How do you find vertical, horizontal and oblique asymptotes for #f(x) =( x^3 - 4x^2 + 5x + 5)/(x - 1)#?

1 Answer
Dec 21, 2017

Answer:

#f(x)# has one vertical asymptote #x=1# and no other linear asymptotes.

Explanation:

#f(x) = (x^3-4x^2+5x+5)/(x-1)#

#color(white)(f(x)) = (x^3-x^2-3x^2+3x+2x-2+7)/(x-1)#

#color(white)(f(x)) = x^2-3x+2 + 7/(x-1)#

This function has a vertical asymptote #x=1#.

As #x->+-oo# we have #7/(x-1)->0#, so #f(x)# is asymptotic to #x^2-3x+2#, which is not a line.

So there are no other linear asymptotes.

graph{(x^3-4x^2+5x+5)/(x-1) [-10, 10, -200, 200]}