Question

How do you find any asymptotes of `f(x)=(x+3)/(x^2-4)`?

Answer 1

There are two vertical asymptotes at `x=-2` and `x=2`
There is a horizontal asymptote at 0 as `x->oo`
There is a horizontal asymptote at 0 as `x->-oo`

Explanation:

`f(x)=(x+3)/(x^2-4)`
`:. f(x)=(x+3)/(x^2-2^2)`
`:. f(x)=(x+3)/((x+2)(x-2))`

Vertical Asymptotes
These occur when the denominator is zero
`:. (x+2)(x-2) = 0`
`x=+-2`#

Horizontal Asymptotes
We also need to examine the behaviour as `x->+-oo`
Now for large `x`,
`x+3 ~~ x` and `x+2 ~~ x` and `x-3 ~~ x`
We have, `f(x)=(x+3)/((x+2)(x-2))`
and so, for large `x`, `f(x) ~~ x/x^2 ~~ 1/x`
so, as `x-> -oo => f(x) -> 0^-`
and, as `x-> oo => f(x) -> 0^+`

Summary
There are two vertical asymptotes at `x=-2` and `x=2`
There is a horizontal asymptote at 0 as `x->oo`
There is a horizontal asymptote at 0 as `x->-oo`

The graph validates this;
graph{(x+3)/((x+2)(x-2)) [-10, 10, -5, 5]}