Question

How do you identify the oblique asymptote of `f(x) = (x^3-6x^2+12x-2)/(x^2-2x+2)`?

Answer 1

Divide numerator by denominator to get a polynomial quotient and a remainder. The quotient is the oblique asymptote `x - 4`

Explanation:

`(x^3-6x^2+12x-2)/(x^2-2x+2)`

`=((x^3-2x^2+2)-(4x^2-8x+8)+(4x+4))/(x^2-2x+2)`

`=((x-4)(x^2-2x+2)+4(x+1))/(x^2-2x+2)`

`=(x-4)+(4(x+1))/(x^2-2x+2)`

It's a little nicer to use synthetic division to do this, but it's not easy to typeset clearly.

Then:

`(4(x+1))/(x^2-2x+2) -> 0` as `x->+-oo`

So the oblique asymptote is `x-4`

graph{(y - (x^3-6x^2+12x-2)/(x^2-2x+2))(y - x + 4) = 0 [-17.75, 22.25, -10.96, 9.04]}