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.