Question

How do you find vertical, horizontal and oblique asymptotes for `(x^2-x+1)/(x-3)`?

Answer 1

The vertical asymptote is `x=3`
No horizontal asymptote.
The oblique asymptote is `y=x+2`

Explanation:

As you cannot divide by `0`, `x!=3`

Therefore,

The vertical asymptote is `x=3`

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

Let's do a long division

`color(white)(aaaa)` `x^2-x+1` `color(white)(aaaa)` `∣` `x-3`

`color(white)(aaaa)` `x^2-3x` `color(white)(aaaaaaa)` `∣` `x+2`

`color(white)(aaaaa)` `0+2x+1`

`color(white)(aaaaaaa)` `+2x-6`

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

Therefore,

`(x^2-x+1)/(x-3)=x+2+7/(x-3)`

The oblique asymptote is `y=x+2`

There is no horizontal asymptote. When `x->+-oo`,

`f(x)->+-oo`

graph{(y-(x^2-x+1)/(x-3))(y-x-2)=0 [-41.1, 41.1, -20.56, 20.56]}