# How do you find the vertical, horizontal and slant asymptotes of:  f(x)=(2x - 2 )/( (x-1)(x^2 + x - 1))?

Oct 31, 2016

The vertical asymptotes are $x = \frac{- 1 - \sqrt{5}}{2}$
and $x = \frac{- 1 + \sqrt{5}}{2}$
The horizontal asymptote is $y = 0$
There is no slant symptote

#### Explanation:

You can start by simplifying $f \left(x\right)$
$f \left(x\right) = \frac{2 x - 2}{\left(x - 1\right) \left({x}^{2} + x - 1\right)} = \frac{2 \left(\cancel{x - 1}\right)}{\cancel{x - 1} \left({x}^{2} + x - 1\right)}$
So $f \left(x\right) = \frac{2}{{x}^{2} + x - 1}$
As we cannot divide by $0$, so ${x}^{2} + x - 1 \ne 0$
The roots of this equation is $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
So the roots are $\frac{- 1 \pm \sqrt{1 + 5}}{2} = \frac{- 1 \pm \sqrt{5}}{2}$
So the vertical asymptotes are $x = \frac{- 1 - \sqrt{5}}{2}$
and $x = \frac{- 1 + \sqrt{5}}{2}$

And the limit of $f \left(x\right)$ as $x \to \infty$ is $0$
so the horizontal asymptote is $y = 0$
graph{2/(x^2+x-1) [-7.474, 8.33, -4.27, 3.63]}