AP®MathsCollege BoardPrecalculusRevision NotesPolynomial & Rational FunctionsRational FunctionsVertical Asymptotes of Rational Functions

Vertical Asymptotes of Rational Functions (College Board AP® Precalculus): Revision Note

Written by: Roger B

Reviewed by: Mark Curtis

Updated on 9 April 2026

Vertical asymptotes of rational functions

What is a vertical asymptote of a rational function?

A vertical asymptote is often represented by a dashed vertical line on a graph
For a rational function, it corresponds to an input value for which the function is not defined
- However the graph of the function approaches closer and closer to the line from either side
- It is a boundary for the graph of the function, rather than a part of the graph itself

Graph showing a rational function curve with a vertical asymptote at a positive value of x, labelled "vertical asymptote". The curve approaches this line from both sides.

Where do vertical asymptotes occur in rational functions?

Let $r (x) = \frac{p (x)}{q (x)}$ be a rational function
- i.e. where $p (x)$ and $q (x)$ are both polynomial functions
If $a$ is a real zero of $q$ , i.e. if $q (a) = 0$
- and if $a$ is not a real zero of $p$ , i.e. if $p (a) \neq 0$
- then the graph of $r$ has a vertical asymptote at $x = a$
  - E.g. $\frac{x + 4}{x - 3}$ has a vertical asymptote at $x = 3$
If $a$ is also a real zero of $p$ , there can still be a vertical asymptote at $x = a$
- if the multiplicity of $a$ as a real zero of $q$ is greater than its multiplicity as a real zero of $p$
  - E.g. $\frac{(x + 4) {(x - 3)}^{5}}{{(x - 3)}^{6}}$ has a vertical asymptote at $x = 3$
    - because the multiplicity of $x = 3$ in the denominator (6) is greater than its multiplicity in the numerator (5)
    - Note that the expression $\frac{(x + 4) {(x - 3)}^{5}}{{(x - 3)}^{6}}$ can be simplified to give $\frac{(x + 4) {(x - 3)}^{5}}{(x - 3) {(x - 3)}^{5}} = \frac{(x + 4)}{(x - 3)}$
- If the multiplicity of $a$ as a real zero of $q$ is less than or equal to its multiplicity as a real zero of $p$ , then there is a hole at $x = a$

Examiner Tips and Tricks

A vertical asymptote can only occur at an input value where the denominator of a rational function becomes zero.

But be careful, not every input value where the denominator becomes zero corresponds with a vertical asymptote.

How do I express the behavior of rational functions near vertical asymptotes using limit notation?

If $x = a$ is a vertical asymptote of a rational function $r$
- then near the vertical asymptote the values of the function's denominator become arbitrarily close to zero
- This means that near the vertical asymptote the values of $r$ increase or decrease without bound
E.g. consider the behavior of $\frac{x + 4}{x - 3}$ close to $x = 3$

$x$	$\frac{x + 4}{x - 3}$
2.9	-69
2.99	-699
2.999	-6999
2.9999	-69999
3	undefined
3.0001	70001
3.001	7001
3.01	701
3.1	71

For input values near $a$ and greater than $a$ , the corresponding limit notation is

$\lim_{x \to a^{+}} r (x) = \infty$ or $\lim_{x \to a^{+}} r (x) = - \infty$

For input values near $a$ and less than $a$ , the corresponding limit notation is

$\lim_{x \to a^{-}} r (x) = \infty$ or $\lim_{x \to a^{-}} r (x) = - \infty$

$\lim_{x \to a^{+}}$ can be read as "the limit as $x$ approaches $a$ from above (or from the right)"
- and $\lim_{x \to a^{-}}$ can be read as "the limit as $x$ approaches $a$ from below (or from the left)"
Whether a particular limit is $\infty$ or $- \infty$ will depend on the precise nature of the function $r$

Graph showing a rational function with vertical asymptotes at x = a and x = b, depicting limits approaching positive and negative infinity. — **Examples of the behavior of a rational function r(x) near vertical asymptotes**

How do I decide whether a limit goes to ∞ or -∞?

To decide whether the limit is $\infty$ or $- \infty$ , look at values of the function near to the vertical asymptote
- E.g. $s (x) = \frac{x + 4}{x - 3}$ has a vertical asymptote at $x = 3$
  - At $x = 2$ , $s (2) = \frac{2 + 4}{2 - 3} = \frac{6}{- 1} = - 6$
    - As $x$ increases towards 3, the numerator stays positive
    - and the denominator stays negative (while getting closer and closer to 0)
    - so $\lim_{x \to 3^{-}} s (x) = - \infty$
  - At $x = 4$ , $s (4) = \frac{4 + 4}{4 - 3} = \frac{8}{1} = 8$
    - As $x$ decreases towards 3, the numerator stays positive
    - and the denominator stays positive (while getting closer and closer to 0)
    - so $\lim_{x \to 3^{+}} s (x) = \infty$

Worked Example

In the $x y$ -plane, the graph of a rational function $f$ has a vertical asymptote at $x = 3$ . Which of the following could be an expression for $f (x)$ ?

(A) $\frac{(x - 3) (x + 3)}{(x + 7) (x - 3)}$

(B) $\frac{(x - 3) (x + 5)}{(x - 2) (x + 3)}$

(D) $\frac{(x - 3) (x + 3)}{4 (x + 3)}$

Answer:

A vertical asymptote can occur where an input value makes the denominator of a rational function zero

When $x = 3$ , $(x - 3) = (3 - 3) = 0$
So options (A) and (C) will both have denominators equal to zero when $x = 3$

But a vertical asymptote does not necessarily occur if the same input value also makes the numerator zero

In option (A), $\frac{(x - 3) (x + 3)}{(x + 7) (x - 3)}$ becomes $\frac{0}{0}$ when $x = 3$
- and the expression can be simplified to $\frac{x + 3}{x + 7}$
- so there is no vertical asymptote at $x = 3$

In option (C), the numerator is not equal to zero when $x = 3$

so (C) has a vertical asymptote at $x = 3$

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