Modeling Using Variation

A pre-owned car dealer has just offered their best candidate, Nicole, a position in sales. The position offers 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. As she considers the offer, she takes into account the typical price of the dealer's cars, the overall market, and how many she can reasonably expect to sell. In this section, we will look at relationships, such as this one, between earnings, sales, and commission rate.

Solving Direct Variation Problems

In the example above, Nicole’s earnings can be found by multiplying her sales by her commission. The formula $e = 0.16 s$ 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.

Table 1 ..
$s$ , sales prices	$e = 0.16 s$	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.

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.

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 $y = k x^{n}$ 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 $n = 1.$

Graph of y=(0.16)x where the horizontal axis is labeled, “s, Sales Price in Dollars”, and the vertical axis is labeled, “e, Earnings, $”. — Figure 1

Example 1

Solving a Direct Variation Problem

The quantity $y$ varies directly with the cube of $x .$ If $y = 25$ when $x = 2,$ find $y$ when $x$ is 6.

Solution

The general formula for direct variation with a cube is $y = k x^{3} .$ The constant can be found by dividing $y$ by the cube of $x .$

\begin{matrix} \begin{array}{l} k = \frac{y}{x^{3}} \end{array} \\ = \frac{25}{2^{3}} \\ = \frac{25}{8} \end{matrix}

Now use the constant to write an equation that represents this relationship.

y = \frac{25}{8} x^{3}

Substitute $x = 6$ and solve for $y .$

\begin{array}{l} y = \frac{25}{8} {(6)}^{3} \\ = 675 \end{array}

Analysis

The graph of this equation is a simple cubic, as shown in Figure 2.

Graph of y=25/8(x^3) with the labeled points (2, 25) and (6, 675). — Figure 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 feet, the formula $T = \frac{14,000}{d}$ gives us the temperature in degrees Fahrenheit at a depth in feet below Earth’s 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.

Table 2 ..
$d,$ depth	$T = \frac{14,000}{d}$	Interpretation
500 ft	$\frac{14,000}{500} = 28$	At a depth of 500 ft, the water temperature is 28° F.
350 ft	$\frac{14,000}{350} = 40$	At a depth of 350 ft, the water temperature is 40° F.
250 ft	$\frac{14,000}{250} = 56$	At a depth of 250 ft, the water temperature is 56° F.

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 $y = \frac{k}{x}$ for inverse variation in this case uses $k = 14,000.$

Graph of y=(14000)/x where the horizontal axis is labeled, “Depth, d (ft)”, and the vertical axis is labeled, “Temperature, T (Degrees Fahrenheit)”. — Figure 3

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 $v t = distance .$ Because the distance is fixed at 100 miles, $v t = 100.$ Solving this relationship for the time gives us our function.

\begin{array}{l} t (v) = \frac{100}{v} \\ = 100 v^{- 1} \end{array}

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.

Example 3

Solving an Inverse Variation Problem

A quantity $y$ varies inversely with the cube of $x .$ If $y = 25$ when $x = 2,$ find $y$ when $x$ is 6.

Solution

The general formula for inverse variation with a cube is $y = \frac{k}{x^{3}} .$ The constant can be found by multiplying $y$ by the cube of $x .$

\begin{array}{l} k = x^{3} y \\ = 2^{3} \cdot 25 \\ = 200 \end{array}

Now we use the constant to write an equation that represents this relationship.

\begin{array}{l} y = \frac{k}{x^{3}}, k = 200 \\ y = \frac{200}{x^{3}} \end{array}

Substitute $x = 6$ and solve for $y .$

\begin{array}{l} y = \frac{200}{6^{3}} \\ = \frac{25}{27} \end{array}

Analysis

The graph of this equation is a rational function, as shown in Figure 4.

Graph of y=25/(x^3) with the labeled points (2, 25) and (6, 25/27). — Figure 4

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, $d .$

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 $z = 8,$ find $x$ when $y = 1$ and $z = 27.$

Solution

Begin by writing an equation to show the relationship between the variables.

x = \frac{k y^{2}}{\sqrt[3]{z}}

Substitute $x = 6,$ $y = 2,$ and $z = 8$ to find the value of the constant $k .$

\begin{array}{l} \begin{array}{l} 6 = \frac{k 2^{2}}{\sqrt[3]{8}} \end{array} \\ 6 = \frac{4 k}{2} \\ 3 = k \end{array}

Now we can substitute the value of the constant into the equation for the relationship.

x = \frac{3 y^{2}}{\sqrt[3]{z}}

To find $x$ when $y = 1$ and $z = 27,$ we will substitute values for $y$ and $z$ into our equation.

\begin{array}{l} x = \frac{3 {(1)}^{2}}{\sqrt[3]{27}} \\ = 1 \end{array}

Key Equations

..
Direct variation	$y = k x^{n}, k is a nonzero constant .$
Inverse variation	$y = \frac{k}{x^{n}}, k is a nonzero constant .$

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.

