f (x) = 2x4 + 3x3 − 5x2 + 7
Exercise 84
f (x) = 4x3 − 6x2 + 2
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(Solution on p. 25.)
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Exercise 85
f (x) = 2x2 1 + 3x − x2
15
7.4 Graphs of Polynomial Functions
For the following exercises, nd all zeros of the polynomial function, noting multiplicities.
Exercise 86
2
f (x) = (x + 3) (2x − 1) (x + 1)
3
(Solution on p. 25.)
Exercise 87
f (x) = x5 + 4x4 + 4x3
Exercise 88
(Solution on p. 25.)
f (x) = x3 − 4x2 + x − 4
For the following exercises, based on the given graph, determine the zeros of the function and note multiplicity.
Exercise 89
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Exercise 90
16
(Solution on p. 25.)
Exercise 91
Use the Intermediate Value Theorem to show that at least one zero lies between 2 and 3 for the
function f (x) = x3 − 5x + 1
7.5 Dividing Polynomials
For the following exercises, use long division to nd the quotient and remainder.
Exercise 92
(Solution on p. 25.)
x3 −2x2 +4x+4
x−2
Exercise 93
3x4 −4x2 +4x+8
x+1
For the following exercises, use synthetic division to nd the quotient. If the divisor is a factor, then write
the factored form.
Exercise 94
x3 −2x2 +5x−1
x+3
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(Solution on p. 25.)
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Exercise 95
x3 +4x+10
x−3
Exercise 96
(Solution on p. 26.)
2x3 +6x2 −11x−12
x+4
Exercise 97
3x4 +3x3 +2x+2
x+1
7.6 Zeros of Polynomial Functions
For the following exercises, use the Rational Zero Theorem to help you solve the polynomial equation.
Exercise 98
(Solution on p. 26.)
2x3 − 3x2 − 18x − 8 = 0
Exercise 99
3x3 + 11x2 + 8x − 4 = 0
Exercise 100
(Solution on p. 26.)
2x4 − 17x3 + 46x2 − 43x + 12 = 0
Exercise 101
4x4 + 8x3 + 19x2 + 32x + 12 = 0
For the following exercises, use Descartes' Rule of Signs to nd the possible number of positive and negative
solutions.
Exercise 102
(Solution on p. 26.)
x3 − 3x2 − 2x + 4 = 0
Exercise 103
2x4 − x3 + 4x2 − 5x + 1 = 0
7.7 Rational Functions
For the following rational functions, nd the intercepts and the vertical and horizontal asymptotes, and then
use them to sketch a graph.
Exercise 104
f (x) =
(Solution on p. 26.)
x+2
x−5
Exercise 105
f (x) =
x2 +1
x2 −4
Exercise 106
f (x) =
(Solution on p. 26.)
3x2 −27
x2 +x−2
Exercise 107
f (x) =
x+2
x2 −9
For the following exercises, nd the slant asymptote.
Exercise 108
f (x) =
x2 −1
x+2
Exercise 109
f (x) =
2x3 −x2 +4
x2 +1
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(Solution on p. 27.)
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7.8 Inverses and Radical Functions
For the following exercises, nd the inverse of the function with the domain given.
Exercise 110
2
(Solution on p. 27.)
f (x) = (x − 2) , x ≥ 2
Exercise 111
2
f (x) = (x + 4) − 3, x ≥ −4
Exercise 112
(Solution on p. 27.)
f (x) = x2 + 6x − 2, x ≥ −3
Exercise 113
f (x) = 2x3 − 3
Exercise√114
f (x) =
(Solution on p. 27.)
4x + 5 − 3
Exercise 115
f (x) =
x−3
2x+1
7.9 Modeling Using Variation (Section )
For the following exercises, nd the unknown value.
Exercise 116
y varies directly as the square of x. If when x = 3, y = 36, nd y if x = 4.
(Solution on p. 27.)
