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Precalculus Study Guide

Chapter 4: Graphing Rational Functions

Introduction

A rational function is the ratio (or quotient ) of two polynomials. At a zero of the denominator, the rational function is undefined. Such a zero is not in the domain of the rational function.

At a real zero of the denominator of the rational function, there may or may not be a vertical asymptote. If the numerator does not vanish (become zero) at that point, then there is a vertical asymptote. If the numerator does become zero at that point, then the behavior of the function can be determined by factoring both the numerator and the denominator into its linear factors, and reducing the expression to lowest terms. The vanishing or nonvanishing of the numerator of the factored and simplified fraction then determines if there is not or is a vertical asymptote.

As with polynomials, real zeros of the numerator that are not themselves zeros of the denominator are $x$ -intercepts on the graph of the rational function.

Rational functions can also have horizontal asymptotes, horizontal lines to which the function tends as $|x|$ gets arbitrarily large.

Rational functions can even have oblique (slanted) asymptotes, slanted lines to which the function tends as $|x|$ gets arbitrarily large.

Chapter Glossary

The following terms in Chapter 4 are linked to the Maple Math Dictionary.

asymptote

divisor

domain

even (function)

hyperbola

infinity

intercept

limit

negative

odd (function)

polynomial

positive

quotient

ratio

rational function

real number

reciprocal

remainder

Typical Problems

For the rational functions given in Problems 1 - 11, sketch a graph and determine the equations of all asymptotes, horizontal, vertical and oblique. In addition, determine the $y$ - and $x$ -intercepts.

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

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

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

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

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

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

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

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

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

4.10. $f (x) = \frac{x^{2} - 9}{x^{2} - x - 2}$ = $\frac{(x + 3) (x - 3)}{(x + 1) (x - 2)}$

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

Maple Initializations

Before accessing the Maple version of the solutions to the typical problems stated above, initialize Maple by pressing the button provided on the right.

Solutions

Problem 4.1

4.1 - Mathematical Solution

The rational function $f (x) = \frac{1}{x}$ has no $y$ -intercept because $f (0)$ is undefined. It has no $x$ -intercept because the equation $f (x) = \frac{1}{x}$ = 0 has no solutions.

Near $x = 0$ with $x > 0$ , the values of $f (x)$ are large and positive. Near $x = 0$ with $x < 0$ , the values of $f (x)$ are large in magnitude, but negative. These two observations are expressed mathematically with the notation

$lim_{x \to 0^{+}} f (x) = \infty$

and

$lim_{x \to 0^{-}} f (x) = - \infty$

Figure 4.1.1 Graph of $f (x) = \frac{1}{x}$

respectively. Consequently, the $y$ -axis is a vertical asymptote.

For $x$ both large and positive, $f (x)$ is positive but very close to zero. For $x$ negative but large in absolute value, $f (x)$ is negative but very close to zero. These two observations are expressed mathematically with the notation

$lim_{x \to \infty} f (x) = 0$

and

$lim_{x \to - \infty} f (x) = 0$

Consequently, the $x$ -axis is a horizontal asymptote.

A Maple-produced graph of this function is the hyperbola shown in Figure 4.1.1.

4.1 - Maplet Solution

A graph of the rational function

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

and the equations of any of its asymptotes can be found with the Rational Function Tutor .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 4.1.2.

The intercepts can be inferred from the graph and/or the function itself.

To use this tutor, enter the numerator and denominator of the rational function in the boxes provided.

Figure 4.1.2 Thumbnail image of the Rational Function Tutor

The Graph button provides a graph of $f (x)$ . Pressing the button labeled Horizontal adds to the graph (as red lines) any horizontal asymptotes. Pressing the button labeled Vertical adds to the graph (as green lines) any vertical asymptotes. Pressing the button labeled Oblique adds to the graph (as blue lines) any oblique asymptotes. (These actions are not cumulative. Pressing the button labeled Show All adds all possible asymptotes to the graph.)

Asymptotes coincident with an axis are drawn more thickly so that they can be seen in the graph.

In addition to modifying the graph, the buttons for asymptotes provide the equations of the respective asymptotes.

It should be clear from the graph of this function that there are no intercepts.

To launch the Rational Function Tutor, click the following link: Rational Function Tutor

4.1 - Interactive Solution

•	Enter the expression for the given rational function.

•	Context Panel: Assign to a Name≻ $f$

Task Template

(The following task template is built into Maple itself. It is designed to call the built-in Rational Function tutor from the Student Precalculus package. The built-in Rational Function tutor has a slightly different appearance than the maplet designed for this study guide. By calling it from the task template, the information about asymptotes does not disappear when the tutor is closed.)

Tools≻Tasks≻Browse: Algebra≻Rational Function - Graph and Asymptotes

Rational Function Tutor

Enter a rational function $\frac{P (x)}{Q (x)} &colon;$

Asymptotes

Horizontal

Oblique

Vertical

Plot

Asymptotes

Vertical Asymptotes

•	Type $f$ and press the Enter key.

•	Context Panel: Denominator

•	Context Panel: Solve

Horizontal Asymptotes

Behavior as $x \to \infty$

•	Type $f$ and press the Enter key.

•	Context Panel: Constructions≻Limit≻ $x$ ≻infinity

•	Context Panel: Evaluate (from inert)

Behavior as $x \to - \infty$

•	Type $f$ and press the Enter key.

•	Context Panel: Constructions≻Limit≻ $x$ ≻-infinity

•	Context Panel: Evaluate (from inert)

Graph

•	Type $f$ and press the Enter key.

•	Context Panel: Plots≻Plot Builder

$- 1 \leq x \leq 1$

        Options:
                Range: $- 1 \leq y \leq 1$
                Find Discontinuities

4.1 - Programmatic Solution

A graph of the rational function

>	$f ≔ \frac{1}{x}$

appears in Figure 4.1.1. It can be created with the command

>	$plot (f, x = - 5 .. 5, y = - 5 .. 5)$

The graph of $f (x)$ crosses neither the $y$ -axis nor the $x$ -axis, so there are no intercepts.

The horizontal line $y = 0$ is a horizontal asymptote. It is discovered asking (and answering) the question

"To what finite value does $f (x)$ approach as $x$ moves far to the left or right on the number line?"

The mathematically precise way of posing and answering this question is via the limit concept. Since Maple implements this with its limit command, we will make use of it in this chapter. Thus, the horizontal asymptote is found to be

>	$y = limit (f, x = \infty) &semi; y = limit (f, x = - \infty)$

It is possible for $f (x)$ to have one horizontal asymptote for large $x$ , and a different horizontal asymptote for small $x$ . Here, the line $y = 0$ is a horizontal asymptote at both extremities.

The vertical line $x = 0$ is a vertical asymptote. A vertical asymptote can occur at points where the denominator of $f (x)$ becomes zero but where the numerator is nonzero.

As $x$ gets close to zero through small positive numbers, $f (x)$ , the reciprocal of these small positive numbers, returns large positive numbers. Thus, as $x$ approaches zero from the right, $f (x)$ tends to large positive values, a phenomenon we will describe by " $\infty$ ", the infinity symbol.

Alternatively, as $x$ gets close to zero through negative numbers of small absolute value, $f (x)$ , the reciprocal of these negative numbers, returns negative numbers of large absolute value. Thus, as $x$ approaches zero from the left, $f (x)$ tends to negative numbers of large absolute value, a phenomenon we will describe by " $- \infty$ ", the negative infinity symbol.

Again, the formal mathematical language for discussing such behaviors is the limit concept, implemented in Maple as

>	$Limit (f, x = 0, right) = limit (f, x = 0, right) &semi; Limit (f, x = 0, left) = limit (f, x = 0, left)$

where the notation indicates that the approach to $x = 0$ is made either from the right, as in the first computation, or from the left, as in the second.

