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anonymous

  • one year ago

Using the completing-the-square method, find the vertex of the function f(x) = –2x2 + 12x + 5 and indicate whether it is a minimum or a maximum and at what point.

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  1. anonymous
    • one year ago
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    it is a maximum because the leading coefficient|dw:1434844163162:dw| is negative so the parabola opens down

  2. campbell_st
    • one year ago
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    |dw:1434843957618:dw|

  3. anonymous
    • one year ago
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    Ok:)

  4. anonymous
    • one year ago
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    \[ f(x) = –2x^2 + 12x + 5 \]to find the vertex, the first coordinate of the vertex is always \[-\frac{b}{2a}\] which in your case is \[-\frac{12}{2\times (-2)}\]

  5. anonymous
    • one year ago
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    the second coordinate of the vertex is what you get when you replace \(x\) by the first coordinate

  6. anonymous
    • one year ago
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    3

  7. dmndlife24
    • one year ago
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    Graph opens downward so it would be a maximum point Vertex formula|dw:1434844392512:dw|

  8. anonymous
    • one year ago
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    yeah that is the first coordinate the second coordinate if \(f(3)\)

  9. anonymous
    • one year ago
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    Now what do I do?

  10. anonymous
    • one year ago
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    have a beer you are done

  11. campbell_st
    • one year ago
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    well completing the square \[f(x) = (-2x^2 + 12x) + 5\] factor out -2 \[f(x) = -2(x^2 - 6x) + 5\] now complete the square in x

  12. anonymous
    • one year ago
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    So is it maximum or minimum??

  13. anonymous
    • one year ago
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    , find the vertex of the function \[(3,23)\] it is a max have a nice day

  14. anonymous
    • one year ago
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    Solve x2 + 12x + 6 = 0 using the completing-the-square method.

  15. anonymous
    • one year ago
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    Thanks friend. I also need help with that one

  16. anonymous
    • one year ago
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    |dw:1434844578090:dw|

  17. campbell_st
    • one year ago
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    so inside the brackets you need to add (-6/2)^2 so the function is \[f(x) = -2(x^2 - 6x + 9) + 5 -18\] which becomes f(x) = -2(x -3)^2 - 13 now you have completed the square have the equation in vertex form and can answer the question using the required method

  18. campbell_st
    • one year ago
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    |dw:1434844393236:dw|

  19. anonymous
    • one year ago
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    \[x^2 + 12x + 6 = 0 \\ x^2+12x=-6\] is a start

  20. anonymous
    • one year ago
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    hahahah you're funny @satellite73 !!!

  21. anonymous
    • one year ago
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    I see, I see

  22. anonymous
    • one year ago
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    |dw:1434844713212:dw|

  23. anonymous
    • one year ago
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    Ok answer the other question please because I'm confused

  24. campbell_st
    • one year ago
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    |dw:1434844498447:dw|

  25. anonymous
    • one year ago
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    \[x^2+12x=-6\]takes two steps to complete the square what is half of 12?

  26. anonymous
    • one year ago
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    6

  27. anonymous
    • one year ago
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    right, and what is \(6^2\_?

  28. anonymous
    • one year ago
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    oops what is \(6^2\)?

  29. anonymous
    • one year ago
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    36

  30. anonymous
    • one year ago
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    so go right from \[x^2+12x=-6\] to \[(x+6)^2=-6+36\] or \[(x+6)^2=30\]

  31. anonymous
    • one year ago
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    then take the square root of both sides, don't forget the \(\pm\) and get \[x+6=\pm\sqrt{30}\]

  32. anonymous
    • one year ago
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    subtract 6 and you are done

  33. anonymous
    • one year ago
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    so is it 6 plus or negative radical 30?

  34. anonymous
    • one year ago
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    no

  35. anonymous
    • one year ago
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    SUBTRACT 6 to solve for \(x\)

  36. anonymous
    • one year ago
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    |dw:1434844993983:dw|

  37. anonymous
    • one year ago
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    nope

  38. anonymous
    • one year ago
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    |dw:1434845032667:dw|

  39. anonymous
    • one year ago
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    |dw:1434845026665:dw|

  40. anonymous
    • one year ago
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    yeah that one

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