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anonymous

  • one year ago

PLEASE HELP!!! f(x) = x^2 + 4x − 1 in vertex form

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  1. anonymous
    • one year ago
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    @Nurali

  2. butterflydreamer
    • one year ago
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    hint: you want to use completing the square ^_^

  3. anonymous
    • one year ago
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    I don't know how to do that.

  4. butterflydreamer
    • one year ago
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    |dw:1439905794934:dw| So now you want to plug b = 4 into the formula: \[(\frac{ b }{ 2})^{2} \] \[= (\frac{ 4}{ 2 }) ^2 = 2^2 = 4 \] Therefore to complete the square we have to add 4 \[f(x) = x^2 + 4x - 1 \] \[= (x^2 + 4x + 4) - 1 \] \[= (x + 2)^2 -1\]

  5. butterflydreamer
    • one year ago
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    oh oops. i made an error

  6. butterflydreamer
    • one year ago
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    Correction: f(x)=x2+4x−1 =(x2+4x+4)−1 - 4 We have to subtract 4 since we added 4 to complete the square =(x+2)^2−1-4 = (x + 2)^2 - 5

  7. anonymous
    • one year ago
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    lol nice, you pretty much got the concept down right though butterfly.

  8. anonymous
    • one year ago
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    thank you so much for explaining that for me :)

  9. anonymous
    • one year ago
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    can you help me with another question?

  10. anonymous
    • one year ago
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    Which of the following values "completes the square," or creates a perfect square trinomial, for x^2 + 6x + ___? a. 1 b.3 c. 6 d.9

  11. butterflydreamer
    • one year ago
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    use the formula i showed you above :) |dw:1439906543104:dw|

  12. anonymous
    • one year ago
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    9

  13. butterflydreamer
    • one year ago
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    excellent :)

  14. anonymous
    • one year ago
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    so the final result would be x^2 + 6x +9?

  15. butterflydreamer
    • one year ago
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    yyeess

  16. anonymous
    • one year ago
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    thank you so much! :)

  17. butterflydreamer
    • one year ago
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    no problemo~ ^_^

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