Question

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

Answer 1

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]}