- Topic
- Limits
- Difficulty
- Introductory
- Suggested time
- 3 min
- Marks
- 2
Limits practice question 8
Which statement gives every vertical asymptote of the graph of the function below?
Not every zero of a denominator gives a vertical asymptote. At each zero of the bottom, what is the top doing?
Factor the top and the bottom before deciding anything: .
A factor that cancels leaves a hole. A zero of the denominator that survives the cancelling, where the numerator is not zero, gives a vertical asymptote.
Theory UBC Math 100 §2.2CLP-1 §1.3Key results
Worked solution
Strategy
Factor first. A zero of the denominator that cancels with the numerator leaves a hole, because the function has a finite limit there; a zero that survives makes the function unbounded, which is a vertical asymptote.
Working
Factor the top and the bottom. Both contain x − 2.
Cancel x − 2. The limit at 2 is finite, so the graph has a hole there, not an asymptote.
The top stays near −3 while the bottom shrinks to 0, so f is unbounded near −6.
x + 6 is negative just left of −6 and positive just right of it, so x = −6 is a vertical asymptote.
Answer
The answer is The only vertical asymptote is x = −6; at x = 2 the graph has a hole..
Check
Near x = 2 the values stay bounded, f(2.01) ≈ 0.625, while near −6 they blow up: f(−5.99) = −299.
Where this goes wrong
Calling every zero of the denominator a vertical asymptote. At x = 2 the numerator is zero too, the factor x − 2 cancels, and f approaches 5/8 there: a hole.
Leans on
Factoring a quadratic, from Functions, graphs and inverses
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