A rational function is undefined wherever its denominator is zero. What happens there depends on the numerator.
Step 1: Let's Learn
Read it, or press Listen and follow the words.
A vertical asymptote
If the factor survives after cancelling, the outputs run off to infinity there.
A removable hole
If the same factor appears above and below and cancels, the graph is a normal curve with one point missing.
Cancelling does not restore the point
The function is still undefined there. Cancelling changes the picture, not the domain.
Two things a zero denominator can mean
If the factor cancels with the numerator, the graph has a hole — one missing point. If it does not, there is a vertical asymptote. Factoring both parts is what distinguishes the cases.
Describing behaviour at an asymptote
The function grows without bound, and the sign can differ on the two sides. AP answers must say which: "as x approaches 2 from the left, f decreases without bound" is the expected form.
A hole still has a limit
At a removable discontinuity the limit exists even though the function does not. Substituting into the cancelled expression gives the y-value of the hole, which is what a graph should mark with an open circle.
The domain excludes both
Holes and asymptotes alike are excluded from the domain, even when a factor cancels. Cancelling changes the formula, not the original function's domain, and stating that restriction is part of a complete answer.
Step 2: Try It Yourself
Tap and try it out.
Step 3: Watch an Example
One step at a time.
Watch Sam Tell Them Apart
Sam examines f(x) = (x − 2)(x + 1) / ((x − 2)(x − 5)).
- Step 1
The denominator is zero at x = 2 and x = 5.
Step 4: Your Turn
Practice makes it stick.
The Denominator
Problem 1 of 2
f(x) = 1 / (x − 4). At what x is there a vertical asymptote?
The Cancelling Factor
Problem 2 of 2
f(x) = (x − 3) / ((x − 3)(x + 1)). At what x is there a hole?
Hole or Asymptote
1 of 8
1 / (x + 6). Vertical asymptote at which x?
2 of 8
(x − 1)/((x − 1)(x + 4)). Hole at which x?
3 of 8
Same function. Asymptote at which x?
4 of 8
Match each feature to what causes it.
Tap a card on the left to start.
5 of 8
Is the function defined at a hole? 1 for yes, 0 for no.
6 of 8
x/(x² − 9). How many vertical asymptotes?
7 of 8
Same function. What is the x-intercept?
8 of 8
1/(x² + 1). How many vertical asymptotes?
Step 5: Quick Check
Show what you know.
Question 1 of 1
(x + 2)/((x + 2)(x − 7)). At what x is the hole?
What You Learned
- A rational function is undefined wherever the denominator is zero.
- A factor that cancels leaves a hole; one that survives gives a vertical asymptote.
- Cancelling changes the picture but never restores the missing point.