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anonymous

  • 5 years ago

is dere any easy way i can do standard form to vertex form?

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  1. amistre64
    • 5 years ago
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    probably; can you give an example?

  2. anonymous
    • 5 years ago
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    f(x)= ax^2+bx+c to f(x)=a(x-h)^2+k

  3. anonymous
    • 5 years ago
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    by completing the square

  4. amistre64
    • 5 years ago
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    yeah... you can find the "x" value for the vertex by taking the -(middle#)/2(first#) that gives you your "h" value plug that in for equation to ge the "y" value for it....which will be the "k" does that make sense?

  5. amistre64
    • 5 years ago
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    x^2 -12x + 45.... vertex = 12/2 = 6 = h (6)^2 -12(6) +45 = k

  6. amistre64
    • 5 years ago
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    y = (x+6)^2 + ...the # you find for k

  7. amistre64
    • 5 years ago
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    x-6...... that is

  8. anonymous
    • 5 years ago
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    f(x)=-4x^2-7x-2 ?? can u show me this one?

  9. amistre64
    • 5 years ago
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    i gave you a bad "equation"...lol...but the steps are good

  10. amistre64
    • 5 years ago
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    h = --7/2(-4) = 7/-8 = -7/8

  11. amistre64
    • 5 years ago
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    k = -4(-7/8)^2 -7(-7/8) -2

  12. amistre64
    • 5 years ago
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    k = 460/64 reduce as needed if I did it right in me head ;)

  13. amistre64
    • 5 years ago
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    thats the "shortest" way I know...

  14. anonymous
    • 5 years ago
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    im still kinda confuse but thanks :)

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