The Vertex Form of a quadratic equation will give the vertex of the parabola. To get this form, you must complete the square.
y = x^2 - 6x + 3 y - 3 = x^2 - 6x (isolate the x terms) y - 3 + 9 = x^2 - 6x + 9 (take half of -6 and square it to complete the square; add to both sides of the equation). y + 6 = (x - 3)^2 (simplify both sides; write the trinomial as a binomial squared)
This is vertex form. The vertex is the point (3, -6), found by taking the opposite of the values with x and y. In general, vertex form is y - k = a(x - h)^2, where (h, k) is the vertex. The "a" value would occur if the x^2 term in the original equation had a coefficient other than 1, but it does not effect the vertex.
This method certainly works, but substituting -b/2a is much easier and faster.
"Ken Starks" <email@example.com> wrote in message news:firstname.lastname@example.org... > > 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... >