1.3: Rational Functions

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Learning Objectives

Use arrow notation.
Solve applied problems involving rational functions.
Find the domains of rational functions.
Identify vertical asymptotes.
Identify horizontal asymptotes.

Suppose we know that the cost of making a product is dependent on the number of items, $x$, produced. This is given by the equation $C (x)=15,000x−0.1x^2+1000.$ If we want to know the average cost for producing $x$ items, we would divide the cost function by the number of items, $x$. The average cost function, which yields the average cost per item for $x$ items produced, is

\[f(x)=\dfrac{15,000x−0.1x^2+1000}{x} \nonumber\]

Many other application problems require finding an average value in a similar way, giving us variables in the denominator. Written without a variable in the denominator, this function will contain a negative integer power.

In the last few sections, we have worked with polynomial functions, which are functions with non-negative integers for exponents. In this section, we explore rational functions, which have variables in the denominator.

Using Arrow Notation

We have seen the graphs of the basic reciprocal function and the squared reciprocal function from our study of toolkit functions. Examine these graphs, as shown in Figure $\PageIndex{1}$, and notice some of their features.

Graphs of f(x)=1/x and f(x)=1/x^2 — Figure $\PageIndex{1}$

Several things are apparent if we examine the graph of $f(x)=\frac{1}{x}$.

On the left branch of the graph, the curve approaches the $x$-axis $(y=0)$ as $x\rightarrow -\infty$.
As the graph approaches $x = 0$ from the left, the curve drops, but as we approach zero from the right, the curve rises.
Finally, on the right branch of the graph, the curves approaches the $x$-axis $(y=0) $ as $x\rightarrow \infty$.

To summarize, we use arrow notation to show that $x$ or $f (x)$ is approaching a particular value (Table $\PageIndex{1}$).

**Table $\PageIndex{1}$**
Symbol	Meaning
$x\rightarrow a^-$	$x$ approaches a from the left ($x<a$ but close to $a$ )
$x\rightarrow a^+$	$x$ approaches a from the right ($x>a$ but close to $a$ )
$x\rightarrow \infty$	$x$ approaches infinity ($x$ increases without bound)
$x\rightarrow −\infty$	$x$ approaches negative infinity ($x$ decreases without bound)
$f(x)\rightarrow \infty$	the output approaches infinity (the output increases without bound)
$f(x)\rightarrow −\infty$	the output approaches negative infinity (the output decreases without bound)
$f(x)\rightarrow a$	the output approaches $a$

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

Let’s begin by looking at the reciprocal function, $f(x)=\frac{1}{x}$. We cannot divide by zero, which means the function is undefined at $x=0$; so zero is not in the domain. As the input values approach zero from the left side (becoming very small, negative values), the function values decrease without bound (in other words, they approach negative infinity). We can see this behavior in Table $\PageIndex{2}$.

Table $\PageIndex{2}$
$x$	–0.1	–0.01	–0.001	–0.0001
$f(x)=\frac{1}{x}$	–10	–100	–1000	–10,000

We write in arrow notation

as $x\rightarrow 0^−,f(x)\rightarrow −\infty$

As the input values approach zero from the right side (becoming very small, positive values), the function values increase without bound (approaching infinity). We can see this behavior in Table $\PageIndex{3}$.

Table $\PageIndex{3}$
$x$	0.1	0.01	0.001	0.0001
$f(x)=\frac{1}{x}$	10	100	1000	10,000

We write in arrow notation

As $x\rightarrow 0^+, f(x)\rightarrow \infty$.

See Figure $\PageIndex{2}$.

Graph of f(x)=1/x which denotes the end behavior. As x goes to negative infinity, f(x) goes to 0, and as x goes to 0^-, f(x) goes to negative infinity. As x goes to positive infinity, f(x) goes to 0, and as x goes to 0^+, f(x) goes to positive infinity. — Figure $\PageIndex{2}$.

This behavior creates a vertical asymptote, which is a vertical line that the graph approaches but never crosses. In this case, the graph is approaching the vertical line $x=0$ as the input becomes close to zero (Figure $\PageIndex{3}$).

Graph of f(x)=1/x with its vertical asymptote at x=0. — Figure $\PageIndex{3}$.

Definition: VERTICAL ASYMPTOTE

A vertical asymptote of a graph is a vertical line $x=a$ where the graph tends toward positive or negative infinity as the inputs approach $a$. We write

As $x\rightarrow a$, $f(x)\rightarrow \infty$, or as $x\rightarrow a$, $f(x)\rightarrow −\infty$.

