What are the asymptotes for #G(x) = (x-1)/(x-x^3)#?

2 Answers
Jun 21, 2017

Answer:

You may start by factorizing the denominator:

Explanation:

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

#x=1# leaves us with the undefined #0/0#

So #x=0andx=-1# are the vertical asympotes.

If we realize that #x-1=-(1-x)# we can simplify, as long as #x!=1#

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

And this goes to zero if #x# gets large enough (either pos or neg)

So #y=0# is the horizontal asymptote.
graph{(x-1)/(x-x^3) [-16.6, 15.44, -7.7, 8.33]}

Jun 21, 2017

Answer:

#"vertical asymptotes at " x=0" and " x=-1#
#"horizontal asymptote at " y=0#

Explanation:

#"simplifying " g(x)#

#g(x)=(x-1)/(x(1-x^2))=-(cancel((x-1)))/(xcancel((x-1))(x+1))#

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

#"the removal of " (x-1)" from the numerator/denominator"#
#"means there is a hole at x = 1"#

#"the graph of " -1/(x(x+1))" is the same as " (x-1)/(x-x^3)#
#"but without the hole"#

The denominator of g(x) cannot be zero as this would make g(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

#"solve " x(x+1)=0rArrx=0,x=-1#

#rArrx=0" and " x=-1" are the asymptotes"#

#"Horizontal asymptotes occur as"#

#lim_(xto+-oo),g(x)toc" ( a constant)"#

divide terms on numerator/denominator by the highest power of x, that is #x^2#

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

as #xto+-oo,g(x)to0/(1+0)#

#rArry=0" is the asymptote"#
graph{(x-1)/(x-x^3) [-10, 10, -5, 5]}