Question 1039269
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If you have a 2nd-degree polynomial function of the form *[tex \Large y\ =\ ax^2\ +\ bx\ +\ c], and the lead coefficient is negative, then the graph is parabola that opens downward.  The vertex of the parabola represents the maximum of the function.  The *[tex \Large x]-coordinate of the vertex is given by *[tex \Large x_v\ =\ -\frac{b}{2a}] and the *[tex \Large y]-coordinate of the vertex, which is the maximum value of the function, is given by *[tex \Large y_v\ =\ a(x_v)^2\ +\ b(x_v)\ +\ c].


For your function, *[tex \Large t] is the independent variable, *[tex \Large C(t)] is the dependent variable (the value of the function), 0.08 is the "b" value, and 0.0004 is the "a" value.


Just do the arithmetic.
 

John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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