# Graphing a Quadratic Function Given in General Form

• Mar 31st 2012, 06:58 AM
Trypanosoma
Graphing a Quadratic Function Given in General Form
If I understand correctly, the formula is:

$\displaystyle -\frac{b}{2a}, \frac{4ac-b^2}{4a}$

Here's what I have:

$\displaystyle 33x^2 -2x +15$

First:

$\displaystyle -\frac{2}{2(33)}$ = $\displaystyle \frac{1}{33}$

Easy enough. Now though...

$\displaystyle \frac{4(33)(15)}\frac{4(33)}$

Doing it that way comes up with wildly incorrect answers. I'd type it out but it's just all wrong. The answer in the book is...

$\displaystyle \frac{494}{33}$

However, doing...

$\displaystyle 15-\frac{-2^2}{4(33)}$ does get me

$\displaystyle \frac{494}{33}$

So that's really $\displaystyle C-\frac{(-b)^2}{4(a)}$

So my question is: Why? I'm missing something in the translation and I'm hoping to get some kind of answer. Thanks for any and all help.
• Mar 31st 2012, 10:55 AM
emakarov
Re: Graphing a Quadratic Function Given in General Form
Quote:

Originally Posted by Trypanosoma
If I understand correctly, the formula is:

$\displaystyle -\frac{b}{2a}, \frac{4ac-b^2}{4a}$

Being psychic, I know that you meant the formula for the vertex coordinates of the parabola ax^2+bx+c (Smile).

Quote:

Originally Posted by Trypanosoma
$\displaystyle 33x^2 -2x +15$

First:

$\displaystyle -\frac{2}{2(33)}$ = $\displaystyle \frac{1}{33}$

Should be $\displaystyle -\frac{-2}{2(33)}$

Quote:

Originally Posted by Trypanosoma
$\displaystyle \frac{4(33)(15)}\frac{4(33)}$

Should be $\displaystyle \frac{4(33)(15)-4}\frac{4(33)}$

Quote:

Originally Posted by Trypanosoma
Doing it that way comes up with wildly incorrect answers.

Not wildly incorrect; the difference between this and the correct answer is only 1/33.

Quote:

Originally Posted by Trypanosoma
So that's really $\displaystyle C-\frac{(-b)^2}{4(a)}$

So my question is: Why?

The y-coordinate of the vertex is $\displaystyle \frac{4ac-b^2}{4a}=c-\frac{b^2}{4a}$.