How do you find the asymptotes for #(x^3-x)/(x^3-4x)#?

1 Answer
Dec 11, 2015

Answer:

Horizontal asymptote: #y= 1#

Vertical asymptote: # x = 2# and #x= -2#

Discontinuity at #(0, 1/4)#

Explanation:

Given #f(x) = (x^3 -x)/(x^3 -4x)#

Let's start by factoring the function #f(x) = (x(x^2-1))/(x(x^2-4) ) => (x(x-1)(x+1))/((x)(x-2)(x+2))#

If you notice, there is a same factor of #x# on numerator and denominator, which can reduce the function to

#f(x) =( (x-1)(x+1))/((x-2)(x+2)) , x!=0 #

Discontinuity aka hole on the graph at
#f(0) = [(0-1)(0+1)]/[(0-2)(0+2)] = 1/4#

Discontinuity at #(0, 1/4)

Horizontal asymptote : #y= 1#

Since degree and coefficient of the numerator and denominator are the same, hence the horizontal asymptote is #y= 1#

Vertical asymptote

Set the denominator of the reduce function equal to zero

#x-2 = 0 ; x + 2 = 0 #
#x = 2 ; and x= -2#
graph{(x^3-x)/(x^3-4x) [-7.024, 7.024, -3.51, 3.51]}