How do I solve a quadratic bezier curve for x and y individually? I'm looking at specific cases of quadratic bezier curves defined by three points: p1 = [0,0],
p2 = [i,m],
p3 = [d,m], where i,d, and m are positive real numbers and i < d. p1 and p3 are the start and end points, p2 is the control point, which does not fall on the curve.
How can I derive the equations to solve for x and y individually? I understand that not all bezier curves are aligned to the x and y axis, so might not follow the usual pattern of a quadratic equation or formula, but intuitively it should be possible to state y = f(x) and x = f(y) because we're only concerned with one root and we know which direction it grows.
 A: A linear Bézier between two points $P_1$ and $P_2$ is given by the linear transition $$b_{1,2}(t) = (1-t)P_1 + tP_2\tag 1$$ from $P_1$ to $P_2$ as $t$ runs from $0$ to $1$.
For a quadratic Bézier, we take two linear ones and interpolate them just like $(1)$:
$$\begin{align}
b_{1,2,3}(t) &= (1-t)b_{1,2}(t) + tb_{2,3}(t) \\
&= (1-t)((1-t)P_1+tP_2) + t((1-t)P_2 + tP_3) \\
&= (1-t)^2P_1+2t(1-t)P_2 + t^2 P_3 \tag 2\\
%&= P_1 - 2tP_1 + t^2P_1 + 2tP_2 - 2t^2P_2 + t^2P_3 \\
%&= t^2 (P_1-2P_2+P_3) + 2t(P_2-P_1)  + P_1  \tag 3\\
\end{align}$$
In terms of $x$ and $y$-coordinates $(2)$ is
$$\begin{align}
\binom xy &= (1-t)^2\binom{x_1}{y_1} + 2t(1-t)\binom{x_2}{y_2}+ t^2\binom{x_3}{y_3} \\
&=  2t(1-t)\binom{i}{m}+ t^2\binom{d}{m} \tag 3\\
\end{align}$$
The final step is left for you: In order to get a function in terms of $x$, i.e. $y(x)$, express $t$ as a function of $x$.  This is just solving a quadratic in $t$.  Then plug in the resulting $t(x)$ into the $y$-coordinate.  The quadratic equation for $x$ is:
$$x = t^2(d-2i)+2ti \tag 4$$
Solve $(4)$ for $t$ and then plug $t$ back into $(3)$.  As $x$ ranges from $0$ to $d$, $t$ will range from $0$ to $1$.

As some comment asked for which branch of the square-root to take: Solving $(4)$, which is the $x$-component of $(3)$, for $t$ yields:
$$t = \frac{-i\pm\sqrt{i^2+(d-2i)x}}{d-2i}\tag 5$$
We take the "$+$" branch because it maps $x=0\mapsto t=0$ and $x=d\mapsto t=1$ due to $d>i>0$.  The $y$-component of $(3)$ is
$$y = mt(2-t)\tag 6$$

In the special case of $d=2i$, equation $(4)$ decays to a linear equation and $t=x/(2i)$. In order to avoid handling different cases, extend $(5)$ with $\sqrt{\cdots}+i$ and rewrite as
$$\begin{align}
t &= \frac{\sqrt{i^2+(d-2i)x}-i}{d-2i} \\
&= \frac{i^2+(d-2i)x-i^2}{(d-2i)(\sqrt{i^2+(d-2i)x}+i)} \\
&= \frac{x}{\sqrt{i^2+(d-2i)x}+i} \tag 7\\
\end{align}$$
so that $d=2i$ is no more problem.  Notice that the denominator of $(7)$ is always positive.
Here is a Desmos plot showing the formulae and values.  Notice that the sliders are such that $d =2i=6$.
A: Your 3 points do define the quadratic relation: $$y=\frac{m\cdot x\cdot (d+i-x)}{d\cdot i}$$
but they don't define a quadratic bezier curve, because you haven't specified which parameter value $0<t_2<1$ that has to correspond to $p_2$. A particular choice of $t_2$ could easily give you a parabola that is rotated in a way so that neither $y = f(x)$ nor $x = f(y)$ would hold for all $x,y$.
