AP®MathsCollege BoardPrecalculusStudy GuidesPolynomial & Rational FunctionsRational FunctionsHoles of Rational Functions

Holes of Rational Functions (College Board AP® Precalculus): Study Guide

Written by: Roger B

Reviewed by: Mark Curtis

Updated on 9 April 2026

Holes of rational functions

What is a hole of a rational function?

A hole is sometimes represented by an open circle on a graph
For a rational function, it corresponds to an input value for which the function is not defined
- Unlike at a vertical asymptote, however, the function does not display unbounded behavior on either side of a hole

Graph depicting a curve on an x-y axis with a hole on the upward slope of the curve, indicated by an open circle with an arrow pointing to it from a label saying "hole". — **Example of a hole in a rational function**

Examiner Tips and Tricks

Be careful when using a graphing calculator, as holes in a rational function will probably not be displayed or be otherwise visually obvious. This is different from a vertical asymptote, where the unbounded behavior of the function on either side of the asymptote is obvious.

Where do holes 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 $c$ is a real zero of both $p$ and $q$ , i.e. if $p (c) = 0$ and $q (c) = 0$
- and if the multiplicity of $c$ as a real zero of $p$ is greater than or equal to its multiplicity as a real zero of $q$
- then the graph of $r$ has a hole at $x = c$
  - E.g. $\frac{(x + 4) {(x - 3)}^{4}}{(x + 1) {(x - 3)}^{4}}$ and $\frac{(x + 4) {(x - 3)}^{7}}{(x + 1) {(x - 3)}^{4}}$ both have holes at $x = 3$
    - because in both cases the numerator and denominator become zero when $x = 3$
    - and the multiplicity of $x = 3$ in the numerator (4 or 7) is greater than or equal to its multiplicity in the denominator (4)
  - Note that the expression $\frac{(x + 4) {(x - 3)}^{4}}{(x + 1) {(x - 3)}^{4}}$ can be simplified to give $\frac{(x + 4) {(x - 3)}^{4}}{(x + 1) {(x - 3)}^{4}} = \frac{x + 4}{x + 1}$
    - and the expression $\frac{(x + 4) {(x - 3)}^{7}}{(x + 1) {(x - 3)}^{4}}$ can be simplified to give $\frac{(x + 4) {(x - 3)}^{3} {(x - 3)}^{4}}{(x + 1) {(x - 3)}^{4}} = \frac{(x + 4) {(x - 3)}^{3}}{x + 1}$
- If the multiplicity of $c$ as a real zero of $p$ is less than its multiplicity as a real zero of $q$ , then there is a vertical asymptote at $x = c$

Examiner Tips and Tricks

The expression for a rational function with a hole at $x = c$ will always be able to be simplified to an expression where the denominator is not equal to zero when $x = c$ . This is a simple way to distinguish a hole from a vertical asymptote.

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

If a rational function $r$ has a hole at $x = c$
- then for input values near the hole the output values of the function will converge towards a particular value
- The value they converge to determines the $y$ -coordinate of the hole
E.g. consider the behavior of $\frac{(x + 4) {(x - 3)}^{4}}{(x + 1) {(x - 3)}^{4}}$ close to $x = 3$
- The output value gets closer and closer to 1.75 as $x$ gets closer and closer to 3

$x$	$\frac{(x + 4) {(x - 3)}^{4}}{(x + 1) {(x - 3)}^{4}}$
2.9	1.76923076923
2.99	1.75187969925
2.999	1.75018754689
2.9999	1.75001875047
3	undefined
3.0001	1.74998125047
3.001	1.74981254686
3.01	1.74812967581
3.1	1.73170731707

If output values of $r$ become arbitrarily close to some number $L$ as input values become sufficiently close to $c$
- then the hole occurs at the point $(c, L)$
- and the corresponding limit notation is

$\lim_{x \to c} r (x) = L$

$\lim_{x \to c}$ can be read as "the limit as $x$ approaches $c$ "

Examiner Tips and Tricks

Note that in the case of a hole at $(c, L)$

$\lim_{x \to c^{-}} r (x) = \lim_{x \to c^{+}} r (x) = \lim_{x \to c} r (x) = L$

I.e. the limits 'from the left' and 'from the right' are both equal to the same number $L$

How can I determine the limit of a rational function at a hole?

