Question

How do you find all the asymptotes for `(x^2 - 9)/(3x - 6)`?

Answer 1

Examine where the denominator is zero to find the vertical asymptote. Long divide to get a quotient linear polynomial that represents the oblique asymptote.

Explanation:

Let `f(x) = (x^2-9)/(3x-6) = ((x-3)(x+3))/(3(x-2))`

The denominator is zero when `x=2`, but the numerator is non-zero. So `f(x)` has a vertical asymptote `x=2`.

Long divide to get a quotient and remainder:

`(x^2-9)/(3x-6) = x/3+2/3-5/(3x-6)`

Then as `x->+-oo` we have `-5/(3x-6) -> 0`

So the oblique asymptote is `y=x/3+2/3`

graph{(y-(x^2-9)/(3x-6))(y-(x/3+2/3))=0 [-10, 10, -5, 5]}