How do you graph #y=(2x^2-3)/(x+2)# using asymptotes, intercepts, end behavior?

1 Answer
Nov 5, 2016

Answer:

The vertical asymptote is #x=-2#
The oblique asymptote is #y=2x+4#

Explanation:

The vertical asymptote is #x=-2#
As the degree of numerator is #># the degree of denominator, we expect an oblique asymptote.
Therefore, we do a long division
#2x^2##color(white)(aaaa)##-3##color(white)(aaaa)##∣#x+2
#2x^2+4x##color(white)(aa)####color(white)(aaaa)##∣#2x-4
#color(white)(a)##0-4x-3#
#color(white)(a)##0-4x-8#
#color(white)(aaaa)##-0+5#
So, #(2x^3-3)/(x+2)=2x-4+5/(x+2)#
So the oblique asymptote is #y=2x-4#
Limit #y=2x^2/x=2x=+-oo#
#x->+-oo#
graph{(y-(2x^2-3)/(x+2))(y-2x+4)=0 [-33.56, 31.36, -22.34, 10.15]}