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Section 3.6 Rational Functions

Before Starting this Section, Review

1 Domain of a function (Section 2.4 , page 216)
2 Zeros of a function (Section 3.2 , page 337)
3 Quadratic formula (Section 1.3 , page 111)
4 Dividing polynomials (Section 3.3 , page 355)

Objectives

1 Define a rational function.
2 Define vertical and horizontal asymptotes.
3 Graph translations of f(x)=1x. $f (x) = \frac{1}{x} .$
4 Find vertical and horizontal asymptotes (if any).
5 Graph rational functions.
6 Graph rational functions with oblique asymptotes.
7 Graph a revenue curve.

Federal Taxes and Revenues

Federal governments levy income taxes on their citizens and corporations to pay for defense, infrastructure, and social services. The relationship between tax rates and tax revenues shown in Figure 3.22 is a “revenue curve.”

All economists agree that a zero tax rate produces no tax revenues and that no one would bother to work with a 100% tax rate, so tax revenue would be zero. It then follows that in a given economy, there is some optimal tax rate T between zero and 100% at which people are willing to maximize their output and still pay their taxes. This optimal rate brings in the most revenue for the government. However, because no one knows the actual shape of the revenue curve, it is impossible to find the exact value of T.

Notice in Figure 3.22 that between the two extreme rates are two rates (such as a and b in the figure) that will collect the same amount of revenue. In a democracy, politicians can argue that taxes are currently too high (at some point b in Figure 3.22) and should therefore be reduced to encourage incentives and harder work (this is supply-side economics); at the same time, the government will generate more revenue. Others can argue that we are well to the left of T (at some point a in Figure 3.22), so the tax rate should be raised (for the rich) to generate more revenue. In Example 10, we sketch the graph of a revenue curve for a hypothetical economy.

Rational Functions

1 Define a rational function.

Recall that a sum, difference, or product of two polynomial functions is also a polynomial function. However, the quotient of two polynomial functions is generally not a polynomial function. In this case, we call the quotient of two polynomial functions a rational function.

Examples of rational functions are

$f (x) = \frac{x - 1}{2 x + 1}, g (y) = \frac{5 y}{y^{2} + 1}, h (t) = \frac{3 t^{2} - 4 t + 1}{t - 2}, and F (z) = 3 z^{4} - 5 z^{2} + 1 .$

( $F (z)$ has the constant polynomial function $D (z) = 1$ as its denominator.) By contrast, $S (x) = \frac{\sqrt{1 - x^{2}}}{x + 1}$ and $G (x) = \frac{| x |}{x}$ are not rational functions.

Example 1 Finding the Domain of a Rational Function

Find the domain of each rational function.

$f (x) = \frac{3 x^{2} - 12}{x - 1}$
$g (x) = \frac{x}{x^{2} - 6 x + 8}$
$h (x) = \frac{x^{2} - 4}{x - 2}$

Solution

We must eliminate all x for which $D (x) = 0,$ so we first solve the equation $D (x) = 0 .$

$D (x) = x - 1 = 0,$ if $x = 1 .$ The domain of $f (x) = \frac{3 x^{2} - 12}{x - 1}$ is the set of all real numbers x except $x = 1$ , or in interval notation, $(- \infty, 1) \cup (1, \infty) .$
The domain of g is the set of all real numbers x for which the denominator D(x) is nonzero.

$\begin{aligned} D (x) = x^{2} - 6 x + 8 = 0 & Set D (x) = 0. \\ (x - 2) (x - 4) = 0 & Factor D (x) . \\ x - 2 = 0 or x - 4 = 0 & Zero-product property \\ x = 2 or x = 4 & Solve for x . \end{aligned}$

The domain of g is the set of all real numbers x except 2 and 4, or in interval notation, $(- \infty, 2) \cup (2, 4) \cup (4, \infty) .$
$D (x) = x - 2 = 0,$ if $x = 2 .$ The domain of $h (x) = \frac{x^{2} - 4}{x - 2}$ is the set of all real numbers x except 2, or in interval notation, $(- \infty, 2) \cup (2, \infty) .$

Practice Problem 1

Find the domain of the rational function $f (x) = \frac{x - 3}{x^{2} - 4 x - 5} .$

In Example 1, part c, note that the functions $h (x) = \frac{x^{2} - 4}{x - 2} = \frac{(x - 2) (x + 2)}{x - 2}$ and $H (x) = x + 2$ are not equal: The domain of H(x) is the set of all real numbers, but the domain of h(x) is the set of all real numbers x except 2. The graph of $y = H (x)$ is a line with slope 1 and y-intercept 2. The graph of $y = h (x)$ is the same line as $y = H (x),$ except that the point (2, 4) is missing. See Figure 3.23.

Figure 3.23

Functions `H`(`x`) and `h`(`x`) have different graphs

As with the quotient of integers, if the polynomials N(x) and D(x) have no common factors, then the rational function $f (x) = \frac{N (x)}{D (x)}$ is said to be in lowest terms. The zeros of N(x) and D(x) will play an important role in graphing the rational function f.

Vertical and Horizontal Asymptotes

2 Define vertical and horizontal asymptotes.

The graph of a rational function may consist of several pieces. Each piece, however, is a continuous and smooth curve. By convention, the pieces are separated by dashed lines through the points that are not in its domain. It is important to study the behavior of a rational function near the points that are excluded from its domain.

Consider the function $f (x) = \frac{1}{x}$ . The domain of this function is $(- \infty, 0) \cup (0, \infty)$ . The function f does not assign any value to $x = 0,$ as f(0) is undefined. However, we more closely examine the behavior of f(x) near the excluded value $x = 0 .$ We start by evaluating f(x) for values of x that are near 0.

Side Note

$\frac{1}{Small Number} = Large Number$

We see that as x is getting closer and closer to 0 from the right (that is, with $x > 0$ ), the values of f(x) increase without bound. In symbols, we write “ $as x \to 0^{+}, f (x) \to \infty .$ ” This statement is read “as x approaches 0 from the right, f(x) approaches infinity.”

Similarly, from the table (with $x < 0$ ), we conclude that “ $as x \to 0^{-}, f (x) \to - \infty$ ” and say that “as x approaches 0 from the left, f(x) approaches negative infinity.”

The graph of $f (x) = \frac{1}{x}$ is the reciprocal function from Chapter 2 (see Figure 3.24). The vertical line $x = 0$ (the y-axis) is a vertical asymptote.

