Finding the Equation of a Parabola

Date: 04/28/2001 at 17:57:42
From: Rakesh
Subject: Finding equation of a parabola

This is the question:

The zeros of a quadratic relation are 0 and 6. The relation has a 
minimum value of -9. Find the equation of the parabola.

I have already graphed the parabola, but I do not know how to find the 
equation of a parabola. All of the questions I've done already provide 
you with the equation. How can you do it?

Date: 04/28/2001 at 21:53:05
From: Doctor Scott
Subject: Re: Finding equation of a parabola

Hi Rakesh.

This is a very interesting problem. Remember that the zeros of the 
quadratic equation are the values that make the quadratic function 
equal to zero. But, by the Factor Theorem, if 6 is a zero of a 
polynomial function, then (x-6) is a factor of the polynomial. So, 
since your zeros are 0 and 6, that means that (x-0) and (x-6) are two 
factors of the quadratic function.

So, if P(x) is the polynomial function, in factored form it would look 
like this: a(x-0)(x-6) = P(x). This is just a(x^2-6x) = P(x). We need 
the a because the parabola could be "stretched" without changing its 
zeros.

So, we now have P(x) = ax^2 - 6ax + 0 as our polynomial. The 
information given says that the vertex occurs when y = -9 (since 
that's the minimum value). So, since the minimum is at the vertex, we 
know that the x-coordinate of the vertex of a quadratic function of 
the form ax^2 + bx + c is given by -b/(2a). So, for our quadratic, the 
x-coordinate of the vertex is at (6a/2a) or x = 3. So, the point 
(3,-9) must lie on the function:

     -9 = a(3^2) - 6a(3) + 0
     -9 = 9a - 18a
     -9 = -9a
so
      a = 1

So, the quadratic must be y = x^2 - 6x + 0

- Doctor Scott, The Math Forum
  http://mathforum.org/dr.math/