How do I find the asymptotes of #f(x)= (- 4x-x^2)/(2 +2x- x^3)#?

1 Answer
Dec 16, 2015

Answer:

asymptotes:
#x=2#
#x=sqrt(2)#
#x=-sqrt(2)#

Explanation:

Start by simplifying the function:

#f(x)=(-4x-x^2)/(2+2x-x^3)#

#f(x)=(-x(4+x))/(-x^3+2x+2)#

#f(x)=(-x(x+4))/(-x(x^2-2)+2)#

#f(x)=(-x(x+4))/((-x+2)(x^2-2))#

#f(x)=(-x(x+4))/(-(x-2)(x^2-2))#

#f(x)=(color(red)cancelcolor(black)-x(x+4))/(color(red)cancelcolor(black)-(x-2)(x^2-2))#

#f(x)=(x(x+4))/((x-2)(x^2-2))#

Take each bracketed polynomial in the denominator, set it to cannot equal to #0#, and solve for #x#.

Finding the asymptotes

#1. x-2!=0#
#color(white)(ixxxx)x!=2#

#2. x^2-2!=0#
#color(white)(xxxx)x^2!=2#
#color(white)(xxxxx)x!=+-sqrt(2)#

The asymptotes are also the values which cannot be substituted into the equation such that the denominator would be #0#.

#:.#, the asymptotes are #x=2#, #x=sqrt(2)#, and #x=-sqrt(2)#.