What are the values and types of the critical points, if any, of #f(x) = (x-5)/(x-3)^2#?

1 Answer
Narad T.
Nov 11, 2016

The vertical asymptote is #x=3#
The horizontal asymptote is #y=0#
The intercepts are #(0,-5/9)# and #(5,0)#

Explanation:

As we cannot divide by #0#, the vertical asymptote is #x=3#

The degree of the numerator is #<# the dergee of the denominator, there is no oblique asymptote.
#lim_(x->-oo)f(x)=lim_(x->-oo)x/x^2=lim_(x->-oo)1/x=0^(-)#
#lim_(x->+oo)f(x)=lim_(x->+oo)x/x^2=lim_(x->+oo)1/x=0^(+)#
The horizontal asymptote is #y=0#

The intercepts are
on the y-axis, #x=0##=>##f(0)=-5/9#
on the x-axis, #y=0##=>##0=x-5##=>##x=5#
graph{(x-5)/(x-3)^2 [-10, 10, -5.005, 4.995]}

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