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

##### 1 Answer

Feb 14, 2016

#### Answer:

Re-express as

#### 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

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

So this quadratic has no Real zeros.

The numerator

Therefore

So the only asymptote is a horizontal one:

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