Search in book...
Toggle Font Controls
Create new playlist

Name your new playlist

Playlist description (optional)
Sign In

Email address

Password

Forgot Password?

or

Continue with Facebook

Continue with Google
Sign Up

Full Name

Email address

Confirm Email Address

Password

or

Continue with Facebook

Continue with Google

4.5 Rational Functions

For a rational function, find the domain and graph the function, identifying all of the asymptotes.
Solve applied problems involving rational functions.

Now we turn our attention to functions that represent the quotient of two polynomials. Whereas the sum, the difference, or the product of two polynomials is a polynomial, in general the quotient of two polynomials is not itself a polynomial.

A rational number can be expressed as the quotient of two integers, $p / q$ , where $q \neq 0$ . A rational function is formed by the quotient of two polynomials, $p (x) / q (x)$ , where $q (x) \neq 0$ . Here are some examples of rational functions and their graphs.

The Domain of a Rational Function

Example 1

Consider

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

Find the domain and graph f.

Solution

When the denominator $x - 3$ is 0, we have $x = 3$ , so the only input that results in a denominator of 0 is 3. Thus the domain is

${x | x \neq 3}, or (- \infty, 3) \cup (3, \infty) .$

The graph of this function is the graph of $y = 1 / x$ translated right 3 units.

Example 2

Determine the domain of each of the functions illustrated at the beginning of this section.

Solution

The domain of each rational function will be the set of all real numbers except those values that make the denominator 0. To determine those exceptions, we set the denominator equal to 0 and solve for x.

FUNCTION	DOMAIN
$f (x) = \frac{1}{x}$ $f (x) = \frac{1}{x^{2}}$ $f (x) = \frac{x - 3}{x^{2} + x - 2} = \frac{x - 3}{(x + 2) (x - 1)}$ $f (x) = \frac{2 x + 5}{2 x - 6} = \frac{2 x + 5}{2 (x - 3)}$ $f (x) = \frac{x^{2} + 2 x - 3}{x^{2} - x - 2} = \frac{x^{2} + 2 x - 3}{(x + 1) (x - 2)}$ $f (x) = \frac{- x^{2}}{x + 1}$	${x \| x \neq 0}$ , or $(- \infty, 0) \cup (0, \infty)$ ${x \| x \neq 0}$ , or $(- \infty, 0) \cup (0, \infty)$ ${x \| x \neq - 2$ and $x \neq 1}$ , or $(- \infty, - 2) \cup (- 2, 1) \cup (1, \infty)$ ${x \| x \neq 3}$ , or $(- \infty, 3) \cup (3, \infty)$ ${x \| x \neq - 1$ and $x \neq 2}$ , or $(- \infty, - 1) \cup (- 1, 2) \cup (2, \infty)$ ${x \| x \neq - 1}$ , or $(- \infty, - 1) \cup (- 1, \infty)$

FUNCTION

DOMAIN

$f (x) = \frac{1}{x}$

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

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

$f (x) = \frac{2 x + 5}{2 x - 6} = \frac{2 x + 5}{2 (x - 3)}$

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

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

${x | x \neq 0}$ , or $(- \infty, 0) \cup (0, \infty)$

${x | x \neq - 2$ and $x \neq 1}$ , or $(- \infty, - 2) \cup (- 2, 1) \cup (1, \infty)$

${x | x \neq 3}$ , or $(- \infty, 3) \cup (3, \infty)$

${x | x \neq - 1$ and $x \neq 2}$ , or $(- \infty, - 1) \cup (- 1, 2) \cup (2, \infty)$

${x | x \neq - 1}$ , or $(- \infty, - 1) \cup (- 1, \infty)$

As a partial check of the domains, we can observe the discontinuities (breaks) in the graphs of these functions. (See p. 267.)

Asymptotes

Vertical Asymptotes

Look at the graph of $f (x) = 1 / (x - 3)$ , shown at left. (Also see Example 1.) Let’s explore what happens as x-values get closer and closer to 3 from the left. We then explore what happens as x-values get closer and closer to 3 from the right.

From the left:

From the right:

We see that as x-values get closer and closer to 3 from the left, the function values (y-values) decrease without bound (that is, they approach negative infinity, $- \infty$ ). Similarly, as the x-values approach 3 from the right, the function values increase without bound (that is, they approach positive infinity, $\infty$ ). We write this as

$f (x) \to - \infty as x \to 3^{-} and f (x) \to \infty as x \to 3^{+} .$

We read $“ f (x) \to - \infty$ as $x \to 3^{-} ”$ as $“ f (x)$ decreases without bound as x approaches 3 from the left.” We read $“ f (x) \to \infty$ as $x \to 3^{+}$ ” as $“ f (x)$ increases without bound as x approaches 3 from the right.” The notation $x \to 3$ means that x gets as close to 3 as possible without being equal to 3. The vertical line $x = 3$ is said to be a vertical asymptote for this curve.

