Question

How do you find the vertical, horizontal or slant asymptotes for `((x^2)+3)/((9x^2)-80x-9)`?

Answer 1

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

Explanation:

We can factorise the denominator

`9x^2-80x-9=(9x+1)(x-9)`

Let, `f(x)=(x^2+3)/(9x^2-80x-9)=(x^2+3)/((9x+1)(x-9))`

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

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

The vertical asymptotes are `x=-1/9` and `x=9`

As the degree of the numerator is `=` to the degree of the denominator, there is no slant asymptote

Now, we calculate the limits of `f(x)` as `x->+-oo`

`lim_(x->+-oo)f(x)=lim_(x->+-oo)x^2/(9x^2)=1/9`

The horizontal asymptote is `y=1/9`

graph{(y-(x^2+3)/(9x^2-80x-9))(y-1/9)=0 [-52.1, 51.93, -26, 25.94]}