Figure 3.24

Reciprocal function, $f (x) = \frac{1}{x}$

This definition says that if the line $x = a$ is a vertical asymptote of the graph of a function f, then the graph of f near $x = a$ behaves like one of the graphs in Figure 3.25.

Figure 3.25

Behavior of a function near a vertical asymptote

Geometrically speaking, a rational function f(x) has a vertical asymptote $x = a$ if its graph approaches the graph of $x = a$ from the left or right side of a.

If $x = a$ is a vertical asymptote for a rational function f, then a is not in its domain; so the graph of f does not cross the line $x = a .$

Just as with polynomials, we study the end behavior of rational functions. Consider again the function $f (x) = \frac{1}{x}$ . We can examine the behavior of f for very large values of $| x |;$ that is, as $x \to \infty$ or $x \to - \infty$ .

Side Note

$\frac{1}{Large Number} = Small Number$

We can see that the larger x is, the closer $f (x) = \frac{1}{x}$ is to 0. We conclude that as $x \to \infty, f (x) \to 0$ . This statement is read “as x approaches $\infty$ , f(x) approaches 0.” Similarly as $x \to - \infty, f (x) \to 0$ . That means, geometrically speaking, that for very large values of $| x |$ the graph of $y = \frac{1}{x}$ approaches the graph of the horizontal line $y = 0 .$

The x-axis, with equation $y = 0$ , is a horizontal asymptote of the graph of $f (x) = \frac{1}{x}$ (see page 258).

If the line $y = k$ is a horizontal asymptote of the graph of a rational function f, then the end behavior of the graph of f as $x \to \infty$ is similar to that of one of the graphs in Figure 3.26. We can reflect the graphs shown in Figure 3.26 with respect to the y-axis to depict the end behavior of a rational function f as $x \to - \infty$ .

Figure 3.26

End behavior of a function near a horizontal asymptote

Geometrically speaking, a rational function f(x) has horizontal asymptote $y = a$ if its graph approaches the graph of $y = a$ as $x \to - \infty$ or as $x \to \infty$

The graph of a function $y = f (x)$ can cross its horizontal asymptote (in contrast to a vertical asymptote). See Example 7, page 394.

We summarize the symbolic notation that we used in the following table.

Symbol	Read As	Meaning
$x \to a^{-}$	`x` approaches a from the left	`x` is getting arbitrarily close to a with $x < a$
$x \to a^{+}$	`x` approaches a from the right	`x` is getting arbitrarily close to a with $x > a$
$x \to - \infty$	`x` approaches negative infinity	`x` decreases without bound
$x \to \infty$	`x` approaches infinity	`x` increases without bound

Symbol	Read As	Meaning
$f (x) \to L$	`f`(`x`) approaches `L`	`f`(`x`) is getting arbitrarily close to `L`
$f (x) \to - \infty$	`f`(`x`) approaches negative infinity	`f`(`x`) decreases without bound
$f (x) \to \infty$	`f`(`x`) approaches infinity	`f`(`x`) increases without bound

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

3 Graph translations of $f (x) = \frac{1}{x} .$

We can graph any function of the form

$g (x) = \frac{a x + b}{c x + d}$

starting with the graph of $f (x) = \frac{1}{x}$ and using the techniques of vertical stretching and compressing, shifting, and/or reflecting.

Example 2 Graphing Rational Functions Using Translations

Graph each rational function, identify the vertical and horizontal asymptotes, and state the domain and range.

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

Solution

Let $f (x) = \frac{1}{x} .$ We can write g(x) in terms of f(x).

$\begin{array}{rcll} g (x) & = & \frac{- 2}{x + 1} \\ = & - 2 (\frac{1}{x + 1}) \\ = & - 2 f (x + 1) & Replace x by x + 1 in f (x) = \frac{1}{x} . \end{array}$

The graph of $y = f (x + 1)$ is the graph of $y = f (x)$ shifted 1 unit to the left. This moves the vertical asymptote $x = 0,$ 1 unit to the left. The graph of $y = - 2 f (x + 1)$ is the graph of $y = f (x + 1)$ stretched vertically 2 units and then reflected about the x-axis. The graph is shown in Figure 3.27. The domain of g is $(- \infty, - 1) \cup (- 1, \infty)$ , and the range is $(- \infty, 0) \cup (0, \infty) .$ The graph has vertical asymptote $x = - 1$ and horizontal asymptote $y = 0$ (the x-axis).

Figure 3.27

Graph of $g (x) = \frac{- 2}{x + 1}$
Using long division, we have

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

Then $h (x) = f (x - 1) + 3 Replace x with x - 1 in f (x) = \frac{1}{x} .$

We see that the graph of $y = h (x)$ is the graph of $f (x) = \frac{1}{x}$ shifted horizontally 1 unit to the right and then vertically up 3 units. The graph is shown in Figure 3.28. The domain of h is $(- \infty, 1) \cup (1, \infty)$ , and the range is $(- \infty, 3) \cup (3, \infty) .$ The graph of h has vertical asymptote $x = 1$ and horizontal asymptote $y = 3$ .

Figure 3.28

Graph of $h (x) = \frac{3 x - 2}{x - 1}$

Practice Problem 2

Repeat Example 2 with
1. $g (x) = \frac{3}{x - 2} .$
2. $h (x) = \frac{2 x + 5}{x + 1} .$

4 Find vertical and horizontal asymptotes (if any).

This means that the vertical asymptotes (if any) are found by locating the real zeros of the denominator.

Example 3 Finding Vertical Asymptotes

Find all vertical asymptotes of the graph of each rational function.

$f (x) = \frac{1}{x - 1}$
$g (x) = \frac{1}{x^{2} - 9}$
$h (x) = \frac{1}{x^{2} + 1}$

Solution

There are no common factors in the numerator and denominator of $f (x) = \frac{1}{x - 1},$ and the only zero of the denominator is 1. Therefore, $x = 1$ is a vertical asymptote of f(x).
The rational function g(x) is in lowest terms. Factoring $x^{2} - 9 = (x + 3) (x - 3),$ we see that the zeros of the denominator are $- 3$ and 3. Therefore, the lines $x = - 3$ and $x = 3$ are the two vertical asymptotes of g(x).
Because the denominator $x^{2} + 1$ has no real zeros, the graph of h(x) has no vertical asymptotes.

