Precalculus Study Guide
Copyright Maplesoft, a division of Waterloo Maple Inc., 2023
Chapter 4: Graphing Rational Functions
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Introduction


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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.
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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 $\leftx\right$ gets arbitrarily large.
Rational functions can even have oblique (slanted) asymptotes, slanted lines to which the function tends as $\leftx\right$ gets arbitrarily large.
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Chapter Glossary


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The following terms in Chapter 4 are linked to the Maple Math Dictionary.
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asymptote
divisor
domain
even (function)
hyperbola
infinity

intercept
limit
negative
odd (function)
polynomial
positive

quotient
ratio
rational function
real number
reciprocal
remainder



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Typical Problems


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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.
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4.1. $f\left(x\right)\=\frac{1}{x}$
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4.2. $f\left(x\right)\=\frac{1}{{x}^{2}}$
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4.3. $f\left(x\right)\=\frac{1}{{x}^{2}\+1}$
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4.4. $f\left(x\right)\=\frac{x}{{x}^{2}\+1}$
4.5. $f\left(x\right)\=\frac{{x}^{2}}{{x}^{2}\+1}$
4.6. $f\left(x\right)\=\frac{{x}^{2}1}{{x}^{2}\+1}$
4.7. $f\left(x\right)\=\frac{1}{{x}^{2}x2}$ = $\frac{1}{\left(x\+1\right)\left(x2\right)}$
4.8. $f\left(x\right)\=\frac{x}{{x}^{2}x2}$ = $\frac{x}{\left(x\+1\right)\left(x2\right)}$
4.9. $f\left(x\right)\=\frac{{x}^{2}}{{x}^{2}x2}$ = $\frac{{x}^{2}}{\left(x\+1\right)\left(x2\right)}$
4.10. $f\left(x\right)\=\frac{{x}^{2}9}{{x}^{2}x2}$ = $\frac{\left(x\+3\right)\left(x3\right)}{\left(x\+1\right)\left(x2\right)}$
4.11. $f\left(x\right)\=\frac{{x}^{3}}{{x}^{2}\+1}$
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Maple Initializations


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




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Solutions



Problem 4.1



4.1  Mathematical Solution


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The rational function $f\left(x\right)\=\frac{1}{x}$ has no $y$intercept because $f\left(0\right)$ is undefined. It has no $x$intercept because the equation $f\left(x\right)\=\frac{1}{x}$ = 0 has no solutions.
Near $x\=0$ with $x\>0$, the values of $f\left(x\right)$ are large and positive. Near $x\=0$ with $x<0$, the values of $f\left(x\right)$ are large in magnitude, but negative. These two observations are expressed mathematically with the notation
$\underset{x\to {0}^{\+}}{lim}\phantom{\rule[0.0ex]{0.4em}{0.0ex}}f\left(x\right)\=\mathrm{\infty}$
and
$\underset{x\to {0}^{}}{lim}\phantom{\rule[0.0ex]{0.4em}{0.0ex}}f\left(x\right)\=\mathrm{\infty}$


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




respectively. Consequently, the $y$axis is a vertical asymptote.
For $x$ both large and positive, $f\left(x\right)$ is positive but very close to zero. For $x$ negative but large in absolute value, $f\left(x\right)$ is negative but very close to zero. These two observations are expressed mathematically with the notation
$\underset{x\to \mathrm{\infty}}{lim}\phantom{\rule[0.0ex]{0.4em}{0.0ex}}f\left(x\right)\=0$
and
$\underset{x\to \mathrm{\infty}}{lim}\phantom{\rule[0.0ex]{0.4em}{0.0ex}}f\left(x\right)\=0$
Consequently, the $x$axis is a horizontal asymptote.
A Mapleproduced graph of this function is the hyperbola shown in Figure 4.1.1.



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4.1  Maplet Solution


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A graph of the rational function
$f\left(x\right)\=\frac{1}{x}$
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 $f\left(x\right)$. 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



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4.1  Interactive Solution


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•

Enter the expression for the given rational function.

