Question 1028268
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The answer depends on what you mean by standard form.  Some people call *[tex \Large y\ =\ ax^2\ +\ bx\ +\ c] standard form and *[tex \Large y\ =\ a(x\ -\ h)^2\ +\ k] vertex form.  And some people refer to *[tex \Large y\ =\ a(x\ -\ h)^2\ +\ k] as standard form.


Either way there are two ways to go about this.


Since you are given the vertex and one other point, you can find a third point on the graph by considering symmetry.  Since the point (-4,1) is two units to the right of the axis of symmetry, then the function value two units to the left of the axis must be the same, hence the point (-8,1) must also be on the graph.


Assuming *[tex \Large y\ =\ ax^2\ +\ bx\ +\ c] as standard form, if (-4,1) is on the graph, then it must be true that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  16a\ -\ 4b\ +\ 1\ =\ 1]


Likewise, considering the other two points,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  36a\ -\ 6b\ +\ 1\ =\ -4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  64a\ -\ 8b\ +\ 1\ =\ 1]


The solution to the 3X3 system of equations yields the standard form coefficients.


I'll leave it to you to verify that *[tex \Large a\ =\ \frac{5}{4}], *[tex \Large b\ =\ 15], and *[tex \Large c\ =\ 41].


On the other hand, if by standard form you mean what most folks call the vertex form, you can proceed as follows:


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


Substituting the vertex coordinates:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ a(x\ +\ 6)^2\ -\ 4]


Then, since you know that the point *[tex \Large (-4,1)] is on the graph,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(-4\ +\ 6)^2\ -\ 4\ =\ 1]


And all you need to do is solve for *[tex \Large a].  Any doubt in your mind that the result will be *[tex \Large a\ =\ \frac{5}{4}]?


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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