Question

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

Answer 1

The vertical asymptote is `x=-2`
The oblique asymptote is `y=-3x+6`
No horizontal asymptote

Explanation:

Let `f(x)=(-3x^2+2)/(x+2)`

The domain of `f(x)` is `D_f(x)=RR-{-2}`

As we cannot divide by `0`, `x!=-2`

So, the vertical asymptote is `x=-2`

The degree of the numerator is `>` the degree of the denominator, we expect an oblique asymptote.

Let's do a long division

`color(white)(aaaa)` `-3x^2+2` `color(white)(aaaa)` `∣` `x+2`

`color(white)(aaaa)` `-3x^2-6x` `color(white)(aaa)` `∣` `-3x+6`

`color(white)(aaaa)` `-0+6x+2`

`color(white)(aaaaaaa)` `+6x+12`

`color(white)(aaaaaaaa)` `+0-10`

Therefore,

`(-3x^2+2)/(x+2)=(-3x+6)-10/(x+2)`

So, the oblique asymptote is `y=-3x+6`

To calculate the limits to `oo`, we take the terms of highest degree in the numerator and the denominator.

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

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

graph{(y-(-3x^2+2)/(x+2))(y+3x-6)=0 [-74.04, 74.05, -37, 37]}