What is 11x^2-28x+4 in vertex form
1 answer:
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Answer:
Now we can set up the derivate equal to 0 and we have:
And solving for t we got:
For the second derivate respect the time we got:
So then we can conclude that t = 4.206 is a maximum for the function.
And the corresponding height would be:
So the maximum occurs at t = 4.206 s and with a height of 135.468 m
Step-by-step explanation:
For this case we have the following function:
In order to maximize this function we need to take the first derivate respect the time and we have:
Now we can set up the derivate equal to 0 and we have:
And solving for t we got:
For the second derivate respect the time we got:
So then we can conclude that t = 4.206 is a maximum for the function.
And the corresponding height would be:
So the maximum occurs at t = 4.206 s and with a height of 135.468 m