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Topic: Vertex formula?
Replies: 1   Last Post: Jun 27, 2005 5:31 PM

 Jeri Heth Posts: 2 Registered: 6/29/05
Re: Vertex formula?
Posted: Jun 27, 2005 5:31 PM

The Vertex Form of a quadratic equation will give the vertex of the
parabola.
To get this form, you must complete the square.

Example:

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
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" <ken@lampsacos.demon.co.uk> wrote in message
news:d9mo7a\$690\$1\$8302bc10@news.demon.co.uk...
>
> 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:
>
> Subtract P
> Multiply by m
>
> 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
>
> Subtract Q
> Divide by m
> Take positive root
> ( optionally change sign here )
>
> 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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