Exercise 117
y varies inversely as the square root of x If when x = 25, y = 2, nd y if x = 4.
Exercise 118
(Solution on p. 27.)
y varies jointly as the cube of x and as z. If when x = 1 and z = 2, y = 6, nd y if x = 2 and z = 3.
Exercise 119
y varies jointly as x and the square of z and inversely as the cube of w. If when x = 3, z = 4, and w =
2, y = 48, nd y if x = 4, z = 5, and w = 3.
For the following exercises, solve the application problem.
Exercise 120
(Solution on p. 27.)
The weight of an object above the surface of the earth varies inversely with the distance from the
center of the earth. If a person weighs 150 pounds when he is on the surface of the earth (3,960
miles from center), nd the weight of the person if he is 20 miles above the surface.
Exercise 121
The volume V of an ideal gas varies directly with the temperature T and inversely with the pressure
P. A cylinder contains oxygen at a temperature of 310 degrees K and a pressure of 18 atmospheres in
a volume of 120 liters. Find the pressure if the volume is decreased to 100 liters and the temperature
is increased to 320 degrees K.
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8 Chapter Test
Perform the indicated operation or solve the equation.
Exercise 122
(Solution on p. 27.)
(3 − 4i) (4 + 2i)
Exercise 123
1−4i
3+4i
Exercise 124
(Solution on p. 27.)
x2 − 4x + 13 = 0
Give the degree and leading coecient of the following polynomial function.
Exercise 125
f (x) = x3 3 − 6x2 − 2x2
Determine the end behavior of the polynomial function.
Exercise 126
(Solution on p. 27.)
f (x) = 8x3 − 3x2 + 2x − 4
Exercise 127
f (x) = −2x2 4 − 3x − 5x2
Write the quadratic function in standard form. Determine the vertex and axes intercepts and graph the
function.
Exercise 128
(Solution on p. 27.)
f (x) = x2 + 2x − 8
Given information about the graph of a quadratic function, nd its equation.
Exercise 129
Vertex (2, 0) and point on graph (4, 12) .
Solve the following application problem.
Exercise 130
(Solution on p. 28.)
A rectangular eld is to be enclosed by fencing. In addition to the enclosing fence, another fence is
to divide the eld into two parts, running parallel to two sides. If 1,200 feet of fencing is available,
nd the maximum area that can be enclosed.
Find all zeros of the following polynomial functions, noting multiplicities.
Exercise 131
3
f (x) = (x − 3) (3x − 1) (x − 1)
2
Exercise 132
f (x) = 2x6 − 6x5 + 18x4
Based on the graph, determine the zeros of the function and multiplicities.
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(Solution on p. 28.)
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Exercise 133
Use long division to nd the quotient.
Exercise 134
(Solution on p. 28.)
2x3 +3x−4
x+2
Use synthetic division to nd the quotient. If the divisor is a factor, write the factored form.
Exercise 135
x4 +3x2 −4
x−2
Exercise 136
(Solution on p. 28.)
2x3 +5x2 −7x−12
x+3
Use the Rational Zero Theorem to help you nd the zeros of the polynomial functions.
Exercise 137
f (x) = 2x3 + 5x2 − 6x − 9
Exercise 138
(Solution on p. 28.)
f (x) = 4x4 + 8x3 + 21x2 + 17x + 4
Exercise 139
f (x) = 4x4 + 16x3 + 13x2 − 15x − 18
Exercise 140
(Solution on p. 28.)
f (x) = x5 + 6x4 + 13x3 + 14x2 + 12x + 8
Given the following information about a polynomial function, nd the function.
Exercise 141
y
It has a double zero at x = 3 and zeroes at x = 1 and x = −2 . It's -intercept is (0, 12) .
Exercise 142
It has a zero of multiplicity 3 at x =
(Solution on p. 28.)
1
2
and another zero at x = −3 . It contains the point (1, 8) .
Use Descartes' Rule of Signs to determine the possible number of positive and negative solutions.
