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marigirl

  • one year ago

Please help me form a quadratic equation

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  1. marigirl
    • one year ago
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    |dw:1440828489889:dw|

  2. marigirl
    • one year ago
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    \[y=a (\frac{ b }{ 2 }+x)(\frac{ b }{ 2}-x)\] Is that a good start?

  3. ganeshie8
    • one year ago
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    Looks good, plugin the point \((0, h)\) and find the value of \(a\)

  4. marigirl
    • one year ago
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    -a

  5. ganeshie8
    • one year ago
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    what do you mean?

  6. marigirl
    • one year ago
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    y= -a (b/2+x) (b/2-x) Or y= a (b/2+x) (b/2-x) Should I do -a or a I should do -a because it's a negative quadratic

  7. ganeshie8
    • one year ago
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    It doesn't matter

  8. ganeshie8
    • one year ago
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    \[y=a (\frac{ b }{ 2 }+x)(\frac{ b }{ 2}-x)\] plugin the point \((0, h)\) and find the value of \(a\)

  9. marigirl
    • one year ago
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    Ok I got a= 4h/b^2 Shouldn't be a negative value?

  10. ganeshie8
    • one year ago
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    so the equation is \[y=\dfrac{4h}{b^2} (\frac{ b }{ 2 }+x)(\frac{ b }{ 2}-x)\] simplify if you want to

  11. Elise_a18
    • one year ago
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    dang. I learned this last year and already forget how

  12. marigirl
    • one year ago
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    Ahhhhhhhhh yes y=(-4h-x^2/b^2)+h

  13. ganeshie8
    • one year ago
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    do you mean y = -(4h/b^2)x^2+h ?

  14. Elise_a18
    • one year ago
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    can you help me with my geometry? ;~;

  15. marigirl
    • one year ago
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    Wait I stuffed up

  16. Elise_a18
    • one year ago
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    im really sorry

  17. marigirl
    • one year ago
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    I expanded the bracket and simplified to y= h - (4hx^2/b^2)

  18. marigirl
    • one year ago
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    Thanks!

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