Question 262271
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Plotting the given points looks like it might be a cubic, so let's presume a 3rd degree polynomial and see what happens.


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


So if the point (-1,1) is on the graph, then:


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


And if the point (0,-2) is on the graph, then:


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


And if the point (1,0) is on the graph, then:


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


And if the point (2,-1) is on the graph, then:


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


Note that the second equation reduces to


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ -2]


Using that information and simplifying we get:


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


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8a\ +\ 4b\ +\ 2c\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ -2]


The solution to the above system is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(-\frac{4}{3},\,\frac{5}{2},\,\frac{5}{6},\,-2\right)]


Verification of the solution of the linear system is left as an exercise for the student.


Hence the desired polynomial is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P(x)\ =\ -\frac{4}{3}x^3\ +\ \frac{5}{2}x^2\ +\ \frac{5}{6}x\ -\ 2]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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