How do you find the Vertical, Horizontal, and Oblique Asymptote given #f(x)= (2x)/(x^2+16)#?

Question

Sophie Adams · Accepted Answer

horizontal asymptote at y = 0

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]}

Caleb Alexander · Answer

undefined To find the vertical asymptotes of $ f(x) = \frac{2x}{x^2 + 16} $, set the denominator equal to zero and solve for $ x $. You'll find $ x = \pm 4i $ which means there are no real vertical asymptotes.

To find horizontal asymptotes, compare the degrees of the numerator and denominator. Since the degree of the numerator (1) is less than the degree of the denominator (2), there is a horizontal asymptote at $ y = 0 $.

To find oblique asymptotes, divide the numerator by the denominator using polynomial long division or synthetic division. The quotient will represent the equation of the oblique asymptote. In this case, it's $ y = 0 $.