What are the asymptotes of #f(x)=-x/((x-1)(x^2+x)) #?

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Answer 1 · 2016-12-20T10:37:44

There is a hole at #x=0#
The vertical asymptotes are #x=-1# and #x=1#
No slant asymptote.
The horizontal asymptote is #y=0#

Let's factorise the denominator

#(x-1)(x^2+x)=x(x-1)(x+1)#

Therefore,

#f(x)=cancelx/(cancelx(x+1)(x-1))#

#=1/((x+1)(x-1))#

There is a hole at #x=0#

The domain of #f(x)# is #D_f(x)=RR-{-1,1} #

As you cannot divide by #0#, #x!=-1# and #x!=1#

The vertical asymptotes are #x=-1# and #x=1#

The degree of the numerator is #<# than the degree of the denominator, there is no slant asymptote.

#lim_(x->+-oo)f(x)=lim_(x->+-oo)1/x^2=0^(+)#

The horizontal asymptote is #y=0#

graph{1/((x+1)(x-1)) [-14.24, 14.24, -7.11, 7.14]}