Finding cubic function from points?
The cubic function $f(x)$ such that $f'(a)=f'(b)=0$ and $f(a)=0$ with $a\not=b$, is given by $$f(x)=A\int_{a}^x(t-b)(t-a)\,dt=A\int_{0}^{x-a}(s+a-b)s\,ds\\ =A\left[\frac{s^3}{3}+\frac{a-b}{2}s^2\right]_{0}^{x-a} =\frac{A}{6}(2x+a-3b)(x-a)^2$$ where $A$ is a constant to be found by imposing that $f(b)=\frac{A}{6}\cdot (a-b)^3$, that is $A=\frac{6f(b)}{(a-b)^3}$.
For $a=100$, $b=50$ and $f(b)=30$ we get $A=\frac{9}{6250}$ and $$f(x)=\frac{3}{6250}(x-25)(x-100)^2.$$