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What are the asymptote(s) and hole(s), if any, of # f(x) =1/((x-3)(x^3-x^2-x+1))#?

Answer 1

Hazel Anglin

asymptotes:
#x=3, -1, 1#
#y=0#

holes:
none

#f(x)=1/((x-3)(x^3-x^2-x+1))#

#f(x)=1/((x-3)(x^2(x-1)-1(x-1))#

#f(x)=1/((x-3)(x^2-1)(x-1))#

#f(x)=1/((x-3)(x+1)(x-1)(x-1))#; #x!=3,-1,1;y!=0#

There are no holes for this function since there are no common bracketed polynomials which appear in the numerator and denominator. There are only restrictions which must be stated for each bracketed polynomial in the denominator. These restrictions are the vertical asymptotes. Keep in mind that there is also a horizontal asymptote of #y=0#.

#:.#, the asymptotes are #x=3#, #x=-1#, #x=1#, and #y=0#.

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Answer 2

Abigail Allen

The function f(x) = 1/((x-3)(x^3-x^2-x+1)) has a vertical asymptote at x = 3 and no horizontal asymptotes. There are no holes in the graph of this function.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.