How do you find the vertical, horizontal or slant asymptotes for #((x-1)(x-3))/(x(x-2) )#?

1 Answer
Apr 15, 2016

Answer:

vertical asymptotes x = 0 , x = 2
horizontal asymptote y = 1

Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation/s set the denominator equal to zero.

solve : x(x - 2) = 0 → x = 0 , x = 2

#rArr x = 0" and " x = 2" are the asymptotes " #

Horizontal asymptotes occur as #lim_(xto+-oo) f(x) to 0 #

now #((x-1)(x-3))/(x(x-2)) = (x^2-4x+3)/(x^2-2x)#

divide terms on numerator/denominator by #x^2 #

#(x^2/x^2 -(4x)/x^2+3/x^2)/(x^2/x^2 -(2x)/x^2)= (1-4/x+3/x^2)/(1-2/x)#

as #x to+-oo , 4/x , 2/x" and " 3/x^2 to 0 #

# y = (1-0+0)/(1-0) = 1/1 = 1 #

#rArr y = 1 " 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.

This is the graph of the function.
graph{((x-1)(x-3))/(x(x-2)) [-10, 10, -5, 5]}