Question 1027749
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A parabola is described by a quadratic function of the form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y(x)\ =\ ax^2\ +\ bx\ c]


Any point on the parabola must be of the form:  *[tex \Large \left(x,\,y(x)\right)]


So if *[tex \Large (-1,10)] is on the graph, it must be true that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(-1)^2\ +\ b(-1)\ +\ c\ =\ 10]


Which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ -\ b\ +\ c\ =\ 10]


Similarly:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4a\ +\ 2b\ +\ c\ =\ -5]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9a\ +\ 3b\ +\ c\ =\ -18]


Solve the 3X3 system to find the coefficients of the desired function.


Once you have the function, the vertex is at the point *[tex \Large \left(-\frac{b}{2a},\,y\left(-\frac{b}{2a}\right)\right)]


If *[tex \Large a\ >\ 0], the parabola opens upward and the vertex is a minimum.  If *[tex \Large a\ <\ 0], the parabola opens downward and the vertex is a maximum.  If *[tex \Large a] IS zero, then you don't have a parabola.  Check the solution to your 3X3 system, because you have made an error in this case.


Given that the vertex is at *[tex \Large (h,k)], the focus is at *[tex \Large \left(h,k\ +\ \frac{1}{4a}\right)], *[tex \Large a] being the lead coefficient of your function.


For the *[tex \Large x]-intercepts, set your function equal to zero and solve for the two roots.  Given correct algebra and arithmetic, your quadratic equation should factor, though it might be easier to just use the quadratic formula.


Good luck.


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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