Question

How do you find the asymptotes for `(2x^2 + x + 2) / (x + 1)`?

Answer 1

The vertical asymptote is `x=-1`
The slant asymptote is `y=2x-1`
No horizontal asymptote

Explanation:

Let `f(x)=(2x^2+x+2)/(x+1)`

The domain of `f(x)` is `D_f(x)=RR-{-1}`

As you cannot divide by `0`, `x!=-1`

The vertical asymptote is `x=-1`

As we the degree of the numerator is `>` than the degree of the denominator, there is a slant asymptote.

Let's do a long division

`color(white)(aaaa)` `2x^2+x+2` `color(white)(aaaa)` `∣` `x+1`

`color(white)(aaaa)` `2x^2+2x` `color(white)(aaaaaaa)` `∣` `2x-1`

`color(white)(aaaaa)` `0-x+2`

`color(white)(aaaaaaa)` `-x-1`

`color(white)(aaaaaaaaa)` `0+3`

Therefore,

`f(x)=(2x-1)+(3)/(x+1)`

So,

`lim_(x->-oo)(f(x)-(2x-1))=lim_(x->-oo)3/(x+1)=0^-`

`lim_(x->+oo)(f(x)-(2x-1))=lim_(x->+oo)3/(x+1)=0^+`

The slant asymptote is `y=2x-1`

graph{(y-(2x^2+x+2)/(x+1))(y-2x+1)(y-50x-50)=0 [-25.64, 25.67, -12.83, 12.84]}