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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

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  1. anonymous
    • one year ago
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    Same

  2. anonymous
    • one year ago
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  3. anonymous
    • one year ago
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    that what the link was

  4. anonymous
    • one year ago
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    standard or vertex?

  5. anonymous
    • one year ago
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    i think standard

  6. anonymous
    • one year ago
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    it might not actually matter which one

  7. anonymous
    • one year ago
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    ice what does that even say

  8. anonymous
    • one year ago
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    ahh okay O.o

  9. ybarrap
    • one year ago
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    You need the equation of a parabola, the vertex location (x,y)=(h,k) and one point where the parabola crosses the x-axis. The equation: $$ y=a(x-h)^2+k $$ The Vertex: (h,k) is your vertex, you can get that from your plot, it is the top-most 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 x-intercept: Next, to find a you need to set y=0, set x equal to the point where the parabola crosses the x-axis. 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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