In general, the line $x = a$ is a vertical asymptote for the graph of f if any of the following is true:

$\begin{array}{l} f (x) \to \infty as x \to a^{-}, or f (x) \to - \infty as x \to a^{-}, or \\ f (x) \to \infty as x \to a^{+}, or f (x) \to - \infty as x \to a^{+} . \end{array}$

The following figures show the four ways in which a vertical asymptote can occur.

The vertical asymptotes of a rational function $f (x) = p (x) / q (x)$ are found by determining the zeros of $q (x)$ that are not also zeros of $p (x)$ . If $p (x)$ and $q (x)$ are polynomials with no common factors other than constants, we need determine only the zeros of the denominator $q (x)$ .

Example 3

Determine the vertical asymptotes for the graph of each of the following functions.

$f (x) = \frac{2 x - 11}{x^{2} + 2 x - 8}$
$h (x) = \frac{x^{2} - 4 x}{x^{3} - x}$
$g (x) = \frac{x - 2}{x^{3} - 5 x}$

Solution

First, we factor the denominator:

$f (x) = \frac{2 x - 11}{x^{2} + 2 x - 8} = \frac{2 x - 11}{(x + 4) (x - 2)} .$

The numerator and the denominator have no common factors. The zeros of the denominator are −4 and 2. Thus the vertical asymptotes for the graph of $f (x)$ are the lines $x = - 4$ and $x = 2$ . (See Fig. 1.)

Figure 1.
We factor the numerator and the denominator:

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

The domain of the function is ${x | x \neq - 1$ and $x \neq 0$ and $x \neq 1}$ , or $(- \infty, - 1) \cup (- 1, 0) \cup (0, 1) \cup (1, \infty)$ . Note that the numerator and the denominator share a common factor, x. The vertical asymptotes of $h (x)$ are found by determining the zeros of the denominator, $x (x + 1) (x - 1)$ , that are not also zeros of the numerator, $x (x - 4)$ . The zeros of $x (x + 1)$ $(x - 1)$ are 0, −1, and 1. The zeros of $x (x - 4)$ are 0 and 4. Thus, although the denominator has three zeros, the graph of $h (x)$ has only two vertical asymptotes, $x = - 1$ and $x = 1$ . (See Fig. 2.)

Figure 2.

The rational expression $[x (x - 4)] / [x (x + 1) (x - 1)]$ can be simplified. Thus,

$h (x) = \frac{x (x - 4)}{x (x + 1) (x - 1)} = \frac{x - 4}{(x + 1) (x - 1)},$

JUST IN TIME 18

where $x \neq 0, x \neq - 1$ , and $x \neq 1$ . The graph of $h (x)$ is the graph of

$h (x) = \frac{x - 4}{(x + 1) (x - 1)}$

with the point where $x = 0$ missing. To determine the y-coordinate of the hole, we substitute 0 for x:

$h (0) = \frac{0 - 4}{(0 + 1) (0 - 1)} = \frac{- 4}{1 \cdot (- 1)} = 4.$

Thus the hole is located at (0, 4).
We factor the denominator:

$g (x) = \frac{x - 2}{x^{3} - 5 x} = \frac{x - 2}{x (x^{2} - 5)} .$

The numerator and the denominator have no common factors. We find the zeros of the denominator, $x (x^{2} - 5)$ . Solving $x (x^{2} - 5) = 0$ , we get

$\begin{array}{rclcrcl} x & = & 0 & or & x^{2} - 5 & = & 0 \\ x & = & 0 & or & x^{2} & = & 5 \\ x & = & 0 & or & x & = & \pm \sqrt{5} . \end{array}$

The zeros of the denominator are 0, $\sqrt{5}$ , and $- \sqrt{5}$ . Thus the vertical asymptotes are the lines $x = 0, x = \sqrt{5}$ , and $x = - \sqrt{5}$ . (See Fig. 3.)

Figure 3.

Now Try Exercises 15 and 19.

Horizontal Asymptotes

Looking again at the graph of $f (x) = 1 / (x - 3)$ , shown at left (also see Example 1), let’s explore what happens to $f (x) = 1 / (x - 3)$ as x increases without bound (approaches positive infinity, $\infty$ ) and as x decreases without bound (approaches negative infinity, $- \infty$ ).

x increases without bound:

x decreases without bound:

We see that

$\frac{1}{x - 3} \to 0 as x \to \infty and \frac{1}{x - 3} \to 0 as x \to - \infty .$

Since $y = 0$ is the equation of the x-axis, we say that the curve approaches the x-axis asymptotically and that the x-axis is a horizontal asymptote for the curve.

