If you know for a fact that your points lie on a polynomial of degree 5, you can use finite differences. This is especially easy when the $x$-values are in arithmetic progression, as in your case. Write the $y$-values:
175 125 100 125 0 -125 -100 -125 -175
Then write the difference between each number and the next:
-50 -25 25 -125 -125 25 -25 -50
Again:
25 50 -150 0 150 -50 -25
Again:
25 -200 150 150 -200 25
Again:
-225 350 0 -350 225
And once more:
575 -350 -350 575
Now if your points were really from a polynomial of degree 5, that last line would have been constant, but it's not, so they're not. But after all, you said they were estimated points - they still might be close to some polynomial of degree 5. To find the polynomial of degree 5 that comes closest to your points, there is a method called least squares, and there are many expositions of it on the web.
EDIT: Here's a start on least squares:
You want a polynomial $p(x)=ax^5+bx^4+cx^3+dx^2+ex+f$ such that $p(800)=175$, $p(600)=125$, etc. We already know that's impossible, so we settle for making all the numbers $p(800)-175$, $p(600)-125$, etc., small. In fact, we form the quantity $$(p(800)-175)^2+(p(600)-125)^2+\cdots+(p(-800)-175)^2$$ and we try to minimize it. This quantity is a function of the 6 variables $a,b,c,d,e,f$, and there are standard calculus techniques for minimizing such a function. It gets very messy, but fortunately you don't actually have to do it; someone has done it for you, in the most general case, and found that you can write down a very simple matrix equation which you can solve for the unknowns $a,b,c,d,e,f$. And that's what you'll find if you search for least squares polynomial fit.
Let us suppose that you have this general problem when you need the quadratic function which goes through three points $[x_i,y_i]$ and let, as written by dinosaur, $$y=f(x)=ax^2+bx+c$$ So, the three equations are $$y_1 = a {x_1}^2+b {x_1}+c$$ $$y_2 = a {x_2}^2+b {x_2}+c$$ $$y_3 = a {x_3}^2+b {x_3}+c$$ Subtracting the first to the second and the second from the third already eliminates $c$ and your are left with two linear equations for two unknowns. You could even eliminate from the first difference $b$ and plug it in the second difference and solve a linear equation in $a$. When $a$ is obtained, go backwards for getting $b$ and then $c$.
If you do the above, you will end with $$a=\frac{{x_1} ({y_3}-{y_2})+{x_2} ({y_1}-{y_3})+{x_3}
({y_2}-{y_1})}{({x_1}-{x_2}) ({x_1}-{x_3})
({x_2}-{x_3})}$$ $$b=\frac{y_2-y_1}{x_2-x_1}-a (x_1+x_2)$$ $$c = y_1-a {x_1}^2-b {x_1}$$ Simple, isn't ?
Best Answer
You can do this most easily with matrices, and a quick Google search shows that matrix multiplication can be done in Excel, but it can be a bit of a pain, especially with so many points.
Let $A$ be the matrix created from the system of equations that result from plugging each point into $f(x)$ and $y$ be the matrix containing the values of $f(x)$. So given a set of points $\{(x_1, y_1),(x_2, y_2),...,(x_n, y_n)\}$, if you want a polynomial fit of degree $N$ (with $N>n$), you'd have the following matrices:
$A = \begin{bmatrix} x_1^{N} & x_1^{N-1} & x_1^{N-2} & ... & 1 \\ x_2^{N} & x_2^{N-1} & x_2^{N-2} & ... & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x_n^{N} & x_n^{N-1} & x_n^{N-2} & ... & 1 \end{bmatrix}$
$Y = \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{bmatrix}$
Now take set the matrix $\beta$ as
$$\beta = (A^T A)^{-1} A^T Y$$ whose elements are the coefficients of the $N$th degree polynomial corresponding to the set of points.