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

As the values of $x$ approach infinity, the function values approach $0$. As the values of $x$ approach negative infinity, the function values approach $0$ (Figure $\PageIndex{4}$). Symbolically, using arrow notation

As $x\rightarrow \infty$, $f(x)\rightarrow 0$,and as $x\rightarrow −\infty$, $f(x)\rightarrow 0$.

Graph of f(x)=1/x which highlights the segments of the turning points to denote their end behavior. — Figure $\PageIndex{4}$.

Based on this overall behavior and the graph, we can see that the function approaches 0 but never actually reaches 0; it seems to level off as the inputs become large. This behavior creates a horizontal asymptote, a horizontal line that the graph approaches as the input increases or decreases without bound. In this case, the graph is approaching the horizontal line $y=0$. See Figure $\PageIndex{5}$.

Graph of f(x)=1/x with its vertical asymptote at x=0 and its horizontal asymptote at y=0. — Figure $\PageIndex{5}$.

Definition: HORIZONTAL ASYMPTOTE

A horizontal asymptote of a graph is a horizontal line $y=b$ where the graph approaches the line as the inputs increase or decrease without bound. We write

As $x\rightarrow \infty \text{ or } x\rightarrow −\infty$, $f(x)\rightarrow b$.

Example $\PageIndex{1}$: Using Arrow Notation.

Use arrow notation to describe the end behavior and local behavior of the function graphed in Figure $\PageIndex{6}$.

Graph of f(x)=1/(x-2)+4 with its vertical asymptote at x=2 and its horizontal asymptote at y=4. — Figure $\PageIndex{6}$.

Solution

Notice that the graph is showing a vertical asymptote at $x=2$, which tells us that the function is undefined at $x=2$.

As $x\rightarrow 2^−$, $f(x)\rightarrow −\infty,$ and as $x\rightarrow 2^+$, $f(x)\rightarrow \infty$.

And as the inputs decrease without bound, the graph appears to be leveling off at output values of $4$, indicating a horizontal asymptote at $y=4$. As the inputs increase without bound, the graph levels off at $4$.

As $x\rightarrow \infty$, $f(x)\rightarrow 4$ and as $x\rightarrow −\infty$, $f(x)\rightarrow 4$.

Exercises $\PageIndex{1}$

Use arrow notation to describe the end behavior and local behavior for the reciprocal squared function.

Answer:

End behavior: as $x\rightarrow \pm \infty$, $f(x)\rightarrow 0$;

Local behavior: as $x\rightarrow 0$, $f(x)\rightarrow \infty$ (there are no x- or y-intercepts)

Example $\PageIndex{2}$: Using Transformations to Graph a Rational Function.

Sketch a graph of the reciprocal function shifted two units to the left and up three units. Identify the horizontal and vertical asymptotes of the graph, if any.

Solution

Shifting the graph left 2 and up 3 would result in the function

\[f(x)=\dfrac{1}{x+2}+3\]

or equivalently, by giving the terms a common denominator,

\[f(x)=\dfrac{3x+7}{x+2}\]

The graph of the shifted function is displayed in Figure $\PageIndex{7}$.

Graph of f(x)=1/(x+2)+3 with its vertical asymptote at x=-2 and its horizontal asymptote at y=3. — Figure $\PageIndex{7}$.

Notice that this function is undefined at $x=−2$, and the graph also is showing a vertical asymptote at $x=−2$.

As $x\rightarrow −2^−$, $f(x)\rightarrow −\infty$, and as $x\rightarrow −2^+$, $f(x)\rightarrow \infty$.

As the inputs increase and decrease without bound, the graph appears to be leveling off at output values of 3, indicating a horizontal asymptote at $y=3$.

As $x\rightarrow \pm \infty$, $f(x)\rightarrow 3$.

Analysis

Notice that horizontal and vertical asymptotes are shifted left 2 and up 3 along with the function.

Exercise $\PageIndex{2}$

Sketch the graph, and find the horizontal and vertical asymptotes of the reciprocal squared function that has been shifted right 3 units and down 4 units.

Solution

Graph of f(x)=1/(x-3)^2-4 with its vertical asymptote at x=3 and its horizontal asymptote at y=-4. — Figure $\PageIndex{8}$.

The function and the asymptotes are shifted 3 units right and 4 units down. As $x\rightarrow 3$, $f(x)\rightarrow \infty$, and as $x\rightarrow \pm \infty$, $f(x)\rightarrow −4$.

