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burhan101

  • 3 years ago

Increasing and decreasing intervals of 4x^2+4x-1

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  1. etemplin
    • 3 years ago
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    do you know derivatives?

  2. burhan101
    • 3 years ago
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    we have to complete the square

  3. burhan101
    • 3 years ago
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    would it be \[\huge 4x^2+4x-4+4-1 \] for the first step of completing the square ?

  4. etemplin
    • 3 years ago
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    yes

  5. burhan101
    • 3 years ago
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    then \[\huge 4 (x-1)^2 -1\] ? :S

  6. etemplin
    • 3 years ago
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    when you multiply that out it doesnt give you 4x^2+4x-4+4-1

  7. etemplin
    • 3 years ago
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    try this one 4(x+(1/2))^2 -2

  8. burhan101
    • 3 years ago
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    i dont get where the 1/2 comes from :S

  9. etemplin
    • 3 years ago
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    i found the error: the leading coefficient must be 1 to complete the square try setting it equal to 0, moving the -1 over, and dividing by 4 then complete the square

  10. burhan101
    • 3 years ago
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    why does it have to be 1 ?

  11. etemplin
    • 3 years ago
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    because thats what the formula says

  12. burhan101
    • 3 years ago
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    oh so it applies at all equations ?

  13. etemplin
    • 3 years ago
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    it applies to quadratics with a leading coefficient of 1

  14. burhan101
    • 3 years ago
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    |dw:1351571054966:dw|

  15. etemplin
    • 3 years ago
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    that works move the -1/4 over divide the middle term by 2 and square it add that answer to both sides factor

  16. etemplin
    • 3 years ago
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    *by middle term, i meant the coefficient with the x^1

  17. burhan101
    • 3 years ago
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    what do you mean by move the -1/4 over ? like why

  18. etemplin
    • 3 years ago
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    because thats what the formula says to do you'll end up with x^2+x=(1/4) the term with x is 1 (1/2)^2 = 1/4 add (1/4) to both sides end up with x^2+x+(1/4) = (1/2) factor into (x+(1/2))^2 = (1/2) solve for x

  19. burhan101
    • 3 years ago
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    |dw:1351571511476:dw|

  20. etemplin
    • 3 years ago
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    yes now factor the left side

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