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/