How do you find the asymptotes for #y=(x-3)/(2x+5)#?

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Answer 1 · 2016-03-20T07:15:45

Vertical Asymptote: #x=-5/2#
Horizontal Asymptote: #y=1/2#

For the vertical asymptote:

Equate the denominator to zero, then solve for x
#2x+5=0#
#2x=-5#
#x=-5/2#

For the horizontal asymptote:

Take the limit of the function

#lim_(xrarr oo) y=lim_(xrarr oo) (x-3)/(2x+5)=1/2#

and therefore

#y=1/2# is a horizontal asymptote

graph{(y- (x-3)/(2x+5))(y-1/2)=0[-20,20,-10,10]}

God bless....I hope the explanation is useful.