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anonymous
 one year ago
could someone help me create an equation from this proabola
http://learn.flvs.net/webdav/educator_algebraII_v14/module06/images/06_05_assessment/06_05_06_01.gif
anonymous
 one year ago
could someone help me create an equation from this proabola http://learn.flvs.net/webdav/educator_algebraII_v14/module06/images/06_05_assessment/06_05_06_01.gif

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that what the link was

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it might not actually matter which one

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ice what does that even say

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.0You need the equation of a parabola, the vertex location (x,y)=(h,k) and one point where the parabola crosses the xaxis. The equation: $$ y=a(xh)^2+k $$ The Vertex: (h,k) is your vertex, you can get that from your plot, it is the topmost point. h represents how for left or right you need to move the vertex and k represents how far up or down it is The xintercept: Next, to find a you need to set y=0, set x equal to the point where the parabola crosses the xaxis. Solve for a. Put it all together: Once you have a,h and k, put back into the equation above and you will have y as a function of x, your equation of a parabola that models your graph. That's it! Note that a should be negative because the parabola goes more and more negative when x gets more and more positive.
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