Writing in Standard Form

**BeSweeet** · September 11th 2009, 02:16 PM

Ok, I'm fixing to go crazy over here. The professor didn't explain these things, and the book DEFINITELY doesn't explain anything like this.

My problems go like this:
"Write the standard form of the equation of the parabola that has the indicated vertex and whose graph passes through the given point."

Vertex: (2, -1) Point: (5, 17)

I know how to stick the vertex into standard form, but I don't know what the crap to do with the point!

I don't freaking get it! The internet isn't any help at all either.

**skeeter** · September 11th 2009, 02:20 PM

Originally Posted by BeSweeet

Ok, I'm fixing to go crazy over here. The professor didn't explain these things, and the book DEFINITELY doesn't explain anything like this.

My problems go like this:
[i][b]"Write the standard form of the equation of the parabola that has the indicated vertex and whose graph passes through the given point."

Vertex: (2, -1) Point: (5, 17)

$y = a(x - h)^2 + k$ where $(h,k)$ are the coordinates of the vertex.

$y = a(x-2)^2 - 1$

substitute $x = 5$ and $y = 17$ ... solve for $a$ and you have it.

**VonNemo19** · September 11th 2009, 02:21 PM

Originally Posted by BeSweeet

I know how to stick the vertex into standard form, but I don't know what the crap to do with the point!

You need that point to get the equation. Put 5 where x goes, and put 17 where y goes.

PS, you have just found the place on the internet where you will finally get some help.

**BeSweeet** · September 11th 2009, 02:25 PM

Do I just replace the variables with the proper numbers? Because this is how it wants me to answer the question:

And I'm supposed to write it in standard form, not solve for a.

**VonNemo19** · September 11th 2009, 02:31 PM

Originally Posted by BeSweeet

And I'm supposed to write it in standard form, not solve for a.

You have to solve for a to get the equation of the parabola. Otherwise, you only have generality. The question gives specifications.

**BeSweeet** · September 11th 2009, 02:35 PM

OH OK!!! I got ya. I'll tell you if what I got was right.

Well I got it wrong... Here's my answer:

And here's my work:

Not sure what to do now... Was excited for a moment.

**VonNemo19** · September 11th 2009, 02:55 PM

Originally Posted by BeSweeet

Not sure what to do now... Was excited for a moment.

$17=a(5-2)^2-1$

$18=a(3)^2$

$18=a(9)$

$\frac{18}{9}=a$

$a=2$

Therefore, the equation is $y=2(x-2)^2-1$

**BeSweeet** · September 11th 2009, 02:56 PM

I feel like an idiot. I divided -1 by both sides instead of adding it.

Reworking it all now.

**VonNemo19** · September 11th 2009, 02:59 PM

Originally Posted by BeSweeet

I feel like an idiot.

Idiots don't care if they get it or not. Therefore, you are not an idiot.

**BeSweeet** · September 11th 2009, 03:13 PM

Awesome! Definitely have this stuff figured out now. Thanks to everyone who helped with those quick responses!!

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