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Re: Vertex formula?
Posted:
Jun 27, 2005 12:11 PM
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Jennifer Toth wrote:
>
> Is there a fornula which can give me the y coordinate of the vertex of a
> parabolilc curve?
>
> I am trying to find a different way other than plugging the value from
> -b/2a back into the equation.
>
>
Let us think backwards, using a geometrical transformation of
the parabola, and a chain of basic functions.
If your parabola has its vertex at (P,Q) and is stretched in the
y direction by factor m (negative for 'upside down'), the
quadratic function can be represented by the CHAIN of simple
functions:
Start with x
Subtract P
Square that answer
Multiply by m
Add Q
.... thats your y
Now relate these values to what you already know:
P = -b/2a
m = a
Therefore you can substitute ANY value into the
'chain of functions' method and the more usual
'2nd order polynomial' formula, and find Q.
Note that the 'Chain of functions' version is VERY easy
to find the inverse of (taking one root at a time,
obviously) by
Start with y
Subtract Q
Divide by m
Take positive root
( optionally change sign here )
Add P
.... thats your x
It is an algorithm that is very easily performed on a
calculator, not needing any brackets or stored values.
So you can solve f(x) = K for lots of values of K
In my opinion the 'Chain of functions' contains SO MUCH
of the personality of the quadratic that it should replace
the usual formula as a way of solving quadratics. But then
I always did like to think in terms of graphs...
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