Question 1176690
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Since y is a function of x, we are dealing with a graph that is a parabola with a vertical axis of symmetry.  Since the vertex is the point *[tex \Large (-2,3)], the equation of the axis of symmetry is *[tex \Large x\ =\ -2].  The axis of symmetry intersects the *[tex \Large x]-axis at the point *[tex \Large (-2,0)].  The distance from the other given point, *[tex \Large (-5,0)], to the axis is *[tex \Large |-5\ -\ (-2)|\ =\ 3].  Hence, because of symmetry, another point on the graph must exist on the *[tex \Large x]-axis 3 units from the intersection of the axis of symmetry and the *[tex \Large x]-axis, to wit, the point *[tex \Large (1,0)]


The standard form of a quadratic function is *[tex \Large y\ =\ ax^2\ +\ bx\ +\ c].


Since the *[tex \Large x]-value of *[tex \Large -5] in the desired function must yield the *[tex \Large y]-value *[tex \Large 0], we can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(-5)^2\ +\ b(-5)\ +\ c\ =\ 0]


And, using the data from the other two points:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(-2)^2\ +\ b(-2)\ +\ c\ =\ 3]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(1)^2\ +\ b(1)\ +\ c\ =\ 0]



Simplifying:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 25a\ -\ 5b\ +\ c\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4a\ -\ 2b\ +\ c\ =\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ +\ b\ +\ c\ =\ 0]


Solve the 3X3 linear system for the a, b, c, coefficients of your desired function.


<b>Alternate solution</b>


Start with the vertex form of a parabola:


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


Insert the coordinates of the vertex for *[tex \Large (h,k)]


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


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


Now, insert the coordinates of the other given point, *[tex \Large (-5,0)], and solve for *[tex \Large a]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(-5\,+\,2)^2\ +\ 3\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9a\ +\ 3\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ -\frac{1}{3}]


Back to our vertex form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -\frac{1}{3}(x\,+\,2)^2\ +\ 3]


All that is left is expanding the squared binomial and collecting like terms to produce the desired standard form.

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

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