Problem 4.2

4.2 - Mathematical Solution

The rational function $f (x) = \frac{1}{x^{2}}$ has no $y$ -intercept because $f (0)$ is undefined. It has no $x$ -intercept because the equation $f (x) = \frac{1}{x^{2}}$ = 0 has no solutions.

Near $x = 0$ with $x > 0$ or with $x < 0$ , the values of $f (x)$ are large and positive. These two observations are expressed mathematically with the notation

$lim_{x \to 0^{+}} f (x) = \infty$

and

$lim_{x \to 0^{-}} f (x) = \infty$

Figure 4.2.1 Graph of $f (x) = \frac{1}{x^{2}}$

respectively. Consequently, the $y$ -axis is a vertical asymptote.

For $x$ large in absolute value, $f (x)$ is positive but very close to zero. This observation is expressed mathematically with the notation

$lim_{x \to \infty} f (x) = 0$

and

$lim_{x \to - \infty} f (x) = 0$

Consequently, the $x$ -axis is a horizontal asymptote.

A Maple-produced graph of this function is shown in Figure 4.2.1.

4.2 - Maplet Solution

A graph of the rational function

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

and the equations of any of its asymptotes can be found with the Rational Function Tutor .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 4.2.2.

The intercepts can be inferred from the graph and/or the function itself.

To use this tutor, enter the numerator and denominator of the rational function in the boxes provided.

Figure 4.2.2 Thumbnail image of the Rational Function Tutor

Asymptotes coincident with an axis are drawn more thickly so that they can be seen in the graph.

In addition to modifying the graph, the buttons for asymptotes provide the equations of the respective asymptotes.

It should be clear from the graph of this function that there are no intercepts.

To launch the Rational Function Tutor, click the following link: Rational Function Tutor

4.2 - Interactive Solution

•	Enter the expression for the given rational function.

•	Context Panel: Assign to a Name≻ $f$

Task Template

Tools≻Tasks≻Browse: Algebra≻Rational Function - Graph and Asymptotes

Rational Function Tutor

Enter a rational function $\frac{P (x)}{Q (x)} &colon;$

Asymptotes

Horizontal

Oblique

Vertical

Plot

Asymptotes

Vertical Asymptotes

•	Type $f$ and press the Enter key.

•	Context Panel: Denominator

•	Context Panel: Solve

Horizontal Asymptotes

Behavior as $x \to \infty$

•	Type $f$ and press the Enter key.

•	Context Panel: Constructions≻Limit≻ $x$ ≻infinity

•	Context Panel: Evaluate (from inert)

Behavior as $x \to - \infty$

•	Type $f$ and press the Enter key.

•	Context Panel: Constructions≻Limit≻ $x$ ≻-infinity

•	Context Panel: Evaluate (from inert)

Graph

•	Type $f$ and press the Enter key.

•	Context Panel: Plots≻Plot Builder

$- 1 \leq x \leq 1$

        Options:
                Range: $- 1 \leq y \leq 1$
                Find Discontinuities

4.2 - Programmatic Solution

A graph of the rational function

>	$f ≔ \frac{1}{x^{2}}$

appears in Figure 4.2.1. It can be created with the command

>	$plot (f, x = - 5 .. 5, y = - 1 .. 4)$

The graph of $f (x)$ crosses neither the $y$ -axis nor the $x$ -axis, so there are no intercepts.

The horizontal line $y = 0$ is a horizontal asymptote. It is discovered asking (and answering) the question

"To what finite value does $f (x)$ approach as $x$ moves far to the left or right on the number line?"

The mathematically precise way of posing and answering this question is via the limit concept. Since Maple implements this with it's limit command, we will make use of it in this chapter. Thus, the horizontal asymptote is found to be

>	$y = limit (f, x = \infty) &semi; y = limit (f, x = - \infty)$

It is possible for $f (x)$ to have one horizontal asymptote for large $x$ , and a different horizontal asymptote for small $x$ . Here, the line $y = 0$ is a horizontal asymptote at both extremities.

The vertical line $x = 0$ is a vertical asymptote. A vertical asymptote can occur at points where the denominator of $f (x)$ becomes zero but where the numerator is nonzero.

As $x$ gets close to zero through small positive numbers, $f (x)$ , the square of the reciprocal of these small positive numbers, returns large positive numbers. Thus, as $x$ approaches zero from the right, $f (x)$ tends to large positive values, a phenomenon we will describe by " $\infty$ ", the infinity symbol.

Alternatively, as $x$ gets close to zero through negative numbers of small absolute value, $f (x)$ , the square of the reciprocal of these negative numbers, returns large positive numbers. Thus, as $x$ approaches zero from the left, $f (x)$ tends to large positive numbers.

Again, the formal mathematical language for discussing such behaviors is the limit concept, implemented in Maple as

>	$Limit (f, x = 0, right) = limit (f, x = 0, right) &semi; Limit (f, x = 0, left) = limit (f, x = 0, left)$

where the notation indicates that the approach to $x = 0$ is made either from the right, as in the first computation, or from the left, as in the second.

Problem 4.3

4.3 - Mathematical Solution

The rational function $f (x) = \frac{1}{x^{2} + 1}$ has 1 as its $y$ -intercept because $f (0)$ = 1. It has no $x$ -intercept because the equation $f (x) = \frac{1}{x^{2} + 1}$ = 0 has no solution.

The denominator, namely, $x^{2} + 1$ , does not equal zero for any real $x$ . Hence, there is no vertical asymptote.

For $x$ large in absolute value, $f (x)$ is positive but very close to zero. This observation is expressed mathematically with the notation

Figure 4.3.1 Graph of $f (x) = \frac{1}{x^{2} + 1}$

$lim_{x \to \infty} f (x) = 0$

and

$lim_{x \to - \infty} f (x) = 0$

Consequently, the $x$ -axis is a horizontal asymptote.

A Maple-produced graph of this function is shown in Figure 4.3.1.

4.3 - Maplet Solution

A graph of the rational function

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

and the equations of any of its asymptotes can be found with the Rational Function Tutor .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 4.3.2.

The intercepts can be inferred from the graph and/or the function itself.

To use this tutor, enter the numerator and denominator of the rational function in the boxes provided.

Figure 4.3.2 Thumbnail image of the Rational Function Tutor

Asymptotes coincident with an axis are drawn more thickly so that they can be seen in the graph.

In addition to modifying the graph, the buttons for asymptotes provide the equations of the respective asymptotes.

It should be clear from the graph of this function that there is just a $y$ -intercept. Since this occurs at $x = 0$ , it is given by $f (0) = 1$ .

To launch the Rational Function Tutor, click the following link: Rational Function Tutor

4.3 - Interactive Solution

•	Enter the expression for the given rational function.

•	Context Panel: Assign to a Name≻ $f$

Task Template

Tools≻Tasks≻Browse: Algebra≻Rational Function - Graph and Asymptotes

Rational Function Tutor

Enter a rational function $\frac{P (x)}{Q (x)} &colon;$

Asymptotes

Horizontal

Oblique

Vertical

Plot

Asymptotes

Vertical Asymptotes

•	Type $f$ and press the Enter key.

•	Context Panel: Denominator

•	Context Panel: Solve

Horizontal Asymptotes

Behavior as $x \to \infty$

•	Type $f$ and press the Enter key.

•	Context Panel: Constructions≻Limit≻ $x$ ≻infinity

•	Context Panel: Evaluate (from inert)

Behavior as $x \to - \infty$

•	Type $f$ and press the Enter key.

•	Context Panel: Constructions≻Limit≻ $x$ ≻-infinity

•	Context Panel: Evaluate (from inert)

Graph

•	Type $f$ and press the Enter key.

