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Asymptote Calculator

Find vertical and horizontal asymptotes with AI-powered step-by-step working
asymptotes of (3x^2 + 2x - 1)/(x^2 - 4)
vertical asymptotes of (x - 3)/(x^2 - 9)
horizontal asymptote of (3x + 1)/sqrt(x^2 + 4)
asymptotes of 1/(x^2 - 5x + 6)

What an Asymptote Is

An asymptote is a line the graph approaches without ever settling on it. Two kinds cover almost every problem.

Vertical asymptote at x=ax = a means the function blows up there:

lim⁡x→a±f(x)=±∞\lim_{x \to a^{\pm}} f(x) = \pm\infty

Horizontal asymptote at y=Ly = L describes the far ends of the graph:

lim⁡x→∞f(x)=Lorlim⁡x→−∞f(x)=L\lim_{x \to \infty} f(x) = L \quad \text{or} \quad \lim_{x \to -\infty} f(x) = L

A function may have at most two horizontal asymptotes — one for each direction — but any number of vertical ones. When a rational function's numerator degree is exactly one more than the denominator's, there is no horizontal asymptote at all; the end behaviour follows a slanted line instead.

How to Find Vertical and Horizontal Asymptotes

Vertical asymptotes

  1. Factor the numerator and denominator fully.
  2. Cancel any common factors first.
  3. Set the remaining denominator to zero and solve. Each surviving root gives a vertical asymptote.
  4. A factor you cancelled produces a removable hole, not an asymptote — this is the distinction most answers get wrong.

Horizontal asymptotes for a rational function

Compare n=deg⁡(numerator)n = \deg(\text{numerator}) with m=deg⁡(denominator)m = \deg(\text{denominator}):

DegreesHorizontal asymptote
n<mn < my=0y = 0
n=mn = my=leading coefficient of numeratorleading coefficient of denominatory = \dfrac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}}
n>mn > mnone (slant when n=m+1n = m+1)

When the function is not rational

Fall back to the limit itself. Divide by the highest power, or reason about growth rates: exe^{x} beats every polynomial, and ln⁡x\ln x loses to all of them. Radicals need care because x2=∣x∣\sqrt{x^2} = |x|, which changes sign as x→−∞x \to -\infty and can give two different horizontal asymptotes.

Common Mistakes to Avoid

  • Not cancelling first: in x−3x2−9\frac{x-3}{x^2-9} the root x=3x = 3 is a hole, not a vertical asymptote. Factor before you conclude anything.
  • Assuming a graph cannot cross its horizontal asymptote: it often does in the middle of the domain. The asymptote only constrains behaviour as x→±∞x \to \pm\infty.
  • Comparing coefficients instead of degrees: the horizontal rule depends on the degrees first; the coefficient ratio matters only when the degrees tie.
  • Forgetting x2=∣x∣\sqrt{x^2} = |x|: skipping the absolute value costs you the second horizontal asymptote as x→−∞x \to -\infty.
  • Reporting "no asymptote" when n>mn > m: there is no horizontal one, but the end behaviour still follows a slant or a polynomial curve.

Examples

Step 1: Factor the denominator: x2−4=(x−2)(x+2)x^2 - 4 = (x-2)(x+2)
Step 2: Check the numerator at those roots: at x=2x = 2 it is 12+4−1=15≠012 + 4 - 1 = 15 \neq 0; at x=−2x = -2 it is 12−4−1=7≠012 - 4 - 1 = 7 \neq 0, so nothing cancels
Step 3: Vertical asymptotes: x=2x = 2 and x=−2x = -2
Step 4: Degrees are equal (n=m=2n = m = 2), so the horizontal asymptote is the ratio of leading coefficients: y=31y = \frac{3}{1}
Answer: Vertical: x=2x = 2, x=−2x = -2; horizontal: y=3y = 3

Step 1: Factor: x−3(x−3)(x+3)\dfrac{x-3}{(x-3)(x+3)}
Step 2: Cancel the common factor x−3x - 3, leaving 1x+3\dfrac{1}{x+3} for x≠3x \neq 3
Step 3: The cancelled factor gives a hole at x=3x = 3, not an asymptote
Step 4: Remaining denominator zero: x=−3x = -3 is the vertical asymptote
Step 5: Degree of numerator 1<1 < degree of denominator 22, so y=0y = 0
Answer: Vertical: x=−3x = -3; horizontal: y=0y = 0; hole at x=3x = 3

Step 1: Divide numerator and denominator by ∣x∣|x|, noting x2+4/∣x∣=1+4/x2\sqrt{x^2+4}/|x| = \sqrt{1 + 4/x^2}
Step 2: As x→+∞x \to +\infty, ∣x∣=x|x| = x: lim⁡3+1/x1+4/x2=31=3\lim \dfrac{3 + 1/x}{\sqrt{1 + 4/x^2}} = \dfrac{3}{1} = 3
Step 3: As x→−∞x \to -\infty, ∣x∣=−x|x| = -x, so the numerator becomes −3−1/x→−3-3 - 1/x \to -3
Step 4: The two directions give different limits
Answer: y=3y = 3 as x→∞x \to \infty and y=−3y = -3 as x→−∞x \to -\infty

Frequently Asked Questions

Yes, as often as it likes in the middle of the domain. A horizontal asymptote only describes the limit as x goes to plus or minus infinity, so crossings at finite x are perfectly normal. A vertical asymptote, by contrast, can never be crossed because the function is undefined there.

Both come from a zero of the denominator. If the factor cancels with the same factor in the numerator, the discontinuity is removable and you get a hole. If it survives cancellation, the function grows without bound and you get a vertical asymptote.

At most two horizontal ones, one for each end of the graph, since each direction has a single limit. There is no limit on vertical asymptotes: tan x, for instance, has infinitely many, at every odd multiple of pi/2.

Because the numerator's degree exceeds the denominator's, so the function grows without bound at the ends. The end behaviour is still describable: if the degrees differ by exactly one it follows a slant line, and if they differ by more it follows a polynomial curve found by division.

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