Question 352781
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First thing:  Since the value of the function for the given point is greater than the value of the function at the vertex, the parabola must open upwards.


Second thing:  By symmetry, since the parabola passes through (8,4) it must also pass through (12,4). (the *[tex \Large x]-value 8 is 2 smaller than the *[tex \Large x]-value of the vertex, so the point with an *[tex \Large x]-value 2 larger than the *[tex \Large x]-value of the vertex must also have a function value of 4.)


Now we have three points and can determine the coefficients of the quadratic function.


We have a quadradic function of the form:


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


And three points on the graph of our function, (10,0), (8,4), and (12,4).


Considering the first point:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(10)^2\ +\ b(10) +\ c\ =\ 0]


which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 100a\ +\ 10b +\ c\ =\ 0]


Likewise, from the coordinates of the other two points we can derive:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 64a\ +\ 8b +\ c\ =\ 4]


And


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 144a\ +\ 12b +\ c\ =\ 4]


Since


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


has a *[tex \Large y]-intercept at *[tex \Large (0,c)], all you need to do is solve the system of equations to determine the value of *[tex \Large c]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 100a\ +\ 10b\ +\ c\ =\ 0]
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \,64a\ +\ \ 8b\ +\ c\ =\ 4]
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 144a\ +\ 12b\ +\ c\ =\ 4]


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