I've been out of school for a long time, and don't remember how to do this. I have some points on a graph with a curved line and I'd like to know an equation for them so I can plot beyond what I've already got.
My current points are $x = 1, y=3000$; $x=2, y=9000$; $x = 3, y = 18000$; $x = 4, y = 30000$
So with the points (1,3000), (2,9000), (3,18000), and (4, 30000), I was able to use Mathematica to find a polynomial to fit the trend. It came to
$y = 1500x+1500x^2$.
You could also use Microsoft Excel if you do not have Mathematica.
If you wanted to do this by hand, then you could use the method of constructing what is known as a Lagrange polynomial. You can read up about this more on the Wikipedia page if needed.
But following the formula on there, you get that
$l_0 = -\frac{(x-2)(x-3)(x-4)}{6}$
$l_1 = \frac{(x-1)(x-3)(x-4)}{2}$
$l_2 = -\frac{(x-1)(x-2)(x-4)}{2}$
$l_3 = \frac{(x-1)(x-2)(x-3)}{6}$
Therefore, our interpolating polynomial $L(x)$ will be
$L(x) = -3000\frac{(x-2)(x-3)(x-4)}{6} + 9000\frac{(x-1)(x-3)(x-4)}{2} - 18000\frac{(x-1)(x-2)(x-4)}{2} + 30000\frac{(x-1)(x-2)(x-3)}{6}$
This also nicely simplifies to
$L(x) = 1500x + 1500x^2$