Section Exercises

Verbal

Exercise 1

What is true of the appearance of graphs that reflect a direct variation between two variables?

Solution

The graph will have the appearance of a power function.

Exercise 2

If two variables vary inversely, what will an equation representing their relationship look like?

Exercise 3

Is there a limit to the number of variables that can jointly vary? Explain.

Solution

No. Multiple variables may jointly vary.

Algebraic

For the following exercises, write an equation describing the relationship of the given variables.

Exercise 4

$y$ varies directly as $x$ and when $x = 6, y = 12.$

Exercise 5

$y$ varies directly as the square of $x$ and when $x = 4, y = 80 .$

Solution

$y = 5 x^{2}$

Exercise 6

$y$ varies directly as the square root of $x$ and when $x = 36, y = 24.$

Exercise 7

$y$ varies directly as the cube of $x$ and when $x = 36, y = 24.$

Solution

$y = \frac{1}{1944} x^{3}$

Exercise 8

$y$ varies directly as the cube root of $x$ and when $x = 27, y = 15.$

Exercise 9

$y$ varies directly as the fourth power of $x$ and when $x = 1, y = 6.$

Solution

$y = 6 x^{4}$

Exercise 10

$y$ varies inversely as $x$ and when $x = 4, y = 2.$

Exercise 11

$y$ varies inversely as the square of $x$ and when $x = 3, y = 2.$

Solution

$y = \frac{18}{x^{2}}$

Exercise 12

$y$ varies inversely as the cube of $x$ and when $x = 2, y = 5.$

Exercise 13

$y$ varies inversely as the fourth power of $x$ and when $x = 3, y = 1.$

Solution

$y = \frac{81}{x^{4}}$

Exercise 14

$y$ varies inversely as the square root of $x$ and when $x = 25, y = 3.$

Exercise 15

$y$ varies inversely as the cube root of $x$ and when $x = 64, y = 5.$

Solution

$y = \frac{20}{\sqrt[3]{x}}$

Exercise 16

$y$ varies jointly with $x$ and $z$ and when $x = 2$ and $z = 3$ , $y = 36.$

Exercise 17

$y$ varies jointly as $x$ , $z$ , and $w$ and when $x = 1$ , $z = 2$ , $w = 5$ , then $y = 100.$

Solution

$y = 10 x z w$

Exercise 18

$y$ varies jointly as the square of $x$ and the square of $z$ and when $x = 3$ and $z = 4$ , then $y = 72.$

Exercise 19

$y$ varies jointly as $x$ and the square root of $z$ and when $x = 2$ and $z = 25$ , then $y = 100.$

Solution

$y = 10 x \sqrt{z}$

Exercise 20

$y$ varies jointly as the square of $x$ the cube of $z$ and the square root of $w .$ When $x = 1$ , $z = 2$ , and $w = 36, then y = 48.$

Exercise 21

$y$ varies jointly as $x$ and $z$ and inversely as $w .$ When $x = 3$ , $z = 5$ , and $w = 6$ , then $y = 10.$

Solution

$y = 4 \frac{x z}{w}$

Exercise 22

$y$ 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 23

$y$ 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 $y = 6.$

Solution

$y = 40 \frac{x z}{\sqrt{w} t^{2}}$

Numeric

For the following exercises, use the given information to find the unknown value.

Exercise 24

$y$ varies directly as $x$ . When $x = 3$ , then $y = 12$ . Find $y$ when $x = 20.$

Exercise 25

$y$ varies directly as the square of $x$ . When $x = 2$ , then $y = 16$ . Find $y$ when $x = 8$ .

Solution

$y = 256$

Exercise 26

$y$ varies directly as the cube of $x$ . When $x = 3$ , then $y = 5$ . Find $y$ when $x = 4$ .

