How to use finite differences to determine an equation of a polynomial given consecutive integer $x$ and corresponding $y$ coordinates of the graph? This chart is given:
for $x=-3$, $y=-9$
for $x=-2$, $y=3$
for $x=-1$, $y=3$
for $x=0$, $y=-3$
for $x=1$, $y=-9$
for $x=2$, $y=-9$
for $x=3$, $y=3$
I found the finite differences to be 6 and degree of the polynomial with the given points to be 3 ($n=3$). But how do I find the polynomial function rule out of this information? Given no factors? Please help with a short solution. Thanks a lot in advance. I also did $y=kx^3$, $-9=k(-3)^3$, $k=1/3$. But of course the rule isn't $y=1/3x^3$. 

Treat your numbers as a sequence with $g(0)$ being the first term, corresponding to $f(-3)$,
$$f(x-3):=g(x) : \color{green}{-9},3,3,-3,-9,-9,3$$
$$\Delta g=g(x+1)-g(x) : \color{green}{12},0,-6,-6,0,12$$
$$\Delta^2 g : \color{green}{-12},-6,0,6,12$$
$$\Delta^3 g : \color{green}{6},6,6$$
Now assume all else is $0$. To get an umbral Taylor series representation,
$$f(x-3)=\color{green}{-9}+\color{green}{12}{x \choose 1}\color{green}{-12}{x \choose 2}+\color{green}{6}{x \choose 3}$$
So that with $x \mapsto x+3$,
$$f(x)=-9+12{{x+3} \choose 1}-12{{x+3} \choose 2}+6{{x+3} \choose 3}$$
$$=x^3-7x-3$$