•	Context Panel: Plots≻Plot Builder

$- 1 \leq x \leq 1$

        Options:
                Range: $- 1 \leq y \leq 1$
                Find Discontinuities

4.3 - Programmatic Solution

A graph of the rational function

>	$f ≔ \frac{1}{x^{2} + 1}$

appears in Figure 4.3.1. It can be created with the command

>	$plot (f, x = - 3 .. 3, y = 0 .. 1, scaling = constrained, tickmarks = [7, 2])$

The graph of $f (x)$ crosses the $y$ -axis but not the $x$ -axis, so there is only a $y$ -intercept, found by computing

>	$eval (f, x = 0)$

The horizontal line $y = 0$ is a horizontal asymptote. It is discovered asking (and answering) the question

"To what finite value does $f (x)$ approach as $x$ moves far to the left or right on the number line?"

>	$y = limit (f, x = \infty) &semi; y = limit (f, x = - \infty)$

It is possible for $f (x)$ to have one horizontal asymptote for large $x$ , and a different horizontal asymptote for small $x$ . Here, the line $y = 0$ is a horizontal asymptote at both extremities.

The denominator, namely,

>	$d ≔ denom (f)$

is never zero, as we see from its graph in Figure 4.3.3.

>	$plot (d, x = - 2 .. 2, - 1 .. 5)$

Figure 4.3.3 Graph of the denominator of $f (x) = \frac{1}{x^{2} + 1}$

Hence, $f (x)$ has no vertical asymptote since vertical asymptotes can occur only at points where the denominator is zero, but the numerator is nonzero.

Problem 4.4

4.4 - Mathematical Solution

The rational function $f (x) = \frac{x}{x^{2} + 1}$ has 0 as its $y$ -intercept because $f (0)$ = 0. It has 0 as its $x$ -intercept because the equation $f (x) = \frac{x}{x^{2} + 1}$ = 0 has the solution $x = 0$ . Consequently, the graph of $f (x)$ passes through the origin.

Because $f (- x) = \frac{(- x)}{{(- x)}^{2} + 1}$ = $- \frac{x}{x^{2} + 1} = - f (x)$ , the function has odd symmetry, just like the function $\sin (x)$ .

The denominator, namely, $x^{2} + 1$ , does not equal zero for any real $x$ . Hence, there is no vertical asymptote.

Figure 4.4.1 Graph of $f (x) = \frac{x}{x^{2} + 1}$

$lim_{x \to \infty} f (x) = 0$ and $lim_{x \to - \infty} f (x) = 0$

Consequently, the $x$ -axis is a horizontal asymptote.

Table 4.4.1 provides a list of values that might help generate a sketch of $f (x)$ if working without the benefits of technology. Figure 4.4.1 is a Maple-generated graph of the function.

$x$	$0$	$\frac{1}{2}$	$1$	$\frac{3}{2}$	$2$	$\frac{5}{2}$	$3$	$\frac{7}{2}$	$4$	$\frac{9}{2}$	$5$
$f (x)$	$0$	$\frac{2}{5}$	$\frac{1}{2}$	$\frac{6}{13}$	$\frac{2}{5}$	$\frac{10}{29}$	$\frac{3}{10}$	$\frac{14}{53}$	$\frac{4}{17}$	$\frac{18}{85}$	$\frac{5}{26}$
$f (x)$	$0$	$0.40$	$0.50$	$0.46$	$0.40$	$0.34$	$0.30$	$0.26$	$0.24$	$0.21$	$0.19$
Table 4.4.1 Values of $f (x)$

4.4 - Maplet Solution

A graph of the rational function

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

and the equations of any of its asymptotes can be found with the Rational Function Tutor .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 4.4.2.

The intercepts can be inferred from the graph and/or the function itself.

To use this tutor, enter the numerator and denominator of the rational function in the boxes provided.

Figure 4.4.2 Thumbnail image of the Rational Function Tutor

Asymptotes coincident with an axis are drawn more thickly so that they can be seen in the graph.

In addition to modifying the graph, the buttons for asymptotes provide the equations of the respective asymptotes.

It should be clear from the graph of this function that the graph of this function passes though the origin. Hence, the $x$ - and $y$ -intercepts are $x = 0$ and $y = 0$ , respectively.

To launch the Rational Function Tutor, click the following link: Rational Function Tutor

4.4 - Interactive Solution

•	Enter the expression for the given rational function.

•	Context Panel: Assign to a Name≻ $f$

Task Template

Tools≻Tasks≻Browse: Algebra≻Rational Function - Graph and Asymptotes

Rational Function Tutor

Enter a rational function $\frac{P (x)}{Q (x)} &colon;$

Asymptotes

Horizontal

Oblique

Vertical

Plot

Asymptotes

Vertical Asymptotes

•	Type $f$ and press the Enter key.

•	Context Panel: Denominator

•	Context Panel: Solve

Horizontal Asymptotes

Behavior as $x \to \infty$

•	Type $f$ and press the Enter key.

•	Context Panel: Constructions≻Limit≻ $x$ ≻infinity

•	Context Panel: Evaluate (from inert)

Behavior as $x \to - \infty$

•	Type $f$ and press the Enter key.

•	Context Panel: Constructions≻Limit≻ $x$ ≻-infinity

•	Context Panel: Evaluate (from inert)

Graph

•	Type $f$ and press the Enter key.

•	Context Panel: Plots≻Plot Builder

$- 1 \leq x \leq 1$

        Options:
                Range: $- 1 \leq y \leq 1$
                Find Discontinuities

4.4 - Programmatic Solution

A graph of the rational function

>	$f ≔ \frac{x}{x^{2} + 1}$

appears in Figure 4.4.1. It can be created with the command

>	$plot (f, x = - 5 .. 5, y = - .5 .. 0.5)$

The graph of $f (x)$ appears to pass through the origin, a hypothesis verified by the calculation

>	$eval (f, x = 0)$

Hence, the origin is both the $x$ -intercept and the $y$ -intercept.

The horizontal line $y = 0$ is a horizontal asymptote. It is discovered asking (and answering) the question

"To what finite value does $f (x)$ approach as $x$ moves far to the left or right on the number line?"

>	$y = limit (f, x = \infty) &semi; y = limit (f, x = - \infty)$

It is possible for $f (x)$ to have one horizontal asymptote for large $x$ , and a different horizontal asymptote for small $x$ . Here, the line $y = 0$ is a horizontal asymptote at both extremities.

The denominator, namely,

>	$d ≔ denom (f)$

is never zero, as we see from its graph in Figure 4.4.3.

>	$plot (d, x = - 2 .. 2, - 1 .. 5)$

Figure 4.4.3 Graph of the denominator of $f (x) = \frac{x}{x^{2} + 1}$

Hence, $f (x)$ has no vertical asymptote since vertical asymptotes can occur only at points where the denominator is zero, but the numerator is nonzero.

Problem 4.5

4.5 - Mathematical Solution

The rational function $f (x) = \frac{x^{2}}{x^{2} + 1}$ has 0 as its $y$ -intercept because $f (0)$ = 0. It has 0 as its $x$ -intercept because the equation $f (x) = \frac{x^{2}}{x^{2} + 1}$ = 0 has the solution $x = 0$ . Consequently, the graph of $f (x)$ passes through the origin.

Because $f (- x) = \frac{{(- x)}^{2}}{{(- x)}^{2} + 1}$ = $\frac{x^{2}}{x^{2} + 1} = f (x)$ , the function has even symmetry, just like the function $\cos (x)$ .

The denominator, namely, $x^{2} + 1$ , does not equal zero for any real $x$ . Hence, there is no vertical asymptote.

Figure 4.5.1 Graph of $f (x) = \frac{x^{2}}{x^{2} + 1}$

For $x$ large in absolute value, $f (x)$ is very close to 1, but remains slightly less than 1. This observation is expressed mathematically with the notation

$lim_{x \to \infty} f (x) = 1$ and $lim_{x \to - \infty} f (x) = 1$

Consequently, the line $y = 1$ is a horizontal asymptote.

