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 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
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" <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:
>
> 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 e-mail to k12math@k12groups.org
private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org
newsgroup website: http://www.thinkspot.net/k12math/
newsgroup charter: http://www.thinkspot.net/k12math/charter.html
|
|
|
|
