Question

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

Answer 1

vertical asymptote at x = -2
oblique asymptote at y = -x + 6

Explanation:

The denominator of the rational function cannot be zero as this would make the function undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve : `x+2=0rArrx=-2" is the asymptote"`

Horizontal asymptotes occur when the degree of the numerator ≤ degree of the denominator. This is not the case here (numerator-degree 2, denominator- degree 1 ). Hence there are no horizontal asymptotes.

Oblique asymptote occur when the degree of the numerator > degree of the denominator. Hence there is an oblique asymptote.
Polynomial division gives.

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

as `xto+-oo,f(x)to-x+6-0`

`rArry=-x+6" is the asymptote"`
graph{(-x^2+4x)/(x+2) [-10, 10, -5, 5]}

Answer 2

The vertical asymptote is `x=-2`
the oblique asymptote is `y=-x+6`
there is no horizontal asymptote

Explanation:

We cannot divide by `0`, so `x!=0`
Therefore `x=-2` is a vertical asymptote
As the degree of the numerator is greater than the degree of the denominator, we expect an oblique asymptote:
so we do a long division
`-x^2+4x` `color(white)(aaaaaaa)` `∣` `x+2`
`-x^2-2x` `color(white)(aaaaaaa)` `∣` `-x+6`
`color(white)(aa)` `0+6x`
`color(white)(aaaaa)` `6x+12`
`color(white)(aaaaaa)` `0-12`
And finally, we have `(-x^2+4x)/(x+2)=-x+6-12/(x+2)`
So the oblique asymptote is `y=-x+6`
graph{(y-(4x-x^2)/(x+2))(y+x-6)=0 [-52, 52.03, -26, 26.04]}