How do you find the Vertical, Horizontal, and Oblique Asymptote given #f(x)= (2x)/(x^2+16)#?
1 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+16=0rArrx^2=-16# This has no real solutions hence 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"# Oblique 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 oblique asymptotes.
graph{(2x)/(x^2+16) [-10, 10, -5, 5]}
