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

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Answer 1 · 2017-02-15T16:57:48

Horizontal asymptote : $larr y = 0 rarr$ .
Indeterminate hole : at x = 1.

$y=(x-1)^2/((x-1)(x^2+1))$

$=((x-1)/(x-1))(x-1)/(x^2+1)$ .

Sans indeterminate hole at x = 1,

$y = (x-1)/(x^2+1)$

As $x to +-oo, y to 0$ .

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.

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