How do you find the asymptotes for #f(x) = (x^2-4x+4) / (x+1)#?

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Answer 1 · 2016-03-02T14:55:02

Oblique Asymptote is #y=x-5#
Vertical Asymptote is #x=-1#

from the given:

#f(x)=(x^2-4x+4)/(x+1)#

perform long division so that the result is

#(x^2-4x+4)/(x+1)=x-5+9/(x+1)#

Notice the part of the quotient

#x-5#

equate this to #y# like as follows

#y=x-5# this is the line which is the Oblique Asymptote

And the divisor #x+1# be equated to zero and that is the Vertical asymptote

#x+1=0# or #x=-1#

You can see the lines #x=-1# and #y=x-5# and the graph of
#f(x)=(x^2-4x+4)/(x+1)#
graph{(y-(x^2-4x+4)/(x+1))(y-x+5)=0[-60,60,-30,30]}

God bless...I hope the explanation is useful..