Exercise 143
8x3 − 21x2 + 6 = 0
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For the following rational functions, nd the intercepts and horizontal and vertical asymptotes, and sketch
a graph.
Exercise 144
f (x) =
(Solution on p. 28.)
x+4
x2 −2x−3
Exercise 145
f (x) =
x2 +2x−3
x2 −4
Find the slant asymptote of the rational function.
Exercise 146
f (x) =
(Solution on p. 29.)
x2 +3x−3
x−1
Find the inverse of the function.
Exercise√147
f (x) =
x−2+4
Exercise 148
(Solution on p. 29.)
f (x) = 3x3 − 4
Exercise 149
f (x) =
2x+3
3x−1
Find the unknown value.
Exercise 150
y varies inversely as the square of x and when x = 3, y = 2. Find y if x = 1.
(Solution on p. 29.)
Exercise 151
y varies jointly with x and the cube root of z. If when x = 2 and z = 27, y = 12, nd y if x =
5 and z = 8.
Solve the following application problem.
Exercise 152
(Solution on p. 29.)
The distance a body falls varies directly as the square of the time it falls. If an object falls 64 feet
in 2 seconds, how long will it take to fall 256 feet?
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Solutions to Exercises in this Module
Solution to Exercise (p. 4)
128
3
Solution to Exercise (p. 7)
9
2
Solution to Exercise (p. 8)
x = 20
Solution to Exercise (p. 9)
The graph will have the appearance of a power function.
Solution to Exercise (p. 9)
No. Multiple variables may jointly vary.
Solution to Exercise (p. 10)
y = 5x2
Solution to Exercise (p. 10)
y = 10x3
Solution to Exercise (p. 10)
y = 6x4
Solution to Exercise (p. 10)
y=
18
x2
y=
81
x4
y=
20
√
3 x
Solution to Exercise (p. 10)
Solution to Exercise (p. 10)
Solution to Exercise (p. 10)
y = 10xzw
Solution√to Exercise (p. 10)
y = 10x z
Solution to Exercise (p. 10)
y = 4 xz
w
Solution to Exercise (p. 10)
y = 40 √xz
wt2
Solution to Exercise (p. 11)
y = 256
Solution to Exercise (p. 11)
y=6
Solution to Exercise (p. 11)
y=6
Solution to Exercise (p. 11)
y = 27
Solution to Exercise (p. 11)
y=3
Solution to Exercise (p. 11)
y = 18
Solution to Exercise (p. 11)
y = 90
Solution to Exercise (p. 11)
y=
81
2
Solution to Exercise (p. 12)
y = 34 x2
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Solution to Ex-
ercise√(p. 12)
y=
1
3
y=
4
x2
x
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Solution to Exercise (p. 12)
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Solution to Ex-
ercise (p. 12)
1.89 years
Solution to Exercise (p. 12)
0.61 years
Solution to Exercise (p. 12)
3 seconds
Solution to Exercise (p. 12)
48 inches
Solution to Exercise (p. 13)
49.75 pounds
Solution to Exercise (p. 13)
33.33 amperes
Solution to Exercise (p. 13)
2.88 inches
Solution to Exercise (p. 13)
2 − 2i
Solution to Exercise (p. 13)
24 + 3i
Solution to Exercise (p. 14)
{2 + i, 2 − i}
Solution to Exercise (p. 14)
2
f (x) = (x − 2) − 9 vertex (2, − − 9) , intercepts (5, 0) ; (− − 1, 0) ; (0, − − 5)
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Solution to Exercise (p. 14)
f (x) =
3
25 (x
2
+ 2) + 3
Solution to Exercise (p. 14)
300 meters by 150 meters, the longer side parallel to river.
