Question

What are the asymptote(s) and hole(s), if any, of `f(x) =1/(x^2+2)`?

Answer 1

`f(x)` has a horizontal asymptote `y=0` and no holes

Explanation:

`x^2 >= 0` for all `x in RR`

So `x^2+2 >= 2 > 0` for all `x in RR`

That is, the denominator is never zero and `f(x)` is well defined for all `x in RR`, but as `x->+-oo`, `f(x) -> 0`. Hence `f(x)` has a horizontal asymptote `y=0`.

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