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

Mar 1, 2016

The asymptotes are $x = 1$ and $y = 0$.

#### Explanation:

${\lim}_{x \to 1} f \left(x\right) = \infty$

${\lim}_{x \to - \infty} f \left(x\right) = 0$

${\lim}_{x \to \infty} f \left(x\right) = 0$

Graph of $y = f \left(x\right)$
graph{1/(x-1)^2 [-10, 10, -5, 5]}

Mar 1, 2016

The asymptotes are $x = 1$ (vertical), $y = 0$ (horizontal).

#### Explanation:

First we can rewrite this as follows

$f \left(x\right) = {\left(\frac{1}{x - 1}\right)}^{2}$ so $f \left(x\right) > 0$ , ${D}_{f} = \mathbb{R} \setminus \setminus \setminus \left\{1\right\}$

Hence for $x \to 1$ ,$f \left(x\right) \to + \infty$

and for $x \to \pm \infty$ ,$f \left(x\right) \to 0$

Hence the asymptotes are $x = 1$ (vertical), $y = 0$ (horizontal).

The graph of $f \left(x\right)$ is