How do you find the asymptotes for #(2x^2)/((-3x-1)^2)#?

1 Answer
Mar 23, 2016

Answer:

#x=-1/3#

Explanation:

Set each factor in the denominator to #0# and solve for #x#. Since the factor, #-3x-1#, appears twice, you only have to solve for #x# in one of the factors.

#-3x-1=0#

#-3x=1#

#color(green)(|bar(ul(color(white)(a/a)x=-1/3color(white)(a/a)|)))#

If you graph the function, you can see that the #x# values approach, but never touch #x=-1/3#.

graph{(2x^2)/(-3x-1)^2 [-7.97, 7.834, -3, 4.9]}