How do you find the vertical, horizontal or slant asymptotes for #(2x)/(x^2+16)#?

1 Answer
Jun 27, 2017

Answer:

#"horizontal asymptote at " y=0#

Explanation:

#f(x)=(2x)/(x^2+16)#

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 they are vertical asymptotes.

#"solve " x^2+16=0rArrx^2=-16#

#"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)=((2x)/x^2)/(x^2/x^2+16/x^2)=(2/x)/(1+16/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{(2x)/(x^2+16) [-10, 10, -5, 5]}