What are the asymptotes and removable discontinuities, if any, of #f(x)=(4)/(x-2)^3 #?

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Answer 1 · 2018-07-13T11:53:10

Vertical asymptote at #x=2# , horizontal asymptote at #y=0# having no removable discontinuity.

#f(x)= 4/(x-2)^3# . Vertical asymptotes are found when

denominator of the function is zero. Here #f(x)# is undefined

when #x=2# . Therefore at #x=2#, we get vertical asymptote.

Since no factor in numerator and denominator cancel each other

there is no removable discontinuity.

Since denominator's degree is greater than that of numerator

, we have a horizontal asymptote at y = 0# (the x-axis).

Vertical asymptote at #x=2# , horizontal asymptote at #y=0#

having no removable discontinuity.

graph{4/(x-2)^3 [-20, 20, -10, 10]} [Ans]