How do you find the vertical, horizontal or slant asymptotes for #(3x-2) /(2x+5)#?

1 Answer
Jun 19, 2016

Answer:

vertical asymptote #x=-5/2#
horizontal asymptote #y=3/2#

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 : 2x + 5 = 0 #rArrx=-5/2#

#rArrx=-5/2" is the asymptote"#

Horizontal asymptotes occur as

#lim_(xto+-oo),ytoc" (a constant)"#

divide terms on numerator/denominator by x

#((3x)/x-2/x)/((2x)/x+5/x)=(3-2/x)/(2+5/x)#

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

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

Slant 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 slant asymptotes.
graph{(3x-2)/(2x+5) [-10, 10, -5, 5]}