How to find the asymptotes of #f(x) =3/(2x-1)# ?

1 Answer
Jan 20, 2016

Answer:

Vertical asymptote: #x=1/2#
Horizontal asymptote: #f(x)=0#

Explanation:

Given
#color(white)("XXX")f(x)=3/(2x-1)#

#f(x)rarr +-oo# as #(2x-1) rarr 0#
#color(white)("XXX")2x-1=0 rarr x=1/2#
giving a vertical asymptote at #x=1/2#

#f(x)=3/(2x-1)#
#rarr 2x-1=3/f(x)#
#rarr x= 3/(2f(x)) +1/2#
#color(white)("XXX")xrarr +-oo# as #f(x)rarr 0#
giving a horizontal asymptote at #f(x)=0#

graph{3/(2x-1) [-12.56, 15.91, -9.15, 5.08]}