Exercise 27

$y$ varies directly as the square root of $x .$ When $x = 16$ , then $y = 4$ . Find $y when x = 36$ .

Solution

$y = 6$

Exercise 28

$y$ varies directly as the cube root of $x .$ When $x = 125$ , then $y = 15$ . Find $y$ when $x = 1$ , $000.$

Exercise 29

$y$ varies inversely with $x .$ When $x = 3$ , then $y = 2$ . Find $y$ when $x = 1$ .

Solution

$y = 6$

Exercise 30

$y$ varies inversely with the square of $x$ . When $x = 4$ , then $y = 3$ . Find $y$ when $x = 2$ .

Exercise 31

$y$ varies inversely with the cube of $x .$ When $x = 3$ , then $y = 1$ . Find $y$ when $x = 1$ .

Solution

$y = 27$

Exercise 32

$y$ varies inversely with the square root of $x .$ When $x = 64,$ then $y = 12.$ Find $y$ when $x = 36.$

Exercise 33

$y$ varies inversely with the cube root of $x .$ When $x = 27,$ then $y = 5.$ Find $y$ when $x = 125.$

Solution

$y = 3$

Exercise 34

$y$ varies jointly as $x and z .$ When $x = 4$ and $z = 2,$ then $y = 16.$ Find $y$ when $x = 3$ and $z = 3.$

Exercise 35

$y$ 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 $w = 3.$

Solution

$y = 18$

Exercise 36

$y$ 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 37

$y$ 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 $z = 25.$

Solution

$y = 90$

Exercise 38

$y$ varies jointly as $x$ and $z$ and inversely as $w .$ When $x = 5,$ $z = 2,$ and $w = 20,$ then $y = 4.$ Find $y$ when $x = 3$ and $z = 8,$ and $w = 48.$

Exercise 39

$y$ 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 $w = 4.$

Solution

$y = \frac{81}{2}$

Exercise 40

$y$ 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 $t = 5.$

Technology

For the following exercises, use a calculator to graph the equation implied by the given variation.

Exercise 41

$y$ varies directly with the square of $x$ and when $x = 2, y = 3.$

Solution

$y = \frac{3}{4} x^{2}$

Exercise 42

$y$ varies directly as the cube of $x$ and when $x = 2, y = 4.$

Exercise 43

$y$ varies directly as the square root of $x$ and when $x = 36, y = 2.$

Solution

$y = \frac{1}{3} \sqrt{x}$

Exercise 44

$y$ varies inversely with $x$ and when $x = 6, y = 2.$

Exercise 45

$y$ varies inversely as the square of $x$ and when $x = 1, y = 4.$

Solution

$y = \frac{4}{x^{2}}$

A graph of a function with a vertical dashed asymptote at x=0 and a horizontal dashed asymptote at y=0. The blue curve is symmetric about the y-axis, approaching infinity near x=0 and zero as x extends horizontally.

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 46

Using the Earth’s time of 1 year and mean distance of 93 million miles, find the equation relating $T$ and $a .$

Exercise 47

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.

Solution

≈ 1.89 years

Exercise 48

Using Earth’s distance of 150 million kilometers, find the equation relating $T$ and $a.$

Exercise 49

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.

Solution

≈ 0.61 years

Exercise 50

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.

Real-World Applications

For the following exercises, use the given information to answer the questions.

Exercise 51

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?

Solution

3 seconds

Exercise 52

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 53

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?

Solution

48 inches

Exercise 54

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 55

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?

Solution

≈ 49.75 pounds

Exercise 56

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 57

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.

Solution

≈ 33.33 amperes

Exercise 58

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 59

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?

Solution

≈ 2.88 inches

Exercise 60

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, find the kinetic energy if the velocity is increased to 20 meters per second.

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.

Complex Numbers

Perform the indicated operation with complex numbers.

$(4 + 3 i) + (- 2 - 5 i)$

Solution

$2 - 2 i$

$(6 - 5 i) - (10 + 3 i)$

$(2 - 3 i) (3 + 6 i)$

Solution

$24 + 3 i$

$\frac{2 - i}{2 + i}$

Solve the following equations over the complex number system.

$x^{2} - 4 x + 5 = 0$

Solution

${2 + i, 2 - i}$

$x^{2} + 2 x + 10 = 0$

Quadratic Functions

For the following exercises, write the quadratic function in standard form. Then, give the vertex and axes intercepts. Finally, graph the function.

