Horizontal Asymptote

Find the horizontal asymptote of a rational function by comparing the degrees and leading coefficients of its numerator and denominator polynomials.

y=anbm when degP=degQy = \frac{a_n}{b_m} \text{ when } \deg P = \deg Q

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What each symbol means

What Horizontal Asymptote takes
PP
QQ
ana_n
bmb_m
yy
Horizontal Asymptote
SymbolMeaning
PPThe polynomial function in the numerator of the rational expression; confusing this with QQ (the denominator) or with its degree degP\deg P leads to applying the formula incorrectly.
QQThe polynomial function in the denominator of the rational expression; the formula only works when degQ=degP\deg Q = \deg P, so mistaking QQ for the numerator breaks the method.
ana_nThe coefficient of the highest-degree term in PP, called the leading coefficient; using any other coefficient like the constant term gives a wrong asymptote.
bmb_mThe coefficient of the highest-degree term in QQ, called the leading coefficient; the formula requires the denominator's leading coefficient, not any other term's coefficient.
yyThe horizontal asymptote value, a constant that the function approaches as x±x \to \pm\infty; mistaking this for a maximum or minimum value misunderstands what an asymptote represents.

When to use it

Apply this to find the horizontal asymptote of a rational function when its numerator and denominator polynomials have equal degree.

Level

Usually taught in: Pre-Calculus

Worked examples

1. Find the horizontal asymptote of a simple rational function

Problem

Find the horizontal asymptote of f(x)=3x2+2x15x2x+4f(x) = \frac{3x^2 + 2x - 1}{5x^2 - x + 4}
  1. degP=2,degQ=2\deg P = 2, \quad \deg Q = 2

    The highest power in the numerator is 2, and the highest power in the denominator is also 2.

  2. an=3,bm=5a_n = 3, \quad b_m = 5

    The leading coefficient of the numerator is 3, and the leading coefficient of the denominator is 5.

  3. y=anbm=35y = \frac{a_n}{b_m} = \frac{3}{5}

    Since the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients.

Answer: y=35y = \frac{3}{5}

When the numerator and denominator have equal degree, the lower-degree terms and constants become negligible for very large values of x|x|, so the function approaches the ratio of its leading coefficients.

2. Find the horizontal asymptote with a negative leading coefficient

Problem

Find the horizontal asymptote of g(x)=4x3+5x22x+12x38x+3g(x) = \frac{-4x^3 + 5x^2 - 2x + 1}{2x^3 - 8x + 3}
  1. degP=3,degQ=3\deg P = 3, \quad \deg Q = 3

    Both polynomials have degree 3; the numerator's leading term is 4x3-4x^3 and the denominator's is 2x32x^3.

  2. an=4,bm=2a_n = -4, \quad b_m = 2

    Identify the leading coefficients, including the sign of the numerator's leading term.

  3. y=anbm=42=2y = \frac{a_n}{b_m} = \frac{-4}{2} = -2

    Apply the equal-degree formula and simplify the fraction to get the final answer.

Answer: y=2y = -2

Even though the denominator might be larger than the numerator for some middle values of xx, the horizontal asymptote depends only on the growth rates of the leading terms. As xx \to \infty or xx \to -\infty, the function approaches y=2y = -2.

3. Video game effectiveness score approaching a limit

Problem

In a new game level, a player's effectiveness score is E(t)=120t2+80t+10040t2+300E(t) = \frac{120t^2 + 80t + 100}{40t^2 + 300} where tt is the time in minutes. What maximum effectiveness rating can the player approach over very long play sessions?
  1. degP=2,degQ=2\deg P = 2, \quad \deg Q = 2

    The numerator has degree 2 (leading term 120t2120t^2) and the denominator has degree 2 (leading term 40t240t^2).

  2. an=120,bm=40a_n = 120, \quad b_m = 40

    The leading coefficient of the numerator is 120 and of the denominator is 40.

  3. y=anbm=12040=3y = \frac{a_n}{b_m} = \frac{120}{40} = 3

    The horizontal asymptote is the ratio of leading coefficients, which simplifies to 3.

Answer: y=3y = 3

The effectiveness score approaches 3.0 as the player spends more time in the level. This tells us that no matter how long someone plays, the score will asymptotically approach—but never exceed—a rating of 3, because the leading terms of the numerator and denominator grow at the same rate.

