How do you find the vertical, horizontal or slant asymptotes for #f(x) = (4x) / (x^2+1)#?

1 Answer
Sep 19, 2016

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 they are vertical asymptotes.

solve : #x^2+1=0rArrx^2=-1#

This has no real solutions hence there are no vertical asymptotes.

Horizontal asymptotes occur as

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

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

#f(x)=((4x)/x^2)/(x^2/x^2+1/x^2)=(4/x)/(1+1/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{(4x)/(x^2+1) [-10, 10, -5, 5]}