I would like to fit a quadratic curve to some data:
Label X Y S 0 105 E 277 85 Min m 72
Min is the minimum of the curve. My aim is to find m. (I suspect m is around 400 to 500)
a) Is this possible given this amount of data?
b) If not, would specifying the slope at S be sufficient?
In either case, how do I find the coefficients of the curve?
This is a real world problem for which my 30 years old school math knowledge is failing me.
You want to adjust a parabola $$y=a+bx+c x^2$$ and you know the value at two specific values of $x$ as well as the value of the minimum which occurs at $x=-\frac b{2c}$ to which corresponds $y=a-\frac{b^2}{4 c}$.
So, the equations are $$105=a \tag 1$$ $$85=a+277b+76729c \tag 2$$ $$72=a-\frac{b^2}{4 c}\tag 3$$ From $(1)$, $a=105$ (simple); replace in $(2)$ and $(3)$ to get $$-20=277b+76729c \tag 4$$ $$33=\frac{b^2}{4 c}\tag 5$$ From $(5)$, $c=\frac{b^2}{132}$. Plug this result in $(4)$ to get $$\frac{76729 }{132}b^2+277 b+20=0 \tag 6$$ Solving the quadratic in $b$, we find two solutions $$b_1=-\frac{2}{277} \left(\sqrt{429}+33\right)\implies c_1=\frac{2 \left(23+\sqrt{429}\right)}{76729}\implies m_1=\frac{277 \left(33+\sqrt{429}\right)}{2 \left(23+\sqrt{429}\right)}\tag 7$$ $$b_2=\frac{2}{277} \left(\sqrt{429}-33\right)\implies c_2=\frac{2 \left(23-\sqrt{429}\right)}{76729}\implies m_2=\frac{277 \left(\sqrt{429}-33\right)}{2 \left(\sqrt{429}-23\right)}\tag 8$$ This means that the minimum is either $m_1\approx 170$ or $m_2\approx 744$ .