How do you find vertical, horizontal and oblique asymptotes for # f(x)= (x^2-4x+4) /( x+1)#?

1 Answer
May 27, 2017

Answer:

Vertical asymptote at #x= -1# , horizontal asymptote is absent.
#f(x)=x-5# is the oblique asymptote.

Explanation:

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

Vertical asymptote: Solving denominator for zero we get #x+1=0 or x = -1# So V.A is at #x= -1#

Since degree of numerator(2) is greater than denominator(1) horizontal asymptote is absent.

Since degree of numerator(2) is greater than denominator(1) by a margin of 1 there is a slant/oblique asymptote , which can be found by long division.

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

So straight line #f(x)=x-5# is the oblique asymptote.

graph{(x^2-4x+4)/(x+1) [-80, 80, -40, 40]} [Ans]