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Horizontal and Slant Asymptotes

A free Precalculus lesson from the “Rational Functions” unit, with a worked example and practice problems including step-by-step solutions.

Horizontal and slant asymptotes describe what a rational graph approaches far left and far right. This lesson is part of Precalculus: Advanced Functions, so the emphasis is on interpreting behavior, choosing the right representation, and explaining the result clearly rather than memorizing isolated algebra moves.

What you'll learn

Why it matters: Rational functions model rates, constraints, efficiency, and quantities that change sharply near restricted inputs.

Worked example

Problem. For r(x) = (3x^2 + 3x + 3)/(x^2 - 3), find the horizontal asymptote's y-value.

  1. The numerator and denominator have the SAME degree (both 2).
  2. For equal degrees the horizontal asymptote is the ratio of leading coefficients: 3/1.
  3. So the asymptote is y = 3, and the y-value is 3.

Answer: 3

Practice problems

1. For r(x) = (3x^2 + 3x + 3)/(x^2 - 3), find the horizontal asymptote's y-value.

Show solution
  1. The numerator and denominator have the SAME degree (both 2).
  2. For equal degrees the horizontal asymptote is the ratio of leading coefficients: 3/1.
  3. So the asymptote is y = 3, and the y-value is 3.

Answer: 3

2. For r(x) = (4x + 4)/(x^2 - 4), the horizontal asymptote is y = ? Enter the y-value.

Show solution
  1. Compare degrees: numerator degree 1 is LESS than denominator degree 2.
  2. When the top degree is smaller, the outputs shrink toward 0 far from the origin.
  3. So the horizontal asymptote is y = 0.

Answer: 0

3. Find the slant asymptote of r(x) = (x^2 + 5)/(x - 1) by long division. Enter it as y = mx + b.

Show solution
  1. The numerator degree (2) is exactly one MORE than the denominator degree (1), so expect a slant asymptote.
  2. Long division: (x - 1) goes into x^2 + 5 a total of x + 1 times, since (x - 1)(x + 1) = x^2 - 1, leaving remainder 6.
  3. The slant asymptote is the quotient line y = x + 1 (the remainder term vanishes far from the origin).

Answer: y = x + 1

4. Find the slant asymptote of r(x) = (x^2 + 3x + 9)/(x + 1) by long division. Enter it as y = mx + b.

Show solution
  1. Degrees differ by one (2 vs 1), so there is a slant asymptote equal to the quotient.
  2. Dividing: x^2 + 3x + 9 = (x + 1)(x + 2) + 7, so the quotient is x + 2 with a nonzero remainder.
  3. The slant asymptote is y = x + 2.

Answer: y = x + 2

5. For r(x) = (x^3 + 2)/(x - 4), which end-behavior asymptote does the graph have?

Choices: neither a horizontal nor a slant asymptote · a horizontal asymptote y = 0 · a horizontal asymptote y = 1 · a slant asymptote y = x

Show solution
  1. The numerator degree (3) exceeds the denominator degree (1) by MORE than one.
  2. A horizontal asymptote needs degrees equal or top-smaller; a slant needs a difference of exactly one.
  3. A gap of 2 gives a polynomial (parabolic) end behavior — no horizontal or slant asymptote.

Answer: neither a horizontal nor a slant asymptote

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