How do you graph #f(x)=(2x^3+x^2-8x-4)/(x^2-3x+2)# using holes, vertical and horizontal asymptotes, x and y intercepts?

1 Answer
Tamir E.
Nov 27, 2017

see below

Explanation:

#f_((x))={x^2(2x+1)-4(2x+1)}/{(x-2)(x-1)}={(x^2-4)(2x+1)}/{(x-2)(x-1)}={cancel((x-2))(x+2)(2x+1)}/{cancel((x-2))(x-1)}=#

hole: #x=2#

#={(x+2)(2x+1)}/{(x-1)}={2x^2+5x+2}/{x-1}=f_((x))#

vertical asymptote: #x=1#

horizontal asymptotes:
#lim_(x rarr +oo)f_((x))=+oo#
#lim_(x rarr -oo)f_((x))=-oo#
#=># none

graph{(2x^3+x^2-8x-4)/(x^2-3x+2) [-200, 200, -150, 150]}

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