Question

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?

Answer 1

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]}