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

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Answer 1 · 2016-03-01T09:41:22

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

#lim_{x to 1}f(x) = oo#

#lim_{x to -oo}f(x) = 0#

#lim_{x to oo}f(x) = 0#

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

Answer 2 · 2016-03-01T11:50:09

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

First we can rewrite this as follows

#f(x)=(1/(x-1))^2# so #f(x)>0# , #D_f = RR \\\ {1}#

Hence for #x->1# ,#f(x)->+oo#

and for #x->+-oo# ,#f(x)->0#

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

The graph of #f(x)# is

enter image source here