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Learn · Pre-Calculus

Rational Functions: Asymptotes, Holes, and What to Do at Each

Vertical asymptote vs. hole — the cancellation rule that decides which. Plus horizontal and slant asymptote shortcuts.

7 minTEKS 3A-3DPre-Calculus

The cancellation rule (most important!)

For a rational function R(x) = p(x)/q(x), fully factor both numerator and denominator. Then:

  • Factor that CANCELS → HOLE at that x-value
  • Factor in denominator that does NOT cancel → VERTICAL ASYMPTOTE there

Example: R(x) = (x² − 4)/(x − 2) = (x − 2)(x + 2)/(x − 2) → cancels to (x + 2) with restriction x ≠ 2. So there is a HOLE at x = 2 (not an asymptote). At the hole, y = 2 + 2 = 4 — the point (2, 4) is removed.

Rational function f(x) = (x − 1)/(x − 2): V.A. at x = 2, H.A. at y = 1V.A. x = 2H.A. y = 1x-int (1, 0)−4−2124621−2V.A. where denominator = 0; H.A. from ratio of leading coefficients (both degree 1 here → y = 1/1 = 1)

Horizontal asymptote rules

For R(x) = p(x)/q(x):

  • deg(p) < deg(q): y = 0
  • deg(p) = deg(q): y = (leading coeff p)/(leading coeff q)
  • deg(p) > deg(q): NO horizontal asymptote (oblique if exactly 1 more)
Hole (removable discontinuity): factor cancels top & bottomhole at (3, ½)f(x) = (x² − 9)/(x − 3) simplifies to x + 3 with a hole at x = 3

Slant (oblique) asymptote

When deg(p) = deg(q) + 1, perform polynomial long division: R(x) = quotient + remainder/q(x). The quotient is the slant asymptote; the remainder vanishes as |x| → ∞.

Horizontal asymptote rules (compare degrees):deg(top) < deg(bot)y = 0bottom grows faster→ f(x) → 0deg(top) = deg(bot)y = a/bratio of leadingcoefficientsdeg(top) > deg(bot)nonemay have slant(oblique) asymptote

Check yourself

📌 Find asymptote
What is the horizontal asymptote of R(x) = (3x² + 1)/(x² − 4)?

Practice with CBE-style practice questions

Pre-Calc Sem A practice for rational function drills.