Table 4.5.1 provides a list of values that might help generate a sketch of $f (x)$ if working without the benefits of technology. Figure 4.5.1 is a Maple-generated graph of the function.

$x$	$0$	$\frac{1}{2}$	$1$	$\frac{3}{2}$	$2$	$\frac{5}{2}$	$3$	$\frac{7}{2}$	$4$	$\frac{9}{2}$	$5$
$f (x)$	$0$	$\frac{1}{5}$	$\frac{1}{2}$	$\frac{9}{13}$	$\frac{4}{5}$	$\frac{25}{29}$	$\frac{9}{10}$	$\frac{49}{53}$	$\frac{16}{17}$	$\frac{81}{85}$	$\frac{25}{26}$
$f (x)$	$0$	$0.20$	$0.50$	$0.69$	$0.80$	$0.86$	$0.90$	$0.92$	$0.94$	$0.95$	$0.96$
Table 4.5.1 Values of $f (x)$

4.5 - Maplet Solution

A graph of the rational function

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

and the equations of any of its asymptotes can be found with the Rational Function Tutor .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 4.5.2.

The intercepts can be inferred from the graph and/or the function itself.

To use this tutor, enter the numerator and denominator of the rational function in the boxes provided.

Figure 4.5.2 Thumbnail image of the Rational Function Tutor

Asymptotes coincident with an axis are drawn more thickly so that they can be seen in the graph.

In addition to modifying the graph, the buttons for asymptotes provide the equations of the respective asymptotes.

It should be clear from the graph of this function that the graph of this function passes though the origin. Hence, the $x$ - and $y$ -intercepts are $x = 0$ and $y = 0$ , respectively.

To launch the Rational Function Tutor, click the following link: Rational Function Tutor

4.5 - Interactive Solution

•	Enter the expression for the given rational function.

•	Context Panel: Assign to a Name≻ $f$

Task Template

Tools≻Tasks≻Browse: Algebra≻Rational Function - Graph and Asymptotes

Rational Function Tutor

Enter a rational function $\frac{P (x)}{Q (x)} &colon;$

Asymptotes

Horizontal

Oblique

Vertical

Plot

Asymptotes

Vertical Asymptotes

•	Type $f$ and press the Enter key.

•	Context Panel: Denominator

•	Context Panel: Solve

Horizontal Asymptotes

Behavior as $x \to \infty$

•	Type $f$ and press the Enter key.

•	Context Panel: Constructions≻Limit≻ $x$ ≻infinity

•	Context Panel: Evaluate (from inert)

Behavior as $x \to - \infty$

•	Type $f$ and press the Enter key.

•	Context Panel: Constructions≻Limit≻ $x$ ≻-infinity

•	Context Panel: Evaluate (from inert)

Graph

•	Type $f$ and press the Enter key.

•	Context Panel: Plots≻Plot Builder

$- 1 \leq x \leq 1$

        Options:
                Range: $- 1 \leq y \leq 1$
                Find Discontinuities

4.5 - Programmatic Solution

A graph of the rational function

>	$f ≔ \frac{x^{2}}{x^{2} + 1}$

appears in Figure 4.5.2. It can be created with the command

>	$plot (f, x = - 7 .. 7, y = 0 .. 1, tickmarks = [7, 4])$

The graph of $f (x)$ appears to pass through the origin, a hypothesis verified by the calculation

>	$eval (f, x = 0)$

Hence, the origin is both the $x$ -intercept and the $y$ -intercept.

The horizontal line $y = 1$ is a horizontal asymptote. It is discovered asking (and answering) the question

"To what finite value does $f (x)$ approach as $x$ moves far to the left or right on the number line?"

>	$y = limit (f, x = \infty) &semi; y = limit (f, x = - \infty)$

Alternatively, rewrite $f (x)$ as

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

via long division, implemented in Maple via its quo command, as follows.

>	$quotient = quo (numer (f), denom (f), x,' r') &semi; remainder = r$

The quotient of $x^{2}$ divided by $x^{2} + 1$ , namely, 1, is returned directly by the quo (quotient) command. The remainder upon executing this division is stored in the variable $r$ , which here, is $- 1$ . Since the remainder must be divided by the divisor, the result follows.

Since we learned in Problem 3 that $\frac{1}{x^{2} + 1}$ tends to zero for $x$ either very large or small, the quotient 1 signals the horizontal asymptote.

Finally, note that it is possible for $f (x)$ to have one horizontal asymptote for large $x$ , and a different horizontal asymptote for small $x$ . Here, the line $y = 1$ is a horizontal asymptote at both extremities.

The denominator, namely,

>	$d ≔ denom (f)$

is never zero, as we see from its graph in Figure 4.5.3.

>	$plot (d, x = - 2 .. 2, - 1 .. 5)$

Figure 4.5.3 Graph of the denominator of $f (x) = \frac{x^{2}}{x^{2} + 1}$

Hence, $f (x)$ has no vertical asymptote since vertical asymptotes can occur only at points where the denominator is zero, but the numerator is nonzero.

Problem 4.6

4.6 - Mathematical Solution

The rational function $f (x) = \frac{x^{2} - 1}{x^{2} + 1}$ has $- 1$ as its $y$ -intercept because $f (0)$ = $- 1$ . It has $x$ = + 1 as its $x$ -intercepts because the equation $f (x) = \frac{x^{2} - 1}{x^{2} + 1}$ = 0 has the solutions $x = - 1$ and 1.

Because $f (- x) = \frac{{(- x)}^{2} - 1}{{(- x)}^{2} + 1}$ = $\frac{x^{2} - 1}{x^{2} + 1} = f (x)$ , the function has even symmetry, just like the function $\cos (x)$ .

The denominator, namely, $x^{2} + 1$ , does not equal zero for any real $x$ . Hence, there is no vertical asymptote.

Figure 4.6.1 Graph of $f (x) = \frac{x^{2} - 1}{x^{2} + 1}$

For $x$ large in absolute value, $f (x)$ is very close to 1, but remains slightly less than 1. This observation is expressed mathematically with the notation

$lim_{x \to \infty} f (x) = 1$ and $lim_{x \to - \infty} f (x) = 1$

Consequently, the line $y = 1$ is a horizontal asymptote.

Table 4.6.1 provides a list of values that might help generate a sketch of $f (x)$ if working without the benefits of technology. Figure 4.6.1 is a Maple-generated graph of the function.

$x$	$0$	$\frac{1}{2}$	$1$	$\frac{3}{2}$	$2$	$\frac{5}{2}$	$3$	$\frac{7}{2}$	$4$	$\frac{9}{2}$	$5$
$f (x)$	$- 1$	$- \frac{3}{5}$	$0$	$\frac{5}{13}$	$\frac{3}{5}$	$\frac{21}{29}$	$\frac{4}{5}$	$\frac{45}{53}$	$\frac{15}{17}$	$\frac{77}{85}$	$\frac{12}{13}$
$f (x)$	$- 1$	$- 0.60$	$0$	$0.38$	$0.60$	$0.72$	$0.80$	$0.85$	$0.88$	$0.91$	$0.92$
Table 4.6.1 Values of $f (x)$

4.6 - Maplet Solution

A graph of the rational function

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

and the equations of any of its asymptotes can be found with the Rational Function Tutor .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 4.6.2.

The intercepts can be inferred from the graph and/or the function itself.

To use this tutor, enter the numerator and denominator of the rational function in the boxes provided.

Figure 4.6.2 Thumbnail image of the Rational Function Tutor

Asymptotes coincident with an axis are drawn more thickly so that they can be seen in the graph.

In addition to modifying the graph, the buttons for asymptotes provide the equations of the respective asymptotes.

It should be clear from the graph of this function that the graph of this function crosses the $x$ -axis twice. The solutions of $f (x) = 0$ are the solutions of $x^{2} - 1 = 0$ . Hence, the $x$ -intercepts are $x$ = + 1. The $y$ -intercept occurs where $x = 0$ , so it is $f (0) = - 1$ .

