How do you find the asymptotes for #G(x) = (x-1)/(x-x^3)#?

1 Answer
Jun 22, 2017

Answer:

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

Explanation:

#"factorise the denominator and simplify"#

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

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

#"since the factor " (x-1)" has been removed from the "#
#"numerator/denominator this 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 denominator 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)to-0/(1+0)#

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