$f (x) = x^{2} - 4 x - 5$

Solution

$f (x) = {(x - 2)}^{2} - 9 vertex (2, –9), intercepts (5, 0); (–1, 0); (0, –5)$

Graph of f(x)=x^2-4x-5.

$f (x) = - 2 x^{2} - 4 x$

For the following problems, find the equation of the quadratic function using the given information.

The vertex is $(- 2, 3)$ and a point on the graph is $(3, 6) .$

Solution

$f (x) = \frac{3}{25} {(x + 2)}^{2} + 3$

The vertex is $(- 3, 6.5)$ and a point on the graph is $(2, 6) .$

Answer the following questions.

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, find the dimensions of the plot to have maximum area.

Solution

300 meters by 150 meters, the longer side parallel to river.

An object projected from the ground at a 45 degree angle with initial velocity of 120 feet per second has height, $h,$ in terms of horizontal distance traveled, $x,$ given by $h (x) = \frac{- 32}{{(120)}^{2}} x^{2} + x .$ Find the maximum height the object attains.

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 coefficient.

$f (x) = 4 x^{5} - 3 x^{3} + 2 x - 1$

Solution

Yes, degree = 5, leading coefficient = 4

$f (x) = 5^{x + 1} - x^{2}$

$f (x) = x^{2} (3 - 6 x + x^{2})$

Solution

Yes, degree = 4, leading coefficient = 1

For the following exercises, determine end behavior of the polynomial function.

$f (x) = 2 x^{4} + 3 x^{3} - 5 x^{2} + 7$

$f (x) = 4 x^{3} - 6 x^{2} + 2$

Solution

$As x \to - \infty, f (x) \to - \infty, as x \to \infty, f (x) \to \infty$

$f (x) = 2 x^{2} (1 + 3 x - x^{2})$

Graphs of Polynomial Functions

For the following exercises, find all zeros of the polynomial function, noting multiplicities.

$f (x) = {(x + 3)}^{2} (2 x - 1) {(x + 1)}^{3}$

Solution

–3 with multiplicity 2, $\frac{1}{2}$ with multiplicity 1, –1 with multiplicity 3

$f (x) = x^{5} + 4 x^{4} + 4 x^{3}$

$f (x) = x^{3} - 4 x^{2} + x - 4$

Solution

4 with multiplicity 1

For the following exercises, based on the given graph, determine the zeros of the function and note multiplicity.

Graph of an odd-degree polynomial with two turning points.

Graph of an even-degree polynomial with two turning points.

Solution

$\frac{1}{2}$ with multiplicity 1, 3 with multiplicity 3

Use the Intermediate Value Theorem to show that at least one zero lies between 2 and 3 for the function $f (x) = x^{3} - 5 x + 1$

Dividing Polynomials

For the following exercises, use long division to find the quotient and remainder.

$\frac{x^{3} - 2 x^{2} + 4 x + 4}{x - 2}$

Solution

$x^{2} + 4$ with remainder 12

$\frac{3 x^{4} - 4 x^{2} + 4 x + 8}{x + 1}$

For the following exercises, use synthetic division to find the quotient. If the divisor is a factor, then write the factored form.

$\frac{x^{3} - 2 x^{2} + 5 x - 1}{x + 3}$

Solution

$x^{2} - 5 x + 20 - \frac{61}{x + 3}$

$\frac{x^{3} + 4 x + 10}{x - 3}$

$\frac{2 x^{3} + 6 x^{2} - 11 x - 12}{x + 4}$

Solution

$2 x^{2} - 2 x - 3$ , so factored form is $(x + 4) (2 x^{2} - 2 x - 3)$

$\frac{3 x^{4} + 3 x^{3} + 2 x + 2}{x + 1}$

Zeros of Polynomial Functions

For the following exercises, use the Rational Zero Theorem to help you solve the polynomial equation.

$2 x^{3} - 3 x^{2} - 18 x - 8 = 0$

Solution

${- 2, 4, - \frac{1}{2}}$

$3 x^{3} + 11 x^{2} + 8 x - 4 = 0$

$2 x^{4} - 17 x^{3} + 46 x^{2} - 43 x + 12 = 0$

Solution

${1, 3, 4, \frac{1}{2}}$

$4 x^{4} + 8 x^{3} + 19 x^{2} + 32 x + 12 = 0$

For the following exercises, use Descartes’ Rule of Signs to find the possible number of positive and negative solutions.

