DWG NO. 2.5 — Lesson 5 of 6

Rational Functions and Asymptotes

Unit 2: Polynomial and Rational Functions · ~25 min

Objective Identify the vertical, horizontal, and slant asymptotes of a rational function, and distinguish an asymptote from a hole.

A rational function is a fraction of two polynomials: f(x) = p\(\frac{x}{q}\)(x). Everywhere a polynomial's graph is one connected curve, a rational function's graph can break apart — because division by zero is never allowed. Understanding where and how it breaks apart is the whole game in this lesson.

Vertical asymptotes and holes

First, factor both the numerator and denominator completely and cancel anything common to both. What's left over decides everything:

Vertical asymptote: at any x-value that still makes the (reduced) denominator 0. The graph shoots toward ±∞ on either side but never touches that vertical line.
Hole: at any x-value that was cancelled out — it made both the original numerator and denominator 0. The function is simply undefined there; a single point is missing from an otherwise continuous graph.

Horizontal asymptotes

Compare the degree of the numerator (n) to the degree of the denominator (m):

Comparison	Horizontal asymptote
n < m	y = 0
n = m	y = (ratio of leading coefficients)
n > m	none (check for a slant asymptote instead)

Slant (oblique) asymptotes

When the numerator's degree is exactly one more than the denominator's (n = m+1), there's no horizontal asymptote — but there is a slant one. Divide the numerator by the denominator (long or synthetic division, from 2.3); the quotient, ignoring the remainder, is the equation of a straight line the graph approaches as x → ±∞.

f(x) = \(\frac{2x+1}{x-1}\) — vertical asymptote x = 1, horizontal asymptote y = 2

Worked Example 1 · Hole vs. vertical asymptote

ProblemFind the holes and vertical asymptotes of f(x) = \(\frac{x+3}{x^{2}-9}\).

1Factor the denominator: x²−9 = (x−3)(x+3). The factor (x+3) appears in both numerator and denominator, so it cancels.

2What's left is \(\frac{1}{x-3}\). The cancelled value x = −3 is a hole; the surviving zero of the denominator, x = 3, is a vertical asymptote.

Hole at x = −3 · Vertical asymptote: x = 3

Worked Example 2 · Horizontal asymptote, equal degrees

ProblemFind the horizontal asymptote of f(x) = \(\frac{2x+1}{x-1}\), the function shown in the diagram above.

1Numerator degree n = 1, denominator degree m = 1. Since n = m, the asymptote is the ratio of leading coefficients.

2Leading coefficients: 2 (numerator) and 1 (denominator). \(\frac{2}{1}\) = 2.

Horizontal asymptote: y = 2

Worked Example 3 · Slant asymptote

ProblemFind the slant asymptote of f(x) = \(\frac{x^{2}-1}{x+2}\).

1Numerator degree 2 is exactly one more than denominator degree 1, so a slant asymptote exists (and there is no horizontal one).

2Divide: x²−1 ÷ (x+2). x²÷x = x; x(x+2) = x²+2x; subtract, bring down: −2x−1. −2x÷x = −2; −2(x+2) = −2x−4; subtract: 3.

3Quotient: x−2, remainder 3. Drop the remainder — it shrinks toward 0 relative to the line as |x| grows.

Slant asymptote: y = x − 2

Guided practice

1.Find any holes and vertical asymptotes of f(x) = \(\frac{x-2}{x^{2}-4}\).
Show answer

x²−4 = (x−2)(x+2); the (x−2) factor cancels. Hole at x = 2 · Vertical asymptote: x = −2
2.Find the horizontal asymptote of f(x) = \(\frac{3x^{2}+1}{2x^{2}-5}\).
Show answer

Equal degrees (both 2); ratio of leading coefficients = 3/2. Horizontal asymptote: y = \(\frac{^{3}}{2}\)
3.Find the horizontal asymptote of f(x) = \(\frac{x+4}{x^{2}-1}\).
Show answer

Numerator degree 1 < denominator degree 2. Horizontal asymptote: y = 0
4.Explain why f(x) = \(\frac{x^{2}+1}{x-3}\) has no horizontal asymptote, and state whether it has a slant asymptote.
Show answer

Numerator degree 2 > denominator degree 1, so there's no horizontal asymptote. Because the numerator's degree is exactly one greater, a slant asymptote does exist — found by dividing the numerator by the denominator.
5.Find the slant asymptote of f(x) = \(\frac{x^{2}+3x+1}{x+1}\).
Show answer

Divide: x²+3x+1 ÷ (x+1) gives quotient x+2 with remainder −1. Slant asymptote: y = x + 2

← 2.4 Finding Real Zeros Unit Overview 2.6 Graphing Rational Functions →