Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
Drexel University or The Math Forum.



Re: Vertex formula?
Posted:
Jun 27, 2005 12:11 PM


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...
 submissions: post to k12.ed.math or email to k12math@k12groups.org private email to the k12.ed.math moderator: kemmoderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html



