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

1 Answer
Nov 4, 2016

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]}