Creating a weighted score I have an audit where there are six criteria, each can be scored Excellent (E), Satisfactory (S), Needs improvement (N) or Unsatisfactory (U).
I know that if someone scores Excellent in all six areas I would like their score to be 100. I also know that if they score 6 Satisfactories, I would like them to score 95. Finally, I also know that if they score unsatusfactory in all area, I would like them to score 0.
I have tried assigning arbitray, evenly distributed, values then plotting these points {$({0,0})$, $({2,95})$, $({3, 100})$} and creating a quadratic line of best fit, using the least squares method, but the resultant equation: $f(x)=\frac{455}{6}x - \frac{85}{6}x^2$ reaches its maximum at $x=91/34$, where $f(x)=41405/408$. As the maximum score is 100 this is no good to me. I cannot simply add a constant either, as I have have the same problem at the intercept, where I would get a negative score. 
Am I approaching this all wrong? Any suggestions as to how I can create a general equation for this and what values I shoudl assign to each criteria?
 A: What if you just assign the following values:
$$E =  {100\over6}, \quad
S =  {95\over6}, \quad
N =  {n\over6}, \quad
U =  0
$$
and choose $n$ as appropriate for your needs?
Also, it would help to have more qualitative information, particularly regarding the relative value of $N$ since that is the one criterion that you haven't used in a formula. Is $N$ very bad, but not terrible, in the same way that $S$ is very good, but not excellent? In other words, if someone scores $6 \times$ Needs Improvement, do they get a score of $5$? Is there such a thing as an overall passing or failing score?
A: Try the following function:
$$f(x) = \frac{5}{18}\big(11x^3-106x^2+339x\big)$$
You can check that $f(0) = 0$, $f(2)=95$, and $f(3)=100$. For "Needs Improvement", the function gives $f(1) = 67.778$. Furthermore, $f$ has a local maximum at $x=3$; you can see that it levels off there nicely:

I found this by setting $f(x)=ax^3+bx^2+cx+d$ and solving for $a,b,c,d$. Your three points at $x=0$, $x=2$, and $x=3$ are not enough to determine these four constants, so based on your description of the problem I added a fourth constraint that $f(x)$ should attain a maximum at $x=3$. (Thus, using calculus, $3ax^2+2bx+c = 0$ when $x=3$.)
Finally, if you plot this against the logistic curve I suggested, you'll see that they are very similar for $x \in [0,3]$.

Edit: further details on the calculation
In looking for a simple polynomial function to fit the data, we can consider there to be four pieces of information:


*

*$f(0) = 0$;

*$f(2) = 95$;

*$f(3) = 100$;

*There should be a maximum at $f(3)$: in other words, the slope of the tangent line there, $f'(3)$, should equal $0$.


Since we have four pieces of information, it makes sense to look at a polynomial of degree $3$, which has four degrees of freedom. Thus $f(x) = ax^3+bx^2+cx+d$ for some $a,b,c,d$. Evaluating this function at $x=0, 2$, and $3$, we have:
$$\begin{align}
d &= 0\\
8a+4b+2c+d&=95\\
27a+9b+3c+d&=100.
\end{align}$$
For the fourth equation, we use the fact that the derivative of $x^n$ is $nx^{n-1}$, which gives $f'(x) = 3ax^2+2bx+c$. Setting this to $0$ at $x=3$, we have:
$$\begin{align}
27a+6b+c&=0.
\end{align}$$
You now have four equations in four unknowns, and can use Excel, or Wolfram Alpha, etc., to solve the system.
