Degree 4 polynomial has 2 points of inflection A&B. The line connecting A&B intersects the function at C&D &→?

Question

Prove that the ratio AB:AC:BD is constant, and what is it?

Answer 1

Let be the polynomial p = x^4 - 6x^2 + 5.

Its second derivative is

p'' = 12 (x^2 - 1), and its roots are +1 and -1.

Then the inflexion point are

A = (-1, 0)
B = (1, 0)

The line connecting A & B is y=0, then the equation for C & D is

x^4 - 6x^2 + 5 = 0

Then

C = (-sqrt(5), 0)
D = (sqrt(5), 0)

The ratios are

(AD) / (AB) = (sqrt(5)+1) / 2

(AB) / (AC) = 2 / (sqrt(5)-1) = (sqrt(5) + 1) / 2

I tried numerically for different 4th degree polynomials, and the ratios are always the golden ratio.

For a general 4th degree polynomial with two inflection points there is rotation and translation followed for an expansion of the coordinates that move the inflexion points to (-1,0) and (1,0). The only difference of this "transformed" polynomial and the one of the example above is that the principal coefficient of p is 1, but in the general case might be any value not null. But this multiplicative factor cannot change the solutions of the homogeneous quartic equation, then the distances between A, B, C and D are the same. The rotation and the translation does not change the distances, and the expansion changes them, but not the ratios, then the result is completely general.

Answer 2

those 2 could have vertically opposite angles and consequently equate those 2 anglles equivalent 17x/6-4=9x/4+3 u gets x as 12 and u gets those 2 angles equivalent to 30 stages.

Answer 3

f(x) = ax^4 + bx^3 + cx^2 + dx + f
f'(x) = 4ax^3 + 3bx^2 + 2cx + d
f''(x) = 12ax^2 + 6bx + 2c

Inflection points are when f''(x) = 0
12ax^2 + 6bx + 2c = 0
x = (-6b +/- SQRT(36b^2 - 96ac)) / (24a)
Inflection points
x_1 = -b/(4a) + SQRT(b^2 - 8ac/3) / (4a)
x_1 = (SQRT(b^2 - 8ac/3) - b) / (4a)
x_2 = -b/(4a) - SQRT(b^2 - 8ac/3) / (4a)
x_2 = (-SQRT(b^2 - 8ac/3) - b) / (4a)

sorry, I have to go, I'll try to come back as soon as I can, this is a long problem, you need to substitute x_1 and x_2 into f(x), get the line equation for AB, where A is the point(x_1,f(x_1)) and B is (x_2,f(x_2)), then you need to equalize that line to f(x) so you can get the intersection points C & D.