Question

How do you find the asymptotes for `f(x)=1/(x^2+4)`?

Answer 1

This function only has a horizontal asymptote: `y = 0`

Explanation:

Notice that `f(x) != 0` for all `x in RR`, since if `f(x) = 0` then `1 = 0(x^2+4) = 0`, which is false.

As `x->+-oo` we find `1/(x^2+4) -> 0`

So we have a horizontal asymptote `y=0`

On the other hand `x^2+4 >= 4 > 0` for all `x in RR`, so the denominator is non-zero for all Real values of `x`.

So there is no vertical asymptote.

graph{1/(x^2+4) [-2.5, 2.5, -1.25, 1.25]}