What are the asymptotes for #(x-4)/(16x-x^3)#?

1 Answer
Jun 18, 2015

Answer:

The vertical asymptotes are #x=0 and x=4 and x=-4#
The horizontal asymptote is the line #y=0#
.

Explanation:

Set the denominator #16x-x^3# equal to zero, and solve for #x#.

#16x-x^3=0#

Factor out #x#.

#x(16-x^2)=0#

#x=0#

#x=0#

#16-x^2=0#

Factor to get

#(4+x)(4-x) =0#

#4+x =0# so #x = -4#

#4-x = 0# so #x=4#

The horizontal asymptote is #y = 0# (The #x#-axis)

because as #x# increases without bound, the value of

#(x-4)/(16x-x^3) = (cancel(x^3) (1/x^2 -4/x^3))/(cancel(x^3)(16/x^2 - 1))# gets closer and closer to #0/1#.