Posted by Jake on Monday, January 7, 2008 at 2:03pm.
If y = f(x) = a x^2 + b x + c
and a, b and c are constants
then you have your parabola.
That IS one.
Now do you want some particular parabola?
for example if I want a parabola that opens up (holds water) and has a vertex at x = 4 and y = -5
Then I can use the form:
(x-h)^2 = 4 g (y-k)
where (h,k) is the vertex and it will open up if g is positive. (by the way g is the distance from the vertex to the focus)
The constant g determines how "steep" it is.
so for our vertex at (4, -5) I have
(x-4)^2 = 4 g (y+5)
for our example let's take g = 1/4 to make it simple
x^2 -8 x +16 = y + 5
y = x^2 -8 x + 11
which is your form that you wanted
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