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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?

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What do you mean by slope? Does your data perfectly fit a quadratic model of that form? or do you need more sophisticated methods for regression? – oldrinb Aug 30 '13 at 1:50
Nowhere in the body of your question is there any mention of the subject of the title. What exactly is your question? Are you trying to determine $p$? – Rick Decker Aug 30 '13 at 1:59
Are you trying to fit a parabola through this set of data? How did you determine that the vertex is (14,25) if all you have is a data set. And what is p? – Betty Mock Aug 30 '13 at 2:17

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 $p = \frac 14, p = \frac 12, p = 1, p = 2$:

enter image description here

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Nice pictures to illustrate the points. +1 – Amzoti Aug 30 '13 at 13:36

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