Question 1125211
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1.  Substitute 2 for *[tex \Large t] and do the arithmetic.


2.  The graph of *[tex \Large y(t)] is a parabola that opens downward because the lead coefficient, -16, is less than zero.  Therefore the maximum value of *[tex \Large y] will be at the vertex of the parabola.  The vertex occurs at the point calculated by dividing the opposite of the linear term by twice the lead coefficient. So for


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


The vertex is at:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (-\frac{b}{2a},f\(-\frac{b}{2a}\)\)]


You need the abscissa of this point, *[tex \LARGE -\frac{b}{2a}], to answer "After how many seconds..." and the ordinate, *[tex \LARGE f\(-\frac{b}{2a}\)], of the point to answer "What is the maximum..."


This might make more sense if you put your function into standard form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y(t)\ =\ -16t^2\ +\ 96t]


so that it is clear that 1: *[tex \Large y] is a function of *[tex \Large t] and 2: the lead coefficient is -16 and the linear term coefficient is 96]


You might also want to note that for this model to be an accurate representation of a physical phenomenon, the object must be touching the ground when it was launched with a 96 feet per second upward initial velocity.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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