•

Context Panel: Assign to a Name≻$f$


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Task Template


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(The following task template is built into Maple itself. It is designed to call the builtin Rational Function tutor from the Student Precalculus package. The builtin 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 $\frac{P\left(x\right)}{Q\left(x\right)}\:$

Asymptotes
Horizontal

Oblique

Vertical




Plot









Asymptotes


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Vertical Asymptotes

•

Type $f$ and press the Enter key.

•

Context Panel: Denominator


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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)


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Behavior as ${x}{\to}{}{\infty}$
•

Type $f$ and press the Enter key.

•

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

•

Context Panel: Evaluate (from inert)


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Graph


${}$${}$
•

Type $f$ and press the Enter key.

•

Context Panel: Plots≻Plot Builder

$1\le x\le 1$
Options:
Range: $1\le y\le 1$
Find Discontinuities

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${}$




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4.1  Programmatic Solution


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A graph of the rational function
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appears in Figure 4.1.1. It can be created with the command
>

$\mathrm{plot}\left(f\,x\=5..5\,y\=5..5\right)$

The graph of $f\left(x\right)$ 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 $f\left(x\right)$ 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
>

$y\=\mathrm{limit}\left(f\,xequals;\infty \right)semi;\phantom{\rule[0.0ex]{0.0em}{0.0ex}}yequals;\mathrm{limit}\left(fcomma;xequals;\infty \right)$

It is possible for $f\left(x\right)$ 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 $f\left(x\right)$ becomes zero but where the numerator is nonzero.
As $x$ gets close to zero through small positive numbers, $f\left(x\right)$, the reciprocal of these small positive numbers, returns large positive numbers. Thus, as $x$ approaches zero from the right, $f\left(x\right)$ 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, $f\left(x\right)$, the reciprocal of these negative numbers, returns negative numbers of large absolute value. Thus, as $x$ approaches zero from the left, $f\left(x\right)$ 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
>

$\mathrm{Limit}\left(f\,xequals;0comma;\mathrm{right}\right)equals;\mathrm{limit}\left(fcomma;xequals;0comma;\mathrm{right}\right)semi;\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\mathrm{Limit}\left(fcomma;xequals;0comma;\mathrm{left}\right)equals;\mathrm{limit}\left(fcomma;xequals;0comma;\mathrm{left}\right)$

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.
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Problem 4.2



4.2  Mathematical Solution


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The rational function $f\left(x\right)\=\frac{1}{{x}^{2}}$ has no $y$intercept because $f\left(0\right)$ is undefined. It has no $x$intercept because the equation $f\left(x\right)\=\frac{1}{{x}^{2}}$ = 0 has no solutions.
Near $x\=0$ with $x\>0$ or with $x<0$, the values of $f\left(x\right)$ are large and positive. These two observations are expressed mathematically with the notation
$\underset{x\to {0}^{\+}}{lim}\phantom{\rule[0.0ex]{0.4em}{0.0ex}}f\left(x\right)\=\mathrm{\infty}$
and
$\underset{x\to {0}^{}}{lim}\phantom{\rule[0.0ex]{0.4em}{0.0ex}}f\left(x\right)\=\mathrm{\infty}$


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




respectively. Consequently, the $y$axis is a vertical asymptote.
For $x$ large in absolute value, $f\left(x\right)$ is positive but very close to zero. This observation is expressed mathematically with the notation
$\underset{x\to \mathrm{\infty}}{lim}\phantom{\rule[0.0ex]{0.4em}{0.0ex}}f\left(x\right)\=0$
and
$\underset{x\to \mathrm{\infty}}{lim}\phantom{\rule[0.0ex]{0.4em}{0.0ex}}f\left(x\right)\=0$
Consequently, the $x$axis is a horizontal asymptote.
A Mapleproduced graph of this function is shown in Figure 4.2.1.



