Question 1031068
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The graph of your function is a parabola.  Since the lead coefficient is negative the  parabola opens downward.  Therefore, the value of the function at the vertex is a maximum.  The value of the independent variable at the vertex of a parabola described by the function:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ p(x)\ =\ ax^2\ +\ bx\ +\ c]


is given by


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


and the maximum value of the function is then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ p(x_{max})]


The total time of flight is the larger of the two roots of the equation formed by setting your function equal to zero.  (zero assumes that the projectile lands at ground level rather than hitting some higher structure or going into a hole.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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