# How do you find the vertical, horizontal and slant asymptotes of: (x^2-2x+1)/(x^3-x^2+x-1)?

Feb 15, 2017

Horizontal asymptote : $\leftarrow y = 0 \rightarrow$.
Indeterminate hole : at x = 1.

#### Explanation:

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

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

Sans indeterminate hole at x = 1,

$y = \frac{x - 1}{{x}^{2} + 1}$

As $x \to \pm \infty , y \to 0$.

So, y = 0 is the asymptote.

${x}^{2} + 1$ has no zeros. So, there are no vertical asymptotes.

graph{((x-1)/(x^2+1)-y)y((x-1)^2+y^2-.004)=0x^2 [-5, 5, -2.5, 2.5]}

Socratic graph with asymptote and location of indeterminate hole.

The hole is a 0-D point, and so, it is termed indeterminate, in

graphics.