Why Cancelling a Factor Can Leave a Missing Point
A rational function is a quotient of polynomials, so its domain excludes every input that makes its original denominator zero. Factoring and cancelling a common factor can make the quotient easier to evaluate, but cancellation does not put those excluded inputs back into the original domain.
This distinction explains how a graph can follow a simple curve everywhere in its domain and still be missing a point on that curve. When the cancelled factor accounts for a denominator zero that disappears completely from the reduced expression, the missing point is called a hole.
The coordinate \(a\) is the input excluded by the original denominator. The coordinate \(g(a)\) is found by substituting that input into the reduced expression, not into the original quotient. The original quotient has denominator zero at \(a\), so it has no function value there.
How to Locate a Hole
For a rational function given by factored polynomials, use the original denominator to find the excluded inputs before simplifying. Then cancel the shared factors and check whether the reduced denominator is nonzero at each excluded input. Only then evaluate the reduced expression to find the corresponding output coordinate.
Set the original denominator equal to zero. Each resulting input is excluded, even if its factor can later be cancelled.
Cancel only factors shared by the numerator and denominator, and keep the original exclusions recorded.
An excluded input gives a hole when the reduced expression is defined there. If the reduced denominator is still zero there, that input is not a hole of this kind.
Substitute the excluded input into the reduced expression. The result gives the hole’s output coordinate.
The check in Step 3 matters when a denominator factor occurs more times than the corresponding numerator factor. Cancelling the shared copies may leave another copy in the reduced denominator. That input remains a zero of the reduced denominator, so it cannot be evaluated there to locate a hole. This is the distinction between a factor that disappears completely and a denominator factor that remains uncancelled.
Worked Example: A Single Cancelled Factor
Worked Example: A Single Cancelled Factor
Find the hole, if any, of \(f(x)=\frac{x^2-9}{x-3}\).
Step 1: Find the original restriction. The denominator is zero when \(x-3=0\), so \(x=3\) is excluded from the domain.
Step 2: Factor and cancel. The numerator is a difference of squares:
Thus, for inputs where the original quotient is defined,
Step 3: Check the reduced expression at the excluded input. The reduced expression \(x+3\) is defined at \(x=3\).
Step 4: Find the output coordinate.
Therefore, the graph has a hole at \((3,6)\).
Check: At \(x=3\), the original denominator is \(3-3=0\), confirming that \(f(3)\) is undefined. The reduced expression gives \(3+3=6\), so \((3,6)\) is the missing point on the line described by the reduced expression.
Why the Reduced Expression Gives the Hole’s Height
Cancellation is justified for inputs where the cancelled factor is not zero. At those inputs, the original function and reduced expression have the same value. At the root of the cancelled factor, however, the original quotient is undefined. If the reduced denominator is nonzero there, the reduced expression still has a well-defined value.
This argument explains both parts of the hole: why the original function has no value at the input and why the reduced expression supplies the missing point’s output coordinate. It relies on \(Q_1(a)\ne 0\). If that condition fails, the reduced expression is not defined at \(a\), so this calculation does not locate a hole there.
Worked Example: Two Cancelled Factors
Worked Example: Two Cancelled Factors
Find all holes of \(f(x)=\frac{(x-1)(x+2)(x-4)}{(x-1)(x+2)(x+3)}\).
Step 1: Record the original exclusions. The original denominator is zero at \(x=1\), \(x=-2\), and \(x=-3\). All three inputs are excluded from the domain.
Step 2: Cancel common factors. Cancelling \(x-1\) and \(x+2\) gives
The restrictions remain those of the original denominator, including the two inputs whose factors were cancelled.
Step 3: Check which excluded inputs are holes. The reduced denominator \(x+3\) is nonzero at \(x=1\) and \(x=-2\). At \(x=-3\), it is zero, so the reduced expression is not defined there. Therefore, the cancelled factors produce holes at \(x=1\) and \(x=-2\).
Step 4: Evaluate the reduced expression at each hole input.
The holes are \(\left(1,-\frac{3}{4}\right)\) and \((-2,-6)\).
Check: The original denominator equals zero at both \(x=1\) and \(x=-2\), so the original function has no value at either input. The reduced denominator equals \(1+3=4\) at \(x=1\) and \(-2+3=1\) at \(x=-2\), so both reduced-expression evaluations are valid. At \(x=-3\), the reduced denominator is \(-3+3=0\), confirming that this excluded input is not a hole found by substitution into the reduced expression.
Worked Example: A Cancelled Factor and a Remaining Denominator Factor
Worked Example: A Cancelled Factor and a Remaining Denominator Factor
Determine the hole of \(f(x)=\frac{(x-2)^2(x+1)}{(x-2)(x+1)(x+4)}\), and identify which excluded input does not give a hole.
Step 1: Find the original exclusions. The original denominator is zero at \(x=2\), \(x=-1\), and \(x=-4\). These inputs are all excluded from the original domain.
Step 2: Cancel the common factors. Cancel one factor of \(x-2\) and the factor \(x+1\):
Step 3: Check the reduced denominator at the excluded inputs. At \(x=2\), the reduced denominator is \(2+4=6\), so the reduced expression is defined. At \(x=-1\), it is \(-1+4=3\), so the reduced expression is also defined. At \(x=-4\), it is \(-4+4=0\), so the reduced expression is undefined there.
Step 4: Calculate the hole coordinate. The cancelled factors give holes at \(x=2\) and \(x=-1\). Substitution into the reduced expression gives
Thus, the holes are \((2,0)\) and \((-1,-1)\). The excluded input \(x=-4\) does not give a hole because its factor remains in the reduced denominator.
Check: The original denominator is zero at all three excluded inputs. The reduced denominator is nonzero at \(2\) and \(-1\), validating the two substituted values, but it is zero at \(-4\). Therefore, the reduced expression supplies hole coordinates for the first two inputs only.
A Common Mistake: Forgetting the Original Domain
A frequent error is to cancel a factor and then treat the simplified formula as though it were the original function at every input. For instance, consider \(\frac{x^2-9}{x-3}\). A mistaken conclusion is
This statement is false at \(x=3\): the left side is undefined there, while the right side equals \(6\). The correct equality includes the original restriction:
The correction is to write down the original domain restrictions before cancelling, retain them after simplifying, and use the reduced expression at an excluded input only to locate the hole. This keeps the formula and the domain of the function distinct.
Why Holes Matter
A simplified formula alone does not always describe the entire original function, because it does not record restrictions lost through cancellation. Keeping track of holes is therefore essential when interpreting a rational function: it preserves the difference between an expression that gives a value at an input and the original function, which may be undefined there.
When sketching a graph, a hole is commonly shown as an open circle at the point obtained from the reduced expression. The open circle marks a location on the reduced curve that is absent from the original function’s graph. If an excluded input remains a zero of the reduced denominator, do not use this hole procedure there; the reduced expression is still undefined at that input.
Check Your Understanding
For each question, keep the original domain restrictions separate from the simplified expression.
- For \(f(x)=\frac{x^2-16}{x-4}\), what input is excluded, and what are the coordinates of the hole?
- For \(f(x)=\frac{(x+1)(x-3)}{(x+1)(x+2)}\), identify the hole and the excluded input where the reduced expression is undefined.
- Why does cancelling a common factor not make the original function defined at the root of that factor?
- For \(f(x)=\frac{(x-1)^2}{(x-1)(x+4)}\), simplify the expression and determine the hole coordinate, if any.
- What condition must the reduced denominator satisfy at an excluded input for substitution into the reduced expression to locate a hole?
