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

1 Answer
Jul 9, 2018

Answer:

#"vertical asymptote at "x=-2#
#"horizontal asymptote at "y=0#

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 "x+2=0rArrx=-2" is the asymptote"#

#"Horizontal asymptotes occur as"#

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

#"divide terms on numerator/denominator by "x#

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

#"as "xto+-oo,yto0/(1+0)#

#y=0" os the asymptote"#
graph{3/(x+2) [-10, 10, -5, 5]}