One way to determine the limit at a point where a rational function has a hole
- is to calculate output values of the function nearer and nearer to the hole
  - and see what value they seem to converge towards
- For example, see the table of values for $\frac{(x + 4) {(x - 3)}^{4}}{(x + 1) {(x - 3)}^{4}}$ above
However a simpler method can be to algebraically simplify the expression for the function
- and then find the value of the simplified expression at the input value where the hole occurs
E.g. you can see from the table above that $\lim_{x \to 3} \frac{(x + 4) {(x - 3)}^{4}}{(x + 1) {(x - 3)}^{4}} = 1.75$
- Instead, start by simplifying the expression
  - $\frac{(x + 4) {(x - 3)}^{4}}{(x + 1) {(x - 3)}^{4}} = \frac{x + 4}{x + 1}$
- Substitute $x = 3$ into the simplified expression
  - $\frac{3 + 4}{3 + 1} = \frac{7}{4} (= 1.75)$
- That can be a much quicker way to find the limit (especially on a non-calculator section of the exam)
  - unless a question specifies that you must use another method

Examiner Tips and Tricks

Be careful here. The expression $\frac{(x + 4) {(x - 3)}^{4}}{(x + 1) {(x - 3)}^{4}}$ can be simplified algebraically to $\frac{(x + 4)}{(x + 1)}$ . However

$f (x) = \frac{(x + 4) {(x - 3)}^{4}}{(x + 1) {(x - 3)}^{4}}$ and $g (x) = \frac{(x + 4)}{(x + 1)}$

are not 'the same function'!

Their behaviors differ at the point where $x = 3$
- $g$ has a well-defined value at $x = 3$ , i.e. $g (3) = \frac{7}{4}$
- $f$ is undefined at $x = 3$ , and has a hole at $(3, \frac{7}{4})$

Worked Example

In the $x y$ -plane, the graph of a rational function $f$ has a hole at $x = - 5$ . Which of the following could be an expression for $f (x)$ ?

(A) $\frac{(x - 5) (x + 3)}{(x + 7) (x - 5)}$

(B) $\frac{(x + 5) (x + 7)}{(x - 3) (x + 5)}$

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

Answer:

A hole can occur where an input value makes both the numerator and denominator of a rational function zero

When $x = - 5$ , $(x + 5) = (- 5 + 5) = 0$
So options (B) and (D) will both have numerators and denominators equal to zero when $x = - 5$

But it is only a hole if simplifying the rational expression can make the zero in the denominator go away

In option (D), $\frac{(x - 3) (x + 5)}{3 {(x + 5)}^{2}} = \frac{(x - 3) (x + 5)}{3 (x + 5) (x + 5)} = \frac{x - 3}{3 (x + 5)}$
- That simplified form still has a denominator equal to zero when $x = - 5$
- but the numerator is no longer equal to zero
- so $x = - 5$ is a vertical asymptote, not a hole

In option (B), the rational expression can be simplified to $\frac{(x + 7)}{(x - 3)}$

The denominator of the simplified form is not equal to zero when $x = - 5$
So $x = - 5$ is a hole of $\frac{(x + 5) (x + 7)}{(x - 3) (x + 5)}$

(B) $\frac{(x + 5) (x + 7)}{(x - 3) (x + 5)}$

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

I would just like to say a massive thank you for putting together such a brilliant, easy to use website.I really think using this site helped me secure my top gradesin science and maths. You really did save my exams! Thank you.

Beth
IGCSE Student

This website is soooo useful and I can’t ever thank you enough for organising questions by topic like this. Furthermore, the name of the website could not have been more appropriate as it literally did SAVE MY EXAMS!

Fathima
A Level Student

Incredible! SO worth my money, the revision notes have everything I need to know and are so easy to understand. I actually enjoy revising! It makes me feel a lot more confident for my GCSEs in a few months.

Kate
GCSE Student

Absolutely brilliant, both my girls used it for A levels and GCSE. It's saves on paper copies, also beneficial exam questions ranked from easy to hard. It's removed a lot of stress from the exams.

Sameera
Parent

Just to say that your resources are the best I have seen and I have been teaching chemistry at different levels for about 40 years

Mark
Chemistry Teacher

Excellent

Holes of Rational Functions (College Board AP® Precalculus): Study Guide

Holes of rational functions

What is a hole of a rational function?

Examiner Tips and Tricks

Where do holes occur in rational functions?

Examiner Tips and Tricks

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

Examiner Tips and Tricks

How can I determine the limit of a rational function at a hole?

Examiner Tips and Tricks

Worked Example

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Roger B

Reviewer: Mark Curtis