Question 762427
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Presume that 0 maps to 12, 1 maps to 8, 2 to 14 and 3 to 18, then with 4 points you can uniquely define a 3rd degree polynomial.


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


So if the point *[tex \LARGE (0, 12)] is on the graph, then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \gamma(0)\ =\ a(0)^3\ +\ b(0)^2\ +\ c(0)\ +\ d\ =\ 12]


Which is to say *[tex \LARGE d\ =\ 12]


Then, if the point *[tex \LARGE (1, 8)] is on the graph:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \gamma(1)\ =\ a(1)^3\ +\ b(1)^2\ +\ c(1)\ +\ d\ =\ 8]


or simplified:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ +\ b\ +\ c\ +\ d\ =\ 8]


Using similar reasoning, write the next two linear equations:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8a\ +\ 4b\ +\ 2c\ +\ d\ =\ 14]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 27a\ +\ 9b\ +\ 3c\ +\ d\ =\ 18]


Since we know the value of *[tex \LARGE d], we can simplify this to a 3X3 system:


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8a\ +\ 4b\ +\ 2c\ =\ 2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 27a\ +\ 9b\ +\ 3c\ =\ 6]


Solve the 3X3 linear system to determine the values of the coefficients for 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \gamma(x)\ =\ ax^3\ +\ bx^2\ +\ cx\ +\ 12]


Then evaluate *[tex \LARGE \gamma(4)] to determine the next number in the sequence.


By the way, next time you post show us a little respect by using the English language correctly instead of of using your nearly unintelligible teenager text-speak.


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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