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

1 Answer
Sep 21, 2016

horizontal asymptote at y = 0


The denominator of the function cannot be zero as this would make the function 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 they 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#


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 (numerator-degree 1 , denominator-degree 2 ) Hence there are no slant asymptotes.
graph{(3x)/(x^2+2) [-10, 10, -5, 5]}