Help on Cubic Polynomials?

p(x) = ax³+bx²+cx+d

we have to minimize the sum of the squared differences :

(p(1)-2)²+(p(2)-6)²+(p(5)-10)²+(p(10)-12)²+

(p(15)-14)²+(p(20)-15)²+(p(25)-18)²+(p(30)-21)²

so we derive with respect to a, b, c, d

dp/da = 2(p(1)-2)*1³+2(p(2)-6)*2³+...

= linear in a,b,c,d

dp/db = ...

dp/dc =...

dp/dd = ...

so we obtain a linear system of 4 equations in 4 variables (a,b,c,d)

solve it and you get your cubic equation.

The rest is straightforward.

The calculation of the matrix of that system is a lot of work !

This can hardly be done manually. I would write a computer program for it.

I wrote a computer program and i get for the system matrix :

1049546940 38128158 1421892 55134 | 1028800

38128158 1421892 55134 2280 | 40776

1421892 55134 2280 108 | 1774

55134 2280 108 8 | 98

This yields

a = 0.002037

b = -0.104522

c = 1.966206

d = 1.455756

Remark : the model is very bad because for bigger values of the time in seconds the number of vegetables grows fast while this is not the case in reality.

So a cubic equation is not appropriate here as extrapolation.

The question p(40) yields 43.237 vegetables, while the person will not find more vegetables after a while. So the model is not good to describe the reality !

that's unquestionably an fairly elementary problem to sparkling up. bear in mind that an n-th order polynomial has n zeros (besides the shown fact that not unavoidably n unique zeros, because of the fact we are able to have double, triple, and better multiplicity roots). For our cubic (order 3) polynomial, we are given all 3 zeros. this suggests, in factored type, our polynomial sounds like: (x + 3)(x + a million)(x - 2) = 0 enable's advance this returned out: (x^2 + 4x + 3)(x - 2) = 0 x^3 + 2x^2 - 5x - 6 = 0 P(x) = x^3 + 2x^2 - 5x - 6 At this factor, although, we see we've an issue. P(0) is meant to equivalent 6, yet for sure it equals -6. for this reason, we could consistently multiply the whole undertaking with the aid of with the aid of -a million. this won't exchange any of the zeros, because of the fact -a million(0) = 0, even though it is going to alter the -6 to a +6: P(x) = -x^3 - 2x^2 + 5x + 6