Question

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

Answer 1

`"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]}