To launch the Rational Function Tutor, click the following link: Rational Function Tutor

4.6 - Interactive Solution

•	Enter the expression for the given rational function.

•	Context Panel: Assign to a Name≻ $f$

Task Template

Tools≻Tasks≻Browse: Algebra≻Rational Function - Graph and Asymptotes

Rational Function Tutor

Enter a rational function $\frac{P (x)}{Q (x)} &colon;$

Asymptotes

Horizontal

Oblique

Vertical

Plot

Asymptotes

Vertical Asymptotes

•	Type $f$ and press the Enter key.

•	Context Panel: Denominator

•	Context Panel: Solve

Horizontal Asymptotes

Behavior as $x \to \infty$

•	Type $f$ and press the Enter key.

•	Context Panel: Constructions≻Limit≻ $x$ ≻infinity

•	Context Panel: Evaluate (from inert)

Behavior as $x \to - \infty$

•	Type $f$ and press the Enter key.

•	Context Panel: Constructions≻Limit≻ $x$ ≻-infinity

•	Context Panel: Evaluate (from inert)

Graph

•	Type $f$ and press the Enter key.

•	Context Panel: Plots≻Plot Builder

$- 1 \leq x \leq 1$

        Options:
                Range: $- 1 \leq y \leq 1$
                Find Discontinuities

4.6 - Programmatic Solution

A graph of the rational function

>	$f := \frac{x^{2} - 1}{x^{2} + 1}$

appears in Figure 4.6.1. It can be created with the command

>	$plot (f, x = - 5 .. 5, y = - 1 .. 1, scaling = constrained, tickmarks = [7, 2])$

The graph of $f (x)$ crosses the $y$ -axis as well as the $x$ -axis, so not only is there is a $y$ -intercept, found by computing

>	$eval (f, x = 0)$

but there are also two $x$ -intercepts, found by solving the equation $f (x) = 0$ for $x$ . Using Maple, we find

>	$solve (f = 0, x)$

that is, the two $x$ -intercepts are $x$ = + 1.

The horizontal line $y = 1$ is a horizontal asymptote. It is discovered asking (and answering) the question

"To what finite value does $f (x)$ approach as $x$ moves far to the left or right on the number line?"

>	$y = limit (f, x = \infty) &semi; y = limit (f, x = - \infty)$

Alternatively, rewrite $f (x)$ as

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

via long division, implemented in Maple via its quo command, as follows.

>	$quotient = quo (numer (f), denom (f), x,' r') &semi;$ $remainder = r$

The quotient of $x^{2} - 1$ divided by $x^{2} + 1$ , namely, 1, is returned directly by the quo (quotient) command. The remainder upon executing this division is stored in the variable $r$ , which here, is $- 2$ . Since the remainder must be divided by the divisor, the result follows.

Since we learned in Problem 3 that $\frac{1}{x^{2} + 1}$ tends to zero for $x$ either very large or small, the quotient 1 signals the horizontal asymptote.

The denominator, namely,

>	$d := denom (f)$

is never zero, as we see from its graph in Figure 4.6.3.

>	$plot (d, x = - 2 .. 2, - 1 .. 5)$

Figure 4.6.3 Graph of the denominator of $f (x) = \frac{x^{2} - 1}{x^{2} + 1}$

Hence, $f (x)$ has no vertical asymptote since vertical asymptotes can occur only at points where the denominator is zero, but the numerator is nonzero.

Problem 4.7

4.7 - Mathematical Solution

The rational function $f (x) = \frac{1}{x^{2} - x - 2}$ = $\frac{1}{(x + 1) (x - 2)}$ has $- \frac{1}{2}$ as its $y$ -intercept because $f (0) = - \frac{1}{2}$ .

It has no $x$ -intercept because the equation $f (x) = 0$ has no solution.

The denominator, namely, $x^{2} - x - 2 = (x + 1) (x - 2)$ has two zeros, namely, $x = - 1$ and $x = 2$ . Because the numerator is 1, there are then two vertical asymptotes, namely, the vertical lines $x = - 1$ and $x = 2$ .

Slightly to the left of $x = - 1$ , the function assumes large positive values, but slightly to the right of this asymptote, the function assumes negative values with large absolute values. These observations are expressed mathematically in the notation

Figure 4.7.1 Graph of $f (x) = \frac{1}{x^{2} - x - 2}$

$lim_{x \to {- 1}^{-}} f (x) = \infty$ and $lim_{x \to {- 1}^{+}} f (x) = - \infty$

Slightly to the left of $x = 2$ , the function assumes negative values with large absolute values, but slightly to the right of this asymptote, the function assumes large positive values. These observations are expressed mathematically in the notation

$lim_{x \to 2^{-}} f (x) = - \infty$ and $lim_{x \to 2^{+}} f (x) = \infty$

For large absolute values of $x$ , the values of the function are positive, but very close to zero. This observation is expressed mathematically in the notation

$lim_{x \to - \infty} f (x) = 0$ and $lim_{x \to \infty} f (x) = 0$

Thus, the $x$ -axis is a horizontal asymptote for the function.

A Maple-produced graph of this function is shown in Figure 4.7.1.

4.7 - Maplet Solution

A graph of the rational function

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

and the equations of any of its asymptotes can be found with the Rational Function Tutor .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 4.7.2.

The intercepts can be inferred from the graph and/or the function itself.

To use this tutor, enter the numerator and denominator of the rational function in the boxes provided.

Note: When entering an expression in a tutor window, use * for multiplication. Thus, for this example the denominator is entered as (x+1)*(x-2).

Figure 4.7.2 Thumbnail image of the Rational Function Tutor

Asymptotes coincident with an axis are drawn more thickly so that they can be seen in the graph.

In addition to modifying the graph, the buttons for asymptotes provide the equations of the respective asymptotes.

It should be clear from the graph of this function that the graph of this function does not cross the $x$ -axis. Hence, there are no $x$ -intercepts. However, the $y$ -intercept occurs where $x = 0$ , so it is given by $f (0) = - \frac{1}{2}$ .

To launch the Rational Function Tutor, click the following link: Rational Function Tutor

4.7 - Interactive Solution

•	Enter the expression for the given rational function.

•	Context Panel: Assign to a Name≻ $f$

Task Template

Tools≻Tasks≻Browse: Algebra≻Rational Function - Graph and Asymptotes

Rational Function Tutor

Enter a rational function $\frac{P (x)}{Q (x)} &colon;$

Asymptotes

Horizontal

Oblique

Vertical

Plot

Asymptotes

Vertical Asymptotes

•	Type $f$ and press the Enter key.

•	Context Panel: Denominator

•	Context Panel: Solve

Horizontal Asymptotes

Behavior as $x \to \infty$

•	Type $f$ and press the Enter key.

•	Context Panel: Constructions≻Limit≻ $x$ ≻infinity

•	Context Panel: Evaluate (from inert)

Behavior as $x \to - \infty$

•	Type $f$ and press the Enter key.

•	Context Panel: Constructions≻Limit≻ $x$ ≻-infinity

•	Context Panel: Evaluate (from inert)

Graph

•	Type $f$ and press the Enter key.

•	Context Panel: Plots≻Plot Builder

$- 1 \leq x \leq 1$

        Options:
                Range: $- 1 \leq y \leq 1$
                Find Discontinuities

4.7 - Programmatic Solution

A graph of the rational function

>	$f := \frac{1}{x^{2} - x - 2}$

appears in Figure 4.7.1. It can be created with the command

>	$plot (f, x = - 5 .. 5, y = - 3 .. 3)$

The graph of $f (x)$ crosses the $y$ -axis but not the $x$ -axis, so there is a $y$ -intercept found in Maple with

>	$y = eval (f, x = 0)$

The horizontal line $y = 0$ is a horizontal asymptote. It is discovered asking (and answering) the question

"To what finite value does $f (x)$ approach as $x$ moves far to the left or right on the number line?"