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4.2  Maplet Solution


${}$
A graph of the rational function
$f\left(x\right)\=\frac{1}{{x}^{2}}$
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




The Graph button provides a graph of $f\left(x\right)$. 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



${}$


4.2  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 builtin Rational Function tutor from the Student Precalculus package. The builtin 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 $\frac{P\left(x\right)}{Q\left(x\right)}\:$

Asymptotes
Horizontal

Oblique

Vertical




Plot









Asymptotes


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Vertical Asymptotes

•

Type $f$ and press the Enter key.

•

Context Panel: Denominator


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${}$

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

$1\le x\le 1$
Options:
Range: $1\le y\le 1$
Find Discontinuities

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${}$




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4.2  Programmatic Solution


${}$
A graph of the rational function
${}$
>

$f\u2254\frac{1}{{x}^{2}}$

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

$\mathrm{plot}\left(f\,x\=5..5\,y\=1..4\right)$

The graph of $f\left(x\right)$ 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 $f\left(x\right)$ 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
${}$
>

$y\=\mathrm{limit}\left(f\,x\=\infty \right)\;\phantom{\rule[0.0ex]{0.0em}{0.0ex}}y\=\mathrm{limit}\left(f\,x\=\infty \right)$

It is possible for $f\left(x\right)$ 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 $f\left(x\right)$ becomes zero but where the numerator is nonzero.
As $x$ gets close to zero through small positive numbers, $f\left(x\right)$, the square of the reciprocal of these small positive numbers, returns large positive numbers. Thus, as $x$ approaches zero from the right, $f\left(x\right)$ 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, $f\left(x\right)$, the square of the reciprocal of these negative numbers, returns large positive numbers. Thus, as $x$ approaches zero from the left, $f\left(x\right)$ tends to large positive numbers.
Again, the formal mathematical language for discussing such behaviors is the limit concept, implemented in Maple as
>

$\mathrm{Limit}\left(f\,x\=0\,\mathrm{right}\right)\=\mathrm{limit}\left(f\,x\=0\,\mathrm{right}\right)\;\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\mathrm{Limit}\left(f\,x\=0\,\mathrm{left}\right)\=\mathrm{limit}\left(f\,x\=0\,\mathrm{left}\right)$

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.
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Problem 4.3



4.3  Mathematical Solution


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The rational function $f\left(x\right)\=\frac{1}{{x}^{2}\+1}$ has 1 as its $y$intercept because $f\left(0\right)$ = 1. It has no $x$intercept because the equation $f\left(x\right)\=\frac{1}{{x}^{2}\+1}$ = 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, $f\left(x\right)$ is positive but very close to zero. This observation is expressed mathematically with the notation


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




$\underset{x\to \mathrm{\infty}}{lim}\phantom{\rule[0.0ex]{0.4em}{0.0ex}}f\left(x\right)\=0$
and
$\underset{x\to \mathrm{\infty}}{lim}\phantom{\rule[0.0ex]{0.4em}{0.0ex}}f\left(x\right)\=0$
Consequently, the $x$axis is a horizontal asymptote.
A Mapleproduced graph of this function is shown in Figure 4.3.1.



${}$


4.3  Maplet Solution


${}$
A graph of the rational function
$f\left(x\right)\=\frac{1}{{x}^{2}\+1}$
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




The Graph button provides a graph of $f\left(x\right)$. 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 is just a $y$intercept. Since this occurs at $x\=0$, it is given by $f\left(0\right)\=1$.
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


(The following task template is built into Maple itself. It is designed to call the builtin Rational Function tutor from the Student Precalculus package. The builtin 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 $\frac{P\left(x\right)}{Q\left(x\right)}\:$

Asymptotes
Horizontal

Oblique

Vertical




Plot









Asymptotes


${}$
Vertical Asymptotes

•

Type $f$ and press the Enter key.

