Question

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

Answer 1

The oblique asymptote is `y=x`
No vertical and horizontal asymptotes

Explanation:

The domain of `f(x)` is `D_f(x)=RR`

The denominator `x^2+9>0`, `AAx in RR`

So there are no vertical asymptotes.

As the degree of the numerator is `>` degree of the denominator, there is an oblique asymptote.

Let's do a long division

`color(white)(aaaa)` `x^3+8` `color(white)(aaaa)` `∣` `x^2+9`

`color(white)(aaaa)` `x^3+9x` `color(white)(aaa)` `∣` `x`

`color(white)(aaaa)` `0-9x+8`

So,

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

So, the oblique asymptote is `y=x`

To calculate the limits as `x-+-oo`, we take the terms of highest degree in the numerator and the denominator

`lim_(x->+-oo)f(x)=lim_(x->+-oo)x^3/x^2=lim_(x->+-oo)x=+-oo`

graph{(y-(x^3+8)/(x^2+9))(y-x)=0 [-5.546, 5.55, -2.773, 2.774]}