# Vertical Asymptotes of Rational Functions

*This post is part of a series.*

Unlike polynomials, rational functions can “blow up” to positive or negative infinity even for relatively small input values. Such input values are called **vertical asymptotes**, because they represent vertical lines that the function approaches but never quite reaches.

## Demonstration

For example, consider the rational function $r(x)=\frac{3x+2}{x-5}$. As we input numbers closer and closer to $5$ while staying greater than $5$, the function output blows up to positive infinity.

On the other hand, as we input numbers closer and closer to $5$ while staying less than $5$, the function output blows up to negative infinity.

As a result, we say the function has a vertical asymptote at $x=5$.

To understand why this happens, notice that as our inputs become closer and closer to $5$, the denominator becomes closer and closer to $0$, while the numerator becomes closer and closer to $17$.

As a result, we end up dividing a fairly constant numerator by a smaller and smaller denominator, which yields a bigger and bigger result.

When the input is greater than $5$, the denominator is positive, which makes the result positive. When the input is less than $5$, the denominator is negative, which makes the result negative.

## Case of Multiple Vertical Asymptotes

In general, vertical asymptotes occur when the denominator is zero and the numerator is nonzero. In the above example, when we input $5$, the denominator is $0$, but the numerator is $17$.

There can also be multiple vertical asymptotes – for example, in the rational function $r(x)=\frac{2x+1}{x^2-4}$, inputting $2$ makes the denominator $0$ while the numerator is $5$, and inputting $-2$ makes the denominator $0$ while the numerator is $-3$.

We confirm that $x=2$ and $x=-2$ are indeed asymptotes by evaluating the function.

## Case of No Vertical Asymptote

On the other hand, if the denominator is zero and the numerator is also zero, then the input is not necessarily a vertical asymptote of the function.

For example, inputting $-1$ to $r(x) = \frac{x^2-x-2}{x+1}$ makes the denominator $0$, but it also makes the numerator $0$, and the result is that the fraction does not blow up to infinity.

To understand this behavior, notice that provided $x$ is not equal to $-1$, the function can simplify.

When we input an $x$ that is not equal to $1$, the $x+1$ factors in the numerator and denominator cancel each other out, and we are left with $x-2$.

As a result, the graph of $r$ is just the graph of $y=x-2$ with a hole at $x=-1$ (where it is undefined).

## Exercises

Find the vertical asymptote(s), if any, of each rational function. (You can view the solution by clicking on the problem.)

## $1) \hspace{.5cm} r(x)=\dfrac{2x}{x-4}$

*Solution:*

$x=4$

## $2) \hspace{.5cm} r(x)=\dfrac{x^2-1}{3x+5}$

*Solution:*

$x=-\frac{5}{3}$

## $3) \hspace{.5cm} r(x)=\dfrac{2x^2-8}{x-2}$

*Solution:*

$\text{none}$

## $4) \hspace{.5cm} r(x)=\dfrac{x+3}{x^2+3x+2}$

*Solution:*

$x=-1,-2$

## $5) \hspace{.5cm} r(x)=\dfrac{x^2+x-2}{x^2-x-6}$

*Solution:*

$x=3$

## $6) \hspace{.5cm} r(x)=\dfrac{x^2+2x-15}{(x^2-9)(x^2+x-12)}$

*Solution:*

$x=-4,-3,3$

## $7) \hspace{.5cm} r(x)=\dfrac{x^4-5x^2+1}{x^2-3x+2}$

*Solution:*

$x=1,2$

## $8) \hspace{.5cm} r(x)=\dfrac{x^2+7x+12}{x^3+x^2-16x-16}$

*Solution:*

$x=-1,4$

## $9) \hspace{.5cm} r(x)=\dfrac{x^2-9}{x^3-3x^2+9x-27}$

*Solution:*

$\text{none}$

## $10) \hspace{.5cm} r(x)=\dfrac{2x-3}{(3x-2)(8x^2-2x-3)}$

*Solution:*

$x=-\frac{1}{2}, \frac{2}{3}, \frac{3}{4}$

*This post is part of a series.*