1. ## Graphing a Quadratic Function Given in General Form

If I understand correctly, the formula is:

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

Here's what I have:

$33x^2 -2x +15$

First:

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

Easy enough. Now though...

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

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...

$\frac{494}{33}$

However, doing...

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

$\frac{494}{33}$

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

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.

2. ## Re: Graphing a Quadratic Function Given in General Form

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

$-\frac{b}{2a}, \frac{4ac-b^2}{4a}$
Being psychic, I know that you meant the formula for the vertex coordinates of the parabola ax^2+bx+c .

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

First:

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

Originally Posted by Trypanosoma
$\frac{4(33)(15)}\frac{4(33)}$
Should be $\frac{4(33)(15)-4}\frac{4(33)}$

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.

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

So my question is: Why?
The y-coordinate of the vertex is $\frac{4ac-b^2}{4a}=c-\frac{b^2}{4a}$.