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

1 Answer
Apr 13, 2016

Answer:

vertical asymptote # x = -1/3 #
horizontal asymptote # y = 2/3#

Explanation:

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

solve : # 3x + 1= 0 rArr x = -1/3 #

#rArr x = -1/3" is the asymptote " #

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

divide terms on numerator/denominator by x

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

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

#rArr y = 2/3 " is the asymptote " #

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