•

Context Panel: Denominator


${}$
${}$

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

$1\le x\le 1$
Options:
Range: $1\le y\le 1$
Find Discontinuities

${}$



${}$




${}$


4.3  Programmatic Solution


${}$
A graph of the rational function
${}$
>

$f\u2254\frac{1}{{x}^{2}\+1}$

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

$\mathrm{plot}\left(f\,xequals;3..3comma;yequals;0..1comma;\mathrm{scaling}equals;\mathrm{constrained}comma;\mathrm{tickmarks}equals;\left[7comma;2\right]\right)$

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

$\mathrm{eval}\left(f\,x\=0\right)$

The horizontal line $y\=0$ is a horizontal asymptote. It is discovered asking (and answering) the question
"To what finite value does $f\left(x\right)$ 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
>

$yequals;\mathrm{limit}\left(fcomma;xequals;\infty \right)semi;\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\phantom{\rule[0.0ex]{0.0em}{0.0ex}}yequals;\mathrm{limit}\left(fcomma;xequals;\infty \right)\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\phantom{\rule[0.0ex]{0.0em}{0.0ex}}$

It is possible for $f\left(x\right)$ 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,
>

$d\u2254\mathrm{denom}\left(f\right)$

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

$\mathrm{plot}\left(d\,xequals;2..2comma;1..5\right)$


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



Hence, $f\left(x\right)$ 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 $f\left(x\right)\=\frac{x}{{x}^{2}\+1}$ has 0 as its $y$intercept because $f\left(0\right)$ = 0. It has 0 as its $x$intercept because the equation $f\left(x\right)\=\frac{x}{{x}^{2}\+1}$ = 0 has the solution $x\=0$. Consequently, the graph of $f\left(x\right)$ passes through the origin.
Because $f\left(x\right)\=\frac{\left(x\right)}{{\left(x\right)}^{2}\+1}$ = $\frac{x}{{x}^{2}\+1}\=f\left(x\right)$, the function has odd symmetry, just like the function $\mathrm{sin}\left(x\right)$.
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 $f\left(x\right)\=\frac{x}{{x}^{2}\+1}$




For $x$ both large and positive, $f\left(x\right)$ is positive but very close to zero. For $x$ negative but large in absolute value, $f\left(x\right)$ is negative but very close to zero. These two observations are expressed mathematically with the notation
$\underset{x\to \mathrm{\infty}}{lim}\phantom{\rule[0.0ex]{0.4em}{0.0ex}}f\left(x\right)\=0$ and $\underset{x\to \mathrm{\infty}}{lim}\phantom{\rule[0.0ex]{0.4em}{0.0ex}}f\left(x\right)\=0$
Consequently, the $x$axis is a horizontal asymptote.
Table 4.4.1 provides a list of values that might help generate a sketch of $f\left(x\right)$ if working without the benefits of technology. Figure 4.4.1 is a Maplegenerated graph of the function.

$x$

$0$

$\frac{1}{2}$

$1$

$\frac{3}{2}$

$2$

$\frac{5}{2}$

$3$

$\frac{7}{2}$

$4$

$\frac{9}{2}$

$5$

$f\left(x\right)$

$0$

$\frac{2}{5}$

$\frac{1}{2}$

$\frac{6}{13}$

$\frac{2}{5}$

$\frac{10}{29}$

$\frac{3}{10}$

$\frac{14}{53}$

$\frac{4}{17}$

$\frac{18}{85}$

$\frac{5}{26}$

$f\left(x\right)$

$0$

$0.40$

$0.50$

$0.46$

$0.40$

$0.34$

$0.30$

$0.26$

$0.24$

$0.21$

$0.19$

Table 4.4.1 Values of $f\left(x\right)$






${}$


4.4  Maplet Solution


${}$
A graph of the rational function
$f\left(x\right)\=\frac{x}{{x}^{2}\+1}$
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




The Graph button provides a graph of $f\left(x\right)$. 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 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


(The following task template is built into Maple itself. It is designed to call the builtin Rational Function tutor from the Student Precalculus package. The builtin 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 $\frac{P\left(x\right)}{Q\left(x\right)}\:$

Asymptotes
Horizontal

Oblique

Vertical




Plot









Asymptotes


${}$
Vertical Asymptotes

•

Type $f$ and press the Enter key.