Practice Problem 3

Find the vertical asymptotes of the graph of

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

The next example illustrates that the graph of a rational function may have gaps (missing points) with or without vertical asymptotes.

Example 4 Rational Functions Whose Graphs Have a Hole

Find all vertical asymptotes of the graph of each rational function.

$h (x) = \frac{x^{2} - 9}{x - 3}$
$g (x) = \frac{x + 2}{x^{2} - 4}$

Solution

$\begin{array}{rcll} h (x) & = & \frac{x^{2} - 9}{x - 3} & Given function \\ = & \frac{(x + 3) (x - 3)}{x - 3} & Factor the numerator: x^{2} - 9 = (x + 3) (x - 3) \\ = & x + 3, if x \neq 3 & Simplify . \end{array}$

The graph of h(x) is the line $y = x + 3$ with a gap (or hole) at $x = 3 .$ See Figure 3.29.

The graph of h(x) has no vertical asymptote at $x = 3,$ the zero of the denominator, because the numerator and the denominator have the common factor $(x - 3),$ both with multiplicity 1.

Figure 3.29

Graph with a hole
$\begin{array}{rcll} g (x) & = & \frac{x + 2}{x^{2} - 4} & Given function \\ = & \frac{x + 2}{(x + 2) (x - 2)} & Factor the denominator . \\ = & \frac{1}{x - 2}, x \neq - 2 & Simplify . \end{array}$

The graph of g(x) has a hole at $x = - 2$ and a vertical asymptote $x = 2 .$ See Figure 3.30.

Figure 3.30

Graph with a hole and an asymptote

Practice Problem 4

Find all vertical asymptotes of the graph of $f (x) = \frac{3 - x}{x^{2} - 9} .$

Based on the above discussion, we have

$g (x) = \frac{x + 2}{x^{2} - 4} = \frac{1}{x - 2}, x \neq - 2,$

and we can also describe the behavior of the function g(x) near the hole at $x = - 2 .$ Replacing x by $- 2$ in the last equation $\frac{1}{x - 2}$ gives the value $\frac{1}{- 2 - 2} = - \frac{1}{4}$ . We conclude that

$g (x) \to - \frac{1}{4}, as x \to - 2 .$

The graph of $g (x) = \frac{x + 2}{x^{2} - 4},$ shown in Figure 3.30, has a horizontal asymptote: $y = 0 .$ This can be verified algebraically. Divide the numerator and the denominator of g by $x^{2},$ the highest power of x in the denominator.

As $x \to \pm \infty$ , the expressions $\frac{1}{x}, \frac{2}{x^{2}},$ and $\frac{4}{x^{2}}$ all approach 0; so

$g (x) \to \frac{0 + 0}{1 - 0} = \frac{0}{1} = 0 .$

Because $g (x) \to 0$ as $x \to \pm \infty$ , the line $y = 0$ (the x-axis) is a horizontal asymptote.

Note that although a rational function can have more than one vertical asymptote (see Example 6), it can have, at most, one horizontal asymptote. We can find the horizontal asymptote (if any) of a rational function by dividing the numerator and denominator by the highest power of x that appears in the denominator and investigating the resulting expression as $x \to \pm \infty .$

Alternatively, one may use the following rules.

Example 5 Finding the Horizontal Asymptote

Find the horizontal asymptote (if any) of the graph of each rational function.

$f (x) = \frac{5 x + 2}{1 - 3 x}$
$g (x) = \frac{2 x}{x^{2} + 1}$
$h (x) = \frac{3 x^{2} - 1}{x + 2}$

Solution

The numerator and denominator of $f (x) = \frac{5 x + 2}{1 - 3 x}$ are both of degree 1. The leading coefficient of the numerator is 5, and that of the denominator is $- 3 .$ By Rule 2, the line $y = \frac{5}{- 3} = - \frac{5}{3}$ is the horizontal asymptote of the graph of f.
For the function $g (x) = \frac{2 x}{x^{2} + 1},$ the degree of the numerator is 1 and that of the denominator is 2. By Rule 1, the line $y = 0$ (the x-axis) is the horizontal asymptote.
The degree of the numerator, 2, of h(x) is greater than the degree of its denominator, 1. By Rule 3, the graph of h has no horizontal asymptote.

Practice Problem 5

Find the horizontal asymptote (if any) of the graph of each function.
1. $f (x) = \frac{2 x - 5}{3 x + 4}$
2. $g (x) = \frac{x^{2} + 3}{x - 1}$
3. $h (x) = \frac{100 x + 57}{0.01 x^{3} + 8 x - 9}$

Graphing Rational Functions

5 Graph rational functions.

In the next procedure, we assume the rational function is in lowest terms. When this is not the case, reduce the function to the lowest terms, and plot the “hole” for the factor (or factors) that are divided out. Recall that for a rational expression to equal zero, its numerator must equal zero.

