How to Find Holes in a Function: Complete Guide

A hole in a function is a single point where the graph is undefined, even though the surrounding curve is smooth. You find it by factoring the numerator and denominator of a rational function, canceling any common factor, and then substituting the excluded x-value into the simplified expression to get the y-coordinate. In practice, finding holes comes down to three steps: factor, cancel, and substitute. A hole differs from a vertical asymptote because the common factor cancels out, leaving a gap at one isolated point rather than a line the graph approaches but never touches.

What Are Holes?

A hole is a removable discontinuity: a single missing point on an otherwise continuous curve. It appears in a rational function — a fraction with polynomials in the numerator and denominator — when the numerator and denominator share a common factor that includes the variable.

Suppose you have $f(x) = \frac{(x-2)(x+3)}{(x-2)(x-1)}$. The factor $(x-2)$ appears in both parts. At $x = 2$, both the numerator and denominator equal zero, so the function is undefined there. But everywhere else, the $(x-2)$ terms divide out, so the graph looks exactly like $y = \frac{x+3}{x-1}$ with a single point punched out at $x = 2$. That missing point is the hole.

It helps to separate the three features that beginners often confuse:

• Zeros occur where the numerator equals zero after canceling — these are x-intercepts, points the graph actually passes through.
• Vertical asymptotes occur where a factor remains in the denominator after canceling — the graph shoots toward positive or negative infinity.
• Holes occur where a factor cancels completely from both numerator and denominator — the graph has a gap at one point.

So the deciding question is simple: does the factor vanish from the denominator after simplifying? If yes, you have a hole. If a copy of it survives in the denominator, you have an asymptote.

Identify Through Factoring

Factoring is the whole game. A hole can only exist if the numerator and denominator share a common polynomial factor, and you cannot see that shared factor until both are written as products.

Work through these steps:

1. Factor the numerator completely into linear or irreducible factors.
2. Factor the denominator completely the same way.
3. Compare the factor lists and circle anything they have in common.

For example, take $g(x) = \frac{x^2 - 9}{x^2 + x - 12}$. The numerator is a difference of squares: $x^2 - 9 = (x-3)(x+3)$. The denominator factors as $(x+4)(x-3)$. Now the shared factor $(x-3)$ jumps out.

A few factoring reminders that matter here:

• Always pull out a greatest common factor first; it simplifies everything after.
• Look for difference of squares ($a^2 - b^2$), perfect square trinomials, and simple quadratics.
• If a factor appears more than once in the denominator but not the numerator, it still produces an asymptote, not a hole.

If, after full factoring, no factor is shared, then the function has no holes at all — any discontinuity present must be an asymptote.

Find the Hole’s Coordinates

Canceling the common factor gives you the simplified function, and from there the coordinates fall out in two moves.

Step 1 — Find the excluded x-value. Set the canceled factor equal to zero and solve. In the example above, the canceled factor is $(x-3)$, so $x = 3$ is excluded from the domain. This is the x-coordinate of the hole.

Step 2 — Substitute into the simplified function. Cancel $(x-3)$ from numerator and denominator to get $g(x) = \frac{x+3}{x+4}$. Now plug in $x = 3$: $\frac{3+3}{3+4} = \frac{6}{7}$. So the hole sits at the point $\left(3, \tfrac{6}{7}\right)$.

That is the entire method, and it is the same whether the numbers are clean or ugly. Two cautions are worth remembering:

• You must substitute into the simplified form, never the original. Plugging $x = 3$ into $\frac{x^2-9}{x^2+x-12}$ gives $\frac{0}{0}$, which is undefined — that indeterminacy is exactly what signals the hole, but it tells you nothing about the y-value.
• There is at most one hole per canceled factor, but a function can have several holes if several distinct factors cancel.

Your two best references for extra practice are Paul’s Online Math Notes and Khan Academy, both of which cover rational functions and removable discontinuities in depth.

A quick reference for what each factor does

Worked Example

Find the hole in $f(x) = \frac{x^2 + 5x + 6}{x^2 - 4}$.

Numerator: $(x+2)(x+3)$. Denominator: $(x-2)(x+2)$. The shared factor is $(x+2)$, so $x = -2$ is excluded. The simplified function is $\frac{x+3}{x-2}$. Substituting $x = -2$ gives $\frac{-2+3}{-2-2} = \frac{1}{-4} = -\frac{1}{4}$. Hole at $\left(-2, -\tfrac{1}{4}\right)$, plus a vertical asymptote at $x = 2$.

Common pitfalls to avoid:

• Substituting the excluded value into the original fraction and stopping when you get $\frac{0}{0}$.
• Forgetting to cancel before evaluating.
• Confusing a canceled factor (hole) with a surviving denominator factor (asymptote).
• Missing holes when the common factor is a bare $x$ rather than a parenthetical term.

When a problem asks only for holes, you can usually skip sketching the whole graph — factor, cancel, substitute, and you are done.

Key Takeaways

• A hole is a removable discontinuity: one missing point where the graph is otherwise continuous.
• Find holes by factoring numerator and denominator and identifying common factors that contain the variable.
• Take each canceled factor, set it to zero to get the excluded x-value, then substitute into the simplified function for the y-coordinate.
• A factor that cancels gives a hole; a factor left in the denominator gives a vertical asymptote; a numerator-only zero after simplifying gives an x-intercept.
• Never substitute into the unsimplified fraction — that just returns the unhelpful form $\frac{0}{0}$.
• A single function can have multiple holes if multiple distinct factors cancel.