How do you find the Vertical, Horizontal, and Oblique Asymptote given #2 /(x^2 + x -1)#?

1 Answer
Nov 4, 2016

Answer:

The vertical asymptotes are #x=(-1+sqrt5)/2# and #x=(-1-sqrt5)/2#
The horizontal asymptote is #y=0#

Explanation:

Let's find the roots of the denominator
#x^2+x-1=0#
#Delta=1-(4*1*-1)=5#
So the roots are #(-1+-sqrt5)/2#
As we cannot divide by #0#, so the vertical asymptotes are
#x=(-1+sqrt5)/2# and #x=(-1-sqrt5)/2#
The limit of the expression as #x->+-oo# is #=0#
So the horizontal asymptote is #y=0#
And as the degree of the numerator is less than the degree of the denominator, ther is no oblique asymptote
graph{2/(x^2+x-1) [-11.25, 11.25, -5.62, 5.62]}