In general, the line $y = b$ is a horizontal asymptote for the graph of f if either or both of the following are true:

$f (x) \to b as x \to \infty or f (x) \to b as x \to - \infty .$

The following figures illustrate four ways in which horizontal asymptotes can occur. In each case, the curve gets close to the line $y = b$ either as $x \to \infty$ or as $x \to - \infty$ . Keep in mind that the symbols $\infty$ and $- \infty$ convey the idea of increasing without bound and decreasing without bound, respectively.

How can we determine a horizontal asymptote? As x gets very large or very small, the value of a polynomial function $p (x)$ is dominated by the function’s leading term. Because of this, if $p (x)$ and $q (x)$ have the same degree, the value of $p (x) / q (x)$ as $x \to \infty$ or as $x \to - \infty$ is dominated by the ratio of the numerator’s leading coefficient to the denominator’s leading coefficient.

For $f (x) = (3 x^{2} + 2 x - 4) / (2 x^{2} - x + 1)$ , we see that the numerator, $3 x^{2} + 2 x - 4$ , is dominated by $3 x^{2}$ and the denominator, $2 x^{2} - x + 1$ , is dominated by $2 x^{2}$ , so $f (x)$ approaches $3 x^{2} / 2 x^{2}$ , or $3 / 2$ , as x gets very large or very small:

$\begin{array}{l} \frac{3 x^{2} + 2 x - 4}{2 x^{2} - x + 1} \to \frac{3}{2}, or 1.5, as x \to \infty, and \\ \frac{3 x^{2} + 2 x - 4}{2 x^{2} - x + 1} \to \frac{3}{2}, or 1.5, as x \to - \infty . \end{array}$

We say that the curve approaches the horizontal line $y = \frac{3}{2}$ asymptotically and that $y = \frac{3}{2}$ is a horizontal asymptote for the curve.

It follows that when the numerator and the denominator of a rational function have the same degree, the line $y = a / b$ is the horizontal asymptote, where a and b are the leading coefficients of the numerator and the denominator, respectively.

Example 4

Find the horizontal asymptote: $f (x) = \frac{- 7 x^{4} - 10 x^{2} + 1}{11 x^{4} + x - 2}$ .

Solution

The numerator and the denominator have the same degree. The ratio of the leading coefficients is $- \frac{7}{11}$ , so the line $y = - \frac{7}{11}$ , or $- 0. \bar{63}$ , is the horizontal asymptote.

Now Try Exercise 21.

To check Example 4, we could evaluate the function for a very large and a very small value of x. Another check, one that is useful in calculus, is to multiply by 1, using $(1 / x^{4}) / (1 / x^{4}) :$

$\begin{matrix} f (x) & = & \frac{- 7 x^{4} - 10 x^{2} + 1}{11 x^{4} + x - 2} \cdot \frac{\frac{1}{x^{4}}}{\frac{1}{x^{4}}} & = & \frac{\frac{- 7 x^{4}}{x^{4}} - \frac{10 x^{2}}{x^{4}} + \frac{1}{x^{4}}}{\frac{11 x^{4}}{x^{4}} + \frac{x}{x^{4}} - \frac{2}{x^{4}}} \\ = & \frac{- 7 - \frac{10}{x^{2}} + \frac{1}{x^{4}}}{11 + \frac{1}{x^{3}} - \frac{2}{x^{4}}} . \end{matrix}$

As $| x |$ becomes very large, each expression whose denominator is a power of x tends toward 0. Specifically, as $x \to \infty$ or as $x \to - \infty$ , we have

$f (x) \to \frac{- 7 - 0 + 0}{11 + 0 - 0}, or f (x) \to - \frac{7}{11} .$

The horizontal asymptote is $y = - \frac{7}{11}$ , or $- 0. \bar{63}$ .

We now investigate the occurrence of a horizontal asymptote when the degree of the numerator is less than the degree of the denominator.

Example 5

Find the horizontal asymptote: $f (x) = \frac{2 x + 3}{x^{3} - 2 x^{2} + 4}$ .

Solution

Let $p (x) = 2 x + 3, q (x) = x^{3} - 2 x^{2} + 4$ , and $f (x) =$ $p (x) / q (x)$ . Note that as $x \to \infty$ , the value of $q (x)$ grows much faster than the value of $p (x)$ . Because of this, the ratio $p (x) / q (x)$ shrinks toward 0. As $x \to - \infty$ , the ratio $p (x) / q (x)$ behaves in a similar manner. The horizontal asymptote is $y = 0$ , the x-axis. This is the case for all rational functions for which the degree of the numerator is less than the degree of the denominator. Note in Example 1 that $y = 0$ , the x-axis, is the horizontal asymptote of $f (x) = 1 / (x - 3)$ .

Now Try Exercise 23.

