Question 649759
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You have three fully described non-collinear points.  I'm going to presume that you were just being sloppy with the hyphen where you started the first row because if the first input value is -1, then the answer comes out with sloppy decimal fractions.


Since you have three non-collinear points, you can uniquely describe either a circle or a parabola, but since this looks like you want a function, I'm going to go with a parabola.


A general function that graphs to a parabola is:


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


Substituting the values for the first three points:


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


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(4)^2\ +\ b(4)\ +\ c\ =\ 21]


Which leads to the following 3X3 linear system:


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4a\ +\ 2b\ +\ c\ =\ 9]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 16a\ +\ 4b\ +\ c\ =\ 21]


You have some work to do:


Step 1:  Solve the 3X3 linear system to get your function coefficients.


Step 2:  Evaluate your function at 5, 8, and 12.


Step 3:  Set your function equal to 57 and solve for *[tex \LARGE x]


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