Question

How do you find the vertical, horizontal or slant asymptotes for `(4x^2+11) /( x^2+8x−9)`?

Answer 1

The vertical asymptotes are `x=-9` and `x=1`
No slant asymptote.
The horizontal asymptote is `y=4`

Explanation:

Let's factorise the denominator

`x^2+8x-9=(x-1)(x+9)`

Let, `f(x)=(4x^2+11)/(x^2+8x-9)`

The domain of `f(x)` is `D_f(x)=RR-{-9,1}`

As we cannot divide by `0`, `x!=-9` and `x!=1`

So, the vertical asymptotes are `x=-9` and `x=1`

The degree of the numerator is `=` to the degree of the denominator, we don't have a slant asymptote.

To find the horizontal asymptote, we compute the limits of `f(x)` as `x->+-oo`

`lim_(x->+-oo)f(x)=lim_(x-+-oo)4x^2/x^2=4`

The horizontal asymptote is `y=4`

graph{(y-(4x^2+11)/(x^2+8x-9))(y-4)(x-1)(x+9)=0 [-52, 52, -26, 26]}