I have 3 points where I would like inflection points. $(t0,f0),(t1,f0),(t2,f2)$
The roots give the following equation for the derivate. $y'=(x^3-(t0+t1+t2)*x^2+(t0*t1+t1*t2+t2*t0)*x-(t0*t1*t2))*g$ where g is the constant multiplied by the first 4 coefficients in case the resulting polynomial is not monic. If monic, g would be 1.
That equation results from multiplying out the following. $y'=g(x-t0)*(x-t1)*(x-t2)$
The integral gives the quartic equation with the roots at t0, t1 and t2. $y=(g/4)*x^4-(g/3)*(t0+t1+t2)*x^3+(g/2)*(t0*t1+t1*t2+t2*t0)*x^2-g*(t0*t1*t2)*x+e$
If I plug in two points, I can get g, by plugging e from one equation into the other. Then I can plug in a value to the equation to get e. I used $(t0,f0) and (t1,f1)$ to get g, and $(t0,f0)$ to get e.
I ran into some problems with this though and can not understand what is wrong.
The first problem is how do I get the equation with the third point included? When I solve the equation with the 1st two points, the third point $(t2, f2)$ isn't included. The inflection point is at t2, but f2 is wrong.
The second problem is if $t2-t1=t1-t0$ is not true, then the first inflection point $(t0,f0)$ has the wrong f0 in addition to the third point being wrong. How can that happen if the equation was derived from that point specifically?
I used the same method for cubics and two inflection points and was able to get it to work with this method, so I'm at a loss.
Is it safe to assume that the quartic can only be made from two points and a third root?
I'm sure the second problem has to do with symmetry, but I don't know why it would render the y value of the first inflection point wrong when the equation was derived from that point.
Best Answer
It works right if you add a third variable and use gaussian elimination.
I used the form $y'=g*(x-t0)*(x-t1)*(x-t2)+h$. Took the integral, plugged in values for x and y, and solved for g,h, and e.