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

1 Answer
Jan 6, 2017

Answer:

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]}