Question

What are the asymptote(s) and hole(s), if any, of `f(x) = (x*(x-2))/(x^2-2x+1)`?

Answer 1

`x=1" "` is the vertical asymptote of `f (x)`.
`" "`
`y=1" "` is the horizantal asymptote of `f (x)`

Explanation:

This rational equation has a vertical and horizantal asymptote .
`" "`
Vertical asymptote is determined by factorizing the denominator :
`" "`
`x^2-2x+1`
`" "`
`=x^2-2 (1)(x)+1^2`
`" "`
`=(x-1)^2`
`" "`
Then,`" "x=1" "`is a vertical asymptote.
`" "`
Let us find the horizantal asymptote :
`" "`
As it is known we have To check both degrees of the
`" '`
numerator and denominator .
`" "`
Here , the degree of the numerator is `2` and that of the
`" "`
denominator is `2` too .
`" "`
If `(ax^2+bx+c)/(a_1x^2+b_1x+c_1)`then the horizantal asymptote is `color (blue)(a/(a_1))`
`" "`
In `f (x)=(x. (x-2))/(x^2-2x+1)=(x^2-2x)/(x^2-2x+1)`
`" "`
Same degree in the numerator and denominator then horizantal
`" "`
asymptote is `y=color (blue)(1/1)=1`
`" "`
`therefore x=1 and y=1 " "` are the asymptotes of `f (x)`.