The following statements describe the two ways in which a horizontal asymptote occurs.

Example 6

Graph

$g (x) = \frac{2 x^{2} + 1}{x^{2}} .$

Include and label all asymptotes.

Solution

Since 0 is the zero of the denominator and is not a zero of the numerator, the y-axis, $x = 0$ , is the vertical asymptote. Note also that the degree of the numerator is the same as the degree of the denominator. Thus, $y = 2 / 1$ , or 2, is the horizontal asymptote.

To draw the graph, we first draw the asymptotes with dashed lines. Then we compute and plot some ordered pairs and draw the two branches of the curve.

`x`	$g (x)$
−2	2.25
−1.5	$2. \bar{4}$
−1	3
−0.5	6
0.5	6
1	3
1.5	$2. \bar{4}$
2	2.25

Now Try Exercise 41.

Oblique Asymptotes

Sometimes a line that is neither horizontal nor vertical is an asymptote. Such a line is called an oblique asymptote, or a slant asymptote.

Example 7

Find all the asymptotes of

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

Solution

The line $x = 2$ is the vertical asymptote because 2 is the zero of the denominator and is not a zero of the numerator. There is no horizontal asymptote because the degree of the numerator is greater than the degree of the denominator. When the degree of the numerator is 1 greater than the degree of the denominator, we divide to find an equivalent expression:

$\begin{array}{l} \frac{2 x^{2} - 3 x - 1}{x - 2} = (2 x + 1) + \frac{1}{x - 2} . & \begin{array}{r} 2 x + 1 \\ x - 2 2 x^{2} - 3 x - 1 \\ \underline{2 x^{2} - 4 x} \\ x - 1 \\ \underline{x - 2} \\ 1 \end{array} \end{array}$

Now we see that when $x \to \infty$ or $x \to - \infty, 1 / (x - 2) \to 0$ and the value of $f (x) \to 2 x + 1$ . This means that as $| x |$ becomes very large, the graph of $f (x)$ gets very close to the graph of $y = 2 x + 1$ . Thus the line $y = 2 x + 1$ is the oblique asymptote.

Now Try Exercise 59.

The following statements are also true.

Shown below is an outline of a procedure that we can follow to create accurate graphs of rational functions.

Example 8

Graph: $f (x) = \frac{2 x + 3}{3 x^{2} + 7 x - 6}$ .

Solution

We find the zeros of the denominator by solving $3 x^{2} + 7 x - 6 = 0$ . Since

$3 x^{2} + 7 x - 6 = (3 x - 2) (x + 3),$

the zeros are $\frac{2}{3}$ and −3. Thus the domain excludes $\frac{2}{3}$ and −3 and is

$(- \infty, - 3) \cup (- 3, \frac{2}{3}) \cup (\frac{2}{3}, \infty) .$

Since neither zero of the denominator is a zero of the numerator, the graph has vertical asymptotes $x = - 3$ and $x = \frac{2}{3}$ . We sketch these as dashed lines.
Because the degree of the numerator is less than the degree of the denominator, the x-axis, $y = 0$ , is the horizontal asymptote.
To find the zeros of the numerator, we solve $2 x + 3 = 0$ and get $x = - \frac{3}{2}$ . Thus, $- \frac{3}{2}$ is the zero of the function, and the pair $(- \frac{3}{2}, 0)$ is the x-intercept.
We find $f (0) :$

$\begin{array}{rcl} f (0) & = & \frac{2 \cdot 0 + 3}{3 \cdot 0^{2} + 7 \cdot 0 - 6} \\ = & \frac{3}{- 6} = - \frac{1}{2} . \end{array}$

The point $(0, - \frac{1}{2})$ is the y-intercept.
We find other function values to determine the general shape. We choose values in each interval of the domain as shown in the table below and then draw the graph. Note that the graph of this function crosses its horizontal asymptote at $x = - \frac{3}{2}$ .

x y

−4.5 −0.26

−3.25 −1.19

−2.5 0.42

−0.5 −0.23

0.5 −2.29

0.75 4.8

1.5 0.53

3.5 0.18

Now Try Exercise 65.

Example 9

Graph: $g (x) = \frac{x^{2} - 1}{x^{2} + x - 6}$ .

