I am given a large set of data where I have determined the vertex to be $(14,25)$ and so I have formed the equation $(x-14)^2=4p(y-25)$. I am just now learning about parabolas and am stuck. Where do I go from here?
Let's just stick with the information provided: I will assume you need to solve for $p$, given a parabola defined by $$(x-14)^2=4p(y-25)$$ (whose vertex must therefore be $(14, 25))$.
Take any one of your data points $(x_i, y_i)$ (any point "on" the parabola, other than the vertex $(14, 25))$, substitute $x = x_i$, $y = y_i$, and solve for $p$. Repeat this for a handful of data points to test for a consistent "approximation" for the value of $p$.
Note that for smaller values of $p$, the corresponding parabola is "thinner" or more narrow (one might say "steeper"), and likewise, the larger the value of $p$, the "wider" (less "steep") the parabola. See, e.g., these graphs produced by Wolfram Alpha, which correspond to (inner graph to outer graph, color coded) values $p = \frac 14, p = \frac 12, p = 1, p = 2$: