Question

How do you find vertical, horizontal and oblique asymptotes for `(x^2-2x)/(x+1)`?

Answer 1

The vertical asymptote is `x=-1`
The oblique asymptote is `y=x`
No horizontal asymptote

Explanation:

Let `f(x)=(x^2-2x)/(x+1)`

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

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

So, the vertical asymptote is `x=-1`

The degree of the numerator is `>` to the degree of the denominator, so we expect an oblique asymptote.

Let's do a long division

`color(white)(aaaa)` `x^2-2x` `color(white)(aaaa)` `∣` `x+1`

`color(white)(aaaa)` `x^2+x` `color(white)(aaaaa)` `∣` `x`

`color(white)(aaaaa)` `0-3x`

So,

`(x^2-2x)/(x+1)=x-(3x)/(x+1)`

Therefore

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

graph{(y-(x^2-x)/(x+1))(y-x)=0 [-36.53, 36.57, -18.26, 18.27]}