Vertical Asymptotes and Holes

Two different things a denominator of zero can mean.

11 problems · about 22 minutes · AP Precalculus 1.10, 1.11

What this lesson teaches

The student distinguishes a vertical asymptote from a removable hole by comparing factors.

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.

Straightforward practice. Get the method working first.

4 problems

(x + 2)/((x + 2)(x − 7)). At what x is the hole?
Answer -2
Why x = −2.
1 / (x + 6). Vertical asymptote at which x?
Answer -6
Why Set the denominator to zero.
(x − 1)/((x − 1)(x + 4)). Hole at which x?
Answer 1
Why The cancelling factor.
Same function. Asymptote at which x?
Answer -4
Why The surviving factor.

The same ideas with more to keep track of.

3 problems

Match each feature to what causes it.
Answer Vertical asymptote → A denominator factor that does not cancel; Removable hole → A factor shared by numerator and denominator; x-intercept → A numerator factor that does not cancel
Why Ask which side the surviving factor is on.
Is the function defined at a hole? 1 for yes, 0 for no.
Answer 0
Why Cancelling does not restore the point.
x/(x² − 9). How many vertical asymptotes?
Answer 2
Why x² − 9 factors into two.

Mixed problems. Work out what kind of question it is before you start.

4 problems

Same function. What is the x-intercept?
Answer 0
Why Where the numerator is zero.
1/(x² + 1). How many vertical asymptotes?
Answer 0
Why Can x² + 1 ever be zero?
The Denominator: f(x) = 1 / (x − 4). At what x is there a vertical asymptote?
Answer 4
Why x = 4.
The Cancelling Factor: f(x) = (x − 3) / ((x − 3)(x + 1)). At what x is there a hole?
Answer 3
Why x = 3.