$x^{3} - 3 x^{2} - 2 x + 4 = 0$

Solution

0 or 2 positive, 1 negative

$2 x^{4} - x^{3} + 4 x^{2} - 5 x + 1 = 0$

Rational Functions

For the following rational functions, find the intercepts and the vertical and horizontal asymptotes, and then use them to sketch a graph.

$f (x) = \frac{x + 2}{x - 5}$

Solution

Intercepts $(–2, 0) and (0, - \frac{2}{5})$ , Asymptotes $x = 5$ and $y = 1.$

Graph of f(x)=(x+1)/(x-5).

$f (x) = \frac{x^{2} + 1}{x^{2} - 4}$

$f (x) = \frac{3 x^{2} - 27}{x^{2} + x - 2}$

Solution

Intercepts (3, 0), (-3, 0), and $(0, \frac{27}{2})$ , Asymptotes $x = 1, x = - 2, y = 3.$

Graph of f(x)=(3x^2-27)/(x^2+x-2).

$f (x) = \frac{x + 2}{x^{2} - 9}$

For the following exercises, find the slant asymptote.

$f (x) = \frac{x^{2} - 1}{x + 2}$

Solution

$y = x - 2$

$f (x) = \frac{2 x^{3} - x^{2} + 4}{x^{2} + 1}$

Inverses and Radical Functions

For the following exercises, find the inverse of the function with the domain given.

$f (x) = {(x - 2)}^{2}, x \geq 2$

Solution

$f^{- 1} (x) = \sqrt{x} + 2$

$f (x) = {(x + 4)}^{2} - 3, x \geq - 4$

$f (x) = x^{2} + 6 x - 2, x \geq - 3$

Solution

$f^{- 1} (x) = \sqrt{x + 11} - 3$

$f (x) = 2 x^{3} - 3$

$f (x) = \sqrt{4 x + 5} - 3$

Solution

$f^{- 1} (x) = \frac{{(x + 3)}^{2} - 5}{4}, x \geq - 3$

$f (x) = \frac{x - 3}{2 x + 1}$

Modeling Using Variation

For the following exercises, find the unknown value.

$y$ varies directly as the square of $x .$ If when $x = 3, y = 36,$ find $y$ if $x = 4.$

Solution

$y = 64$

$y$ varies inversely as the square root of $x$ If when $x = 25, y = 2,$ find $y$ if $x = 4.$

$y$ varies jointly as the cube of $x$ and as $z .$ If when $x = 1$ and $z = 2,$ $y = 6,$ find $y$ if $x = 2$ and $z = 3.$

Solution

$y = 72$

$y$ varies jointly as $x$ and the square of $z$ and inversely as the cube of $w .$ If when $x = 3,$ $z = 4,$ and $w = 2,$ $y = 48,$ find $y$ if $x = 4,$ $z = 5,$ and $w = 3.$

For the following exercises, solve the application problem.

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 person weighs 150 pounds when he is on the surface of the earth (3,960 miles from center), find the weight of the person if he is 20 miles above the surface.

Solution

148.5 pounds

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.

Chapter Test

Perform the indicated operation or solve the equation.

$(3 - 4 i) (4 + 2 i)$

Solution

$20 - 10 i$

$\frac{1 - 4 i}{3 + 4 i}$

$x^{2} - 4 x + 13 = 0$

Solution

${2 + 3 i, 2 - 3 i}$

Give the degree and leading coefficient of the following polynomial function.

$f (x) = x^{3} (3 - 6 x - 2 x^{2})$

Determine the end behavior of the polynomial function.

$f (x) = 8 x^{3} - 3 x^{2} + 2 x - 4$

Solution

$A s x \to - \infty, f (x) \to - \infty, a s x \to \infty, f (x) \to \infty$

$f (x) = - 2 x^{2} (4 - 3 x - 5 x^{2})$

Write the quadratic function in standard form. Determine the vertex and axes intercepts and graph the function.

$f (x) = x^{2} + 2 x - 8$

Solution

$f (x) = {(x + 1)}^{2} - 9$ , vertex $(−1, −9)$ , intercepts $(2, 0); (−4, 0); (0, −8)$

Graph of f(x)=x^2+2x-8.

Given information about the graph of a quadratic function, find its equation.

Vertex $(2, 0)$ and point on graph $(4, 12) .$

Solve the following application problem.