Solution to Exercise (p. 14)
Yes, degree = 5, leading coecient = 4
Solution to Exercise (p. 14)
Yes, degree = 4, leading coecient = 1
Solution to Exercise (p. 14)
As x → −∞, f (x) → −∞, as x → ∞, f (x) → ∞
Solution to Exercise (p. 15)
3 with multiplicity 2,− 12 with multiplicity 1, 1 with multiplicity 3
Solution to Exercise (p. 15)
4 with multiplicity 1
Solution to Exercise (p. 16)
1
2
with multiplicity 1, 3 with multiplicity 3
Solution to Exercise (p. 16)
x2 + 4 with remainder 12
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Solution to Exercise (p. 16)
x2 − 5x + 20 −
61
x+3
Solution to Exercise (p. 17)
2x2 − 2x − 3, so factored form is (x + 4) 2x2 − 2x − 3
Solution to Exercise (p. 17)
{−2, 4, − 12 }
Solution to Exercise (p. 17)
{1, 3, 4,
1
2}
Solution to Exercise (p. 17)
0 or 2 positive, 1 negative
Solution to Exercise (p. 17)
Intercepts (− − 2, 0) and 0, − 25 , Asymptotes x = 5 and y = 1.
Solution to Exercise (p. 17)
, Asymptotes x = 1, x = − − 2, y = 3.
Intercepts (3, 0), (-3, 0), and 0, 27
2
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Solution to Exercise (p. 17)
y = x−2
Solution to
√ Exercise (p. 18)
f −1 (x) =
x+2
f −1 (x) =
x + 11 − 3
Solution to
√ Exercise (p. 18)
Solution to Exercise (p. 18)
f −1 (x) =
(x+3)2 −5
,
4
x ≥ −3
Solution to Exercise (p. 18)
y = 64
Solution to Exercise (p. 18)
y = 72
Solution to Exercise (p. 18)
148.5 pounds
Solution to Exercise (p. 19)
20 − 10i
Solution to Exercise (p. 19)
{2 + 3i, 2 − 3i}
Solution to Exercise (p. 19)
As x → −∞, f (x) → −∞, as x → ∞, f (x) → ∞
Solution to Exercise (p. 19)
2
f (x) = (x + 1) − 9, vertex (−1, −9), intercepts (2, 0) ; (−4, 0) ; (0, −8)
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Solution to Exercise (p. 19)
60,000 square feet
Solution to Exercise (p. 19)
0 with multiplicity 4, 3 with multiplicity 2
Solution to Exercise (p. 20)
2x2 − 4x + 11 −
26
x+2
Solution to Exercise (p. 20)
2x2 − x − 4. So factored form is (x + 3) 2x2 − x − 4
Solution to Exercise (p. 20)
√
− 21 (has multiplicity 2), −1±i2
15
Solution to Exercise (p. 20)
− 2 (has multiplicity 3), ± i
Solution to Exercise (p. 20)
3
f (x) = 2(2x − 1) (x + 3)
Solution to Exercise (p. 21)
Intercepts (−4, 0) , 0, − 43 , Asymptotes x = 3, x = −1, y = 0.
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OpenStax-CNX module: m49353
Solution to Exercise (p. 21)
y =x+4
Solution to
qExercise (p. 21)
f −1 (x) =
3
x+4
3
Solution to Exercise (p. 21)
y = 18
Solution to Exercise (p. 21)
4 seconds
Glossary
Denition 1: constant of variation
the non-zero value k that helps dene the relationship between variables in direct or inverse variation
Denition 2: direct variation
the relationship between two variables that are a constant multiple of each other; as one quantity
increases, so does the other
Denition 3: inverse variation
the relationship between two variables in which the product of the variables is a constant
Denition 4: inversely proportional
a relationship where one quantity is a constant divided by the other quantity; as one quantity
increases, the other decreases
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Denition 5: joint variation
a relationship where a variable varies directly or inversely with multiple variables
Denition 6: varies directly
a relationship where one quantity is a constant multiplied by the other quantity
Denition 7: varies inversely
a relationship where one quantity is a constant divided by the other quantity
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