Question

How do you find all the asymptotes for function `f(x)= (x+3)/(x^2-9)`?

Answer 1

`f(x) = (x+3)/(x^2-9) = (x+3)/((x-3)(x+3)) = 1/(x-3)`

with exclusion `x != -3`.

Hence the asymptotes are `y = 0` and `x = 3`.

Explanation:

Use the difference of squares identity: `a^2-b^2 = (a-b)(a+b)`

with `a=x` and `b=3` to find:

`(x^2-9) = (x^2-3^2) = (x-3)(x+3)`

So:

`f(x) = (x+3)/(x^2-9) = (x+3)/((x-3)(x+3)) = 1/(x-3)`

with exclusion `x != -3`.

As `x->+-oo` we find `f(x) -> 0`, so `y=0` is a horizontal asymptote.

There is a vertical asymptote at `x=3` since the denominator becomes zero and the numerator is non-zero.