Unit 1: Polynomial & Rational Functions

Rational Asymptotes & Holes

Identifying vertical asymptotes and removable discontinuities (holes) of a rational function.

Advanced25 min lesson3 min readUpdated August 12, 2026Author not yet attributed

Prerequisites

Polynomial End Behavior & Zeros

Where a Rational Function Breaks — and How

Factoring a polynomial's numerator just revealed its zeros directly. Now put a polynomial in a denominator too, and ask: what happens at the input values that make the denominator zero — and does it matter whether that same factor also appears in the numerator?

Definition — Rational Function

A rational function is a quotient of two polynomials, f(x) = p(x)/q(x). Its domain excludes every input where q(x) = 0, since division by zero is undefined there — but not every excluded point behaves the same way on the graph.

The distinction comes down to whether a factor of q(x) also divides out of p(x). If a factor appears in both the numerator and denominator, it cancels algebraically — but the original function was still undefined at the input that makes that factor zero, so the graph shows a hole (a single missing point) rather than unbounded behavior there. If a factor of q(x) has no matching factor in p(x), it can't cancel, and the function's output grows without bound near that input: a vertical asymptote. Zeros of the simplified function come from whatever numerator factors remain uncanceled.

Worked Example — Finding a Hole, an Asymptote, and a Zero

Analyze f(x) = (x² − 4)/(x² − x − 6). Factor both: numerator (x − 2)(x + 2), denominator (x − 3)(x + 2). The factor (x + 2) is common, so it cancels, leaving (x − 2)/(x − 3) — but x = −2 is still excluded from the domain, producing a hole there. Its height: plug x = −2 into the simplified form, (−2 − 2)/(−2 − 3) = −4/−5 = 4/5, so the hole sits at (−2, 4/5). The uncanceled denominator factor (x − 3) gives a vertical asymptote at x = 3. The uncanceled numerator factor (x − 2) gives a zero at x = 2.

\frac{x^2-4}{x^2-x-6} = \frac{(x-2)(x+2)}{(x-3)(x+2)} = \frac{x-2}{x-3}, \quad x \ne -2

Worked Example — A Rational Function with No Holes

Analyze g(x) = (x + 1)/(x² − 9). Factor the denominator: (x − 3)(x + 3). Neither factor matches anything in the numerator, so nothing cancels — there's no hole. Both denominator factors produce vertical asymptotes, at x = 3 and x = −3, and the numerator's own zero gives a zero at x = −1.

Function Explorer

f(x) =

sin(x)

Transform: g(x) = a·f(b(x − h)) + k

a — vertical stretch/reflect1

b — horizontal stretch/reflect1

h — horizontal shift0

k — vertical shift0

Composition

Compose with another function

Trace mode

Show inverse view

Analysis (of the transformed function, in view)

y-intercept: (0, 0)
x-intercepts: (-9.42, 0), (-6.28, 0), (-3.14, 0), (0, 0), (3.14, 0), (6.28, 0), (9.42, 0)
Extrema: local min at (-7.85, -1); local max at (-4.71, 1); local min at (-1.57, -1); local max at (1.57, 1); local min at (4.71, -1); local max at (7.85, 1)
Inflection points: (-9.42, 0), (-6.28, 0), (-3.14, 0), (0, 0), (3.14, 0), (6.28, 0), (9.42, 0)
Vertical asymptotes: none found in view
Horizontal asymptotes: none found
Domain: all real numbers in view
Range (estimated): approximately [-1, 1]

Tip

Always factor first, before deciding anything about zeros, asymptotes, or holes — an unfactored rational expression hides exactly which excluded points are holes and which are asymptotes.

Common Mistakes

Treating every input that makes the original denominator zero as a vertical asymptote.
First cancel any common factors. An input where a cancelled factor was zero produces a hole, not an asymptote — only the surviving, uncancelled denominator factors give vertical asymptotes.
Forgetting that the domain restriction from a cancelled factor still applies after simplifying.
The simplified expression (x − 2)/(x − 3) is only equal to the original function where x ≠ −2 — that excluded point has to be carried along even though it no longer appears algebraically.

Key Takeaways

A rational function's domain excludes every zero of its denominator, but those excluded points aren't all the same kind of break.
A factor common to numerator and denominator cancels and produces a hole; an uncancelled denominator factor produces a vertical asymptote.
Zeros of a rational function come from the numerator's uncancelled factors.

Summary

Factoring reveals a rational function's zeros, holes, and vertical asymptotes all at once. The next lesson pushes the same idea further: rewriting an expression in a different, equivalent form to reveal a feature that its original form hides entirely — most importantly, what the function does far from the origin.

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