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

1 Answer
Sep 25, 2017

Answer:

#"horizontal asymptote at "y=0#

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then thet are vertical asymptotes.

#"solve "x^2+2=0rArrx^2=-2#

#"this has no real solutions hence there are no vertical"#
#"asymptotes"#

#"horizontal asymptotes occur as"#

#lim_(xto+-oo),f(x)toc" ( a constant)"#

#"divide terms on numerator/denominator by the highest"#
#"power of x that is "x^2#

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

as #xto+-oo,f(x)to0/(1+0)#

#rArry=0" is the asymptote"#

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there are no slant asymptotes.
graph{(3x)/(x^2+2) [-10, 10, -5, 5]}