Procedure in Action

Example 6 Graphing a Rational Function

OBJECTIVE	EXAMPLE
Graph $f (x) = \frac{N (x)}{D (x)} .$	Sketch the graph of $f (x) = \frac{2 x^{2} - 2}{x^{2} - 9} .$
Step 0 Plot a hole. Remove any common factor(s) such as $x - a$ of `N`(`x`) and `D`(`x`).	Because $2 x^{2} - 2$ is not zero when $x = \pm 3,$ the Factor Theorem guarantees that there is no common factor of the form $x - a$ for $2 x^{2} - 2$ and $x^{2} - 9 .$
Step 1 Find the intercepts. Because `f` is in lowest terms, the `x`-intercepts are found by solving the equation $N (x) = 0 .$ The `y`-intercept, if there is one, is `f`(0).	$\begin{array}{rcll} 2 x^{2} - 2 & = & 0 & Set N (x) = 0. \\ 2 (x - 1) (x + 1) & = & 0 & Factor . \\ x - 1 = 0 or x + 1 & = & 0 & Zero-product \\ property \\ x = 1 or x & = & - 1 & Solve for x . \end{array}$ The `x`-intercepts are $- 1$ and 1, so the graph passes through the points $(- 1, 0)$ and (1, 0). $\begin{array}{rcll} f (0) & = & \frac{2 {(0)}^{2} - 2}{{(0)}^{2} - 9} & Replace x with 0 in f (x) . \\ = & \frac{2}{9} & Simplify . \end{array}$ The `y`-intercept is $\frac{2}{9},$ so the graph of `f` passes through the point $(0, \frac{2}{9}) .$
Step 2 Find the vertical asymptotes (if any). Solve $D (x) = 0$ to find the vertical asymptotes of the graph. Sketch the vertical asymptotes.	$\begin{array}{rcll} x^{2} - 9 & = & 0 & Set D (x) = 0. \\ (x - 3) (x + 3) & = & 0 & Factor . \\ x - 3 = 0 or x + 3 & = & 0 & Zero-product property \\ x = 3 or x & = & - 3 & Solve for x . \end{array}$ The vertical asymptotes of the graph of `f` are the lines $x = - 3$ and $x = 3 .$ See the figure in step 5.
Step 3 Find the horizontal asymptote (if any). Use the rules on page 391 for finding the horizontal asymptote.	Because $n = m = 2,$ by Rule 2, the horizontal asymptote is $y = \frac{Leading coefficient of N (x)}{Leading coefficient of D (x)} = \frac{2}{1} = 2 .$
Step 4 Locate the graph relative to the horizontal asymptote (if any). If $y = k$ is a horizontal asymptote, divide `N`(`x`) by `D`(`x`) and write $f (x) = \frac{N (x)}{D (x)} = k + \frac{R (x)}{D (x)} .$ Use a sign graph for $f (x) - k = \frac{R (x)}{D (x)}$ and test numbers associated with the zeros of `R`(`x`) and `D`(`x`) to determine intervals where the graph of `f` is above the asymptote $y = k, (f (x) - k is positive)$ and where it is below $y = k, (f (x) - k is negative) .$ The graph of `f` intersects the line $y = k$ if $f (x) - k = 0 .$	So $f (x) = \frac{2 x^{2} - 2}{x^{2} - 9}$ $\begin{array}{r} 2 \\ Using long division: x^{2} - 9 2 x^{2} - 2 \\ \underline{2 x^{2} - 18} \\ 16 \end{array}$ $= 2 + \frac{16}{x^{2} - 9}; R (x) = 16$ has no zeros, and `D`(`x`) has zeros $- 3$ and 3. These zeros divide the `x`-axis into three intervals. See the figure. We choose test points $- 4, 0,$ and 4 to determine the sign of $\frac{16}{x^{2} - 9} .$
Step 5 Sketch the graph. Plot some points and graph the asymptotes found in Steps 1–4. Plot “holes” (if any) in the graph at $x = a$ from Step 0.

Practice Problem 6

Sketch the graph of $f (x) = \frac{2 x}{x^{2} - 1} .$

Notice that the rational function $g (x) = \frac{2 x^{3} - 2 x}{x^{3} - 9 x}$ reduces to f(x) of Example 6. The graph of g is identical to the graph of f with one exception: The graph of g has a “hole” at (0, 2/9), because 0 is not in the domain of g.

Example 7 Graphing a Rational Function

Sketch the graph of $f (x) = \frac{x^{2} + 2}{(x + 2) (x - 1)} .$

Solution

Step 0 f(x) is in lowest terms.
Step 1 Because $x^{2} + 2 > 0,$ the graph has no x-intercepts.

$\begin{array}{rcll} f (0) & = & \frac{0^{2} + 2}{(0 + 2) (0 - 1)} & Replace x with 0 in f (x) . \\ = & - 1 & Simplify . \end{array}$

The y-intercept is $- 1 .$
Step 2 Set $(x + 2) (x - 1) = 0 .$ Solving for x, we have $x = - 2$ or $x = 1 .$ The vertical asymptotes are the lines $x = - 2$ and $x = 1 .$
Step 3 We have $(x + 2) (x - 1) = x^{2} + x - 2,$ so $f (x) = \frac{x^{2} + 2}{x^{2} + x - 2} .$ By Rule 2, page 391, the horizontal asymptote is $y = \frac{1}{1} = 1,$ because the leading coefficient for both the numerator and the denominator is 1.
Step 4 By long division (see margin), $f (x) = \frac{x^{2} + 2}{(x + 2) (x - 1)} = 1 + \frac{4 - x}{(x + 2) (x - 1)};$ $R (x) = 4 - x$ has one zero, 4, and $D (x) = (x + 2) (x - 1)$ has two zeros, $- 2$ and 1. These three zeros divide the x-axis into four intervals. See the figure. We choose test points $- 3, 0, 2,$ and 5 to determine the sign of $f (x) - 1 = \frac{4 - x}{(x + 2) (x - 1)} .$
Step 5 The graph of f is shown in Figure 3.31 .

Figure 3.31

Graph crossing horizontal asymptote

Practice Problem 7

Sketch the graph of $f (x) = \frac{2 x^{2} - 1}{2 x^{2} + x - 3} .$

Graphing a rational function is more complicated than graphing a polynomial function. Plotting a larger number of points and checking to see if the graph of the function crosses its horizontal asymptote (if any) gives additional detail. Notice that in Figure 3.31 the graph of f crosses the horizontal asymptote $y = 1 .$ To find this point of intersection, set $f (x) = 1$ and solve for x.

$\begin{array}{rcll} \frac{x^{2} + 2}{x^{2} + x - 2} & = & 1 & Set f (x) = 1; (x + 2) (x - 1) = x^{2} + x - 2 \\ x^{2} + 2 & = & x^{2} + x - 2 & Multiply both sides by x^{2} + x - 2. \\ x & = & 4 & Solve for x . \end{array}$

The graph of f crosses the horizontal asymptote at the point (4, 1).

Side Note

To determine whether the graph of a rational function with horizontal asymptote $y = k$ intersects the asymptote, solve the equation $f (x) = k .$ Any solution is the first coordinate of a point of intersection.

We know that the graphs of polynomial functions are continuous (all in one piece). The next example shows that the graph of a rational function (other than a polynomial function) can also be continuous.

Example 8 Graphing a Rational Function

Sketch a graph of $f (x) = \frac{x^{2}}{x^{2} + 1} .$

Solution

Step 0 f(x) is in lowest terms.
Step 1 Now $f (0) = 0$ and solving $f (x) = 0$ gives $x = 0 .$ Therefore, 0 is both the x-intercept and the y-intercept for the graph of f.
Step 2 Because $x^{2} + 1 > 0$ for all x, there are no real zeros for the denominator, so there are no vertical asymptotes.
Step 3 By Rule 2 for locating horizontal asymptotes, the horizontal asymptote is $y = 1 .$
Step 4 By long division (see margin), $f (x) = \frac{x^{2}}{x^{2} + 1} = 1 + \frac{- 1}{x^{2} + 1} .$ Neither $R (x) = - 1$ nor $D (x) = x^{2} + 1$ has real zeros. Because $\frac{- 1}{x^{2} + 1}$ is negative for all values of x, the graph of f is always below the line $y = 1 .$
Step 5 The graph of $y = f (x)$ is shown in Figure 3.32 .

