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

##### 2 Answers

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]}

The vertical asymptote is

the oblique asymptote is

there is no horizontal asymptote

#### Explanation:

We cannot divide by

Therefore

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

And finally, we have

So the oblique asymptote is

graph{(y-(4x-x^2)/(x+2))(y+x-6)=0 [-52, 52.03, -26, 26.04]}