Question 636143
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A parabola is represented by a general form quadratic function:  *[tex \LARGE \rho(x)\ =\ ax^2\ +\ bx\ +\ c].


If it passes through a point *[tex \LARGE \left(x_i,y_i\right)], then the following will hold:


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


For the three points you were given:


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(3)^2\ +\ b(3)\ +\ c\ =\ -2]


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


From which the following 3X3 system can be derived:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ c\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9a\ +\ 3b\ +\ c\ =\ -2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4a\ -\ 2b\ +\ c\ =\ 2]


Substituting the first into the other two gives us a 2X2 linear system:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9a\ +\ 3b\ =\ -2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4a\ -\ 2b\ =\ 2]


Solve the 2X2 by whatever means you find convenient and then you will have the coefficients for your function.


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