The function is $f(x)=\frac{1}{{(x−3)}^2}−4$.

Solving Applied Problems Involving Rational Functions

In Example $\PageIndex{2}$, we shifted a toolkit function in a way that resulted in the function $f(x)=\frac{3x+7}{x+2}$. This is an example of a rational function. A rational function is a function that can be written as the quotient of two polynomial functions. Many real-world problems require us to find the ratio of two polynomial functions. Problems involving rates and concentrations often involve rational functions.

Definition: RATIONAL FUNCTION

A rational function is a function that can be written as the quotient of two polynomial functions $P(x)$ and $Q(x)$.

\[f(x)=\dfrac{P(x)}{Q(x)}=\dfrac{a_px^p+a_{p−1}x^{p−1}+...+a_1x+a_0}{b_qx^q+b_{q−1}x^{q−1}+...+b_1x+b_0},\space Q(x)≠0\]

Example $\PageIndex{3}$: Solving an Applied Problem Involving a Rational Function

A large mixing tank currently contains 100 gallons of water into which 5 pounds of sugar have been mixed. A tap will open pouring 10 gallons per minute of water into the tank at the same time sugar is poured into the tank at a rate of 1 pound per minute. Find the concentration (pounds per gallon) of sugar in the tank after 12 minutes. Is that a greater concentration than at the beginning?

Solution

Let t be the number of minutes since the tap opened. Since the water increases at 10 gallons per minute, and the sugar increases at 1 pound per minute, these are constant rates of change. This tells us the amount of water in the tank is changing linearly, as is the amount of sugar in the tank. We can write an equation independently for each:

water: $W(t)=100+10t$ in gallons

sugar: $S(t)=5+1t$ in pounds

The concentration, $C$, will be the ratio of pounds of sugar to gallons of water

\[C(t)=\dfrac{5+t}{100+10t}\]

The concentration after 12 minutes is given by evaluating $C(t)$ at $t= 12$.

\[\begin{align} C(12) & =\dfrac{5+12}{100+10(12)} \\ &= \dfrac{17}{220} \end{align} \]

This means the concentration is 17 pounds of sugar to 220 gallons of water.

At the beginning, the concentration is

\[\begin{align} C(0) & =\dfrac{5+0}{100+10(0)} \\ & =\dfrac{1}{20} \end{align} \]

Since $\frac{17}{220}≈0.08>\frac{1}{20}=0.05$, the concentration is greater after 12 minutes than at the beginning.

Analysis

To find the horizontal asymptote, divide the leading coefficient in the numerator by the leading coefficient in the denominator:

\[\dfrac{1}{10}=0.1\]

Notice the horizontal asymptote is $y= 0.1.$ This means the concentration, $C,$ the ratio of pounds of sugar to gallons of water, will approach 0.1 in the long term.

Exercise $\PageIndex{3}$

There are 1,200 freshmen and 1,500 sophomores at a prep rally at noon. After 12 p.m., 20 freshmen arrive at the rally every five minutes while 15 sophomores leave the rally. Find the ratio of freshmen to sophomores at 1 p.m.

Answer: $\frac{12}{11}$

Finding the Domains of Rational Functions

A vertical asymptote represents a value at which a rational function is undefined, so that value is not in the domain of the function. A reciprocal function cannot have values in its domain that cause the denominator to equal zero. In general, to find the domain of a rational function, we need to determine which inputs would cause division by zero.

Definition: DOMAIN OF A RATIONAL FUNCTION

The domain of a rational function includes all real numbers except those that cause the denominator to equal zero.

HowTO: Given a rational function, find the domain.

Set the denominator equal to zero.
Solve to find the x-values that cause the denominator to equal zero.
The domain is all real numbers except those found in Step 2.

Example $\PageIndex{4}$: Finding the Domain of a Rational Function

Find the domain of $f(x)=\dfrac{x+3}{x^2−9}$.

Solution

Begin by setting the denominator equal to zero and solving.

\[x^2-9=0\]

\[x^2=9\]

\[x=\pm 3\]

The denominator is equal to zero when $x=\pm 3$. The domain of the function is all real numbers except $x=\pm 3$.

Analysis

A graph of this function, as shown in Figure $\PageIndex{9}$, confirms that the function is not defined when $x=\pm 3$.

Graph of f(x)=1/(x-3) with its vertical asymptote at x=3 and its horizontal asymptote at y=0. — Figure $\PageIndex{9}$.

