# 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

The oblique asymptote is

#### Explanation:

The vertical asymptote is

As the degree of numerator is

Therefore, we do a long division

So,

So the oblique asymptote is

Limit

graph{(y-(2x^2-3)/(x+2))(y-2x+4)=0 [-33.56, 31.36, -22.34, 10.15]}