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):
As indicated in amd's comment, with the parabola having an axis of symmetry parallel to the $y$ axis, we have that
For simpler algebra, let $a = \frac{1}{4f}$ and $b = v_1$. If we have $g$ be the amount you wish to gain on point $2$, i.e., $(6,4)$, then $v_2 = 4 + g$. Thus, the parabola equation can then be written as
$$y = a(x - b)^2 + (4 + g) \implies y = ax^2 - 2abx + ab^2 + (4 + g) \tag{1}\label{eq1A}$$
Plugging in your $2$ points gives $2$ equations in $2$ unknowns which we can then solve for $a$ and $b$.
In particular, for $(3, 2)$ with your example, we get
$$2 = 9a - 6ab + ab^2 + (4 + g) \tag{2}\label{eq2A}$$
and, for $(6, 4)$, we have
$$4 = 36a - 12ab + ab^2 + (4 + g) \tag{3}\label{eq3A}$$
Next, \eqref{eq3A} minus \eqref{eq2A} gives
$$2 = 27a - 6ab = a(27 - 6b) \tag{4}\label{eq4A}$$
Multiply both sides of \eqref{eq2A} by $27 - 6b$ and use \eqref{eq4A} gives
$$\begin{equation}\begin{aligned} 2(27 - 6b) & = 9(2) - 6(2)b + (2)b^2 + (4 + g)(27 - 6b) \\ 54 - 12b & = 18 - 12b + 2b^2 + 108 - 24b + 27g - 6gb \\ 0 & = 2b^2 + (-24 - 6g)b + (72 + 27g) \end{aligned}\end{equation}\tag{5}\label{eq5A}$$
This is now a quadratic equation you can solve for $b$ (making sure, if you want the maximum to be between the $2$ points, that the root you choose is such that $3 \lt b \lt 6$), then use this to determine $a$ in \eqref{eq4A} (and ensure that $a \lt 0$ so the parabola vertex is a maximum, rather than a minimum).