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

1 Answer
Aug 19, 2017

Answer:

#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)#.