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Discussion Thread: A problem where the Z coordinate is possibly 'implied'.   [#45674] / General Questions
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The Question by Robt W. Brown  01023450 :
2024-09-24 at 03:37GMT

The Vertex function below is based on the coordinates of the Vertex where: x = h  & y=k  ....

y= a(x - h)^2 + k

stated as a function we have:

f(x) = a(x - h)^2 + k  ...........................the Vertex formula
..............................
Let's begin with a vertex point (0,-1) and we have:

f(x) = a(x - 0)^2 - 1

Let's expand (x -0)(x -0) = x(x) =x^2

f(x) = ax^2 - 1
---------------------------------
Would there be any  rule broken, if I suggest we add another term to the vertex formula, so we have a graphing in 3 dimensions?

f(x) = a(x - h)^2 + k + ix
..............................

Although this extra term is added, the resulting parabola will still have the 'same vertex point on the Y axis & XY plane, here's proof:
We still : h = 0   k = -1
---------------------

,,,,y= a(x - h)^2 + k + ix   

g(x) = a(x - 0)^2 - 1 + ix     ...where f(x) = y + z

g(0) = a(0 - 0)^2 - 1 + ix

g(0) = a(0)^2 - 1 + i(0)

g(0) = -1 + i(0)

y = -1    z= 0      when x = 0   ......vertex of parabola #2   (0, -1, 0)
.........................................
This may seem like a manipulation of numbers, but the parabola of function g(x) will intersect the Y axis at its own vertex. And this is precisely the same point which is the vertex of f(x) .. So, if we try to argue that pt(0, -1)  is not the same as: (0, -1, 0) ......the contra-argument is, in the f(x) graphing, coordinate Z must be implied to be: z = 0
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A Response by Michael :
2024-09-25 at 20:17GMT

Yes. (0,-1) is no longer the vertex of the full quadratic with the ix term, but it is the vertex of the parabola projected back into the XY plane (which gets rid of the ix term).
A Response by Robt W. Brown  01023450 :
2024-09-26 at 04:47GMT

OK...enuff said, Thanks..

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