Rational Functions and Holes — AP® Precalculus Notes

Topic 1.9 found vertical asymptotes at zeros of the denominator, but not at every zero: when a factor cancels, there is no asymptote. This topic is about what is there instead, a single missing point called a hole in the graph.

When a factor cancels

Factor the numerator and the denominator. If $(x - a)$ divides the numerator at least as many times as the denominator, it cancels completely from the denominator. The function is still undefined at $x = a$ (its original denominator is $0$ there), but near $a$ it behaves like the simplified formula, which has a value at $a$ . The graph is the simplified graph with one point removed: a hole.

Worked example 1. Analyze $r(x)$ $= \frac{x^2 - x - 6}{x^2 - 5x + 6}$ .

Factor: $\frac{(x - 3)(x + 2)}{(x - 2)(x - 3)}$ . So $r(x) = \frac{x + 2}{x - 2}$ for $x \ne 3$ .

Hole: $x - 3$ cancels. Height: $\frac{3 + 2}{3 - 2} = 5$ . So $\lim_{x \to 3} r(x) = 5$ , and the hole is at $(3, 5)$ .

Vertical asymptote: $x - 2$ remains below. $\lim_{x \to 2^-} r(x) = -\infty$ , $\lim_{x \to 2^+} r(x) = \infty$ .

Horizontal asymptote: equal degrees, $\lim_{x \to \pm\infty} r(x) = 1$ . Zero: $x = -2$ (not $3$ , which is not in the domain).

The graph of r(x) = (x² + x − 2)/(x² − 1), which equals (x + 2)/(x + 1) for x ≠ 1. It has a hole (open circle) at (1, 1.5), a dashed vertical asymptote x = −1 and a dashed horizontal asymptote y = 1. Its zero is x = −2. — **Figure 1.** An open circle marks the hole at $\left(1, \frac{3}{2}\right)$ ; dashed lines mark the two asymptotes. LevlPrep original diagram.

Hole or asymptote? Compare multiplicities

$(x - a)$ appears	At $x = a$	Example
only in the denominator	vertical asymptote	$\frac{x + 1}{x - 1}$ at $x = 1$
more times below than above	vertical asymptote	$\frac{x - 1}{(x - 1)^2}$ at $x = 1$
at least as many times above	hole	$\frac{(x - 1)^2}{x - 1}$ at $x = 1$ , hole at $(1, 0)$

In the last example the hole sits on the $x$ -axis, at $(1, 0)$ . It is still not a zero: $x = 1$ is not in the domain.

The height of a hole is a limit

A hole at $(a, L)$ means the outputs approach $L$ as $x$ approaches $a$ from both sides, while $r(a)$ does not exist. Write it as $\lim_{x \to a} r(x) = L$ . A table shows the same thing: for $k(x) = \frac{x^2 - 1}{x + 1}$ , the values at $-1.01$ , $-1.001$ , $-0.999$ , $-0.99$ are $-2.01$ , $-2.001$ , $-1.999$ , $-1.99$ , so $\lim_{x \to -1} k(x) = -2$ .

Contrast this with a vertical asymptote, where the table's values grow without bound in size. Both are inputs outside the domain; the limit tells them apart.

Building a function from its features

The same rules run backward. To make a rational function with a hole at $x = 1$ and a vertical asymptote $x = -2$ , put the factor $x - 1$ both above and below (so it cancels) and leave $x + 2$ only below:

$f(x) = \frac{x - 1}{(x - 1)(x + 2)}.$

Check it: for $x \ne 1$ , $f(x) = \frac{1}{x + 2}$ , so the hole is at $\left(1, \frac{1}{3}\right)$ , the vertical asymptote is $x = -2$ , and since the denominator has the higher degree the horizontal asymptote is $y = 0$ . Swapping the roles of the two factors would swap the features.

Common slips

Writing $r(3) = 5$ for a hole at $(3, 5)$ . The function has no value at $3$ ; use $\lim_{x \to 3} r(x) = 5$ .
Calling $\frac{0}{0}$ zero. It has no value; cancel and evaluate.
Listing the hole's input as a zero or as a vertical asymptote.