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anonymous

  • one year ago

Write the quadratic function in the form f(x)=a−(x-h)^ 2+k . Then, give the vertex of its graph.f(x)=-2x^2+8x-5

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  1. zzr0ck3r
    • one year ago
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    \(f(x) = ax^2+bx+c\) is the same thing as \(f(x) =a(x+\dfrac{b}{2a})^2+c-a(\dfrac{b}{2a})^2 \)

  2. anonymous
    • one year ago
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    so i just plug it in?

  3. sepeario
    • one year ago
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    isn't the x point for the vertex -b/2a?

  4. misty1212
    • one year ago
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    hi!!

  5. misty1212
    • one year ago
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    this is not that hard it is \[f(x)=-2x^2+8x-5\]right?

  6. anonymous
    • one year ago
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    yes

  7. misty1212
    • one year ago
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    you want to write it as \[f(x)=a(x-h)^2+k\] all you need is \(h\) and \(k\)

  8. misty1212
    • one year ago
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    since evidently \(a=-2\)

  9. misty1212
    • one year ago
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    to find \(h\) compute \(-\frac{b}{2a}\) which, in your case, is \(-\frac{8}{2\times (-2)}\)

  10. misty1212
    • one year ago
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    what do you get ?

  11. anonymous
    • one year ago
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    is it -8/4x? and simplify so its -2x?

  12. anonymous
    • one year ago
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    or is it 2/x

  13. anonymous
    • one year ago
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    it's just 2 Kimes

  14. anonymous
    • one year ago
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    oh opps i mistaken x and a variable

  15. anonymous
    • one year ago
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    as*

  16. zzr0ck3r
    • one year ago
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    @Kimes yes just plug things :)

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