Question 522226
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Since you are concerned with whether the parabola opens up or down I have to assume that the axis is a vertical line.  You don't say so, and there really is no other clue about the orientation of your parabola.  When you are writing descriptions of things, do not assume anything.  Had you stated that you wanted to derive a <i>function</i> whose graph was a parabola, that would have been a different thing.


The vertex form for a vertical axis parabola is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ a(x\ -\ h)^2\ +\ k]


Where *[tex \Large \left(h,\,k\right)] are the coordinates of the vertex.


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ a(x\ -\ 3)^2\ -\ 2]


Is true for this parabola, and if the point *[tex \Large \left(-2,\,-8\right)] is indeed on the graph, then the following must be true as well:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -8\ =\ a(-2\ -\ 3)^2\ -\ 2]


Which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ -\frac{6}{25}]


And the full vertex form equation becomes


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -\frac{6}{25}(x\ -\ 3)^2\ -\ 2]


Parabolas with a negative lead coefficient open downward.  This makes sense upon examining the two given points.  The vertex of any parabola function is either a minimum or a maximum.  Hence, if the value of the function (the *[tex \Large y]-coordinate) at any point other than the vertex is less than the value of the function at the vertex, the vertex must be a maximum and that means the parabola must open downwards.  -8 is less than -2 QED.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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