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### 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.

T.J.
```

```
Date: 03/14/2002 at 08:35:31
From: Doctor Floor
Subject: Re: Parabolas, -b/2a?

Hi, T.J.,

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/
```
Associated Topics:
High School Conic Sections/Circles
High School Geometry

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