Solution

We find the zeros of the denominator by solving $x^{2} + x - 6 = 0$ . Since

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

the zeros are −3 and 2. Thus the domain excludes the x-values −3 and 2 and is

$(- \infty, - 3) \cup (- 3, 2) \cup (2, \infty) .$

Since neither zero of the denominator is a zero of the numerator, the graph has vertical asymptotes $x = - 3$ and $x = 2$ . We sketch these as dashed lines.
The numerator and the denominator have the same degree, so the horizontal asymptote is determined by the ratio of the leading coefficients: $1 / 1$ , or 1. Thus, $y = 1$ is the horizontal asymptote. We sketch it with a dashed line.
To find the zeros of the numerator, we solve $x^{2} - 1 = 0$ . The solutions are −1 and 1. Thus, −1 and 1 are the zeros of the function and the pairs (−1, 0) and (1, 0) are the x-intercepts.
We find $g (0) :$

$g (0) = \frac{0^{2} - 1}{0^{2} + 0 - 6} = \frac{- 1}{- 6} = \frac{1}{6} .$

Thus, $(0, \frac{1}{6})$ is the y-intercept.
We find other function values to determine the general shape and then draw the graph.

Now Try Exercise 77.

The magnified portion of the graph in Example 9 above shows another situation in which a graph can cross its horizontal asymptote. The point where $g (x)$ crosses $y = 1$ can be found by setting $g (x) = 1$ and solving for x:

$\begin{array}{rcl} \frac{x^{2} - 1}{x^{2} + x - 6} & = & 1 \\ x^{2} - 1 & = & x^{2} + x - 6 \\ - 1 & = & x - 6 Subtracting x^{2} \\ 5 & = & x . Adding 6 \end{array}$

The point of intersection is (5, 1). Note the behavior of the curve after it crosses the horizontal asymptote at $x = 5$ . (See the graph at left.) It continues to decrease for a short interval and then begins to increase, getting closer and closer to $y = 1$ as $x \to \infty$ .

Graphs of rational functions can also cross an oblique asymptote. The graph of

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

shown below crosses its oblique asymptote $y = 2 x$ . Remember, graphs can cross horizontal asymptotes or oblique asymptotes, but they cannot cross vertical asymptotes.

Let’s now graph a rational function $f (x) = p (x) / q (x)$ , where $p (x)$ and $q (x)$ have a common factor, $x - c$ . The graph of such a function has a “hole” in it. We first saw this situation in Example 3(b), where the common factor was x.

Example 10

Graph: $g (x) = \frac{x - 2}{x^{2} - x - 2}$ .

Solution

We first express the denominator in factored form:

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

The domain of the function is ${x | x \neq - 1$ and $x \neq 2}$ , or $(- \infty, - 1) \cup$ $(- 1, 2) \cup (2, \infty)$ . Note that the numerator and the denominator have the common factor $x - 2$ . The zeros of the denominator are −1 and 2, and the zero of the numerator is 2. Since −1 is the only zero of the denominator that is not a zero of the numerator, the graph of the function has $x = - 1$ as its only vertical asymptote. The degree of the numerator is less than the degree of the denominator, so $y = 0$ is the horizontal asymptote. There are no zeros of the function and thus no x-intercepts, because 2 is the only zero of the numerator and 2 is not in the domain of the function. Since $g (0) = 1, (0, 1)$ is the y-intercept. The rational expression

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

can be simplified. Thus,

`x`	`y`
−3	$- \frac{1}{2}$
−2	−1
−1	Not defined
0	1
1	$\frac{1}{2}$
2	Not defined
3	$\frac{1}{4}$

$g (x) = \frac{x - 2}{(x + 1) (x - 2)} = \frac{1}{x + 1}, where x \neq - 1 and x \neq 2.$

The graph of $g (x)$ is the graph of $y = 1 / (x + 1)$ with the point where $x = 2$ missing. To determine the coordinates of the “hole,” we substitute 2 for x in $g (x) = 1 / (x + 1) :$

$g (2) = \frac{1}{2 + 1} = \frac{1}{3} .$

Thus the “hole” is located at $(2, \frac{1}{3})$ . We draw the graph indicating the “hole” when $x = 2$ with an open circle.

Now Try Exercise 49.

Example 11

Graph: $f (x) = \frac{- 2 x^{2} - x + 15}{x^{2} - x - 12}$ .

Solution

We first express the numerator and the denominator in factored form:

$f (x) = \frac{- 2 x^{2} - x + 15}{x^{2} - x - 12} = \frac{- (2 x^{2} + x - 15)}{x^{2} - x - 12} = \frac{- (2 x - 5) (x + 3)}{(x - 4) (x + 3)} .$

The domain of the function is ${x | x \neq - 3$ and $x \neq 4}$ , or $(- \infty, - 3) \cup (- 3, 4) \cup$ $(4, \infty)$ . The numerator and the denominator have the common factor $x + 3$ . The zeros of the denominator are −3 and 4, and the zeros of the numerator are −3 and $\frac{5}{2}$ . Since 4 is the only zero of the denominator that is not a zero of the numerator, the graph of the function has $x = 4$ as its only vertical asymptote.

