Question

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

Answer 1

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]}

Answer 2

`"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]}