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