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

2 Answers
Nov 1, 2016

Answer:

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]}

Nov 1, 2016

Answer:

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]}