A rectangular field is to be enclosed by fencing. In addition to the enclosing fence, another fence is to divide the field into two parts, running parallel to two sides. If 1,200 feet of fencing is available, find the maximum area that can be enclosed.

Solution

60,000 square feet

Find all zeros of the following polynomial functions, noting multiplicities.

$f (x) = {(x - 3)}^{3} (3 x - 1) {(x - 1)}^{2}$

$f (x) = 2 x^{6} - 12 x^{5} + 18 x^{4}$

Solution

0 with multiplicity 4, 3 with multiplicity 2

Based on the graph, determine the zeros of the function and multiplicities.

Use long division to find the quotient.

$\frac{2 x^{3} + 3 x - 4}{x + 2}$

Solution

$2 x^{2} - 4 x + 11 - \frac{26}{x + 2}$

Use synthetic division to find the quotient. If the divisor is a factor, write the factored form.

$\frac{x^{4} + 3 x^{2} - 4}{x - 2}$

$\frac{2 x^{3} + 5 x^{2} - 7 x - 12}{x + 3}$

Solution

$2 x^{2} - x - 4$ . So factored form is $(x + 3) (2 x^{2} - x - 4)$

Use the Rational Zero Theorem to help you find the zeros of the polynomial functions.

$f (x) = 2 x^{3} + 5 x^{2} - 6 x - 9$

$f (x) = 4 x^{4} + 8 x^{3} + 21 x^{2} + 17 x + 4$

Solution

$- \frac{1}{2}$ (has multiplicity 2), $\frac{- 1 \pm i \sqrt{15}}{2}$

$f (x) = 4 x^{4} + 16 x^{3} + 13 x^{2} - 15 x - 18$

$f (x) = x^{5} + 6 x^{4} + 13 x^{3} + 14 x^{2} + 12 x + 8$

Solution

$- 2$ (has multiplicity 3), $\pm i$

Given the following information about a polynomial function, find the function.

It has a double zero at $x = 3$ and zeroes at $x = 1$ and $x = - 2$ . Its y-intercept is $(0, 12) .$

It has a zero of multiplicity 3 at $x = \frac{1}{2}$ and another zero at $x = - 3$ . It contains the point $(1, 8) .$

Solution

$f (x) = 2 {(2 x - 1)}^{3} (x + 3)$

Use Descartes’ Rule of Signs to determine the possible number of positive and negative solutions.

$8 x^{3} - 21 x^{2} + 6 = 0$

For the following rational functions, find the intercepts and horizontal and vertical asymptotes, and sketch a graph.

$f (x) = \frac{x + 4}{x^{2} - 2 x - 3}$

Solution

Intercepts $(- 4, 0), (0, - \frac{4}{3})$ , Asymptotes $x = 3, x = −1, y = 0$ .

Graph of f(x)=(x+4)/(x^2-2x-3).

$f (x) = \frac{x^{2} + 2 x - 3}{x^{2} - 4}$

Find the slant asymptote of the rational function.

$f (x) = \frac{x^{2} + 3 x - 3}{x - 1}$

Solution

$y = x + 4$

Find the inverse of the function.

$f (x) = \sqrt{x - 2} + 4$

$f (x) = 3 x^{3} - 4$

Solution

$f^{- 1} (x) = \sqrt[3]{\frac{x + 4}{3}}$

$f (x) = \frac{2 x + 3}{3 x - 1}$

Find the unknown value.

$y$ varies inversely as the square of $x$ and when $x = 3,$ $y = 2.$ Find $y$ if $x = 1.$

Solution

$y = 18$

$y$ varies jointly with $x$ and the cube root of $z .$ If when $x = 2$ and $z = 27,$ $y = 12,$ find $y$ if $x = 5$ and $z = 8.$

Solve the following application problem.

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?

Solution

4 seconds

constant of variation

the non-zero value

k

that helps define the relationship between variables in direct or inverse variation

direct variation

the relationship between two variables that are a constant multiple of each other; as one quantity increases, so does the other

inverse variation

the relationship between two variables in which the product of the variables is a constant

inversely proportional

a relationship where one quantity is a constant divided by the other quantity; as one quantity increases, the other decreases

joint variation

a relationship where a variable varies directly or inversely with multiple variables

varies directly

a relationship where one quantity is a constant multiplied by the other quantity

varies inversely

a relationship where one quantity is a constant divided by the other quantity