Vertical Asymptotes and Holes

A rational function breaks wherever its denominator is zero — but the graph reacts differently depending on whether that zero also cancels with the numerator. One kind sends the curve to infinity; the other kind just skips a single point.

By the end you'll be able to find every denominator zero of a rational function, classify each one as a vertical asymptote or a hole, and explain why the distinction comes down to cancelling factors.

Predict: what's the vertical asymptote of \(1/(x-3)\)? Set a = 3 on the slider below and check.

This plots \(f(x) = 1/(x-a)\). Drag a and watch the dashed vertical asymptote track it — it always sits exactly where \(x - a = 0\). Then check the box: it turns on a cancelling factor at \(x = 2\), which punches a hole in that same curve instead of a second asymptote, since that denominator zero now cancels with the numerator. Hover the curve or the hole for exact readouts.

f(x) = 1/(x−a); vertical asymptote at x = -1.00. Cancelling factor off — x = 2 plays no role.

f(x) = 1/(x−a) — curve, asymptote at x = a, optional hole at x = 2
f(x) = 1/(x−a) asymptote x = a hole (when cancelled)

Every denominator zero excludes a point from the domain — but factoring tells you whether the graph reacts with a vertical asymptote or just a hole.

Visual

Picture the curve racing toward a dashed vertical line — it gets closer and closer but never touches, shooting off toward \(+\infty\) on one side and \(-\infty\) on the other. A hole looks completely different: the curve is perfectly smooth and continuous, except for one missing point you could patch by hand — like a pixel someone erased from an otherwise unbroken line.

Formal

For \(f(x) = P(x)/Q(x)\), fully factor both P(x) and Q(x), then look at each zero of Q(x):

  • If the factor causing that zero does not cancel with a matching factor in P(x), the graph has a vertical asymptote there — the numerator stays nonzero while the denominator shrinks to zero, so \(|f(x)| \to \infty\).
  • If the factor does cancel, the graph has a hole there instead — the reduced expression is defined there by a limit, even though the original expression is not.

Every denominator zero is excluded from the domain either way. The cancellation test only decides which kind of break shows up on the graph — and you must find those zeros from the original, unreduced denominator, or you'll miss the holes entirely.

Applied

An engineer modeling a system's response near a resonant frequency sees a genuine vertical asymptote — the output really does blow up there. But a formula that simplifies algebraically (say, a rate that cancels out of both a distance and a time term) can look like it breaks down at one input value when really it's just a single patchable gap — the value the formula "should" give is recoverable by simplifying first.

Worked example

Classify the breaks of \(f(x) = \dfrac{x^2-1}{x^2+3x+2}\). Factor: \(f(x) = \dfrac{(x-1)(x+1)}{(x+1)(x+2)}\). Denominator zeros: \(x=-1\) and \(x=-2\) (both excluded from the domain). The factor \((x+1)\) cancels, so \(x=-1\) is a hole — the reduced form \((x-1)/(x+2)\) at \(x=-1\) gives \((-2)/(1) = -2\), so the hole sits at \((-1,-2)\). The factor \((x+2)\) does not cancel, so \(x=-2\) is a vertical asymptote.

Your turn

Classify the breaks of \(h(x) = \dfrac{x^2-4}{x^2+x-6}\). Factor: \(h(x) = \dfrac{(x-2)(x+2)}{(x-2)(x+3)}\). Denominator zeros: \(x=2\) and \(x=-3\). The factor \((x-2)\) cancels, so \(x=2\) is a ____. The reduced form \((x+2)/(x+3)\) at \(x=2\) gives \((4)/(5) = \) ____, so the hole sits at \((2,\) ____\()\). The factor \((x+3)\) does not cancel, so \(x=-3\) is a ____.

Reveal the answer

\(x=2\) is a hole, since \((x-2)\) cancels. Its y-value is \((4)/(5) = \) 0.8, so the hole sits at \((2, 0.8)\). \(x=-3\) is a vertical asymptote, since \((x+3)\) never cancels. Try setting the diagram's slider to \(a=-3\) and checking the cancel box to see a similar pattern — an asymptote at one zero, a hole at another.

More info — why a hole is a limit, not a value

Near a cancelled factor, the original and reduced expressions agree everywhere except at the excluded point itself — so the curve approaches the same y-value from both sides, it just never officially reaches it in the original formula. That approached-but-undefined value is exactly what a limit describes, which is why a hole is also called a removable discontinuity — you can patch it by simplifying first (see simplifying a rational expression in the Dive deeper link below). A vertical asymptote can't be patched this way — there's no finite value the curve approaches from either side.

Check your understanding

Question 1 of 4

What is the vertical asymptote of \(f(x) = \dfrac{1}{x+5}\)?

Question 2 of 4

For \(f(x) = \dfrac{x-4}{(x-4)(x+1)}\), what happens at \(x = 4\)?

Question 3 of 4

A rational function \(f(x) = P(x)/Q(x)\) has a denominator zero at x = 7 that shares no common factor with the numerator. What happens on the graph at x = 7?

Question 4 of 4

Classify the breaks of \(g(x) = \dfrac{x^2-9}{x^2-x-12}\) at x = -3 and x = 4. (Hint: factor both.)

Recap

  • A vertical asymptote sits at every denominator zero of \(f(x)=P(x)/Q(x)\) that doesn't cancel with a matching factor in the numerator.
  • A denominator zero that does cancel produces a hole (removable discontinuity) instead — a single missing point, not a break to infinity.
  • Find denominator zeros from the original, unfactored denominator — reducing first can hide the holes.
  • A hole's y-coordinate comes from evaluating the reduced expression at that x-value.

Dive deeper

Sources

  • Vertical Asymptotes and Holes