Question 609613
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The standard (or general) form of a parabola is


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


Given that (2,4) is on the parabola,


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


Given that (1,1) is on the parabola,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(1)^2\ +\ b(1)\ +\ c\ =\ 1]


and


Given that (3,1) is on the parabola,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(9)^2\ +\ b(3)\ +\ c\ =\ 1]


Giving us the 3X3 system



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ +\ b\ +\ c\ =\ 1]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9a\ +\ 3b\ +\ c\ =\ 1]


Verification of the solution set element *[tex \LARGE (-3,12-8)] is left as an exercise for the student.


Hence the standard (or general, if you like) form of the equation is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -3x^2\ +\ 12x\ -\ 8]


Complete the square to find vertex form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -3(x^2\ -\ 4x)\ -\ 8]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -3(x^2\ -\ 4x\ +\ 4)\ -\ 8\ +\ 12]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -3(x\ -\ 2)^2\ +\ 4]


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