Finding the Equation of a ParabolaDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/