>	$y = limit (f, x = \infty) &semi; y = limit (f, x = - \infty)$

It is possible for $f (x)$ to have one horizontal asymptote for large $x$ , and a different horizontal asymptote for small $x$ . Here, the line $y = 0$ is a horizontal asymptote at both extremities.

The vertical lines $x = - 1$ and $x = 2$ are vertical asymptotes. A vertical asymptote can occur at points where the denominator of $f (x)$ becomes zero but where the numerator is nonzero.

The zeros of the denominator of $f (x)$ are found in Maple as follows. $discont = true$

>	$solve (denom (f) = 0, x)$

The behavior of $f (x)$ near $x = - 1$ is determined by the two limits

>	$Limit (f, x = - 1, right) = limit (f, x = - 1, right) &semi; Limit (f, x = - 1, left) = limit (f, x = - 1, left)$

The behavior of $f (x)$ near $x = 2$ is determined by the two limits

>	$Limit (f, x = 2, right) = limit (f, x = 2, right) &semi; Limit (f, x = 2, left) = limit (f, x = 2, left)$

Problem 4.8

4.8 - Mathematical Solution

The rational function $f (x) = \frac{x}{x^{2} - x - 2}$ = $\frac{x}{(x + 1) (x - 2)}$ has $0$ as its $y$ -intercept because $f (0) = 0$ . It has zero as its $x$ -intercept because the equation $f (x) = 0$ has the solution $x = 0$ . Thus, the graph of the function passes through the origin.

The denominator, namely, $x^{2} - x - 2 = (x + 1) (x - 2)$ has two zeros, namely, $x = - 1$ and $x = 2$ . Because the numerator is not zero at either of these locations, there are then two vertical asymptotes, namely, the vertical lines $x = - 1$ and $x = 2$ .

Slightly to the left of $x = - 1$ , the function assumes negative values with large absolute values, but slightly to the right of this asymptote, the function assumes large positive values. These observations are expressed mathematically in the notation

$lim_{x \to {- 1}^{-}} f (x) = - \infty$ and $lim_{x \to {- 1}^{+}} f (x) = \infty$

Figure 4.8.1 Graph of $f (x) = \frac{x}{x^{2} - x - 2}$

$lim_{x \to 2^{-}} f (x) = - \infty$ and $lim_{x \to 2^{+}} f (x) = \infty$

For large positive values of $x$ , the values of the function are positive, but very close to zero. For negative values of $x$ with large absolute value, the values of the function are negative, but again close to zero.

These observations are expressed mathematically in the notation

$lim_{x \to - \infty} f (x) = 0$ and $lim_{x \to \infty} f (x) = 0$

Thus, the $x$ -axis is a horizontal asymptote for this function.

A Maple-produced graph of this function is shown in Figure 4.8.1.

4.8 - Maplet Solution

A graph of the rational function

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

and the equations of any of its asymptotes can be found with the Rational Function Tutor .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 4.8.2.

The intercepts can be inferred from the graph and/or the function itself.

To use this tutor, enter the numerator and denominator of the rational function in the boxes provided.

Figure 4.8.2 Thumbnail image of the Rational Function Tutor

Asymptotes coincident with an axis are drawn more thickly so that they can be seen in the graph.

In addition to modifying the graph, the buttons for asymptotes provide the equations of the respective asymptotes.

It should be clear from the graph of this function that just one branch the graph of this function passes though the origin. Hence, the $x$ - and $y$ -intercepts are $x = 0$ and $y = 0$ , respectively.

To launch the Rational Function Tutor, click the following link: Rational Function Tutor

4.8 - Interactive Solution

•	Enter the expression for the given rational function.

•	Context Panel: Assign to a Name≻ $f$

Task Template

Tools≻Tasks≻Browse: Algebra≻Rational Function - Graph and Asymptotes

Rational Function Tutor

Enter a rational function $\frac{P (x)}{Q (x)} &colon;$

Asymptotes

Horizontal

Oblique

Vertical

Plot

Asymptotes

Vertical Asymptotes

•	Type $f$ and press the Enter key.

•	Context Panel: Denominator

•	Context Panel: Solve

Horizontal Asymptotes

Behavior as $x \to \infty$

•	Type $f$ and press the Enter key.

•	Context Panel: Constructions≻Limit≻ $x$ ≻infinity

•	Context Panel: Evaluate (from inert)

Behavior as $x \to - \infty$

•	Type $f$ and press the Enter key.

•	Context Panel: Constructions≻Limit≻ $x$ ≻-infinity

•	Context Panel: Evaluate (from inert)

Graph

•	Type $f$ and press the Enter key.

•	Context Panel: Plots≻Plot Builder

$- 1 \leq x \leq 1$

        Options:
                Range: $- 1 \leq y \leq 1$
                Find Discontinuities

4.8 - Programmatic Solution

A graph of the rational function

>	$f := \frac{x}{x^{2} - x - 2}$

appears in Figure 4.8.1. It can be created with the command

>	$plot (f, x = - 5 .. 5, y = - 3 .. 3)$

The graph of $f (x)$ appears to pass through the origin, a hypothesis verified by the calculation

>	$y = eval (f, x = 0)$

Hence, the origin is both the $x$ -intercept and the $y$ -intercept.

The horizontal line $y = 0$ is a horizontal asymptote. It is discovered asking (and answering) the question

"To what finite value does $f (x)$ approach as $x$ moves far to the left or right on the number line?"

>	$y = limit (f, x = \infty) &semi; y = limit (f, x = - \infty)$

It is possible for $f (x)$ to have one horizontal asymptote for large $x$ , and a different horizontal asymptote for small $x$ . Here, the line $y = 0$ is a horizontal asymptote at both extremities.

The vertical lines $x = - 1$ and $x = 2$ are vertical asymptotes. A vertical asymptote can occur at points where the denominator of $f (x)$ becomes zero but where the numerator is nonzero.

The zeros of the denominator of $f (x)$ are found in Maple as follows.

>	$solve (denom (f) = 0, x)$

The behavior of $f (x)$ near $x = - 1$ is determined by the two limits

>	$Limit (f, x = - 1, right) = limit (f, x = - 1, right) &semi; Limit (f, x = - 1, left) = limit (f, x = - 1, left)$

The behavior of $f (x)$ near $x = 2$ is determined by the two limits

>	$Limit (f, x = 2, right) = limit (f, x = 2, right) &semi; Limit (f, x = 2, left) = limit (f, x = 2, left)$

Problem 4.9

4.9 - Mathematical Solution

The rational function $f (x) = \frac{x^{2}}{x^{2} - x - 2}$ = $\frac{x^{2}}{(x + 1) (x - 2)}$ has $0$ as its $y$ -intercept because $f (0) = 0$ . It has zero as its $x$ -intercept because the equation $f (x) = 0$ has the solution $x = 0$ . Thus, the graph of the function passes through the origin.

Slightly to the left of $x = - 1$ , the function assumes large positive values, but slightly to the right of this asymptote, the function assumes negative values with large absolute value. These observations are expressed mathematically in the notation

Figure 4.9.1 Graph of $f (x) = \frac{x^{2}}{x^{2} - x - 2}$

$lim_{x \to {- 1}^{-}} f (x) = \infty$ and $lim_{x \to {- 1}^{+}} f (x) = - \infty$

$lim_{x \to 2^{-}} f (x) = - \infty$ and $lim_{x \to 2^{+}} f (x) = \infty$

For large positive values of $x$ , the values of the function are very close to 1. This is because for such values of $x$ , the function behaves essentially like the fraction $\frac{x^{2}}{x^{2}} = 1$ .