•

Context Panel: Denominator


${}$
${}$

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

$1\le x\le 1$
Options:
Range: $1\le y\le 1$
Find Discontinuities

${}$



${}$




${}$


4.4  Programmatic Solution


${}$
A graph of the rational function
${}$
>

$f\u2254\frac{x}{{x}^{2}\+1}$

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

$\mathrm{plot}\left(f\,x\=5..5\,y\=.5..0.5\right)$

The graph of $f\left(x\right)$ appears to pass through the origin, a hypothesis verified by the calculation
${}$
>

$\mathrm{eval}\left(f\,x\=0\right)$

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 $f\left(x\right)$ 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
>

$yequals;\mathrm{limit}\left(fcomma;xequals;\infty \right)semi;\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\phantom{\rule[0.0ex]{0.0em}{0.0ex}}yequals;\mathrm{limit}\left(fcomma;xequals;\infty \right)\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\phantom{\rule[0.0ex]{0.0em}{0.0ex}}$

It is possible for $f\left(x\right)$ 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,
>

$d\u2254\mathrm{denom}\left(f\right)$

is never zero, as we see from its graph in Figure 4.4.3.
${}$
>

$\mathrm{plot}\left(d\,xequals;2..2comma;1..5\right)$


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



Hence, $f\left(x\right)$ 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 $f\left(x\right)\=\frac{{x}^{2}}{{x}^{2}\+1}$ has 0 as its $y$intercept because $f\left(0\right)$ = 0. It has 0 as its $x$intercept because the equation $f\left(x\right)\=\frac{{x}^{2}}{{x}^{2}\+1}$ = 0 has the solution $x\=0$. Consequently, the graph of $f\left(x\right)$ passes through the origin.
Because $f\left(x\right)\=\frac{{\left(x\right)}^{2}}{{\left(x\right)}^{2}\+1}$ = $\frac{{x}^{2}}{{x}^{2}\+1}\=f\left(x\right)$, the function has even symmetry, just like the function $\mathrm{cos}\left(x\right)$.
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 $f\left(x\right)\=\frac{{x}^{2}}{{x}^{2}\+1}$




For $x$ large in absolute value, $f\left(x\right)$ is very close to 1, but remains slightly less than 1. This observation is expressed mathematically with the notation
$\underset{x\to \mathrm{\infty}}{lim}\phantom{\rule[0.0ex]{0.4em}{0.0ex}}f\left(x\right)\=1$ and $\underset{x\to \mathrm{\infty}}{lim}\phantom{\rule[0.0ex]{0.4em}{0.0ex}}f\left(x\right)\=1$
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 $f\left(x\right)$ if working without the benefits of technology. Figure 4.5.1 is a Maplegenerated graph of the function.

$x$

$0$

$\frac{1}{2}$

$1$

$\frac{3}{2}$

$2$

$\frac{5}{2}$

$3$

$\frac{7}{2}$

$4$

$\frac{9}{2}$

$5$

$f\left(x\right)$

$0$

$\frac{1}{5}$

$\frac{1}{2}$

$\frac{9}{13}$

$\frac{4}{5}$

$\frac{25}{29}$

$\frac{9}{10}$

$\frac{49}{53}$

$\frac{16}{17}$

$\frac{81}{85}$

$\frac{25}{26}$

$f\left(x\right)$

$0$

$0.20$

$0.50$

$0.69$

$0.80$

$0.86$

$0.90$

$0.92$

$0.94$

$0.95$

$0.96$

Table 4.5.1 Values of $f\left(x\right)$






${}$


4.5  Maplet Solution


${}$
A graph of the rational function
$f\left(x\right)\=\frac{{x}^{2}}{{x}^{2}\+1}$
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




The Graph button provides a graph of $f\left(x\right)$. 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 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


(The following task template is built into Maple itself. It is designed to call the builtin Rational Function tutor from the Student Precalculus package. The builtin 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 $\frac{P\left(x\right)}{Q\left(x\right)}\:$

Asymptotes
Horizontal

Oblique

Vertical




Plot









Asymptotes


${}$
Vertical Asymptotes

•

Type $f$ and press the Enter key.

