# How do you find the asymptotes for G(x) = (x-1)/(x-x^3)?

Jun 22, 2017

$\text{vertical asymptotes at " x=0" and } x = - 1$
$\text{horizontal asymptote at } y = 0$

#### Explanation:

$\text{factorise the denominator and simplify}$

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

$\Rightarrow g \left(x\right) = - \frac{1}{x \left(x + 1\right)}$

$\text{since the factor " (x-1)" has been removed from the }$
$\text{numerator/denominator this means there is a hole at}$
x = 1"

$\text{the graph of " -1/(x(x+1))" is the same as } \frac{x - 1}{x - {x}^{3}}$
$\text{but without the hole}$

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

$\text{solve } x \left(x + 1\right) = 0 \Rightarrow x = 0 , x = - 1$

$\Rightarrow x = 0 \text{ and " x=-1" are the asymptotes}$

$\text{horizontal asymptotes occur as}$

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

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

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

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

$\Rightarrow y = 0 \text{ is the asymptote}$
graph{(x-1)/(x-x^3) [-10, 10, -5, 5]}