Question

How do you find the vertical, horizontal or slant asymptotes for `f(x) = (-x^2 + 4x)/(x+2)`?

Answer 1

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

Explanation:

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

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

The vertical asymptote is `x=-2`

The degree of the numerator is `>` than the degree of the denominator, so there is a slant asymptote.

To find the slant, we start by doing a long division

`color(white)(aaaa)` `-x^2+4x` `color(white)(aaaa)` `∣` `x+2`

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

`color(white)(aaaaaaa)` `0+6x`

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

`color(white)(aaaaaaaaaaaaaa)` `-12`

So,

`f(x)=(-4x^2+4x)/(x+2)=-x+6-12/(x+2)`

`lim_(x->-oo)f(x)+(x-6)=lim_(x->-oo)-12/(x+2)=0^+`

`lim_(x->+oo)f(x)+(x-6)=lim_(x->+oo)-12/(x+2)=0^-`

The slant asymptote is `y=-x+6`

No horizontal asymptote

graph{(y-(-x^2+4x)/(x+2))(y+x-6)(y-25x-50)=0 [-35, 38.04, -14.36, 22.2]}