Question

How do you find the asymptotes for `f(x)=(x^2+x-2)/(x^2-x+2)`?

Answer 1

Re-express as `f(x) = 1 + (2x-4)/(x^2-x+2)` to find horizontal asymptote `y=1`

Explanation:

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

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

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

Now `x^2-x+2` is of the form `ax^2+bx+c` with `a=1`, `b=-1` and `c=2`, so has discriminant `Delta` given by the formula:

`Delta = b^2-4ac = (-1)^2-(4*1*2) = 1-8 = -7`

So this quadratic has no Real zeros.

The numerator `(2x-4)` is of lower degree than the denominator, so `(2x-4)/(x^2-x+2) -> 0` as `x->+-oo`

Therefore `f(x)->1+0 = 1` as `x->+-oo`

So the only asymptote is a horizontal one: `y=1`

graph{(x^2+x-2)/(x^2-x+2) [-10, 10, -5, 5]}