How do you find vertical, horizontal and oblique asymptotes for # x / (3x(x-1))#?

Question

Sophie Adams · Accepted Answer

There is a vertical asymptote at #x=1# and a horizontal asymptote at #y=0#

To find all the asymptotes for function #y=x/(3x(x-1))#, we first observe that #x# cancels out from numerator and denominator and the function is primarily #x/(3(x-1))#, but there is a hole at #x=0#

Let us first start with vertical asymptotes, which are given by putting denominator equal to zero or #x-1=0# i.e. #x=1#.

Further as in #y=3/(x-1)#, there is no variable in numerator, we have a horizontal asymptote at #y=0#

graph{x/(3x(x-1)) [-10, 10, -5, 5]}

Elias Ackerman · Answer

undefined To find the vertical asymptotes of the function $ \frac{x}{3x(x-1)} $, we identify the values of $ x $ that make the denominator equal to zero:

Setting $ 3x(x-1) = 0 $ gives us $ x = 0 $ and $ x = 1 $. Therefore, the vertical asymptotes are $ x = 0 $ and $ x = 1 $.

To find horizontal asymptotes, we examine the behavior of the function as $ x $ approaches positive and negative infinity. We compare the degrees of the numerator and denominator:

The degree of the numerator is 1, and the degree of the denominator is 3. Since the degree of the denominator is greater, the horizontal asymptote is at $ y = 0 $.

To determine if there are any oblique asymptotes, we perform polynomial division of the numerator by the denominator. However, since the degree of the numerator is less than the degree of the denominator, there are no oblique asymptotes in this case.