Can someone help me pls
2 answers:
You might be interested in
Answer:
Horizontal asymptote of the graph of the function f(x) = 16x/(3+x^2) is y=0
Step-by-step 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