The degrees of the numerator and the denominator are the same, so the line $y = \frac{- 2}{1} = - 2$ is the horizontal asymptote. The zeros of the numerator are $\frac{5}{2}$ and −3. Because −3 is not in the domain of the function, the only x-intercept is $(\frac{5}{2}, 0)$ . Since $f (0) = - \frac{15}{12} = - \frac{5}{4}$ , then $(0, - \frac{5}{4})$ is the y-intercept. The rational function

$\frac{- (2 x - 5) (x + 3)}{(x - 4) (x + 3)}$

can be simplified.

Thus,

$f (x) = \frac{- (2 x - 5) (x + 3)}{(x - 4) (x + 3)} = \frac{- (2 x - 5)}{x - 4}, where x \neq - 3 and x \neq 4.$

The graph of $f (x)$ is the graph of $y = - (2 x - 5) / (x - 4)$ with the point where $x = - 3$ missing. To determine the coordinates of the hole, we substitute −3 for x in $f (x) = (- 2 x + 5) / (x - 4) :$

$f (- 3) = \frac{- [2 (- 3) - 5]}{- 3 - 4} = \frac{- [- 11]}{- 7} = \frac{11}{- 7} = - \frac{11}{7} .$

Thus the hole is located at $(- 3, - \frac{11}{7})$ . We draw the graph indicating the hole when $x = - 3$ with an open circle.

`x`	`y`
−5	−1.67
−4	−1.63
−3	Not defined
−2	−1.5
−1	−1.4
0	−1.25
1	−1
2	−0.5
3	1
3.5	4
4	Not defined
5	−5
6	−3.5
7	−3
8	−2.75

Now Try Exercise 67.

Applications

Example 12

Temperature During an Illness. A person’s temperature T, in degrees Fahrenheit, during an illness is given by the function

$T (t) = \frac{4 t}{t^{2} + 1} + 98.6,$

where time t is given in hours since the onset of the illness. The graph of this function is shown at left.

Find the temperature at $t = 0, 1, 2, 5, 12, and 24$ .
Find the horizontal asymptote of the graph of $T (t)$ . Complete:

$T (t) \to as t \to \infty .$
Give the meaning of the answer to part (b) in terms of the application.

Solution

We have

$\begin{array}{l} T (0) = 98.6, T (1) = 100.6, T (2) = 100.2, \\ T (5) \approx 99.369, T (12) \approx 98.931, and T (24) \approx 98.766. \end{array}$
Since

$T (t) = \frac{4 t}{t^{2} + 1} + 98.6 = \frac{98.6 t^{2} + 4 t + 98.6}{t^{2} + 1},$

the horizontal asymptote is $y = 98.6 / 1$ , or 98.6. Then it follows that $T (t) \to 98.6$ as $t \to \infty$ .
As time goes on, the temperature returns to “normal,” which is $98.6 ° F$ .

Now Try Exercise 83.

Visualizing the Graph

Match the function with its graph.

$f (x) = - \frac{1}{x^{2}}$
$f (x) = x^{3} - 3 x^{2} + 2 x + 3$
$f (x) = \frac{x^{2} - 4}{x^{2} - x - 6}$
$f (x) = - x^{2} + 4 x - 1$
$f (x) = \frac{x - 3}{x^{2} + x - 6}$
$f (x) = \frac{3}{4} x + 2$
$f (x) = x^{2} - 1$
$f (x) = x^{4} - 2 x^{2} - 5$
$f (x) = \frac{8 x - 4}{3 x + 6}$
$f (x) = 2 x^{2} - 4 x - 1$

Answers on page A-19

4.5 Exercise Set

Determine the domain of the function.

1. $f (x) = \frac{x^{2}}{2 - x}$
2. $f (x) = \frac{1}{x^{3}}$
3. $f (x) = \frac{x + 1}{x^{2} - 6 x + 5}$
4. $f (x) = \frac{{(x + 4)}^{2}}{4 x - 3}$
5. $f (x) = \frac{3 x - 4}{3 x + 15}$
6. $f (x) = \frac{x^{2} + 3 x - 10}{x^{2} + 2 x}$

In Exercises 7–12, use your knowledge of asymptotes and intercepts to match the equation with one of the graphs (a)–(f) that follow. List all asymptotes.

7. $f (x) = \frac{8}{x^{2} - 4}$
8. $f (x) = \frac{8}{x^{2} + 4}$
9. $f (x) = \frac{8 x}{x^{2} - 4}$
10. $f (x) = \frac{8 x^{2}}{x^{2} - 4}$
11. $f (x) = \frac{8 x^{3}}{x^{2} - 4}$
12. $f (x) = \frac{8 x^{3}}{x^{2} + 4}$

Determine the vertical asymptotes of the graph of the function.

