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December 22, 2014

December 22, 2014

Posted by **Jake** on Monday, January 7, 2008 at 2:03pm.

- Math -
**Damon**, Monday, January 7, 2008 at 2:33pmIf 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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