A hole in a rational function is a removable discontinuity. It appears when a value of
\(x\) makes the original denominator zero, but the zero comes from a factor that also appears in
the numerator and can be canceled algebraically. The graph behaves like the simplified function,
except that one point is missing.
1. Rational function form
A rational function has the form
\[
f(x)=\frac{P(x)}{D(x)},
\qquad D(x)\ne0.
\]
The denominator controls where the original function is undefined. Every real solution of
\(D(x)=0\) must be excluded from the original domain.
2. When does a hole occur?
A hole occurs when the numerator and denominator share a common factor. Suppose
\[
P(x)=G(x)P_1(x),
\qquad
D(x)=G(x)D_1(x).
\]
Then
\[
\frac{P(x)}{D(x)}
=
\frac{G(x)P_1(x)}{G(x)D_1(x)}
=
\frac{P_1(x)}{D_1(x)},
\qquad G(x)\ne0.
\]
The simplified expression describes the curve, but the values that make \(G(x)=0\) are still
excluded from the original domain. If the simplified denominator is not zero at such a value,
the missing point is a removable hole.
3. Hole coordinate
If \(x=a\) is a zero of the canceled common factor, and
\[
D_1(a)\ne0,
\]
then the hole is located at
\[
\left(a,\frac{P_1(a)}{D_1(a)}\right).
\]
The original function is undefined at \(x=a\), but the limit exists:
\[
\lim_{x\to a}\frac{P(x)}{D(x)}
=
\frac{P_1(a)}{D_1(a)}.
\]
4. Example
Consider
\[
f(x)=\frac{x^2-4}{x-2}.
\]
Factor the numerator:
\[
x^2-4=(x-2)(x+2).
\]
Then
\[
f(x)=\frac{(x-2)(x+2)}{x-2}.
\]
For \(x\ne2\), this simplifies to
\[
f(x)=x+2.
\]
However, the original denominator is zero at \(x=2\), so \(x=2\) is not in the domain.
The y-value of the missing point comes from the simplified expression:
\[
y=2+2=4.
\]
Therefore, the graph is the line \(y=x+2\) with an open circle at
\[
\boxed{(2,4)}.
\]
5. Hole versus vertical asymptote
A denominator zero does not always create a hole. If the factor does not cancel completely, the
graph may still have a vertical asymptote. The key distinction is what happens after simplification.
6. Domain restrictions remain
A common mistake is to cancel a factor and then forget that the original expression was undefined
there. Cancellation is allowed for simplification, but it does not restore the removed domain value.
For example,
\[
\frac{(x-3)(x+1)}{x-3}=x+1
\]
only when \(x\ne3\). The graph is the line \(y=x+1\) with a hole at \(x=3\).
7. Multiple holes
A rational function can have more than one hole if more than one real factor cancels. For example,
\[
\frac{(x-1)(x+2)(x+4)}{(x-1)(x+2)}
\]
simplifies to \(x+4\), but the original function is undefined at \(x=1\) and \(x=-2\). Therefore,
the graph has two holes:
\[
(1,5)
\qquad\text{and}\qquad
(-2,2).
\]
8. Graph interpretation
On the graph, holes are shown as open circles. The curve approaches the open circle from both
sides, but the point itself is not included. This is different from a vertical asymptote, where
the graph shoots upward or downward without approaching a finite point.
9. Common mistakes
- Do not call every denominator zero a hole. First check whether the factor cancels.
- Do not forget that canceled denominator zeros remain excluded from the original domain.
- Do not use the original expression to evaluate the y-coordinate of a hole. Use the simplified expression.
- Do not confuse a removable hole with an x-intercept. A hole is missing from the graph.
10. Final checklist
- Factor or otherwise identify common factors in \(P(x)\) and \(D(x)\).
- Cancel the common factor to get the simplified expression.
- Keep the canceled x-values excluded from the original domain.
- Evaluate the simplified expression at each canceled real x-value.
- Write each hole as an ordered pair \((a,y)\).
- Check remaining denominator zeros for vertical asymptotes.
- Draw holes as open circles on the graph.
