Question 324748
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(t)\ =\ -16t^2\ +\ 128t\ +\ 15]


Note that the graph of *[tex \Large h] is a parabola.  It opens downward because the lead coefficient is negative, hence the vertex is a maximum.  The value of the *[tex \Large x]-coordinate of the vertex of a parabola which equation is in standard form is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_v\ =\ \frac{-b}{2a}]


for this problem:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_v\ =\ \frac{-128}{2(-16)}\ =\ 4]


hence the maximum height is reached 4 seconds after the flare is fired.  The maximum height reached is therefore *[tex \Large h(4)].


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(4)\ =\ -16(4)^2\ +\ 128(4)\ +\ 15]


You can do your own arithmetic.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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