These observations are expressed mathematically in the notation

$lim_{x \to - \infty} f (x) = 1$ and $lim_{x \to \infty} f (x) = 1$

Thus, the line $y = 1$ is a horizontal asymptote for this function.

A Maple-produced graph of this function is shown in Figure 4.9.1.

4.9 - Maplet Solution

A graph of the rational function

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

and the equations of any of its asymptotes can be found with the Rational Function Tutor .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail in Figure 4.9.2.

The intercepts can be inferred from the graph and/or the function itself.

To use this tutor, enter the numerator and denominator of the rational function in the boxes provided.

Figure 4.9.2 Thumbnail image of the Rational Function Tutor

Asymptotes coincident with an axis are drawn more thickly so that they can be seen in the graph.

In addition to modifying the graph, the buttons for asymptotes provide the equations of the respective asymptotes.

To launch the Rational Function Tutor, click the following link: Rational Function Tutor

4.9 - Interactive Solution

•	Enter the expression for the given rational function.

•	Context Panel: Assign to a Name≻ $f$

Task Template

Tools≻Tasks≻Browse: Algebra≻Rational Function - Graph and Asymptotes

Rational Function Tutor

Enter a rational function $\frac{P (x)}{Q (x)} &colon;$

Asymptotes

Horizontal

Oblique

Vertical

Plot

Asymptotes

Vertical Asymptotes

•	Type $f$ and press the Enter key.

•	Context Panel: Denominator

•	Context Panel: Solve

Horizontal Asymptotes

Behavior as $x \to \infty$

•	Type $f$ and press the Enter key.

•	Context Panel: Constructions≻Limit≻ $x$ ≻infinity

•	Context Panel: Evaluate (from inert)

Behavior as $x \to - \infty$

•	Type $f$ and press the Enter key.

•	Context Panel: Constructions≻Limit≻ $x$ ≻-infinity

•	Context Panel: Evaluate (from inert)

Graph

•	Type $f$ and press the Enter key.

•	Context Panel: Plots≻Plot Builder

$- 1 \leq x \leq 1$

        Options:
                Range: $- 1 \leq y \leq 1$
                Find Discontinuities

4.9 - Programmatic Solution

A graph of the rational function

>	$f := \frac{x^{2}}{x^{2} - x - 2}$

appears in Figure 4.9.1. It can be created with the command

>	$plot (f, x = - 5 .. 7, y = - 3 .. 5)$

The graph of

f (x)

appears to pass through the origin, a hypothesis verified by the calculation

>	$y = eval (f, x = 0)$

Hence, the origin is both the $x$ -intercept and the $y$ -intercept.

The horizontal line $y = 1$ is a horizontal asymptote. It is discovered asking (and answering) the question

"To what finite value does $f (x)$ approach as $x$ moves far to the left or right on the number line?"

>	$y = limit (f, x = \infty) &semi; y = limit (f, x = - \infty)$

It is possible for $f (x)$ to have one horizontal asymptote for large $x$ , and a different horizontal asymptote for small $x$ . Here, the line $y = 1$ is a horizontal asymptote at both extremities.

The vertical lines $x = - 1$ and $x = 2$ are vertical asymptotes. A vertical asymptote can occur at points where the denominator of $f (x)$ becomes zero but where the numerator is nonzero.

The zeros of the denominator of $f (x)$ are found in Maple as follows.

>	$solve (denom (f) = 0, x)$

The behavior of $f (x)$ near $x = - 1$ is determined by the two limits

>	$Limit (f, x = - 1, right) = limit (f, x = - 1, right) &semi; Limit (f, x = - 1, left) = limit (f, x = - 1, left)$

The behavior of $f (x)$ near $x = 2$ is determined by the two limits

>	$Limit (f, x = 2, right) = limit (f, x = 2, right) &semi; Limit (f, x = 2, left) = limit (f, x = 2, left)$

Problem 4.10

4.10 - Mathematical Solution

The rational function $f (x) = \frac{x^{2} - 9}{x^{2} - x - 2}$ = $\frac{x^{2} - 9}{(x + 1) (x - 2)}$ has $\frac{9}{2}$ as its $y$ -intercept because $f (0) = \frac{9}{2}$ . It has $x = - 3$ and $x = 3$ as its $x$ -intercepts because the equation $f (x) = 0$ has the solutions $x = \pm 3$ .

Slightly to the left of $x = - 1$ , the function assumes negative values with large absolute value, but slightly to the right of this asymptote, the function assumes large positive values. These observations are expressed mathematically in the notation

Figure 4.10.1 Graph of $f (x) = \frac{x^{2} - 9}{x^{2} - x - 2}$

$lim_{x \to {- 1}^{-}} f (x) = - \infty$ and $lim_{x \to {- 1}^{+}} f (x) = \infty$

Slightly to the left of $x = 2$ , the function assumes large positive values, but slightly to the right of this asymptote, the function assumes negative values with large absolute value. These observations are expressed mathematically in the notation

$lim_{x \to 2^{-}} f (x) = \infty$ and $lim_{x \to 2^{+}} f (x) = - \infty$

These observations are expressed mathematically in the notation

$lim_{x \to - \infty} f (x) = 1$ and $lim_{x \to \infty} f (x) = 1$

Thus, the line $y = 1$ is a horizontal asymptote for this function.

A Maple-produced graph of this function is shown in Figure 4.10.1.

4.10 - Maplet Solution

A graph of the rational function

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

and the equations of any of its asymptotes can be found with the Rational Function Tutor .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 4.10.2.

The intercepts can be inferred from the graph and/or the function itself.

To use this tutor, enter the numerator and denominator of the rational function in the boxes provided.

Figure 4.10.2 Thumbnail image of the Rational Function Tutor

Asymptotes coincident with an axis are drawn more thickly so that they can be seen in the graph.

In addition to modifying the graph, the buttons for asymptotes provide the equations of the respective asymptotes.

It should be clear from the graph of this function that the graph of this function crosses the $x$ -axis twice, and the $y$ -axis once. The $x$ -intercepts are the solutions of the equation $f (x) = 0$ , or more simply, $x^{2} - 9 = 0$ . Hence, the $x$ -intercepts are given by $x$ = + 3, whereas the $y$ -intercept is given by $f (0) = \frac{9}{2}$ .

To launch the Rational Function Tutor, click the following link: Rational Function Tutor

4.10 - Interactive Solution

•	Enter the expression for the given rational function.

•	Context Panel: Assign to a Name≻ $f$

Task Template

Tools≻Tasks≻Browse: Algebra≻Rational Function - Graph and Asymptotes

Rational Function Tutor

Enter a rational function $\frac{P (x)}{Q (x)} &colon;$

Asymptotes

Horizontal

Oblique

Vertical

Plot

Asymptotes

Vertical Asymptotes

•	Type $f$ and press the Enter key.

•	Context Panel: Denominator

•	Context Panel: Solve

Horizontal Asymptotes

Behavior as $x \to \infty$

•	Type $f$ and press the Enter key.

•	Context Panel: Constructions≻Limit≻ $x$ ≻infinity

•	Context Panel: Evaluate (from inert)

Behavior as $x \to - \infty$

•	Type $f$ and press the Enter key.

•	Context Panel: Constructions≻Limit≻ $x$ ≻-infinity

•	Context Panel: Evaluate (from inert)

Graph

•	Type $f$ and press the Enter key.

•	Context Panel: Plots≻Plot Builder

$- 1 \leq x \leq 1$

        Options:
                Range: $- 1 \leq y \leq 1$
                Find Discontinuities

4.10 - Programmatic Solution

A graph of the rational function

>	$f := \frac{x^{2} - 9}{x^{2} - x - 2}$

appears in Figure 4.10.1. It can created with the command

>	$plot (f, x = - 5 .. 7, y = - 3 .. 7)$

The graph of $f (x)$ crosses both axes. Hence, the $y$ -intercept is given by

>	$y = eval (f, x = 0)$

whereas the $x$ -intercepts are given by $x = c$ , where $c$ is either of the two numbers

>	$solve (f = 0, x)$

The horizontal line $y = 1$ is a horizontal asymptote. It is discovered asking (and answering) the question

"To what finite value does $f (x)$ approach as $x$ moves far to the left or right on the number line?"

