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

Apr 12, 2017

$\text{vertical asymptote at } x = 1$
$\text{horizontal asymptote at } y = 0$

#### Explanation:

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

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

$\text{solve "(1-x)^2=0rArrx=1" is the asymptote}$

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 ${x}^{2}$

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

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

$\Rightarrow y = 0 \text{ is the asymptote}$

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there are no slant asymptotes.
graph{(x)/((1-x)^2) [-10, 10, -5, 5]}