Question

How do you find vertical, horizontal and oblique asymptotes for `(x^3+8) /(x^2+9)`?

Answer 1

`(x^3+8)/(x^2+9)` has an oblique asymptote `y=x` and no other asymptotes.

Explanation:

For all Real values of `x`, the denominator `x^2+9 >= 9 > 0`.

So there are no vertical asymptotes.

`(x^3+8)/(x^2+9) = (x^3+9x-9x+8)/(x^2+9)= (x(x^2+9)-9x+8)/(x^2+9)=x+(-9x+8)/(x^2+9)`

Then since the denominator has higher degree than the numerator:

`(-9x+8)/(x^2+9)->0` as `x->+-oo`

Hence:

`(x^3+8)/(x^2+9) = x+(-9x+8)/(x^2+9)`

is asymptotic to `x` as `x->+-oo`

So `f(x)` has no vertical or horizontal asymptotes, but has one oblique asymptote:

`y = x`

graph{(y-(x^3+8)/(x^2+9))(y-x) = 0 [-20, 20, -10, 10]}