•

Context Panel: Denominator


${}$
${}$

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

$1\le x\le 1$
Options:
Range: $1\le y\le 1$
Find Discontinuities

${}$



${}$




${}$


4.5  Programmatic Solution


${}$
A graph of the rational function
>

$f\u2254\frac{{x}^{2}}{{x}^{2}\+1}$

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

$\mathrm{plot}\left(f\,xequals;7..7comma;yequals;0..1comma;\mathrm{tickmarks}equals;\left[7comma;4\right]\right)$

The graph of $f\left(x\right)$ appears to pass through the origin, a hypothesis verified by the calculation
${}$
>

$\mathrm{eval}\left(f\,x\=0\right)$

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 $f\left(x\right)$ 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
>

$yequals;\mathrm{limit}\left(fcomma;xequals;\infty \right)semi;\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\phantom{\rule[0.0ex]{0.0em}{0.0ex}}yequals;\mathrm{limit}\left(fcomma;xequals;\infty \right)\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\phantom{\rule[0.0ex]{0.0em}{0.0ex}}$

Alternatively, rewrite $f\left(x\right)$ 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.
>

$\mathrm{quotient}equals;\mathrm{quo}\left(\mathrm{numer}\left(f\right)comma;\mathrm{denom}\left(f\right)comma;xcomma;apos;rapos;\right)semi;\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\mathrm{remainder}equals;r\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\phantom{\rule[0.0ex]{0.0em}{0.0ex}}$

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 $f\left(x\right)$ 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,
>

$d\u2254\mathrm{denom}\left(f\right)$

is never zero, as we see from its graph in Figure 4.5.3.
${}$
>

$\mathrm{plot}\left(d\,xequals;2..2comma;1..5\right)$


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



Hence, $f\left(x\right)$ 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 $f\left(x\right)\=\frac{{x}^{2}1}{{x}^{2}\+1}$ has $1$ as its $y$intercept because $f\left(0\right)$ = $1$. It has $x$ = + 1 as its $x$intercepts because the equation $f\left(x\right)\=\frac{{x}^{2}1}{{x}^{2}\+1}$ = 0 has the solutions $x\=1$ and 1.
Because $f\left(x\right)\=\frac{{\left(x\right)}^{2}1}{{\left(x\right)}^{2}\+1}$ = $\frac{{x}^{2}1}{{x}^{2}\+1}\=f\left(x\right)$, the function has even symmetry, just like the function $\mathrm{cos}\left(x\right)$.
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 $f\left(x\right)\=\frac{{x}^{2}1}{{x}^{2}\+1}$




${}$
For $x$ large in absolute value, $f\left(x\right)$ is very close to 1, but remains slightly less than 1. This observation is expressed mathematically with the notation
$\underset{x\to \mathrm{\infty}}{lim}\phantom{\rule[0.0ex]{0.4em}{0.0ex}}f\left(x\right)\=1$ and $\underset{x\to \mathrm{\infty}}{lim}\phantom{\rule[0.0ex]{0.4em}{0.0ex}}f\left(x\right)\=1$
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 $f\left(x\right)$ if working without the benefits of technology. Figure 4.6.1 is a Maplegenerated graph of the function.

$x$

$0$

$\frac{1}{2}$

$1$

$\frac{3}{2}$

$2$

$\frac{5}{2}$

$3$

$\frac{7}{2}$

$4$

$\frac{9}{2}$

$5$

$f\left(x\right)$

$1$

$\frac{3}{5}$

$0$

$\frac{5}{13}$

$\frac{3}{5}$

$\frac{21}{29}$

$\frac{4}{5}$

$\frac{45}{53}$

$\frac{15}{17}$

$\frac{77}{85}$

$\frac{12}{13}$

$f\left(x\right)$

$1$

$0.60$

$0$

$0.38$

$0.60$

$0.72$

$0.80$

$0.85$

$0.88$

$0.91$

$0.92$

Table 4.6.1 Values of $f\left(x\right)$






${}$


4.6  Maplet Solution


${}$
A graph of the rational function
$f\left(x\right)\=\frac{{x}^{2}1}{{x}^{2}\+1}$
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




The Graph button provides a graph of $f\left(x\right)$. 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 the graph of this function crosses the $x$axis twice. The solutions of $f\left(x\right)\=0$ 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 $f\left(0\right)\=1$.
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


(The following task template is built into Maple itself. It is designed to call the builtin Rational Function tutor from the Student Precalculus package. The builtin 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 $\frac{P\left(x\right)}{Q\left(x\right)}\:$

Asymptotes
Horizontal

Oblique

Vertical




Plot









Asymptotes


${}$
Vertical Asymptotes

•

Type $f$ and press the Enter key.

