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Find the equation of the parabola which has the given vertex $V$, which passes through the given point $P$, and which has the specified axis of symmetry.

$V(4,-2), P(2,14)$, vertical axis of symmetry.

The answer is $(x-4)^2=\frac 14(y+2)$, but I do not know how to get this answer. I know it is an upward facing parabola with $x=4$ as the axis of symmetry.

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You are trying to find a parabola of the form $$(x-4)^2 = 4p (y+2)$$ If you replace $x=2$ and $y=14$ en the last equation, you will find $p=\frac{1}{16}$. So that, $$(x-4)^2 = \frac{1}{4} (y+2)$$ is the parabola you are looking for.

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