How do you find a hole in a rational function

Rational functions are quotients of polynomials. When a factor of the denominator also appears in the numerator, the factor can be reduced, leaving the function undefined only at the point where the factor equals zero. That isolated point is called a hole, or removable discontinuity, and it is different from a vertical asymptote, which occurs when the denominator is zero but no cancellation is possible.

Step‑by‑step procedure

Follow these actions to locate holes:

1. 1Factor the numerator and denominator completely.
2. 2Identify any common factors between them.
3. 3Cancel the common factors algebraically.
4. 4Set the cancelled factor equal to zero; that x‑value is the hole.

Consider the function $f(x)=\frac{(x-2)(x+3)}{(x-2)(x-5)}$. First, factor both parts; they are already factored. The factor $(x-2)$ appears in both numerator and denominator, so it cancels, leaving $g(x)=\frac{x+3}{x-5}$ for all $x\neq2$. The cancelled factor $(x-2)$ equals zero at $x=2$, so the original function has a hole at $x=2$. The y‑coordinate of the hole is found by plugging $x=2$ into the simplified function $g(x)$, giving $g(2)=\frac{2+3}{2-5}=\frac{5}{-3}=-\frac{5}{3}$. Thus the hole is the point $(2,-5/3)$.

Common pitfalls when searching for holes:

• Cancelling a factor that is not actually common; this creates a false hole.
• Forgetting to check the domain after cancellation; the cancelled root remains excluded.
• Treating a vertical asymptote as a hole because the numerator also approaches zero.

Factor summary for the example: