Question

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

Answer 1

The oblique asymptote is `y=x+2`
There is no vertical asymptote
No horizontal asymptote

Explanation:

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

The degree of the numerator is greater than the degree of the denominator, so there is an oblique asymptote.

Let's perform a long division

`color(white)(aaaa)` `x^2-2x+2` `color(white)(aa)` `|` `x^3+0x^2+0x+1` `color(white)(aaaa)` `|` `x+2`

`color(white)(aaaaaaaaaaaaaaaaaa)` `x^3-2x^2+2x`

`color(white)(aaaaaaaaaaaaaaaaaaa)` `0+2x^2-2x+1`

`color(white)(aaaaaaaaaaaaaaaaaaaaaaa)` `2x^2+4x+4`

`color(white)(aaaaaaaaaaaaaaaaaaaaaaaaa)` `0+2x-3`

Therefore,

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

`lim_(x->-oo)f(x)-(x+2)=lim_(x->-oo)(2x-3)/(x^2-2x+2)=lim_(x->-oo)(2x)/x^2=O^-`

`lim_(x->+oo)f(x)-(x+2)=lim_(x->+oo)(2x-3)/(x^2-2x+2)=lim_(x->+oo)(2x)/x^2=O^+`

The oblique asymptote is `y=x+2`

The denominator is

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

`AA x in RR, (x-1)^2+1>0`

There is no vertical asymptote
graph{(y-(x^3-1)/(x^2-2x+2))(y-x-2)=0 [-10.97, 11.53, -2.97, 8.28]}