Question

In Exercise,find the horizontal asymptote of the graph of the function.
f(x) = 16x/3+x^2

Answer 1

Horizontal asymptote of the graph of the function f(x) = 16x/(3+x^2) is y=0

Explanation:

I attached the graph of the function. Graphically it can be seen that the horizontal asymptote of the graph of the function is y=0.

When denominator's degree (2) is higher than nominator's degree (1) then the horizontal asymptote is y=0