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
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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
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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
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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
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,,,,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)
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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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