There is a vertical asymptote at $x=3$ and a hole in the graph at $x=−3$. We will discuss these types of holes in greater detail later in this section.

Exercise $\PageIndex{4}$

Find the domain of $f(x)=\frac{4x}{5(x−1)(x−5)}$.

Answer: The domain is all real numbers except $x=1$ and $x=5$.

Identifying Vertical Asymptotes of Rational Functions

By looking at the graph of a rational function, we can investigate its local behavior and easily see whether there are asymptotes. We may even be able to approximate their location. Even without the graph, however, we can still determine whether a given rational function has any asymptotes, and calculate their location.

Vertical Asymptotes

The vertical asymptotes of a rational function may be found by examining the factors of the denominator that are not common to the factors in the numerator. Vertical asymptotes occur at the zeros of such factors.

HowTO: Given a rational function, identify any vertical asymptotes of its graph

Factor the numerator and denominator.
Note any restrictions in the domain of the function.
Reduce the expression by canceling common factors in the numerator and the denominator.
Note any values that cause the denominator to be zero in this simplified version. These are where the vertical asymptotes occur.
Note any restrictions in the domain where asymptotes do not occur. These are removable discontinuities, or “holes.”

Example $\PageIndex{5}$: Identifying Vertical Asymptotes

Find the vertical asymptotes of the graph of $k(x)=\frac{5+2x^2}{2−x−x^2}$.

Solution

First, factor the numerator and denominator.

\[k(x)=\dfrac{5+2x^2}{2−x−x^2}\].

\[=\dfrac{5+2x^2}{(2+x)(1-x)}\]

To find the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero:

\[(2+x)(1−x)=0\]

\[x=−2,1\]

Neither $x=–2$ nor $x=1$ are zeros of the numerator, so the two values indicate two vertical asymptotes. The graph in Figure $\PageIndex{10}$ confirms the location of the two vertical asymptotes.

$Graph of k(x)=(5+2x)^2/(2-x-x^2) with its vertical asymptotes at x=-2 and x=1 and its horizontal asymptote at y=-2.$

Figure $\PageIndex{10}$.

Removable Discontinuities

Occasionally, a graph will contain a hole: a single point where the graph is not defined, indicated by an open circle. We call such a hole a removable discontinuity. For example, the function $f(x)=\frac{x^2−1}{x^2−2x−3}$ may be re-written by factoring the numerator and the denominator.

\[f(x)=\dfrac{(x+1)(x−1)}{(x+1)(x−3)}\]

Notice that $x+1$ is a common factor to the numerator and the denominator. The zero of this factor, $x=−1$, is the location of the removable discontinuity. Notice also that $x–3$ is not a factor in both the numerator and denominator. The zero of this factor, $x=3$, is the vertical asymptote. See Figure $\PageIndex{11}$. [Note that removable discontinuities may not be visible when we use a graphing calculator, depending upon the window selected.]

Graph of f(x)=(x^2-1)/(x^2-2x-3) with its vertical asymptote at x=3 and a removable discontinuity at x=-1. — Figure $\PageIndex{11}$.

REMOVABLE DISCONTINUITIES OF RATIONAL FUNCTIONS

A removable discontinuity occurs in the graph of a rational function at $x=a$ if $a$ is a zero for a factor in the denominator that is common with a factor in the numerator. We factor the numerator and denominator and check for common factors. If we find any, we set the common factor equal to 0 and solve. This is the location of the removable discontinuity. This is true if the multiplicity of this factor is greater than or equal to that in the denominator. If the multiplicity of this factor is greater in the denominator, then there is still an asymptote at that value.

Example $\PageIndex{6}$: Identifying Vertical Asymptotes and Removable Discontinuities for a Graph

Find the vertical asymptotes and removable discontinuities of the graph of $k(x)=\frac{x−2}{x^2−4}$.

Solution

Factor the numerator and the denominator.

\[k(x)=\dfrac{x−2}{(x−2)(x+2)}\]

Notice that there is a common factor in the numerator and the denominator, $x–2$. The zero for this factor is $x=2$. This is the location of the removable discontinuity.

Notice that there is a factor in the denominator that is not in the numerator, $x+2$. The zero for this factor is $x=−2$. The vertical asymptote is $x=−2$. See Figure $\PageIndex{12}$.

Graph of k(x)=(x-2)/(x-2)(x+2) with its vertical asymptote at x=-2 and a removable discontinuity at x=2. — Figure $\PageIndex{12}$.

