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LekaFEV [45]
3 years ago
10

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

Mathematics
1 answer:
matrenka [14]3 years ago
4 0

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

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Step-by-step explanation:

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