13. $g (x) = \frac{1}{x^{2}}$
14. $f (x) = \frac{4 x}{x^{2} + 10 x}$
15. $h (x) = \frac{x + 7}{2 - x}$
16. $g (x) = \frac{x^{4} + 2}{x}$
17. $f (x) = \frac{3 - x}{(x - 4) (x + 6)}$
18. $h (x) = \frac{x^{2} - 4}{x (x + 5) (x - 2)}$
19. $g (x) = \frac{x^{3}}{2 x^{3} - x^{2} - 3 x}$
20. $f (x) = \frac{x + 5}{x^{2} + 4 x - 32}$

Determine the horizontal asymptote of the graph of the function.

21. $f (x) = \frac{3 x^{2} + 5}{4 x^{2} - 3}$
22. $g (x) = \frac{x + 6}{x^{3} + 2 x^{2}}$
23. $h (x) = \frac{x^{2} - 4}{2 x^{4} + 3}$
24. $f (x) = \frac{x^{5}}{x^{5} + x}$
25. $g (x) = \frac{x^{3} - 2 x^{2} + x - 1}{x^{2} - 16}$
26. $h (x) = \frac{8 x^{4} + x - 2}{2 x^{4} - 10}$

Determine the oblique asymptote of the graph of the function.

27. $g (x) = \frac{x^{2} + 4 x - 1}{x + 3}$
28. $f (x) = \frac{x^{2} - 6 x}{x - 5}$
29. $h (x) = \frac{x^{4} - 2}{x^{3} + 1}$
30. $g (x) = \frac{12 x^{3} - x}{6 x^{2} + 4}$
31. $f (x) = \frac{x^{3} - x^{2} + x - 4}{x^{2} + 2 x - 1}$
32. $h (x) = \frac{5 x^{3} - x^{2} + x - 1}{x^{2} - x + 2}$

Graph the function. Be sure to label all the asymptotes. List the domain and the x- and y-intercepts.

33. $f (x) = \frac{1}{x}$
34. $g (x) = \frac{1}{x^{2}}$
35. $h (x) = - \frac{4}{x^{2}}$
36. $f (x) = - \frac{6}{x}$
37. $g (x) = \frac{x^{2} - 4 x + 3}{x + 1}$
38. $h (x) = \frac{2 x^{2} - x - 3}{x - 1}$
39. $f (x) = \frac{- 2}{x - 5}$
40. $f (x) = \frac{1}{x - 5}$
41. $f (x) = \frac{2 x + 1}{x}$
42. $f (x) = \frac{3 x - 1}{x}$
43. $f (x) = \frac{x + 3}{x^{2} - 9}$
44. $f (x) = \frac{x - 1}{x^{2} - 1}$
45. $f (x) = \frac{x}{x^{2} + 3 x}$
46. $f (x) = \frac{3 x}{3 x - x^{2}}$
47. $f (x) = \frac{1}{{(x - 2)}^{2}}$
48. $f (x) = \frac{- 2}{{(x - 3)}^{2}}$
49. $f (x) = \frac{x^{2} + 2 x - 3}{x^{2} + 4 x + 3}$
50. $f (x) = \frac{x^{2} - x - 2}{x^{2} - 5 x - 6}$
51. $f (x) = \frac{1}{x^{2} + 3}$
52. $f (x) = \frac{- 1}{x^{2} + 2}$
53. $f (x) = \frac{x^{2} - 4}{x - 2}$
54. $f (x) = \frac{x^{2} - 9}{x + 3}$
55. $f (x) = \frac{x - 1}{x + 2}$
56. $f (x) = \frac{x - 2}{x + 1}$
57. $f (x) = \frac{x^{2} + 3 x}{2 x^{3} - 5 x^{2} - 3 x}$
58. $f (x) = \frac{3 x}{x^{2} + 5 x + 4}$
59. $f (x) = \frac{x^{2} - 9}{x + 1}$
60. $f (x) = \frac{x^{3} - 4 x}{x^{2} - x}$
61. $f (x) = \frac{x^{2} + x - 2}{2 x^{2} + 1}$
62. $f (x) = \frac{x^{2} - 2 x - 3}{3 x^{2} + 2}$
63. $g (x) = \frac{3 x^{2} - x - 2}{x - 1}$
64. $f (x) = \frac{2 x^{2} - 5 x - 3}{2 x + 1}$
65. $f (x) = \frac{x - 1}{x^{2} - 2 x - 3}$
66. $f (x) = \frac{x + 2}{x^{2} + 2 x - 15}$
67. $f (x) = \frac{3 x^{2} + 11 x - 4}{x^{2} + 2 x - 8}$
68. $f (x) = \frac{2 x^{2} - 3 x - 9}{x^{2} - 2 x - 3}$
69. $f (x) = \frac{x - 3}{{(x + 1)}^{3}}$
70. $f (x) = \frac{x + 2}{{(x - 1)}^{3}}$
71. $f (x) = \frac{x^{3} + 1}{x}$
72. $f (x) = \frac{x^{3} - 1}{x}$
73. $f (x) = \frac{x^{3} + 2 x^{2} - 15 x}{x^{2} - 5 x - 14}$
74. $f (x) = \frac{x^{3} + 2 x^{2} - 3 x}{x^{2} - 25}$
75. $f (x) = \frac{5 x^{4}}{x^{4} + 1}$
76. $f (x) = \frac{x + 1}{x^{2} + x - 6}$
77. $f (x) = \frac{x^{2}}{x^{2} - x - 2}$
78. $f (x) = \frac{x^{2} - x - 2}{x + 2}$

