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

1 Answer
Apr 8, 2016

Answer:

vertical asymptote # x = 1/2 #
horizontal asymptote # y = 1/2 #

Explanation:

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

solve: 2x - 1 = 0 #rArr x = 1/2" is the asymptote " #

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

divide all terms on numerator/denominator by x

#(x/x - 3/x )/((2x)/x - 1/x) = (1 - 3/x)/(2 - 1/x) #

as # x to+-oo , 3/x" and " 1/x to 0 #

#rArr y = 1/2" 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.

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