$\begin{array}{r} 1 \\ x^{2} + 1 x^{2} \\ \underline{x^{2} + 1} \\ - 1 \end{array}$

Figure 3.32

A continuous rational function

Practice Problem 8

Sketch the graph of $f (x) = \frac{x^{2} + 1}{x^{2} + 2} .$

Oblique Asymptotes

6 Graph rational functions with oblique asymptotes.

Suppose

$f (x) = \frac{N (x)}{D (x)}$

and the degree of N(x) is greater than the degree of D(x). We know from Rule 3 on page 391 that the graph of f has no horizontal asymptote. Use either long division or synthetic division to obtain

$f (x) = \frac{N (x)}{D (x)} = Q (x) + \frac{R (x)}{D (x)},$

where the degree of R(x) is less than the degree of D(x). Rule 1 on page 391 tells us that as $x \to \infty$ or as $x \to - \infty,$ the expression $\frac{R (x)}{D (x)} \to 0 .$

Therefore, as $x \to \pm \infty,$

$f (x) \to Q (x) + 0 = Q (x) .$

This means that as $| x |$ gets very large, the graph of f behaves like the graph of the polynomial Q(x). If the degree of N(x) is exactly one more than the degree of D(x), then Q(x) will have the linear form $m x + b .$ In this case, the graph of f is said to have an oblique (or slant) asymptote. Consider, for example, the function

$f (x) = \frac{x^{2} + x}{x - 1} = x + 2 + \frac{2}{x - 1} Use long division; see margin .$

Now as $x \to \pm \infty,$ the expression $\frac{2}{x - 1} \to 0$ so that the graph of f approaches the graph of the oblique asymptote—the line $y = x + 2,$ as shown in Figure 3.33. The graph of f is above the line $y = x + 2$ on $(1, \infty)$ and below it on $(- \infty, 1)$ .

$\begin{array}{r} x + 2 \\ x - 1 x^{2} + x \\ \underline{x^{2} - x} \\ 2 x \\ \underline{2 x - 2} \\ 2 \end{array}$

Figure 3.33

Graph with vertical and oblique asymptotes

Example 9 Graphing a Rational Function with an Oblique Asymptote

Sketch the graph of $f (x) = \frac{x^{2} - 4}{x + 1} .$

Solution

Step 0 f(x) is in lowest terms.
Step 1 Intercepts: $f (0) = \frac{0 - 4}{0 + 1} = - 4,$ so the y-intercept is $- 4 .$ Set $x^{2} - 4 = 0 .$

We have $x = - 2$ and $x = 2,$ so the x-intercepts are $- 2$ and 2.
Step 2 Vertical asymptotes: Set $x + 1 = 0 .$ We get $x = - 1,$ so the line $x = - 1$ is a vertical asymptote.
Step 3 Asymptotes: Since the degree of the numerator is greater than the degree of the denominator, f(x) has no horizontal asymptote. However, by long division (see margin),

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

As $x \to \pm \infty,$ the expression $\frac{3}{x + 1} \to 0;$ so the line $y = x - 1$ is an oblique asymptote: The graph gets close to the line $y = x - 1$ as $x \to \infty$ and as $x \to - \infty .$

$\begin{array}{r} x - 1 \\ x + 1 x^{2} + 0 x - 4 \\ \underline{x^{2} + x} \\ - x - 4 \\ \underline{- x - 1} \\ - 3 \end{array}$
Step 4 Location: Use the intervals determined by the zeros of the numerator and of the denominator of $f (x) - (x - 1) = \frac{x^{2} - 4}{x + 1} - (x - 1) = \frac{- 3}{x + 1}$ to create a sign graph. The numerator, $- 3$ , has no zero, and the denominator, $x + 1$ , has one zero, $- 1 .$ The zero, $- 1$ , divides the x-axis into two intervals. See the figure. We choose test points $- 3$ and 0 to determine the sign of $f (x) - (x - 1) = \frac{- 3}{x + 1} .$

The graph is above the line $y = x - 1$ on $(- \infty, 1)$ and below it on $(- 1, \infty) .$
Step 5 The graph of f is shown in Figure 3.34 .

Figure 3.34

Practice Problem 9

Sketch the graph of $f (x) = \frac{x^{2} + 2}{x - 1} .$

Graph of a Revenue Curve

7 Graph a revenue curve.

Example 10 Graphing a Revenue Curve

The revenue curve for an economy of a country is given by

$R (x) = \frac{x (100 - x)}{x + 10},$

where x is the tax rate in percent and R(x) is the tax revenue in billions of dollars.

Find and interpret R(10), R(20), R(30), R(40), R(50), and R(60).
Sketch the graph of $y = R (x)$ for $0 \leq x \leq 100 .$
Use a graphing calculator to estimate the tax rate that yields the maximum revenue.

Solution

$R (10) = \frac{10 (100 - 10)}{10 + 10} = $ 45$ billion. This means that if the income is taxed at the rate of 10%, the total revenue for the government will be $45 billion.

$\begin{array}{rcl} Similarly, R (20) & \approx & $ 53.3 billion, \\ R (30) & = & $ 52.5 billion, \\ R (40) & = & $ 48 billion, \\ R (50) & \approx & $ 41.7 billion, and \\ R (60) & \approx & $ 34.3 billion . \end{array}$
The graph of the function $y = R (x)$ for $0 \leq x \leq 100$ is shown in Figure 3.35.

Figure 3.35
From the calculator graph of

$Y = \frac{100 x - x^{2}}{x + 10},$

using the TRACE feature, you can see that the tax rate of about 23% produces the maximum tax revenue of about $53.7 billion for the government.

