# How do you find the asymptotes for  f(x) = (2x-2) /( (x-1)(x^2 + x -1))?

May 2, 2017

$\text{vertical asymptotes at "x~~-1.62" and } x \approx 0.62$

$\text{horizontal asymptote at } y = 0$

#### Explanation:

$\text{the first step is to factorise and simplify f(x)}$

$f \left(x\right) = \frac{2 \cancel{\left(x - 1\right)}}{\cancel{\left(x - 1\right)} \left({x}^{2} + x - 1\right)} = \frac{2}{{x}^{2} + x - 1}$

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

$\text{solve " x^2+x-1=0" using the quadratic formula}$

$x = \frac{- 1 \pm \sqrt{1 + 4}}{2}$

$\Rightarrow x = - \frac{1}{2} \pm \frac{1}{2} \sqrt{5}$

$\Rightarrow x \approx - 1.62 \text{ and " x~~0.62" are the asymptotes}$

Horizontal asymptotes occur as

${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ ( a constant)}$

divide terms on numerator/denominator by the highest power of x, that is ${x}^{2}$

$f \left(x\right) = \frac{\frac{2}{x} ^ 2}{{x}^{2} / {x}^{2} + \frac{x}{x} ^ 2 - \frac{1}{x} ^ 2} = \frac{\frac{2}{x} ^ 2}{1 + \frac{1}{x} - \frac{1}{x} ^ 2}$

as $x \to \pm \infty , f \left(x\right) \to \frac{0}{1 + 0 - 0}$

$\Rightarrow y = 0 \text{ is the asymptote}$
graph{2/(x^2+x-1) [-10, 10, -5, 5]}