Question

What are the vertical and horizontal asymptotes of `y = (x+3)/(x^2-9)`?

Answer 1

vertical asymptote at `x=3`

horizontal asymptote at `y=0`

hole at `x=-3`

Explanation:

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

First factor:

`y = ((x+3))/((x+3)(x-3))`

Since the factor `x+3` cancels that is a discontinuity or hole, the factor `x-3` does not cancel so it is a asymptote:

`x-3=0`

vertical asymptote at `x=3`

Now let's cancel out the factors and see what the functions does as x gets really big in the positive or negative:

`x -> +-oo, y ->?`

`y = cancel((x+3))/(cancel((x+3))(x-3))=1/(x-3)`

As you can see the reduced form is just `1` over some number `x`, we can ignore the `-3` because when `x` is huge it is insignificant.

We know that: `x ->+-oo, 1/x -> 0` hence, our original function has the same behavior:

`x ->+-oo, ((x+3))/((x+3)(x-3)) -> 0`

Therefore, the function has a horizontal asymptote at `y=0`

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