Find a rational function that satisfies the given conditions. Answers may vary, but try to give the simplest answer possible.

79. Vertical asymptotes $x = - 4, x = 5$
80. Vertical asymptotes $x = - 4, x = 5;$ x-intercept (−2, 0)
81. Vertical asymptotes $x = - 4, x = 5;$ horizontal asymptote $y = \frac{3}{2};$ x-intercept (−2, 0)
82. Oblique asymptote $y = x - 1$
83. Medical Dosage. The function

$N (t) = \frac{0.8 t + 1000}{5 t + 4}, t \geq 15,$

gives the body concentration $N (t)$ , in parts per million, of a certain dosage of medication after time t, in hours.
1. Find the horizontal asymptote of the graph and complete the following:
  
  $\begin{array}{l} N (t) \to as t \to \infty . & N (t) \to 0.16 as t \to \infty \end{array}$
2. Explain the meaning of the answer to part (a) in terms of the application.
84. Average Cost. The average cost per light, in dollars, for a company to produce x roadside emergency lights is given by the function

$A (x) = \frac{2 x + 100}{x}, x > 0.$
1. Find the horizontal asymptote of the graph and complete the following:
  
  $A (x) \to as x \to \infty .$
2. Explain the meaning of the answer to part (a) in terms of the application.
85. Population Growth. The population P, in thousands, of a resort community is given by

$P (t) = \frac{500 t}{2 t^{2} + 9},$

where t is the time, in months, since the city council raised the property taxes.
1. Find the population at $t = 0, 1, 3$ , and 8 months.
2. Find the horizontal asymptote of the graph and complete the following:
  
  $P (t) \to as t \to \infty .$
3. Explain the meaning of the answer to part (b) in terms of the application.

Skill Maintenance

Vocabulary Reinforcement

In each of Exercises 86–94, fill in the blank with the correct term. Some of the given choices will not be used. Others will be used more than once.

x-intercept
y-intercept
odd function
even function
domain
range
slope
distance formula
midpoint formula
horizontal lines
vertical lines
point–slope equation
slope–intercept equation
difference quotient
$f (x) = f (- x)$
$f (- x) = - f (x)$

86. A function is a correspondence between a first set, called the ______________, and a second set, called the ______________, such that each member of the ______________ corresponds to exactly one member of the ______________. [1.2]
87. The ______________ of a line containing $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is given by $(y_{2} - y_{1}) / (x_{2} - x_{1})$ . [1.3]
88. The ______________ of the line with slope m and y-intercept (0, b) is $y = m x + b$ . [1.3]
89. The ______________ of the line with slope m passing through $(x_{1}, y_{1})$ is $y - y_{1} = m (x - x_{1})$ . [1.4]
90. A(n) ______________ is a point $(a, 0)$ . [1.1]
91. For each x in the domain of an odd function f, $\underline{f (- x) = - f (x)}$ . [2.4]
92. ______________ are given by equations of the type $x = a$ . [1.3]
93. The ______________ is $(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ . [1.1]
94. A(n) ______________ is a point $(0, b)$ . [1.1]

Synthesis

Find the nonlinear asymptote of the function.

95. $f (x) = \frac{x^{5} + 2 x^{3} + 4 x^{2}}{x^{2} + 2}$
96. $f (x) = \frac{x^{4} + 3 x^{2}}{x^{2} + 1}$

Graph the function.

97. $f (x) = \frac{2 x^{3} + x^{2} - 8 x - 4}{x^{3} + x^{2} - 9 x - 9}$
98. $f (x) = \frac{x^{3} + 4 x^{2} + x - 6}{x^{2} - x - 2}$

..................Content has been hidden....................

You can't read the all page of ebook, please click here login for view all page.

`x`	`y`
−4.5	−0.26
−3.25	−1.19
−2.5	0.42
−0.5	−0.23
0.5	−2.29
0.75	4.8
1.5	0.53
3.5	0.18

Table of Contents for 4.5 Rational Functions

Create new playlist

Sign In

Sign Up

Table of Contents for
4.5 Rational Functions