How do you find vertical, horizontal and oblique asymptotes for #(x^3-8)/(x^2-5x+6)#?

1 Answer
Jun 13, 2018

Answer:

the oblique asymptote is #y=x+5# and the horizontal asymptote is #x=3#

Explanation:

#(x^3-8)/(x^2-5x+6)=(cancel(x-2)(x^2+2x+4))/((x-3)cancel(x-2))#
=#(x^2+2x+4)/(x-3)=((x-3)(x+5)+19)/(x-3)=x+5+19/(x-3)#

Therefore, the oblique asymptote is #y=x+5# and the horizontal asymptote is #x=3#

The horizontal asymptote is found by letting the denominator equal to 0.

The oblique asymptote is found by long division