•

Context Panel: Denominator


${}$
${}$

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

$1\le x\le 1$
Options:
Range: $1\le y\le 1$
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
>

$\mathrm{plot}\left(f\,x\=5..5\,y\=1..1\,\mathrm{scaling}equals;\mathrm{constrained}comma;\mathrm{tickmarks}equals;\left[7comma;2\right]\right)$

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

$\mathrm{eval}\left(f\,x\=0\right)$

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

$\mathrm{solve}\left(f\=0\,x\right)$

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 $f\left(x\right)$ 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
>

$yequals;\mathrm{limit}\left(fcomma;xequals;\infty \right)semi;\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\phantom{\rule[0.0ex]{0.0em}{0.0ex}}yequals;\mathrm{limit}\left(fcomma;xequals;\infty \right)\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\phantom{\rule[0.0ex]{0.0em}{0.0ex}}$

Alternatively, rewrite $f\left(x\right)$ 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.
>

$\mathrm{quotient}equals;\mathrm{quo}\left(\mathrm{numer}\left(f\right)comma;\mathrm{denom}\left(f\right)comma;xcomma;apos;rapos;\right)semi;$$\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\mathrm{remainder}equals;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.
Finally, note that it is possible for $f\left(x\right)$ 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,
>

$d:=\mathrm{denom}\left(f\right)$

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

$\mathrm{plot}\left(d\,xequals;2..2comma;1..5\right)$


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



Hence, $f\left(x\right)$ 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 $f\left(x\right)\=\frac{1}{{x}^{2}x2}$ = $\frac{1}{\left(x\+1\right)\left(x2\right)}$ has $\frac{1}{2}$ as its $y$intercept because $f\left(0\right)\=\frac{1}{2}$.
It has no $x$intercept because the equation $f\left(x\right)\=0$ has no solution.
The denominator, namely, ${x}^{2}x2\=\left(x\+1\right)\left(x2\right)$ 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 $f\left(x\right)\=\frac{1}{{x}^{2}x2}$




$\underset{x\to {1}^{}}{lim}\phantom{\rule[0.0ex]{0.4em}{0.0ex}}f\left(x\right)\=\mathrm{\infty}$ and $\underset{x\to {1}^{\+}}{lim}\phantom{\rule[0.0ex]{0.4em}{0.0ex}}f\left(x\right)\=\mathrm{\infty}$
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
$\underset{x\to {2}^{}}{lim}\phantom{\rule[0.0ex]{0.4em}{0.0ex}}f\left(x\right)\=\mathrm{\infty}$ and $\underset{x\to {2}^{\+}}{lim}\phantom{\rule[0.0ex]{0.4em}{0.0ex}}f\left(x\right)\=\mathrm{\infty}$
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
$\underset{x\to \mathrm{\infty}}{lim}\phantom{\rule[0.0ex]{0.4em}{0.0ex}}f\left(x\right)\=0$ and $\underset{x\to \mathrm{\infty}}{lim}\phantom{\rule[0.0ex]{0.4em}{0.0ex}}f\left(x\right)\=0$
Thus, the $x$axis is a horizontal asymptote for the function.
A Mapleproduced graph of this function is shown in Figure 4.7.1.



${}$


4.7  Maplet Solution


${}$
A graph of the rational function
$f\left(x\right)\=\frac{1}{\left(x\+1\right)\left(x2\right)}$
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.


Figure 4.7.2 Thumbnail image of the Rational Function Tutor




The      