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 #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]}