How do you find the vertical, horizontal and slant asymptotes of: #y= (3x-2) /(2x+5)#?

1 Answer
Aug 30, 2016

Answer:

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

Explanation:

The denominator of y cannot be zero as this would make y undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve: #2x+5=0rArrx=-5/2" is the asymptote"#

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

#y=((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) [-20, 20, -10, 10]}