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

Nov 4, 2016

The vertical asymptotes are $x = \frac{- 1 + \sqrt{5}}{2}$ and $x = \frac{- 1 - \sqrt{5}}{2}$
The horizontal asymptote is $y = 0$

Explanation:

Let's find the roots of the denominator
${x}^{2} + x - 1 = 0$
$\Delta = 1 - \left(4 \cdot 1 \cdot - 1\right) = 5$
So the roots are $\frac{- 1 \pm \sqrt{5}}{2}$
As we cannot divide by $0$, so the vertical asymptotes are
$x = \frac{- 1 + \sqrt{5}}{2}$ and $x = \frac{- 1 - \sqrt{5}}{2}$
The limit of the expression as $x \to \pm \infty$ 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]}