Practice Problem 10

Repeat Example 10 for $R (x) = \frac{x (100 - x)}{x + 20} .$

Answers to Practice Problems

1. Domain: $(- \infty, - 1) \cup (- 1, 5) \cup (5, \infty)$
2.
1. Domain: $(- \infty, 2) \cup (2, \infty)$ ; range: $(- \infty, 0) \cup (0, \infty)$ ; Vertical asymptote: $x = 2$ ; horizontal asymptote: $y = 0$
2. Domain: $(- \infty, - 1) \cup (- 1, \infty)$ ; range: $(- \infty, 2) \cup (2, \infty)$ ; vertical asymptote: $x = - 1$ ; horizontal asymptote: $y = 2$
3. $x = - 5$ and $x = 2$
4. $x = - 3$
5.
1. $y = \frac{2}{3}$
2. No horizontal asymptote
3. $y = 0$
6. x-intercept: 0; y-intercept: 0; vertical asymptotes: $x = - 1$ and $x = 1$ ; horizontal asymptote: x-axis. The graph is above the x-axis on $(- 1, 0) \cup (1, \infty)$ and below the x-axis on $(- \infty, - 1) \cup (0, 1) .$
7. x-intercepts: $\pm \frac{\sqrt{2}}{2};$ y-intercept: $\frac{1}{3};$ vertical asymptotes: $x = - \frac{3}{2}$ and $x = 1;$ horizontal asymptote: $y = 1;$ The graph is above the line $y = 1$ on $(- \infty, - \frac{3}{2}) \cup (1, \infty)$ and below it on $(- \frac{3}{2}, 1)$ .

The graph crosses the horizontal (2, 1).
8. No x-intercept; y-intercept: $\frac{1}{2};$ no vertical asymptote; horizontal asymptote: $y = 1 .$ The graph is below the line $y = 1$ on $(- \infty, \infty) .$
9. No x-intercept; y-intercept: $- 2;$ vertical asymptote: $x = 1;$ no horizontal asymptote; $y = x + 1$ is an oblique asymptote. The graph is above the line $y = x + 1$ on $(1, \infty)$ and below it on $(- \infty, 1) .$
10.
1. $R (10) = $ 30$ billion
  
  $R (20) = $ 40$ billion
  
  $R (30) = $ 42$ billion
  
  $R (40) = $ 40$ billion
  
  $R (50) = $ 35.7$ billion
  
  $R (60) = $ 30$ billion
3. 29%

Section 3.6 Exercises

Concepts and Vocabulary

A rational function can be expressed in the form .
The line $x = a$ is a vertical asymptote of f if $| f (x) | \to \infty$ as $x \to \underline{}$ or as $x \to \underline{}$ .
The line $y = k$ is a horizontal asymptote of f if $f (x) \to k$ as $x \to \underline{}$ or as $x \to \underline{}$ .
If an asymptote is neither horizontal nor vertical, then it is called a(n) .
True or False. Every rational function has at least one vertical asymptote.
True or False. Every rational function has, at most, one horizontal asymptote.
True or False. The graph of a rational function may cross its vertical asymptote.
True or False. The graph of a rational function may cross its horizontal asymptote.

Building Skills

In Exercises 9–16, find the domain of each rational function.

$f (x) = \frac{x - 3}{x + 4}$
$f (x) = \frac{x + 1}{x - 1}$
$g (x) = \frac{x - 1}{x^{2} + 1}$
$g (x) = \frac{x + 2}{x^{2} + 4}$
$h (x) = \frac{x - 3}{x^{2} - x - 6}$
$h (x) = \frac{x - 7}{x^{2} - 6 x - 7}$
$F (x) = \frac{2 x + 3}{x^{2} - 6 x + 8}$
$F (x) = \frac{3 x - 2}{x^{2} - 3 x + 2}$

In Exercises 17–26, use the graph of the rational function f(x) to complete each statement.

As $x \to 1^{+}, f (x) \to \underline{}$ .
As $x \to 1^{-}, f (x) \to \underline{}$ .
As $x \to - 2^{+}, f (x) \to \underline{}$ .
As $x \to - 2^{-}, f (x) \to \underline{}$ .
As $x \to \infty, f (x) \to \underline{}$ .
As $x \to - \infty, f (x) \to \underline{}$ .
The domain of f(x) is .
There are vertical asymptotes.
The equations of the vertical asymptotes of the graph are and .
The equation of the horizontal asymptote of the graph is .

In Exercises 27–34, graph each rational function as a translation of $f (x) = \frac{1}{x} .$ Identify the vertical and horizontal asymptotes and state the domain and range.

$f (x) = \frac{3}{x - 4}$
$g (x) = \frac{- 4}{x + 3}$
$f (x) = \frac{- x}{3 x + 1}$
$g (x) = \frac{2 x}{4 x - 2}$
$g (x) = \frac{- 3 x + 2}{x + 2}$
$f (x) = \frac{- x + 1}{x - 3}$
$g (x) = \frac{5 x - 3}{x - 4}$
$f (x) = \frac{2 x + 12}{x + 5}$

In Exercises 35–44, find the vertical asymptotes, if any, of the graph of each rational function.

$f (x) = \frac{x}{x - 1}$
$f (x) = \frac{x + 3}{x - 2}$
$g (x) = \frac{(x + 1) (2 x - 2)}{(x - 3) (x + 4)}$
$g (x) = \frac{(2 x - 1) (x + 2)}{(2 x + 3) (3 x - 4)}$
$h (x) = \frac{x^{2} - 1}{x^{2} + x - 6}$
$h (x) = \frac{x^{2} - 4}{3 x^{2} + x - 4}$
$f (x) = \frac{x^{2} - 6 x + 8}{x^{2} - x - 12}$
$f (x) = \frac{x^{2} - 9}{x^{3} - 4 x}$
$g (x) = \frac{2 x + 1}{x^{2} + x + 1}$
$g (x) = \frac{x^{2} - 36}{x^{2} + 5 x + 9}$

In Exercises 45–52, find the horizontal asymptote, if any, of the graph of each rational function.

$f (x) = \frac{x + 1}{x^{2} + 5}$
$f (x) = \frac{2 x - 1}{x^{2} - 4}$
$g (x) = \frac{2 x - 3}{3 x + 5}$
$g (x) = \frac{3 x + 4}{- 4 x + 5}$
$h (x) = \frac{x^{2} - 49}{x + 7}$
$h (x) = \frac{x + 3}{x^{2} - 9}$
$f (x) = \frac{2 x^{2} - 3 x + 7}{3 x^{3} + 5 x + 11}$
$f (x) = \frac{3 x^{3} + 2}{x^{2} + 5 x + 11}$

In Exercises 53–58, match the rational function with its graph.

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

In Exercises 59–74, use the six-step procedure on pages 392 and 393 to graph each rational function.

