Hard cubic?

Question

Well this is part of a much larger problem involving all sorts of polynomials anyway i am stuck on this one particular part which involves a cubic where you know 2 points and half the stationary point of inflection the points given are as follows:

[45,0] , [32,-20] and the stationary point of inflection occurs at [?, -7]

Basically ur supposed to find the equation for the cubic. is it possible to do this or have i overlooked something? The cubic is part of a much larger diagram involving all sorts of parabolas, semi-circles and ellipses which we need to find the formula for in order to create a computer generated diagram. if need be i can upload a picture of the diagram for u guys to look at if u think i might have missed something.

Answer 1

You have enough information to solve it, it's just really tedious.

y = ax^3 + bx^2 + cx + d
y' = 3ax^2 + 2bx + c
y" = 6ax + 2b

y" = 0 when x = -b/(3a)
y'(-b/3a) = 0, which leads to
c = b^2 / (3a) ...(1)

y(-b/3a) = -7 which leads to
d = 1/27 * b^3/a^2 - 7 ...(2)

y(45) = 0 leads to
(45)^3*a + (45)^2*b + 15*(b^2/a) + 1/27*(b^3/a^2) - 7 = 0 ...(3)

y(32) = -20 leads to
(32)^3*a + (32)^2*b + 32/3*(b^2/a) + 1/27*(b^3/a^2) + 13 = 0 ...(4)

Now we need to solve (3) and (4) simultaneously. Using a second order Newton Raphson method to solve those equations,

a = 0.0353, b = -4.147, c = 162.44, d = -2127.83

FINAL ANSWER:
y = 0.0353*x^3 – 4.147*x^2 + 162.44*x – 2127.83

You can verify that
y(45) = 0
y(32) = -20
y"(39.168) = 0
y’(39.168) = 0
y(39.168) = -7

Answer 2

do the Lagrange polynomial