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

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