$f (x) = \frac{2 x}{x - 3}$
$f (x) = \frac{- x}{x - 1}$
$f (x) = \frac{x}{x^{2} - 4}$
$f (x) = \frac{x}{1 - x^{2}}$
$h (x) = \frac{- 2 x^{2}}{x^{2} - 9}$
$h (x) = \frac{4 - x^{2}}{x^{2}}$
$f (x) = \frac{2}{x^{2} - 2}$
$f (x) = \frac{- 2}{x^{2} - 3}$
$g (x) = \frac{x + 1}{(x - 2) (x + 3)}$
$g (x) = \frac{x - 1}{(x + 1) (x - 2)}$
$h (x) = \frac{x^{2}}{x^{2} + 1}$
$h (x) = \frac{2 x^{2}}{x^{2} + 4}$
$f (x) = \frac{x^{3} - 4 x}{x^{3} - 9 x}$
$f (x) = \frac{x^{3} + 32 x}{x^{3} + 8 x}$
$g (x) = \frac{{(x - 2)}^{2}}{x - 2}$
$g (x) = \frac{{(x - 1)}^{2}}{x - 1}$

In Exercises 75–78, find an equation of a rational function having the given asymptotes, intercepts, and graph.

In Exercises 79–86, find the oblique asymptote and sketch the graph of each rational function.

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

Applying the Concepts

For Exercises 87 and 88, use the following definition: Given a cost function C, the average cost of producing the first x items is found by the formula:

$\bar{C} (x) = \frac{C (x)}{x}, x > 0 .$

Average cost. The Genuine Trinket Co. manufactures “authentic” trinkets for gullible tourists. Fixed daily costs are $2000, and it costs $0.50 to produce each trinket.
1. Write the cost function C for producing x trinkets.
2. Write the average cost $\bar{C}$ of producing x trinkets.
3. Find and interpret $\bar{C} (100), \bar{C} (500),$ and $\bar{C} (1000) .$
4. Find and interpret the horizontal asymptote of the rational function $\bar{C} (x) .$
Average cost. The monthly cost C of producing x portable CD players is given by

$C (x) = - 0.002 x^{2} + 6 x + 7000 .$
1. Write the average cost function $\bar{C} (x) .$
2. Find and interpret $\bar{C} (100), \bar{C} (500),$ and $\bar{C} (1000) .$
3. Find and interpret the oblique asymptote of $\bar{C} (x) .$
Biology: birds collecting seeds. A bird is collecting seed from a field that contains 100 grams of seed. The bird collects x grams of seed in t minutes, where

$t = f (x) = \frac{4 x + 1}{100 - x}, 0 < x < 100 .$
1. Sketch the graph of $t = f (x) .$
2. How long does it take the bird to collect
  1. 50 grams?
  2. 75 grams?
  3. 95 grams?
  4. 99 grams?
3. Complete the following statements if applicable:
  1. As $x \to 100^{-}, f (x) \to \underline{}$ .
  2. As $x \to 100^{+}, f (x) \to \underline{}$ .
4. Does the bird ever collect all of the seed from the field?
Criminology. Suppose Las Vegas decides to be crime-free. The estimated cost of catching and convicting x% of the criminals is given by

$C (x) = \frac{1000}{100 - x} million dollars .$
1. Find and interpret C(50), C(75), C(90), and C(99).
2. Sketch a graph of $C (x), 0 \leq x < 100 .$
3. What happens to C(x) as $x \to 100^{-} ?$
4. What percentage of the criminals can be caught and convicted for $30 million?
Environment. Suppose an environmental agency decides to get rid of the impurities from the water of a polluted river. The estimated cost of removing $x %$ of the impurities is given by

$C (x) = \frac{3 x^{2} + 50}{x (100 - x)} billion dollars .$
1. How much will it cost to remove 50% of the impurities?
2. Sketch the graph of $y = C (x) .$
3. Estimate the percentage of the impurities that can be removed at a cost of $30 billion.
Biology. The growth function g(x) describes the growth rate of organisms as a function of some nutrient concentration x. Suppose

$g (x) = a \frac{x}{k + x}, x \geq 0,$

where a and k are positive constants.
1. Find the horizontal asymptote of the graph of g(x). Use the asymptote to explain why a is called the saturation level of the nutrient.
2. Show that k is the half-saturation constant; that is, show that if $x = k,$ then $g (x) = \frac{a}{2} .$
Population of bacteria. The population P (in thousands) of a colony of bacteria at time t (in hours) is given by

$P (t) = \frac{8 t + 16}{2 t + 1}, t \geq 0 .$
1. Find the initial population of the colony. (Find the population at $t = 0$ hours.)
2. What is the long-term behavior of the population? (What happens when $t \to \infty$ ?)
Drug concentration. The concentration c (in milligrams per liter) of a drug in the bloodstream of a patient at time $t \geq 0$ (in hours since the drug was injected) is given by

$c (t) = \frac{5 t}{t^{2} + 1}, t \geq 0 .$
1. By plotting points or by using a graphing calculator, sketch the graph of $y = c (t) .$
2. Find and interpret the horizontal asymptote of the graph of $y = c (t) .$
3. Find the approximate time when the concentration of drug in the bloodstream is maximal.
4. At what time is the concentration level equal to 2 milligrams per liter?
Book publishing. The printing and binding cost for a college algebra book is $10. The editorial cost is $200,000. The first 2500 books are samples and are given free to professors. Let x be the number of college algebra books produced.
1. Write a function f describing the average cost of saleable books.
2. Find the average cost of a saleable book if 10,000 books are produced.
3. How many books must be produced to bring the average cost of a saleable book under $20?
4. Find the vertical and horizontal asymptotes of the graph of $y = f (x) .$ How are they related to this situation?
A “phony” sale. A jewelry merchant at a local mall wants to have a “p percent off” sale. She marks up the stock by q percent to break even with respect to the prices before the sale.
1. Show that $q = \frac{p}{1 - p}, 0 < p < 1 .$
2. Sketch the graph of q in part (a).
3. How much should she mark up the stock to break even if she wants to have a “25% off” sale?

Beyond the Basics

In Exercises 97–106, use transformations of the graph of $y = \frac{1}{x}$ or $y = \frac{1}{x^{2}}$ to graph each rational function f(x).

