How do you find vertical, horizontal and oblique asymptotes for #(2x+3)/(3x+1) #?

1 Answer
May 24, 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 set the denominator equal to zero.

solve : 3x + 1 = 0 #rArrx=-1/3" is the asymptote"#

Horizontal asymptotes occur as #lim_(xto+-oo) , y to 0#

divide terms on numerator/denominator by x

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

as #xto+-oo ,yto(2+0)/(3+0)#

#rArry=2/3" is the asymptote"#

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both of degree 1). Hence there are no oblique asymptotes.
graph{(2x+3)/(3x+1) [-10, 10, -5, 5]}