Holes, Asymptotes & Removable Discontinuities

You’ll be able to

Distinguish a hole from a vertical asymptote by factoring numerator and denominator
Find the coordinates of a hole, not just its x value
Explain why a graphing tool may not show a hole at all

Two ways for a denominator zero to behave

Every zero of the denominator is excluded from the domain, but not every one produces a vertical asymptote. Factor both numerator and denominator and cancel. A factor that cancels leaves a hole — a single missing point in an otherwise ordinary curve. A factor that survives in the denominator alone gives a vertical asymptote, where the function blows up. Factoring is not a shortcut here; it is the only way to tell the two apart.

Hole vs vertical asymptote

factor cancels → hole at that x · factor remains in denominator → vertical asymptote

A factor appearing twice in the denominator and once in the numerator still leaves one copy behind — so it is an asymptote, not a hole.

A hole has two coordinates

Saying "there is a hole at x = −2" is only half an answer. To get the y-coordinate, cancel the common factor and evaluate the simplified expression at that x. The simplified function agrees with the original everywhere except at the hole, so its value there is exactly the height the missing point would have had. This is called a removable discontinuity precisely because filling in that one point would make the function continuous.

Worked example

Analyze f(x) = (x² − x − 6)/(x² − 4) completely.

1.Factor: numerator x² − x − 6 = (x − 3)(x + 2); denominator x² − 4 = (x − 2)(x + 2).
2.The domain excludes x = 2 and x = −2.
3.The factor (x + 2) cancels, so x = −2 is a hole; (x − 2) remains in the denominator, so x = 2 is a vertical asymptote.
4.Simplified: f(x) = (x − 3)/(x − 2) for x ≠ −2.
5.Height of the hole: substitute x = −2 into the simplified form: (−2 − 3)/(−2 − 2) = −5/−4 = 1.25.
6.Degrees are equal, so the horizontal asymptote is the ratio of leading coefficients, y = 1.

Answer: Domain: all reals except x = −2 and x = 2. There is a hole at (−2, 1.25), a vertical asymptote at x = 2, a horizontal asymptote at y = 1, and an x-intercept at x = 3.

Watch out

Do not substitute the hole's x value into the original expression — you will get 0/0, which is undefined and tells you nothing. The height comes from the simplified form only.

Checkpoint

For g(x) = (x − 5)/(x² − 25), what happens at x = 5 and at x = −5?

Why your calculator hides the hole

A graphing tool samples the function at a finite list of x values and connects them. Unless one of those samples lands exactly on the hole, nothing looks unusual — the curve is drawn straight through the missing point. So a graph is evidence about asymptotes and shape but is never evidence about holes. Holes are found algebraically, by factoring, and that is the only reliable method.

On the exam

A favorite exam setup: a rational function whose x-intercept appears to be at the hole. In the worked example above, x = −2 makes the numerator zero, yet it is not an x-intercept — the point is missing from the graph. Always check candidate intercepts against the domain.

Checkpoint

h(x) = (x + 1)(x − 4)/[(x − 4)²]. What is at x = 4?

Answer the 2 checkpoints as you read.