Question

How do you find the vertical, horizontal and slant asymptotes of: `x/(1-x)^2`?

Answer 1

`"vertical asymptote at " x=1`
`"horizontal asymptote at " y=0`

Explanation:

`"let " f(x)=x/(1-x)^2`

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

`"solve "(1-x)^2=0rArrx=1" is the asymptote"`

Horizontal asymptotes occur as

`lim_(xto+-oo),f(x)toc" (a constant)"`

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

`f(x)=(x/x^2)/(1/x^2-(2x)/x^2+x^2/x^2)=(1/x)/(1/x^2-2/x+1)`

as `xto+-oo,f(x)to0/(0-0+1)`

`rArry=0" is the asymptote"`

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there are no slant asymptotes.
graph{(x)/((1-x)^2) [-10, 10, -5, 5]}