SOLUTION: determine the poynomial whose graph passes through the points (-1,1), (0,-2), (1,0), and (2,-1)
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Question 262271: determine the poynomial whose graph passes through the points (-1,1), (0,-2), (1,0), and (2,-1)
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Plotting the given points looks like it might be a cubic, so let's presume a 3rd degree polynomial and see what happens.
\ =\ ax^3\ +\ bx^2\ +\ cx\ +\ d)
So if the point (-1,1) is on the graph, then:
^3\ +\ b(-1)^2\ +\ c(-1)\ +\ d\ =\ 1)
And if the point (0,-2) is on the graph, then:
^3\ +\ b(0)^2\ +\ c(0)\ +\ d\ =\ -2)
And if the point (1,0) is on the graph, then:
^3\ +\ b(1)^2\ +\ c(1)\ +\ d\ =\ 0)
And if the point (2,-1) is on the graph, then:
^3\ +\ b(2)^2\ +\ c(2)\ +\ d\ =\ -1)
Note that the second equation reduces to
Using that information and simplifying we get:
The solution to the above system is:
)
Verification of the solution of the linear system is left as an exercise for the student.
Hence the desired polynomial is:
\ =\ -\frac{4}{3}x^3\ +\ \frac{5}{2}x^2\ +\ \frac{5}{6}x\ -\ 2)
John
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