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anonymous

  • one year ago

What is the equation of the following graph?

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  1. anonymous
    • one year ago
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    please help me!

  2. anonymous
    • one year ago
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    This is just like the other one, except now you have to use the graph to pick out the values. What is the vertex?

  3. anonymous
    • one year ago
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    (3,-2)?

  4. anonymous
    • one year ago
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    yes. and the roots?

  5. anonymous
    • one year ago
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    i'm not sure what the roots are...

  6. anonymous
    • one year ago
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    the roots are where the graph crosses the x-axis. For example|dw:1434816829118:dw|

  7. anonymous
    • one year ago
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    What is the x where your parabola crosses?

  8. anonymous
    • one year ago
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    1 and 5?

  9. anonymous
    • one year ago
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    yes. Now try plugging the roots into the equation. Remember the form is \[y = a(x-p)(x-q)\] Where p and q are the roots

  10. anonymous
    • one year ago
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    -2=a(3-4)(3-3)?

  11. anonymous
    • one year ago
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    not quite. We're not using the vertex yet or plugging in for x and y... yet. We have to set up the equation first. The root are 1 and 5, so plug in 1 and 5 for p and q to get y = a(x - 1)(x - 5) *It doesn't matter which is p and which is q*

  12. anonymous
    • one year ago
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    NOW you plug in the vertex to solve for a

  13. anonymous
    • one year ago
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    oh, okay. let me try.

  14. anonymous
    • one year ago
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    -2=a(3-1)(3-5)

  15. anonymous
    • one year ago
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    yes, now solve it for a

  16. anonymous
    • one year ago
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    thanks. hold on.

  17. anonymous
    • one year ago
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    a=1/2?

  18. anonymous
    • one year ago
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    yes. can I see your final equation to make sure you got the format right?

  19. anonymous
    • one year ago
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    sure.-2=a(3-1)(3-5) a=1/2

  20. anonymous
    • one year ago
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    The a, p, and q are constants that we used properties to find. They are the same for every point on the parabola. The x and y are variables and change for every point, so we need them to stay. So the equation is y = ½(x -1)(x - 5)

  21. anonymous
    • one year ago
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    so i solve this equation?

  22. anonymous
    • one year ago
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    no, that's the answer

  23. anonymous
    • one year ago
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    thanks again man, this helped.

  24. anonymous
    • one year ago
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    you're welcome

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