Question 571736
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Let the four vertices of the square (starting at the negative *[tex \Large x] axis and going clockwise) be *[tex \Large A: \left(-\alpha,\,0\right)], *[tex \Large B: \left(0,\,\beta\right)], *[tex \Large C: \left(\alpha,\,0\right)], *[tex \Large D: \left(0,\,-\beta\right)].


Since the second degree term variable in the function is *[tex \Large x] we know that the axis of symmetry of the parabola graph is a vertical line.  Furthermore, since we are given that *[tex \Large a\ >\ 0], we know that the parabola opens upward, and the vertex is a minimum local extrema.  That fixes the three points of the square that are in common with the graph of the function as *[tex \Large A], *[tex \Large C], and *[tex \Large D] with *[tex \Large D] as the vertex of the parabola.


The standard form of a parabola is:


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


where *[tex \Large a\ \neq\ 0]


So if the parabola passes through the point *[tex \Large \left(x_i,\,y_i\right)] then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y_i\ =\ a(x_i)^2\ +\ b(x_i)\ +\ c]


In the case of your function at point *[tex \Large D: \left(0,\,-\beta\right)],


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(0)^2\ +\ (0)(0)\ -\ 4\ =\ -\beta]


from which we deduce


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \beta\ =\ 4]


But since the given figure was a square, we know that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \alpha\ =\ \beta\ =\ 4]


Subsituting the coordinates of *[tex \Large C: \left(\alpha,\,0\right)] which we now know to be *[tex \Large C: \left(4,\,0\right)] into your given function:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(4)^2\ +\ (0)(4)\ -\ 4\ =\ 0]


from which we get


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ \frac{1}{4}] 


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