Question 325939
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The equation of a general parabola with a vertical axis of symmetry is:


*[tex \Large \ \ \ \ \ \ \ \ \ \ y\ =\ ax^2\ +\ bx\ +\ c]


Substituting the coordinates of the first given point:


*[tex \Large \ \ \ \ \ \ \ \ \ \ -6\ =\ a(0)^2\ +\ b(0)\ +\ c]


*[tex \Large \ \ \ \ \ \ \ \ \ \ -6\ =\ c]


Substituting the coordinates of the second given point:


*[tex \Large \ \ \ \ \ \ \ \ \ \ -3\ =\ a(1)^2\ +\ b(1)\ +\ c]


*[tex \Large \ \ \ \ \ \ \ \ \ \ -3\ =\ a\ +\ b\ +\ c]


Substituting the coordinates of the third point:


*[tex \Large \ \ \ \ \ \ \ \ \ \ 6\ =\ a(4)^2\ +\ b(2)\ +\ c]


*[tex \Large \ \ \ \ \ \ \ \ \ \ 6\ =\ 4a\ +\ 2b\ +\ c]


Substituting the value we have determined for c:


*[tex \Large \ \ \ \ \ \ \ \ \ \ -3\ =\ a\ +\ b\ -\ 6]


*[tex \Large \ \ \ \ \ \ \ \ \ \ a\ +\ b\ =\ 3]


and


*[tex \Large \ \ \ \ \ \ \ \ \ \ 6\ =\ 4a\ +\ 2b\ -\ 6]


*[tex \Large \ \ \ \ \ \ \ \ \ \ 4a\ +\ 2b\ =\ 12]


Now just solve the 2 X 2 system:


*[tex \Large \ \ \ \ \ \ \ \ \ \ \ \,a\ +\ \ b\ =\ 3]
*[tex \Large \ \ \ \ \ \ \ \ \ \ 4a\ +\ 2b\ =\ 12]


and you will have all three coefficients to fill in the general equation:


*[tex \Large \ \ \ \ \ \ \ \ \ \ y\ =\ ax^2\ +\ bx\ +\ c]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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