Question

How to find the asymptotes of `f(x) = (x+3) /( x^2 + 8x + 15)` ?

Answer 1

Simplify `f(x)` to see that there is a horizontal asymptote `y=0`, vertical asymptote `x=-5` and hole (removable singularity) at `(-3, 1/2)`

Explanation:

`f(x) = (x+3)/(x^2+8x+15)`

`=(x+3)/((x+3)(x+5))`

`=1/(x+5)`

with exclusion `x != -3`

This function has horizontal asymptote `y=0` since `1/(x+5)->0` as x->+-oo#

It has a vertical asymptote `x=-5`, where the denominator is zero, but the numerator non-zero.

It has a hole (removable singularity) at `(-3, 1/2)` since the left and right limits exist at `(-3, 1/2)` and are equal.