A friend of mine im trying to help in math was given a question on an assignment along the lines of: suggest an equation to satisfy the following points:
x F(x)
-3 0
-2 4
-1 0
0 -6
1 -8
2 0
3 24
its obvisously 3rd order and i know the leading coefficient is 1...so have x^3 + bx^2 + cx + d = F(x)
can anyone hint as to how to go about it?
Finding the equation given many points
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stapel_eliz
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stefan88 wrote:its obvisously 3rd order and i know the leading coefficient is 1...so have x^3 + bx^2 + cx + d = F(x)
can anyone hint as to how to go about it?
Just as a quadratic (a polynomial of degree two) is uniquely determined by three points, so also a cubic (a polynomial of degree three) is uniquely determined by four points.
So pick any four of the given points, and plug them into the cubic's equation. I like smaller values, so I'd probably pick (-1, 0), (0, -6), (1, -8), and (2, 0):
. . . . .-1 + b - c + d = 0
. . . . .d = -6
. . . . .1 + b + c + d = -8
. . . . .8 + 4b + 2c + d = 0
Then solve the system of equations for the values of the coefficients.
P.S. Alternatively, you could plug all of the points into a graphing calculator in "STAT" mode, and do the cubic regression. You will discover that your impulse was exactly right: this doesn't just look like a cubic; it is a cubic.
