# Finding cubic function from points?

How can I find a cubic function from two known points $\left(50,30\right)$ and $\left(100,0\right)$ which are turning points, hence the gradient at these points is zero. My final function must be in the form $\,\mathrm{F}\left(x\right) = A\left(x + a\right)\left(x + b\right)\left(x + c\right)$.

I already know that (X+b)= (x-100) (X+c)=(x-100)

So I'm missing A and a

The graph will then be set with a domain so it flows smoothly in this graph

I need very clear steps on finding a graph that flows smoothly and ends up in this form :)

Thank you!

The cubic function $f(x)$ such that $f'(50)=f'(100)=0$ and $f(50)=30$, $f(100)=0$ is given by $$f(x)=A\int_{100}^x(t-50)(t-100)\,dt=A\int_{0}^{x-100}(s+50)s\,ds\\ =A\left[\frac{s^3}{3}+25s^2\right]_{0}^{x-100} =\frac{A}{3}(x-25)(x-100)^2$$ where $A$ is a constant to be found by imposing that $30=f(50)=\frac{A}{3}\cdot 25\cdot 50^2$, that is $A=\frac{9}{6250}$.
2. There is a double root at $x=100$ so factorization will be easy.