Question 1182759
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If *[tex \Large (-2,27)] is a point on the function


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y(x)\ =\ ax^3\ +\ bx^2\ +\ cx\ +\ d]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y(-2)\ =\ -8a\ +\ 4b\ -\ 2c\ +\ d\ =\ 27]


Similarly,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y(1)\ =\ a\ +\ b\ +\ c\ +\ d\ =\ 0]


At the turning points, the first derivative must equal zero, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  y'(-2)\ =\ 3a(-2)^2\ +\ 2b(-2)\ +\ c\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  12a\ -\ 4b\ +\ c\ =\ 0]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  y'(1)\ =\ 3a(1)^2\ +\ 2b(1)\ +\ c\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  3a\ +\ 2b\ +\ c\ =\ 0]


Solve the 4X4 system for *[tex \Large a, b, c, d]

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

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
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