>	$y = limit (f, x = \infty) &semi; y = limit (f, x = - \infty)$

It is possible for $f (x)$ to have one horizontal asymptote for large $x$ , and a different horizontal asymptote for small $x$ . Here, the line $y = 1$ is a horizontal asymptote at both extremities.

The vertical lines $x = - 1$ and $x = 2$ are vertical asymptotes. A vertical asymptote can occur at points where the denominator of $f (x)$ becomes zero but where the numerator is nonzero.

The zeros of the denominator of $f (x)$ are found in Maple as follows.

>	$solve (denom (f) = 0, x)$

The behavior of $f (x)$ near $x = - 1$ is determined by the two limits

>	$Limit (f, x = - 1, right) = limit (f, x = - 1, right) &semi; Limit (f, x = - 1, left) = limit (f, x = - 1, left)$

The behavior of $f (x)$ near $x = 2$ is determined by the two limits

>	$Limit (f, x = 2, right) = limit (f, x = 2, right) &semi; Limit (f, x = 2, left) = limit (f, x = 2, left)$

Problem 4.11

4.11 - Mathematical Solution

The rational function $f (x) = \frac{x^{3}}{x^{2} + 1}$ has $y = 0$ as its $y$ -intercept because $f (0) = 0$ . It has $x = 0$ as its $x$ -intercept because $x^{3} = 0$ implies $x = 0$ .

There is neither a horizontal nor a vertical asymptote. However, there is an oblique (slant) asymptote given by $y = x$ . The equation for the oblique asymptote can be found by long division, as represented in the tableau in Table 4.11.1.

$x$

________

$x^{2} + 1$ ) $x^{3}$

$x^{3} + x$

_______

$- x$

Table 4.11.1 Long division

Consequently, we can write

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

From this representation, it follows that as $|x|$ gets large, the fractional part (i.e., the remainder part) tends to zero. Hence, the function approaches the oblique asymptote $y = x$ .

There are other calculus-based techniques that could be used to determine the oblique asymptote, but they would be beyond the scope of the present discussion.

We end this discussion with the Maple-generated graph in Figure 4.11.1.

Figure 4.11.1 Graph of $f (x) = \frac{x^{3}}{x^{2} + 1}$

4.11 - Maplet Solution

A graph of the rational function

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

and the equations of any of its asymptotes can be found with the Rational Function Tutor .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 4.11.2.

The intercepts can be inferred from the graph and/or the function itself.

To use this tutor, enter the numerator and denominator of the rational function in the boxes provided.

Figure 4.11.2 Thumbnail image of the Rational Function Tutor

Asymptotes coincident with an axis are drawn more thickly so that they can be seen in the graph.

In addition to modifying the graph, the buttons for asymptotes provide the equations of the respective asymptotes.

It should be clear from the graph of this function that the graph of this function passes through the origin. Hence, the $x$ - and $y$ -intercepts are $x = 0$ and $y = 0$ , respectively.

To launch the Rational Function Tutor, click the following link: Rational Function Tutor

4.11 - Interactive Solution

•	Enter the expression for the given rational function.

•	Context Panel: Assign to a Name≻ $f$

Task Template

Tools≻Tasks≻Browse: Algebra≻Rational Function - Graph and Asymptotes

Rational Function Tutor

Enter a rational function $\frac{P (x)}{Q (x)} &colon;$

Asymptotes

Horizontal

Oblique

Vertical

Plot

Asymptotes

Vertical Asymptotes

•	Type $f$ and press the Enter key.

•	Context Panel: Denominator

•	Context Panel: Solve

Horizontal Asymptotes

Behavior as $x \to \infty$

•	Type $f$ and press the Enter key.

•	Context Panel: Constructions≻Limit≻ $x$ ≻infinity

•	Context Panel: Evaluate (from inert)

Behavior as $x \to - \infty$

•	Type $f$ and press the Enter key.

•	Context Panel: Constructions≻Limit≻ $x$ ≻-infinity

•	Context Panel: Evaluate (from inert)

Oblique Asymptotes

•	Type $f$ and press the Enter key.

•	Context Panel: Conversions≻Partial Fractions≻ $x$

Graph

•	Type $f$ and press the Enter key.

•	Context Panel: Plots≻Plot Builder

$- 1 \leq x \leq 1$

        Options:
                Range: $- 1 \leq y \leq 1$
                Find Discontinuities

4.11 - Programmatic Solution

A graph of the rational function

>	$f := \frac{x^{3}}{x^{2} + 1}$

appears in Figure 4.11.1. It can be created with the command

>	$plot (f, x = - 2 .. 2, y = - 2 .. 2)$

The graph of $f (x)$ passes through the origin. Hence, the $y$ -intercept is given by

>	$y = eval (f, x = 0)$

The graph of $f (x)$ passes through the origin. Hence, the $y$ -intercept is given by

>	$y = eval (f, x = 0)$

and the $x$ -intercept is given by $x = 0$ .

There is neither a horizontal nor a vertical asymptote. However, there is an oblique (or slant) asymptote, found algebraically by long dividing the denominator into the numerator. In Maple, this calculation is implemented with the quo (quotient) command. The quotient of the long division is given by

>	$quo (x^{3}, x^{2} + 1, x,' r')$

and the remainder is given by

$r$

Hence, we could write the function $f (x)$ as

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

From this representation, it follows that as $|x|$ gets large, the fractional part (i.e., the remainder part) tends to zero. Hence, the function approaches the oblique asymptote $y = x$ .

Maple has other calculus-based tools that could also be used to determine the oblique asymptote, but they would be beyond the scope of the present discussion.

Exercises - Chapter 4

For the rational functions given in Exercises 4.1 - 4.15, sketch a graph and determine the equations of all asymptotes, both horizontal and vertical. In addition, determine the $y$ - and $x$ -intercepts.

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

4.2. $f (x) = \frac{2 x}{3 x - 4}$

4.3. $f (x) = \frac{5 x + 2}{4 x + 3}$

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

4.5. $f (x) = \frac{3 x + 7}{4 x^{2} - 11 x + 6}$

4.6. $f (x) = \frac{8 x^{2} + 6 x - 9}{3 x^{2} + 11 x + 6}$

4.7. $f (x) = \frac{x^{2} - 4 x + 20}{x^{2} - x - 2}$

4.8. $f (x) = \frac{5}{x^{2} + 6 x + 9}$

4.9. $f (x) = \frac{3}{x^{2} - 4 x + 13}$

4.10. $f (x) = \frac{x - 2}{x^{2} + 8 x + 16}$

4.11. $f (x) = \frac{2 x - 3}{x^{2} + 8 x + 17}$

4.12. $f (x) = \frac{2 x^{2} - x - 15}{x^{2} + 2 x + 1}$

4.13. $f (x) = \frac{x^{2} - 2 x + 5}{x^{2} - 2 x + 10}$

4.14. $f (x) = \frac{3 x^{2} - x - 2}{x^{2} + 6 x + 18}$

4.15. $f (x) = \frac{x^{2} + 2 x + 8}{x^{2} + 10 x + 25}$

Go to Chapter 1 2 3 5 6 7 8 9 10 11

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Robotics

Machine Design & Industrial Automation

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Application Areas

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Purchasing

Institutional Student Licensing

Maplesoft Elite Maintenance (EMP)

Support

Product Training

Online Product Help

Webinars & Events

Publications

Content Hubs

Examples & Applications

Community

About Maplesoft

Media Center

User Community

Contact

Online Help

All Products Maple MapleSim