Question

How do you find vertical, horizontal and oblique asymptotes for `f(x) = (x^2 - 9) / (x - 4)`?

Answer 1

The vertical asymptote is `x=4`
The oblique asymptote is `y=x+4`
No horizontal asymptote.

Explanation:

The domain of `f(x)` is `D_f(x)=RR-{4}`

As you cannot divide by `0`, `x!=4`

The vertical asymptote is `x=4`

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

Let's do a long division

`color(white)(aaaa)` `x^2` `color(white)(aaaaaa)` `-9` `color(white)(aaaa)` `∣` `color(red)(x-4)`

`color(white)(aaaa)` `x^2-4x` `color(white)(aaaaaaaa)` `∣` `color(blue)(x+4)`

`color(white)(aaaaa)` `0+4x-9`

`color(white)(aaaaaaa)` `+4x-16`

`color(white)(aaaaaaaaa)` `+0+7`

Therefore,

`f(x)=(x^2-9)/(x-4)=color(blue)(x+4)+7/(x-4)`

`lim_(x->-oo)(f(x)-(x+4))=lim_(x->-oo)7/x=0^-`

`lim_(x->+oo)(f(x)-(x+4))=lim_(x->+oo)7/x=0^+`

The oblique asymptote is `y=x+4`

graph{(y-(x^2-9)/(x-4))(y-x-4)=0 [-52, 52, -26.03, 26]}