$f (x) = - \frac{2}{x}$
$f (x) = \frac{1}{2 - x}$
$f (x) = \frac{1}{{(x - 2)}^{2}}$
$f (x) = \frac{1}{x^{2} + 2 x + 1}$
$f (x) = \frac{1}{{(x - 1)}^{2}} - 2$
$f (x) = \frac{1}{{(x + 2)}^{2}} + 3$
$f (x) = \frac{1}{x^{2} + 12 x + 36}$
$f (x) = \frac{3 x^{2} + 18 x + 28}{x^{2} + 6 x + 9}$
$f (x) = \frac{x^{2} - 2 x + 2}{x^{2} - 2 x + 1}$
$f (x) = \frac{2 x^{2} + 4 x - 3}{x^{2} + 2 x + 1}$
Sketch and discuss the graphs of the rational functions of the form $y = \frac{a}{x^{n}},$ where a is a nonzero real number and n is a positive integer, in the following cases:
1. $a > 0$ and n is odd.
2. $a < 0$ and n is odd.
3. $a > 0$ and n is even.
4. $a < 0$ and n is even.
Graphing the reciprocal of a polynomial function. Let f(x) be a polynomial function and $g (x) = \frac{1}{f (x)} .$ Justify the following statements about the graphs of f(x) and g(x).
1. If c is a zero of f(x), then $x = c$ is a vertical asymptote of the graph of g(x).
2. If $f (x) > 0,$ then $g (x) > 0,$ and if $f (x) < 0,$ then $g (x) < 0 .$
3. The graphs of f and g intersect for those values (and only those values) of x for which $f (x) = g (x) = \pm 1 .$
4. When f is increasing (decreasing, constant) on an interval of its domain, g is decreasing (increasing, constant) on that interval.
Use Exercise 108 to sketch the graphs of $f (x) = x^{2} - 4$ and $g (x) = \frac{1}{x^{2} - 4}$ on the same coordinate axes.
Use Exercise 108 to sketch the graphs of $f (x) = x^{2} + 1$ and $g (x) = \frac{1}{x^{2} + 1}$ on the same coordinate axes.
Recall that the inverse of a function f(x) is written as $f^{- 1} (x),$ while the reciprocal of f(x) can be written as either $\frac{1}{f (x)}$ or ${[f (x)]}^{- 1} .$ The point of this exercise is to make clear that the reciprocal of a function has nothing to do with the inverse of a function. Let $f (x) = 2 x + 3 .$ Find both ${[f (x)]}^{- 1}$ and $f^{- 1} (x) .$ Compare the two functions. Graph all three functions on the same coordinate axes.
Let $f (x) = \frac{x - 1}{x + 2} .$ Find ${[f (x)]}^{- 1}$ and $f^{- 1} (x) .$ Graph all three functions on the same coordinate axes.
Sketch the graph of $g (x) = \frac{2 x^{3} + 3 x^{2} + 2 x - 4}{x^{2} - 1} .$ Discuss the end behavior of g(x).
Sketch the graph of $f (x) = \frac{x^{3} - 2 x^{2} + 1}{x - 2} .$ Discuss the end behavior of f(x).

In Exercises 115 and 116, the graph of the rational function f(x) crosses the horizontal asymptote. Graph f(x) and find the points of intersection of the curve with its horizontal asymptote.

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

In Exercises 117–120, find an equation of a rational function f satisfying the given conditions.

Vertical asymptote: $x = 3$

Horizontal asymptote: $y = - 1$

x-intercept: 2
Vertical asymptotes: $x = - 1, x = 1$

Horizontal asymptote: $y = 1$

x-intercept: 0
Vertical asymptote: $x = 0$

Oblique asymptote: $y = - x$

x-intercepts: $- 1, 1$
Vertical asymptote: $x = 3$

Oblique asymptote: $y = x + 4$

y-intercept: $2; f (4) = 14$

In Exercises 121–126, write a rational function f(x) that has all of the characteristics given in the exercise.

Has a vertical asymptote at $x = 2,$ a horizontal asymptote at $y = 1,$ and a y-intercept at $(0, - 2)$
Has vertical asymptotes at $x = 2$ and $x = - 1,$ a horizontal asymptote at $y = 0,$ a y-intercept at (0, 2), and an x-intercept at 4
Has $f (\frac{1}{2}) = 0; f (x) \to 4$ as $x \to \pm \infty, f (x) \to \infty$ as $x \to 1^{-},$ and $f (x) \to \infty$ as $x \to 1^{+}$
Has $f (0) = 0; f (x) \to 2$ as $x \to \pm \infty;$ has no vertical asymptotes and is symmetric about the y-axis
Has $y = 3 x + 2$ as an oblique asymptote; has a vertical asymptote at $x = 1$
Is it possible for the graph of a rational function to have both a horizontal and an oblique asymptote? Explain.

Critical Thinking / Discussion / Writing

Give an example of a rational function that intersects its horizontal asymptote at
1. No points.
2. One point.
3. Two points.
Give an example of a rational function that intersects its oblique asymptote at
1. No points.
2. One point.
3. Two points.
Construct a rational function R(x) that has $y = x + 1$ as its oblique asymptote and the graph of R(x) intersects the asymptote at the points (2, R(2)) and (5, R(5)).
Describe a rational function $R (x) = \frac{N (x)}{D (x)}$ that has asymptote $y = a x + b$ and R(x) intersects the asymptote at n points whose x-coordinates are $c_{1}, c_{2}, \dots, c_{n} .$ Discuss the relationship between the degree of D(x) and the number of points of intersection of R(x) and $y = a x + b$ .

Getting Ready for the Next Section

1. Find the equation of the line in slope–intercept form that passes through the point $(- 5, 3)$ and has slope $- \frac{2}{3} .$
2. From the equation in part (a), find y when $x = 1 .$
1. Find the equation of the line in slope–intercept form that passes through the point $(- 2, - 3)$ and is perpendicular to the line $4 x + 5 y = 6 .$
2. From the equation in part (a), find y when $x = 10 .$
Find c if the line $3 x + c y = 11$ passes through the point (1, 2).
Find k if the graph of the equation $y = k x^{2} + 1$ passes through the point $(- 2, 7) .$ Then find y when $x = 2 .$
Find k if the graph of the equation $y = \frac{k}{x^{2}}$ passes through the point $(\frac{1}{2}, 12) .$ Then find y when $x = 2 .$

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Table of Contents for Section 3.6 Rational Functions

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Table of Contents for
Section 3.6 Rational Functions