An asymptote is a line a graph gets closer and closer to without reaching it (or, for a horizontal one, without staying away from it). For a rational function, a polynomial over a polynomial, three rules decide the horizontal asymptote from nothing but the degrees of the top and bottom, and one factoring step finds the vertical ones. The rules are below with the reason for each, which is what turns a memorised list into marks on the harder questions.
Download the asymptote rules sheet (PDF) (the rules with the reasons and the worked examples on page 1, the practice set with answers on page 2; two pages, free).
The horizontal asymptote rules
For f(x) = p(x)/q(x), let n be the degree of the numerator p and m the degree of the denominator q, with leading coefficients a and b.
| Case | Horizontal asymptote | Example |
|---|---|---|
| n < m (bottom-heavy) | y = 0 | (3x + 1)/(x² − 4) → y = 0 |
| n = m (equal degrees) | y = a/b | (2x² − x)/(5x² + 3) → y = 2/5 |
| n > m (top-heavy) | none | (x³ + 1)/(x − 2) → none |
| n = m + 1 exactly | no horizontal, but an oblique asymptote y = mx + c from dividing | (x² + 1)/(x − 1) → y = x + 1 |
Why. For very large x, only the highest power in each polynomial matters: p(x) behaves like ax^n and q(x) like bx^m, so f(x) behaves like (a/b)·x^(n − m). If n < m the power is negative and the value shrinks to 0. If n = m the power is zero and the value settles at a/b. If n > m the power is positive and the value grows without limit, so there is no horizontal line it approaches; when the difference is exactly 1 it grows like a straight line, the oblique asymptote.
A graph can cross its horizontal asymptote in the middle; the asymptote only describes the far left and far right. (x)/(x² + 1) has y = 0 and crosses it at the origin.
How to find vertical asymptotes
- Factor the numerator and the denominator fully.
- Cancel any factor that appears in both. A cancelled factor (x − c) marks a hole at x = c, a single missing point, not an asymptote.
- Set what remains of the denominator equal to zero. Each solution x = c is a vertical asymptote.
- Check the side behaviour if the question asks: plug in a value just left and just right of c to see whether the graph goes up or down on each side.
Example: f(x) = (x² − 1)/(x² − 3x + 2) = (x − 1)(x + 1)/((x − 1)(x − 2)). Cancel (x − 1): hole at x = 1 (at height (1 + 1)/(1 − 2) = −2). Vertical asymptote at x = 2. Degrees equal, so the horizontal asymptote is y = 1/1 = 1.
Oblique (slant) asymptotes
When the top's degree is exactly one more than the bottom's, divide. For (x² + 1)/(x − 1): long division gives x + 1 with remainder 2, so f(x) = x + 1 + 2/(x − 1). As x grows the remainder term vanishes and the graph hugs y = x + 1. If the degrees differ by 2 or more, the graph follows a curve (a parabola for a difference of 2), which is not called an asymptote in school courses.
Asymptotes of other functions
| Function | Vertical asymptote | Horizontal asymptote |
|---|---|---|
| y = a·b^x + c (exponential) | none | y = c |
| y = log_b(x − h) + k | x = h | none |
| y = tan x | x = π/2 + nπ | none |
| y = 1/x | x = 0 | y = 0 |
| y = (ax + b)/(cx + d) | x = −d/c | y = a/c |
The last row is the Algebra 2 workhorse: a linear over a linear always has one vertical asymptote where the bottom is zero and a horizontal one at the ratio of the x coefficients.
Ten worked examples
- (5x + 2)/(x² + 1): n = 1 < m = 2, so y = 0; denominator never zero, so no vertical asymptote.
- (3x² + x)/(x² − 9): equal degrees, y = 3; x² − 9 = (x − 3)(x + 3), so x = 3 and x = −3.
- (x³ − 1)/(x + 2): n = 3 > m = 1 by two, so no horizontal or oblique asymptote; vertical at x = −2.
- (2x² + 3x)/(x − 1): degrees 2 and 1, divide: 2x + 5 remainder 5, so oblique y = 2x + 5; vertical x = 1.
- (x − 2)/(x² − 4) = (x − 2)/((x − 2)(x + 2)) = 1/(x + 2): hole at x = 2, vertical x = −2, horizontal y = 0.
- (4x³)/(2x³ + x): equal degrees, y = 4/2 = 2; bottom x(2x² + 1) = 0 only at x = 0.
- y = 3·2^x − 4: horizontal y = −4, no vertical.
- y = ln(x + 5): vertical x = −5, no horizontal.
- (x² + 2x + 1)/(x + 1) = (x + 1)²/(x + 1) = x + 1: a hole at x = −1, no asymptotes at all; the graph is the line y = x + 1 with one point missing.
- (−x² + 3)/(2x² + 1): equal degrees, y = −1/2; denominator never zero, no vertical asymptote.
The mistakes that lose marks
- Setting the denominator to zero before cancelling: a cancelled factor is a hole, not an asymptote (example 5).
- Reading "no horizontal asymptote" as "no asymptote": a top-heavy function still has its vertical ones, and may have an oblique one.
- Using the constant terms instead of the leading coefficients in the equal-degree case: (2x² − x)/(5x² + 3) has y = 2/5, not y = −1/3.
- Claiming a graph cannot cross a horizontal asymptote. It can; it cannot cross a vertical one.
- Forgetting that y = a·b^x + c has its asymptote at y = c, not y = 0, once the curve is shifted.
Where to get help
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Related pages
- The square root function and its graph
- Exponential growth and decay
- Graphing linear equations, four methods
Questions parents ask
1What are the horizontal asymptote rules?
2How do you find a vertical asymptote?
3Can a graph cross a horizontal asymptote?
4What is the horizontal asymptote of an exponential function?
5What is an oblique asymptote?
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