Common mistakes

Where Horizontal Asymptote usually goes wrong
Answer came out wrong
Writing y=24y = \frac{2}{4} for f(x)=2x2+34x+1f(x) = \frac{2x^2 + 3}{4x + 1}
First check whether the degrees are equal. If deg P>P > deg QQ, there is no horizontal asymptote; if deg Q>Q > deg PP, the asymptote is y=0y = 0.
For f(x)=3x2+15x22f(x) = \frac{3x^2 + 1}{5x^2 - 2}, writing y=53y = \frac{5}{3}
Remember that y=anbmy = \frac{a_n}{b_m} where ana_n is from the numerator and bmb_m is from the denominator, so the correct answer is y=35y = \frac{3}{5}.
For f(x)=3x2+75x22f(x) = \frac{3x^2 + 7}{5x^2 - 2}, computing y=72y = \frac{7}{-2}
Always use the leading coefficients from the highest-degree terms: y=35y = \frac{3}{5}.
The mistakeWhy it is wrongThe fix
Writing y=24y = \frac{2}{4} for f(x)=2x2+34x+1f(x) = \frac{2x^2 + 3}{4x + 1}The numerator has degree 2 and the denominator has degree 1, so they are not equal; the formula y=anbmy = \frac{a_n}{b_m} only applies when degP=degQ\deg P = \deg Q.First check whether the degrees are equal. If deg P>P > deg QQ, there is no horizontal asymptote; if deg Q>Q > deg PP, the asymptote is y=0y = 0.
For f(x)=3x2+15x22f(x) = \frac{3x^2 + 1}{5x^2 - 2}, writing y=53y = \frac{5}{3}This reverses the numerator and denominator leading coefficients, putting the denominator's coefficient on top instead of the numerator's.Remember that y=anbmy = \frac{a_n}{b_m} where ana_n is from the numerator and bmb_m is from the denominator, so the correct answer is y=35y = \frac{3}{5}.
For f(x)=3x2+75x22f(x) = \frac{3x^2 + 7}{5x^2 - 2}, computing y=72y = \frac{7}{-2}The constant terms 7 and 2-2 have no effect on the function's behavior as xx \to \infty; only the coefficients of the highest-degree terms determine the asymptote.Always use the leading coefficients from the highest-degree terms: y=35y = \frac{3}{5}.

Tips and when to use something else

  • Check degree equality first: only use this formula when degP=degQ\deg P = \deg Q. For deg P>P > deg QQ, there's no horizontal asymptote—try Polynomial Long Division to find a slant asymptote instead.
  • The horizontal asymptote tells you the direction the function heads as x±x \to \pm\infty, not a barrier it cannot cross—a function may intersect its asymptote at finite values of xx.
  • Simplify your final fraction before writing your answer: y=84=2y = \frac{-8}{4} = -2, not y=84y = \frac{-8}{4}.
  • Only the leading coefficients matter—ignore all lower-degree terms and constants, because they become negligible compared to the highest-degree terms when x|x| is very large.

Frequently asked questions

What's the difference between horizontal and vertical asymptotes?
A vertical asymptote is a vertical line x=ax = a where the function approaches ±\pm\infty as the denominator approaches zero. A horizontal asymptote is a horizontal line y=by = b that the function approaches as x±x \to \pm\infty. Rational functions can have both types, but you find them using different methods.
What if my rational function doesn't have a horizontal asymptote?
Horizontal asymptotes exist only in certain cases: when degP=degQ\deg P = \deg Q (giving y=anbmy = \frac{a_n}{b_m}), or when deg Q>Q > deg PP (giving y=0y = 0). If deg P>P > deg QQ, the function grows without bound as xx \to \infty, so there is no horizontal asymptote.
Can a function cross its horizontal asymptote?
Yes, a function can cross a horizontal asymptote multiple times at finite values of xx. The asymptote only describes the behavior as x±x \to \pm\infty, not the function's behavior everywhere.
Why do only the leading coefficients matter?
For very large values of x|x|, the leading term of a polynomial grows so much faster than all lower-degree terms that the smaller terms become negligible by comparison. The ratio of the highest-degree terms dominates the overall behavior of the rational function, so only those coefficients determine the horizontal asymptote.

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Reviewed 2026-09-18