Finding parabola (arc) that passes through two points with a given height gain Given that I have two points (3, 2) and (6, 4) as well as a determined desired height gain (say perhaps of 2). How can I construct the equation for a parabola that passes these two points?
My goal is to then take the explicit derivative at point 1 and launch an item at that velocity (at that angle which I'm not sure how I'll get) in order to have it follow the path of the parabola.
Diagram (excuse my inability to use Demos efficiently):

 A: If you let $g$ be the amount you wish to gain on point $2$, i.e., $(6,4)$, note the parabola equation can be written in the form
$$y = a(x - b)^2 + (4 + g) \implies y = ax^2 - 2abx + ab^2 + (4 + g) \tag{1}\label{eq1A}$$
where $b$ is the $x$ value where the vertex occurs. You can plug in your $2$ points to get $2$ equations in $2$ unknowns you can then solve for $a$ and $b$.
In particular with your example, for $(3, 2)$, you will get
$$2 = 9a - 6ab + ab^2 + (4 + g) \tag{2}\label{eq2A}$$
and for $(6, 4)$, you have
$$4 = 36a - 12ab + ab^2 + (4 + g) \tag{3}\label{eq3A}$$
Next, \eqref{eq3A} minus \eqref{eq2A} gives
$$2 = 25a - 6ab = a(25 - 6b) \tag{4}\label{eq4A}$$
Multiplying both sides of \eqref{eq2A} by $25 - 2b$ and using \eqref{eq4A} gives
$$\begin{equation}\begin{aligned}
2(25 - 6b) & = 9(2) - 6(2)b + (2)b^2 + (4 + g)(25 - 6b) \\
50 - 4b & = 18 - 12b + 2b^2 + 100 + 25g - 12b - 6gb \\
0 & = 2b^2 + (-20 - 6g)b + (68 + 25g)
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
This is now a quadratic equation you can solve for $b$ (making sure the root you choose is such that $2 \lt b \lt 4$ so it's in the region you expect), then use that to determine $a$ in \eqref{eq4A}.
