- Topic
- Curve sketching
- Difficulty
- Standard
- Suggested time
- 4 min
- Marks
- 2
Curve sketching practice question 12
The function below has exactly one vertical asymptote. Find its -value.
A zero of the denominator is only a candidate for an asymptote. What would make the graph have a hole there instead?
The denominator vanishes at two points, but one of them is cancelled by the numerator. Factor both.
A cancelled factor gives a hole, not an asymptote.
A whole number.
Theory UBC Math 100 §7CLP-1 §3.6Key results
Worked solution
Strategy
Vertical asymptotes come from zeros of the denominator that survive cancellation. Factor, cancel, and look at what is left.
Working
The factor (x − 2) cancels.
Finite limit: no asymptote at 2.
The surviving zero of the denominator is the vertical asymptote.
Answer
The answer is 3.
Check
f(2.99) = −499 and f(3.01) = 501 blow up, while f(1.99) and f(2.01) are both close to −4.
Where this goes wrong
Reporting both x = 2 and x = 3 as asymptotes. At x = 2 the numerator vanishes too, and the limit is finite: a hole in the graph, not a vertical asymptote.
Leans on
Removable and infinite discontinuities, from Continuity
Try a similar question
Classify the discontinuities of the function below.Saved in this browser only.