AP®MathsCollege BoardPrecalculusRevision NotesPolynomial & Rational FunctionsRational FunctionsHoles of Rational Functions

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

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

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

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