Please help me with this problem
1 answer:
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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
