# Graphing & Identifying Quadratic Functions

• Sep 11th 2009, 03:53 PM
BeSweeet
I have to graph a quadratic function and find the vertex (simple), axis of symmetry (simple), and possible x/y intercepts (simple). I'm just having trouble converting this quadratic function into standard form (not 100% sure if that's what I need to do).

Here's the question:
http://img30.imageshack.us/img30/323...90911at531.png

To my understanding, I have to make the function look like this:
$(2x^2-6x+9)-9+7$

And then I would factor what's inside of the parenthesis, but I don't think it's possible with this problem... Or is it? Please help me on this one. Almost done with this assignment. Been driving me nuts for a couple of days.
• Sep 11th 2009, 04:06 PM
skeeter
Quote:

Originally Posted by BeSweeet
I have to graph a quadratic function and find the vertex (simple), axis of symmetry (simple), and possible x/y intercepts (simple). I'm just having trouble converting this quadratic function into standard form (not 100% sure if that's what I need to do).

Here's the question:
http://img30.imageshack.us/img30/323...90911at531.png

To my understanding, I have to make the function look like this:
$(2x^2-6x+9)-9+7$

And then I would factor what's inside of the parenthesis, but I don't think it's possible with this problem... Or is it? Please help me on this one. Almost done with this assignment. Been driving me nuts for a couple of days.

you have to complete the square ...

$2x^2-6x+7$

$2(x^2 - 3x) + 7$

$2\left(x^2 - 3x + \frac{9}{4}\right) + 7 - \frac{9}{2}
$

$2\left(x - \frac{3}{2}\right)^2 + \frac{5}{2}$
• Sep 11th 2009, 04:39 PM
BeSweeet
Ah, ok. I forgot that you didn't have to factor out something in the whole function.

Now I don't know how to graph the thing.

All these other problems are starting to piss me off...
• Sep 11th 2009, 07:17 PM
BeSweeet
Quote:

Originally Posted by skeeter
you have to complete the square ...

$2x^2-6x+7$

$2(x^2 - 3x) + 7$

$2\left(x^2 - 3x + \frac{9}{4}\right) + 7 - \frac{9}{2}
$

$2\left(x - \frac{3}{2}\right)^2 + \frac{5}{2}$

Where did $-\frac{9}{2}$ come from??
• Sep 11th 2009, 07:24 PM
11rdc11
Quote:

Originally Posted by BeSweeet
Where did $-\frac{9}{2}$ come from??

$2x^2-6x+7= 0$

$2x^2 -6x = -7$

$x^2 -3x = -\frac{7}{2}$

$(x^2 - \frac{3x}{2} +\frac{9}{4}) = -\frac{7}{2} + \frac{9}{4}$

$(x-\frac{3}{2}) + \frac{5}{4}$
• Sep 11th 2009, 07:27 PM
BeSweeet
So what would the axis os symmetry be? And I can't seem to graph it without getting it wrong...

http://img151.imageshack.us/img151/5...90911at927.png
• Sep 11th 2009, 07:31 PM
skeeter
Quote:

Originally Posted by 11rdc11
Skeeter made a mistake it suppose to be $-\frac{9}{4}$

no skeeter didn't ...

$\textcolor{red}{2}\left(x^2 - 3x + \textcolor{red}{\frac{9}{4}}\right) + 7 \textcolor{red}{- \frac{9}{2}}$

$2\cdot \frac{9}{4} = \frac{9}{2}$ ...

$\frac{9}{2}$ was added to the quadratic, not $\frac{9}{4}$ ... $-\frac{9}{2}$ is needed to balance the expression.
• Sep 11th 2009, 07:39 PM
11rdc11
Quote:

Originally Posted by skeeter
no skeeter didn't ...

$\textcolor{red}{2}\left(x^2 - 3x + \textcolor{red}{\frac{9}{4}}\right) + 7 \textcolor{red}{- \frac{9}{2}}$

$2\cdot \frac{9}{4} = \frac{9}{2}$ ...

$\frac{9}{2}$ was added to the quadratic, not $\frac{9}{4}$ ... $-\frac{9}{2}$ is needed to balance the expression.

My apologizes, skeeter is correct