I am trying to come up with a simple equation that fits a curve.?

What a great challenge. I love it, but I'm about to get some kip, so I'll try and add some more tomorrow. Just as a preliminary, I've found a cubic equation that satisfies the "frown" shape only, although it's mirror image about the straight line should be all that's required. No doubt there will be some restrictions on P, but I'll try and work those out later. The equation is : y = ax^3 + bx^2 + 10, where, a = (10P^2 - 10T^2 - RP^2) / [P^2T^2(P - T)] and b = (RP^3 + 10T^3 - 10P^3) / [P^2T^2(P - T)] EDIT: As with Ben and Scythian1950, I did consider trigonometric curves and the very simple circle, but I opted for a polynomial because of the ease of differentiating. As Ben stated, a quadratic won't cut it, so I used the next best thing, a cubic. The cubic above does work between all your restrictions, but as I said I would, I have now calculated the restriction on P. It is that P > T[10/(10 - R)]^(1/3). (I set up the equation so that P will always be one of the positive roots of y.) Note that the curve always intersects (0, 10) and (T, R). Changing P changes the curvature. Example 1: P = 28, T = 20, R = 6 http://www.wolframalpha.com/input/?i=plot+a*x%5e3%2bb*x%5e2%2b10+where+a%3d(10*28%5e2-10*20%5e2-6*28%5e2)%2f(28%5e2*20%5e2*(28-20))%2c+b%3d(6*28%5e3%2b10*20%5e3-10*28%5e3)%2f(28%5e2*20%5e2*(28-20))&incParTime=true Example 2: P = 50, T = 20, R = 6 http://www.wolframalpha.com/input/?i=plot+a*x%5e3%2bb*x%5e2%2b10+where+a%3d(10*50%5e2-10*20%5e2-6*50%5e2)%2f(50%5e2*20%5e2*(50-20))%2c+b%3d(6*50%5e3%2b10*20%5e3-10*50%5e3)%2f(50%5e2*20%5e2*(50-20))&incParTime=true Hopefully, I'll get back with the other half of the problem. EDIT2: I've just seen Scythian's beautiful work, so as mine is not going too well, I'd prefer to desist bashing my head against a brick wall and just give him a TU. EDIT3: Thanks to other answerers for allowing me to see that a quadratic can work. Sometimes I just (enjoyably) muddle through things I've never done before, or don't understand very well, so I'm in awe at learning from such knowledgeable people.

Of course there are many ways to do this. If we want to stick with a quadratic function (y=ax^2+bx+c), then everything is straightforward until we get to your last restriction that the y-values be bounded between 10 and R for x between 0 and T. This essentially puts a bound on the concavity (second derivative), so taking a=P won't work. A solution that is simple for the program but potentially ugly would be to only allow certain values of P as input. Unfortunately these restrictions would have to depend on T,R... A more elegant solution (IMO) is to make a transformation. Take a= [2(R-10) / (pi T^2) ] * arctan P. Since arctan is bounded between +/- pi/2, this a is bounded between +/- (R-10)/T^2. This is essentially what we need to achieve your last condition. We of course need c=10 for the curve to pass through (0,10). And to pass through (T,R) requires that b = (R-10-aT^2)/T. In case you're interested in the derivation of what's above: I started solving for c and b, in terms of a. Now to keep our y-values bounded appropriately, it's enough to bound the a-values so that at the extremes, the derivative at 0 and at T are zero (so the parabola has its vertex at the endpoints in the extreme, or outside of the range (0,T) in the intermediate). Taking the derivative and evaluating at 0 and T gives (R-10 +/- aT^2) / T and finding where each of these is zero gives the appropriate extreme values of a. Now we need some way of taking P and producing some value of a in this range, and the arctangent was the first thing to pop to mind. Hopefully that works. The major downside in my eyes would be the requirement to calculate an arctangent of P. It might be interesting to also try an exponential function or the like. EDIT: While working out the formulas above, I hadn't really thought of the value of a as relating to the position of the vertex of the parabola, but it leads to an interesting remark. At the extremal values of a the vertex is at x=0 and x=T, and between those values the vertex is OUTSIDE of that range. So at some point it "goes through infinity", and of course this occurs when our curve is the straight line. Mildly amusing IMO. In response to additions from the other posters, first note that a quadratic DOES work, you just can't take an unlimited range of values of a. (But of course you need to also limit the coefficients of the cubic.) I hadn't thought of circular arcs, but I like the idea. Note that here too you would need to limit the parameter Scythian suggests to fulfill the asker's final condition (you could do a similar arctangent trick as I've done). Finally, I must defer to Scythian's greater knowledge on applications. It does seem likely that a parametric (x(t), y(t)) would be easier to deal with, and Bezier curves are something I recognize (largely) in name only.