Question 797696
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Vertex: *[tex \LARGE (-3,-1)]


Contains: *[tex \LARGE (-2,-3)]


By symmetry must then contain *[tex \LARGE (-4,-3)]


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


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


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \rho(-4)\ =\ a(-4)^2\ +\ (-4)b\ +\ c\ =\ -3]


Which leads to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4a\ -\ 2b\ +\ 1\ =\ -\ 3]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 16a\ -\ 4b\ +\ 1\ =\ -\ 3]


Use any convenient method to solve what you will find to be the independent and consistent 3X3 system of equations for the ordered triple *[tex \Large \left(a,\,b,\,c\right)] which gives you the coefficients to plug in to:


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


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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