# How do you find the vertical, horizontal or slant asymptotes for f(x)=x/(x-1)^2?

Aug 9, 2017

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

#### Explanation:

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.

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

$\text{horizontal asymptotes occur as}$

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

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

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

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

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

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/(x-1)^2 [-10, 10, -5, 5]}