Constraining parameters for a quadratic equation

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This discussion focuses on constraining parameters for a quadratic equation to fit specific criteria derived from experimental data. The quadratic function is defined as y=ax²+bx+c, with constraints that it must pass through the point (1, 0) and remain within the bounds of 0≤y≤1 and y≤-x+1. A key insight is that the slope of the quadratic at the point (1, 0) must be -1 or less, leading to the condition 2a+b < -1. This establishes a clear relationship between the coefficients a, b, and c, which is essential for maintaining the desired fit.

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kungfuscious
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Hi there! I'm working on analyzing some data from an experiment, and you can see a graph of some of the results in the attached .jpg image. I've done both a linear and a quadratic fit on the data points, weighted to error bars (the red lines are the fits). I use percents in the graph, but I'd rather use more general values between 0 and 1.

I'm trying to write out constraints on the parameters of a quadratic fit in more general terms.

Constraints:
1) I know that the quadratic fit needs to pass through the point x=1, y=0.

2) For the domain, 0[itex]\leq[/itex]x[itex]\leq[/itex]1

3) The range is similar: 0[itex]\leq[/itex]y[itex]\leq[/itex]1

4) The quadratic must fit below the line going from the top left to the bottom right. (the line y=-x+1)

Different quadratics could be drawn in here, and I'm trying to generalize. y=ax[itex]^{2}[/itex]+bx+c

What I've figured out so far is that for 1), 0=a+b+c

Then, for 4), I can say that 0[itex]\leq[/itex]y[itex]\leq[/itex]1 , and
y[itex]\leq[/itex]-x+1

I guess that also means I can say that ax[itex]^{2}[/itex]+bx+c[itex]\leq[/itex]-x+1

I can plug in for a or b or c in that equation (using 0=a+b+c), but I just end up running around in a circle.

I get the feeling there's a lot more I can do with this to write out how a and b and c might be related or constrained, but I'm a bit stuck. Can anyone help?

Cheers,

Kungfuscious
 

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If you want the quadratic curve to fit under the -x+1 line even right at the x=1,y=0 point, then you could add an additional constraint where the tangent of the quadratic curve has a slope of -1 or less at that x=1,y=0 point...otherwise, the curve could excursion beyond your -x+1 line

so dy/dx = 2ax + b and so 2ax + b < -1 and at x=1, 2a+b < -1

no?
 
Thanks a lot! I had figured out another few things, but I think your suggestion will work. Thank you so much!

-Kungfuscious
 

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