Parabolas, -b/2a ?Date: 03/13/2002 at 22:38:48 From: T.J. Subject: Parabolas, -b/2a? Hi, I've thought about why the form -b/2a works when trying to graph a parabola, but I just cannot figure it out. I asked my teacher and she said just not to worry about it. I couldn't not worry about it because I only learn how to do the math when I understand why it works. Thanks ahead of time! T.J. Date: 03/14/2002 at 08:35:31 From: Doctor Floor Subject: Re: Parabolas, -b/2a? Hi, T.J., Thanks for your question. Let's consider the parabola y = ax^2 + bx + c. Let's find the intersections with y = c; if we have found these intersections, then we know that the midpoint of these intersections gives the x-coordinate of the vertex of the parabola, because the parabola is symmetric in the vertical line through the vertex. We choose y = c in order to eliminate c from the equation to find the intersections. We have ax^2 + bx + c = c ax^2 + bx = 0 x(ax+b) = 0 x = 0 or ax+b = 0 x = 0 or ax = -b x = 0 or x = -b/a so the intersections are (0,c) and (-b/a,c). The midpoint is (-b/(2a),c) and thus the x-coordinate of the vertex is -b/(2a), just as you have learned. See for an approach using "completing the square" (and a little bit more difficult, but also more complete) from the Dr. Math archive: Graphing Parabolas http://mathforum.org/dr.math/problems/mui1.16.97.html If you have more questions, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
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