The graph of this function will have the vertical asymptote at $x=−2$, but at $x=2$ the graph will have a hole.

Exercise $\PageIndex{5}$

Find the vertical asymptotes and removable discontinuities of the graph of $f(x)=\frac{x^2−25}{x^3−6x^2+5x}$.

Answer

Removable discontinuity at $x=5$.

Vertical asymptotes: $x=0$, $x=1$.

Identifying Horizontal Asymptotes of Rational Functions

While vertical asymptotes describe the behavior of a graph as the output gets very large or very small, horizontal asymptotes help describe the behavior of a graph as the input gets very large or very small. Recall that a polynomial’s end behavior will mirror that of the leading term. Likewise, a rational function’s end behavior will mirror that of the ratio of the function that is the ratio of the leading terms.

There are two distinct outcomes when checking for horizontal asymptotes:

Case 1: If the degree of the denominator > degree of the numerator, there is a horizontal asymptote at $y=0$.

Example: $f(x)=\dfrac{4x+2}{x^2+4x−5}$

In this case, the end behavior is $f(x)≈\frac{4x}{x^2}=\frac{4}{x}$. This tells us that, as the inputs increase or decrease without bound, this function will behave similarly to the function $g(x)=\frac{4}{x}$, and the outputs will approach zero, resulting in a horizontal asymptote at $y=0$. See Figure $\PageIndex{13}$. Note that this graph crosses the horizontal asymptote.

Graph of f(x)=(4x+2)/(x^2+4x-5) with its vertical asymptotes at x=-5 and x=1 and its horizontal asymptote at y=0. — Figure $\PageIndex{13}$: Horizontal asymptote $y=0$ when $f(x)=\dfrac{p(x)}{q(x)}$,$q(x)≠0$where degree of $p$ < degree of $q$.

Case 2: If the degree of the denominator = degree of the numerator, there is a horizontal asymptote at $y=\dfrac{a_n}{b_n}$, where $a_n$ and $b_n$ are respectively the leading coefficients of $p(x)$ and $q(x)$ for $f(x)=\dfrac{p(x)}{q(x)}$, $q(x)≠0$.

Example: $f(x)=\dfrac{3x^2+2}{x^2+4x−5}$

In this case, the end behavior is $f(x)≈\dfrac{3x^2}{x^2}=3$. This tells us that as the inputs grow large, this function will behave like the function $g(x)=3$, which is a horizontal line. As $x\rightarrow \pm \infty$, $f(x)\rightarrow 3$, resulting in a horizontal asymptote at $y=3$. See Figure $\PageIndex{15}$. Note that this graph crosses the horizontal asymptote.

Graph of f(x)=(3x^2+2)/(x^2+4x-5) with its vertical asymptotes at x=-5 and x=1 and its horizontal asymptote at y=3. — Figure $\PageIndex{15}$: Horizontal asymptote when $f(x)=\frac{p(x)}{q(x)}$, $q(x)≠0$ where degree of $p$ =degree of $q$.

Notice that, while the graph of a rational function will never cross a vertical asymptote, the graph may or may not cross a horizontal or slant asymptote. Also, although the graph of a rational function may have many vertical asymptotes, the graph will have at most one horizontal (or slant) asymptote.

It should be noted that, if the degree of the numerator is larger than the degree of the denominator by more than one, the end behavior of the graph will mimic the behavior of the reduced end behavior fraction. For instance, if we had the function

\[f(x)=\dfrac{3x^5−x^2}{x+3}\]

with end behavior

\[f(x)≈\dfrac{3x^5}{x}=3x^4\],

the end behavior of the graph would look similar to that of an even polynomial with a positive leading coefficient.

$x\rightarrow \pm \infty, f(x)\rightarrow \infty$

HORIZONTAL ASYMPTOTES OF RATIONAL FUNCTIONS

The horizontal asymptote of a rational function can be determined by looking at the degrees of the numerator and denominator.

Degree of numerator is less than degree of denominator: horizontal asymptote at $y=0$.
Degree of numerator is greater than degree of denominator by one: no horizontal asymptote; slant asymptote.
Degree of numerator is equal to degree of denominator: horizontal asymptote at ratio of leading coefficients.

Example $\PageIndex{7}$: Identifying Horizontal and Slant Asymptotes

For the functions listed, identify the horizontal or slant asymptote.

