Collinear Points on a Curve

Question

I encountered a problem recently wherein I had to find the value of the coefficient of the second degree term in a 4th degree equation (other coefficients were given) such that there are $4$ collinear points on the curve.
I found a similar question in a problem set wherein the solution included taking the second derivative and solving for the discriminant to be greater than $0$ and obtain an inequality which was the solution but I can't seem to figure out why this works. Please help.

The curve given in the question was $$y=x^4+9x^3+ax^2+2008x+2050$$

Answer 1

For a monic polynomial $P(x) \in \Bbb{R}[x]$ of degree $4$. Pick two points $(a,P(a)),(a+c,P(a)+d)$ on the curve $P(x)-y=0$, let $$h(t) =P(a+tc) - (P(a)+td) = c^4 t(t-1) (t^2+ft+g) $$

$g = (\frac{h'(0)}{c^4}=-\frac{c P'(a)-d}{c^4}$, $\frac{h(2)}{c^4} = 8+4f+2g, f= \frac{\frac{h(2)}{c^4}-8-2g}{4}$

Iff $f^2-4g>0$ then $t^2+ft+g = (t-u)(t-v)$ has two real roots so $h$ has 4 real roots and $(a+uc,P(a+uc)),(a+vc),P(a+vc))$ are two more points of the curve on the line $(a+tc,P(a)+td)$

Answer 2

One sufficient condition would be that there exists 2 local minima at $x_1$ and $x_3$ and a local maxima at $x_2$ such that $$x_1<x_2<x_3$$ $$f(x_1), f(x_2) < f(x_3)$$ Since $\lim_{|x|\rightarrow \infty} f(x) = \infty$, we can guarente a line will intersect $f$ at 4 points above $f(x_1)$ and $f(x_3)$ and below $f(x_2)$. I.e. there are 3 "humps". This means that $$f'(x_1) = f'(x_2) = f'(x_3) = 0 $$ and $$f''(x_1), f''(x_3)>0 \text{ and } f''(x_2) < 0$$ Take $a = -1000$, then $$f'(x) = 4 (-18.6353 + x) (-1.02017 + x) (26.4055 + x)$$ $$f''(x) = 12 (-166.667 + 4.50002 x + x^2)$$ a quick check shows that the above properties are satisfied for $$x_1 = -26.4055, \ x_2 = 1.02017, \ x_3 = 18.6353$$ see this plot of f(x)