Question

How do you find vertical, horizontal and oblique asymptotes for `(3x^2 + 4)/(x+1)`?

Answer 1

The vertical asymptote is `x=-1`
The oblique asymptote is `y=3x-3`

Explanation:

As you canot divide by `0`, the vertical asymptote is `x=-1`
As the degree of the numerator is `>` than the degree of the dedenominator, so we expect an oblique asymptote:
Therefore we do a lond division
`3x^2` `color(white)(aaaaaa)` `+4` `∣` `x+1`
`3x^2+3x` `color(white)(aaaa)` #∣3x-3#
`color(white)(aa)` `0+3x+4` `color(white)(a)`
`color(white)(aaaaa)` `+0-3` `color(white)(a)`
`color(white)(aaaaaaaaa)` `+7` `color(white)(a)`

So, `(3x^2+4)/(x+1)=3x-3+7/(x+1)`
So the oblique asymptote is `y=3x-3`
`lim_(n rarr +-oo) (3x^2+4)/(x+1)=lim_(n rarr +-oo) 3x=+-oo`
So there is no horizontal asymptote
graph{(y-(3x^2)/(x+1))(y-3x+3)=0 [-43.83, 38.34, -24.63, 16.5]}