$g(x)=\dfrac{6x^3−10x}{2x^3+5x^2}$
$h(x)=\dfrac{x^2−4x+1}{x+2}$
$k(x)=\dfrac{x^2+4x}{x^3−8}$

Solution

For these solutions, we will use $f(x)=\dfrac{p(x)}{q(x)},\space q(x)≠0$.

$g(x)=\frac{6x^3−10x}{2x^3+5x^2}$: The degree of $p=$degree of $q=3$, so we can find the horizontal asymptote by taking the ratio of the leading terms. There is a horizontal asymptote at $y =\frac{6}{2}$ or $y=3$.
$h(x)=\frac{x^2−4x+1}{x+2}$: The degree of $p=2$ and degree of $q=1$. Since $p>q$ by 1, there is a slant asymptote found at $\dfrac{x^2−4x+1}{x+2}$.
$clipboard_e5ff4f00a1466e92caa3ddac0dac32f5b.png$
$k(x)=\frac{x^2+4x}{x^3−8}$ : The degree of $p=2$ < degree of $q=3$, so there is a horizontal asymptote $y=0$.

Example $\PageIndex{8}$ Identifying Horizontal Asymptotes

In the sugar concentration problem earlier, we created the equation $C(t)=\frac{5+t}{100+10t}$.

Find the horizontal asymptote and interpret it in context of the problem.

Solution

Both the numerator and denominator are linear (degree 1). Because the degrees are equal, there will be a horizontal asymptote at the ratio of the leading coefficients. In the numerator, the leading term is $t$, with coefficient 1. In the denominator, the leading term is 10t, with coefficient 10. The horizontal asymptote will be at the ratio of these values:

$t\rightarrow \infty,\space C(t)\rightarrow \frac{1}{10}$

This function will have a horizontal asymptote at $y=\frac{1}{10}$.

This tells us that as the values of t increase, the values of $C$ will approach $\frac{1}{10}$. In context, this means that, as more time goes by, the concentration of sugar in the tank will approach one-tenth of a pound of sugar per gallon of water or $\frac{1}{10}$ pounds per gallon.

Example $\PageIndex{9}$: Identifying Horizontal and Vertical Asymptotes

Find the horizontal and vertical asymptotes of the function

$f(x)=\dfrac{(x−2)(x+3)}{(x−1)(x+2)(x−5)}$

Solution

First, note that this function has no common factors, so there are no potential removable discontinuities.

The function will have vertical asymptotes when the denominator is zero, causing the function to be undefined. The denominator will be zero at $x=1,–2,$and $5$, indicating vertical asymptotes at these values.

The numerator has degree $2$, while the denominator has degree 3. Since the degree of the denominator is greater than the degree of the numerator, the denominator will grow faster than the numerator, causing the outputs to tend towards zero as the inputs get large, and so as $x\rightarrow \pm \infty$, $f(x)\rightarrow 0$. This function will have a horizontal asymptote at $y =0.$ See Figure $\PageIndex{16}$.

Graph of f(x)=(x-2)(x+3)/(x-1)(x+2)(x-5) with its vertical asymptotes at x=-2, x=1, and x=5 and its horizontal asymptote at y=0. — Figure $\PageIndex{16}$.

Exercise $\PageIndex{6}$

Find the vertical and horizontal asymptotes of the function:

$f(x)=\dfrac{(2x−1)(2x+1)}{(x−2)(x+3)}$

Answer

Vertical asymptotes at $x=2$ and $x=–3$

horizontal asymptote at $y =4$.

Contributor

Jay Abramson (Arizona State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at https://openstax.org/details/books/precalculus.

Symbol	Meaning
\(x\rightarrow a^-\)	\(x\) approaches a from the left (\(x<a\) but close to \(a\) )
\(x\rightarrow a^+\)	\(x\) approaches a from the right (\(x>a\) but close to \(a\) )
\(x\rightarrow \infty\)	\(x\) approaches infinity (\(x\) increases without bound)
\(x\rightarrow −\infty\)	\(x\) approaches negative infinity (\(x\) decreases without bound)
\(f(x)\rightarrow \infty\)	the output approaches infinity (the output increases without bound)
\(f(x)\rightarrow −\infty\)	the output approaches negative infinity (the output decreases without bound)
\(f(x)\rightarrow a\)	the output approaches \(a\)

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Local Behavior of \(f(x)=\frac{1}{x}\)

End Behavior of \(f(x)=\frac{1}{x}\)

Vertical Asymptotes

\(x\)	–0.1	–0.01	–0.001	–0.0001
\(f(x)=\frac{1}{x}\)	–10	–100	–1000	–10,000