anonymous
  • anonymous
Using the direction, size and vertex of your parabola, think of a quadratic equation in general form, y = a(x - h)2 + k, that would closely match your parabola. Remember (h, k) represents the vertex of the parabola.
Algebra
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chestercat
  • chestercat
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anonymous
  • anonymous
This is the picture of the parabola I am using...it's the inside of the first arch. I can't figure out what numbers to put in the equation
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AravindG
  • AravindG
This is strange ...The only thing I can predict here is the coefficient x^2 is negative. Do you have more data?
anonymous
  • anonymous
No. It's part of a project that I'm doing. I know you're supposed to be using GeoGebra to have a visual representation...I just didn't know how to set up the equation. I'm under the impression that you do your best to guess an equation and see how close it is to the parabola in the picture.

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anonymous
  • anonymous
It's alright if you don't get it. I was hoping to finish the project tonight but I can always ask my teacher on Tuesday...
AravindG
  • AravindG
First thing to do would be fix the origin at a convenient point.
anonymous
  • anonymous
Is the origin the vertex? For the vertex I got the point of (2.47, 5.24)
AravindG
  • AravindG
How did you get that?
anonymous
  • anonymous
I uI used the point finder on GeoGebra...is it totally wrong? I'm so sorry, I have no clue what I'm doing
AravindG
  • AravindG
I just dont have any idea I mean you just got a parabola without a coordinate axes.How are you supposed to give the vertex point?
anonymous
  • anonymous
THe parabola is the McDonald's arch...right? That's what I thought the picture was for
AravindG
  • AravindG
yes it is.
AravindG
  • AravindG
Are you given the dimensions or something?
anonymous
  • anonymous
The dimensions?
AravindG
  • AravindG
like the length of the mcdonald part or sthng.
anonymous
  • anonymous
Ah, no. I put the picture in GeoGebra, which is basically like putting the picture on an actual graph...I could use a different picture?
AravindG
  • AravindG
I see. All the best sorry I dont know how to help you on this.
anonymous
  • anonymous
It's alright. Thank you for trying :)
ybarrap
  • ybarrap
My initial thought was to use GeoGebra. Import the gif you have into GeoGebra and use the Parabola tool to select the point and directrix. Then use the "Focus" command and the "Vertex" command to find the focus and vertex. Then use the "Distance" command to find the distance between the focus and vertex to use for P in the equation below. With the vertex command, you will also have h and k: \(\Large y={(x-h)^2\over 4p}+k\). I get for h, p and k: \(\Large y={(x-2.97)^2\over 4(-.13)}-0.49\) http://en.wikipedia.org/wiki/Parabola http://www.wolframalpha.com/input/?i=y%3D-%28x-2.97%29^2%2F%284*0.13%29+-0.49
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ybarrap
  • ybarrap
Let me know if you have any questions.

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