IA on Quartic Functions

IA's for IB math HL are destroying me greatly, so I'll ask here. I have this information:

Having investigated the function f(x)= (x^4)-8(x^3)+18(x^2)-12x+24, I have found that the ratios of the distances PQ:QR:RS (Q and R being the inflection points on this "W" shaped quartic, and P and S being where a line QR would intersect the original function) to be 1:phi:1 (phi being the golden ratio)

I have found the same ratios to be true in another "W" shaped function f(x)=(x^4)-9(x^3)+23(x^2)--13x+30

The next steps I have to do are

5. Form a conjecture and formally prove it u sing a general quartic function.
6. Extend this investigation to other quartic functions that are not strictly of a "W" shape.

I know that forming a conjecture has something to do with the golden ratio being QR, and that's about it. I need help.

Well I can certainly prove your conjecture, I didn't begin with a general quatric. Make of that what you will...

Any function of a particular "W" shape for which a conjecture might hold can be translated freely without loss of character. It may be translated such that it has inflection points at x=a and x=-a for some a.
Then the second derivative of the quatric is given:

f''(x) = b(x+a)(x-a)
f''(x) = b(x^2-a^2)

Anti-differentiate

f'(x) = b(1/3*x^3 - a^2*x + c)

Anti-differentiate

f(x) = b(1/12*x^4 - a^2/2*x^2 + cx + d)

Solve at the inflection points:

f(a) = b(1/12*(a)^4 - a^2/2*(a)^2 + c(a) + d)
f(-a) = b(1/12*(-a)^4 - a^2/2*(-a)^2 + c(-a) + d)

Solve for the slope (m) of the line connecting the inflection points:

f(a) - f(-a) = 2abc
a - (-a) = 2a
m = bc

Solve for slope/intercept form:

y = mx + i

f(a) = bc(a) + i
b(1/12*(a)^4 - a^2/2*(a)^2 + c(a) + d) = bca + i
i = b/12*(a)^4 - ba^4/2 + bd
i = (-5b/12)(a^4) + bd

y = bcx + (-5b/12)(a^4) + bd

Solve for intersections of line with f(x)

b/12*x^4 - ba^2/2*x^2 + bcx + bd = bcx + (-5b/12)(a^4) + bd
x^4 - 6(a^2)*x^2 = -5(a^4)
x^4 - 6(a^2)*x^2 + 5(a^4) = 0

let w = x^2

w^2 - 6(a^2)w + 5(a^4) = 0

w = 6(a^2)/2 +/- sqrt([-6(a^2)]^2 - 4(5(a^4))]/2
w = 3(a^2) +/- sqrt(36a^4 - 20a^4) / 2
w = 3(a^2) +/- sqrt(16a^4)/2
w = 3(a^2) +/- 2(a^2)
w1 = 5(a^2), w2 = (a^2)

x1 = +sqrt(w1) = sqrt(5)a = Projection of S onto x-axis
x2 = -sqrt(w1) = -sqrt(5)a = Projection of P onto x-axis
x3 = +sqrt(w2) = a = Projection of R onto x-axis
x4 = -sqrt(w2) = -a = Projection of Q onto x-axis

(Note that our original inflection points have fallen out: beauty).
Then there is some constant t such that the distance between points is given:

|PQ| = t|x2x4|
|QR| = t|x4x3|
|RS| = t|x3x1|

|PQ| = ta(sqrt(5) - 1)
|QR| = ta(2)
|RS| = ta(sqrt(5) - 1)

Gives ratios of phi. QED

This proof extends to any Quatric with two distinct real inflection points. We cannot draw a line to intersect a quatric four times if that quatirc has fewer than two distinct real inflection points.

Edit: Made my linear algebra a little more clear (hopefully)...

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