Asymptotes of Rational Functions

You’ll be able to

Locate vertical asymptotes from the denominator of a rational function
Determine the horizontal asymptote by comparing degrees
Distinguish a hole (removable discontinuity) from a vertical asymptote

Vertical asymptotes: the denominator hits zero

A rational function is a ratio of polynomials, p(x)/q(x). A vertical asymptote appears at any x-value where the denominator is zero but the numerator is not — there the function blows up toward ±∞. For f(x) = (3x² + 1)/(x² − 4), the denominator x² − 4 is zero at x = 2 and x = −2, and the numerator is nonzero there, so both are vertical asymptotes.

Horizontal asymptotes: compare the degrees

A horizontal asymptote describes the value f(x) settles toward as x → ±∞, and it comes from comparing the degrees of numerator and denominator. If the numerator’s degree is less, the asymptote is y = 0. If the degrees are equal, the asymptote is the ratio of leading coefficients. If the numerator’s degree is greater, there is no horizontal asymptote (the function grows without bound, possibly along a slant asymptote).

Horizontal-asymptote test

deg(top) < deg(bottom) → y = 0 · equal → y = (lead top)/(lead bottom) · top > bottom → none

This only describes end behavior. The graph may cross a horizontal asymptote in the middle; the rule is about x → ±∞.

Holes: a factor that cancels

When the same factor appears in both numerator and denominator, it cancels and creates a hole — a removable discontinuity — rather than an asymptote. For f(x) = (x + 2)/[(x − 5)(x + 2)], the (x + 2) cancels, leaving a hole at x = −2. The denominator still vanishes at x = 5 (no cancellation there), so x = 5 remains a true vertical asymptote.

Worked example

Find the horizontal asymptote of f(x) = (3x² + 1)/(x² − 4).

1.Compare degrees: the numerator 3x² + 1 has degree 2 and the denominator x² − 4 has degree 2 — they are equal.
2.When the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients.
3.The leading coefficient of the top is 3; of the bottom is 1. Their ratio is 3/1 = 3.
4.So the horizontal asymptote is y = 3.

Answer: The degrees are equal, so the horizontal asymptote is the ratio of leading coefficients, 3/1 = 3. That is, y = 3.

Checkpoint

Find the horizontal asymptote of f(x) = (3x² + 1)/(x² − 4).

Watch out

A vertical asymptote needs the denominator to be zero and the numerator nonzero at that x. If a factor cancels, that x is a hole, not an asymptote — check for common factors before you declare an asymptote.

Checkpoint

For f(x) = (x + 2)/[(x − 5)(x + 2)], what occurs at x = −2?

On the exam

Always factor fully first. The exam loves a rational function whose denominator “looks like” two asymptotes but where one factor cancels into a hole. Simplify, then read vertical asymptotes and holes off the reduced form.

Answer the 2 checkpoints as you read.