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anonymous

  • one year ago

um i dont understand this part...when finding the max/min.vertex etc. P (x) = 16x - x2 P (x) = -x2 + 16x = -(x2 - 16x + 64 - 64) P (x) = -(x - 8)2 + 64

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  1. anonymous
    • one year ago
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    okay

  2. anonymous
    • one year ago
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    How did this = -(x2 - 16x + 64 - 64) turn to this P (x) = -(x - 8)2 + 64

  3. anonymous
    • one year ago
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    thats the only part i dont get

  4. anonymous
    • one year ago
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    you distribute the 2....

  5. anonymous
    • one year ago
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    after the perenthesis

  6. anonymous
    • one year ago
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    hm?

  7. anonymous
    • one year ago
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    P (x) = -(x - 8)2 + 64 ^------this two, you multiply the two *s what ever is inside the ()

  8. anonymous
    • one year ago
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    thats x^2 - 16... but what about the 64?

  9. anonymous
    • one year ago
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    64 minus 64 is 0

  10. anonymous
    • one year ago
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    :(

  11. anonymous
    • one year ago
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    P (x) = -(x - 8)2 + 64, i believe this step is suppose to come before the first equation in the sentence...

  12. anonymous
    • one year ago
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    it would still be -(x - 8)^2 +0

  13. anonymous
    • one year ago
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    help

  14. perl
    • one year ago
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    Do you agree that 16x - x^2 = -(x-8)^2 + 64?

  15. anonymous
    • one year ago
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    no

  16. perl
    • one year ago
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    How about this step, does this make sense $$ \Large { -x^2 + 16x = -(x^2 - 16x + 64 - 64) }$$

  17. anonymous
    • one year ago
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    uhum i guess so

  18. anonymous
    • one year ago
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    well no actually

  19. perl
    • one year ago
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    Let me add a few steps \[ \Large { -x^2 + 16x \\ = -x^2 + 16x + 0 \\ = -(x^2 - 16x + 0) \\ = -(x^2 - 16x + 64 - 64) }\]

  20. anonymous
    • one year ago
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    hm

  21. anonymous
    • one year ago
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    why did u change the + to a minus

  22. perl
    • one year ago
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    we factor out the negative

  23. anonymous
    • one year ago
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    ok?

  24. perl
    • one year ago
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    \[ \Large { -x^2 + 16x \\ = -x^2 + 16x \\ = -(x^2 - 16x ) \\ = -(x^2 - 16 x + 0) \\ = -(x^2 - 16x + 64 - 64) }\]

  25. anonymous
    • one year ago
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    oh ok i see

  26. anonymous
    • one year ago
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    but how does it become a (x +8) ^2 + 64

  27. perl
    • one year ago
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    ok so far you agree with the steps above ?

  28. anonymous
    • one year ago
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    yes

  29. perl
    • one year ago
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    \[ \Large { -x^2 + 16x \\ = -x^2 + 16x \\ = -(x^2 - 16x ) \\ = -(x^2 - 16 x + 0) \\ = -(x^2 - 16x + 64 - 64) \\ = - (x^2 -16x +64) - (-64) }\]

  30. perl
    • one year ago
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    I distributed the negative on the last term -64

  31. anonymous
    • one year ago
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    ok i see but how does it become a (x +8) ^2 + 64

  32. perl
    • one year ago
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    \[ \Large { -x^2 + 16x \\ = -x^2 + 16x \\ = -(x^2 - 16x ) \\ = -(x^2 - 16 x + 0) \\ = -(x^2 - 16x + 64 - 64) \\ = - (x^2 -16x +64) - (-64) \\ = - (x^2 -16x +64) + 64 \\ = - (x-8)^2 +64 }\]

  33. perl
    • one year ago
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    the reason why we added 64 in the first place was to complete the square

  34. anonymous
    • one year ago
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    sorry i couldnt help :(

  35. perl
    • one year ago
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    To complete the square we use the identity \[ \Large x^2 + bx + \left( b/2 \right)^2 = (x+ b/2)^2 \]Here we have \(\large b=-16\) \[ \Large x^2 + (\text{-}16)x + \left( \text{-}16/2 \right)^2 = (x+ (\text{-}16)/2)^2 \]We can simplify \[ \Large x^2 -16x + \left( \text{-}8 \right)^2 = \left( x+ (\text{-}8) \right)^2 \]\[ \Large x^2 -16x + 64 = \left( x - 8 \right)^2 \]

  36. perl
    • one year ago
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    Can you verify that this is always true. $$ \Large x^2 + bx + \left( b/2 \right)^2 = (x+ b/2)^2 $$ Start from the right side and expand

  37. anonymous
    • one year ago
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    @perl oh ok

  38. anonymous
    • one year ago
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    i mean i still dont understand this =−(x2−16x+64−64)=−(x2−16x+64)−(−64)=−(x2−16x+64)+64=−(x−8)2+64

  39. perl
    • one year ago
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    x^2 -16x + 64 = (x-8)^2 so we substituted

  40. anonymous
    • one year ago
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    why do u only multiply the - by 64 nd not all the numbers?

  41. perl
    • one year ago
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    The expressions in red are exactly equal $$\Large −\color{red}{(x^2−16x+64)}+64=−\color{red}{(x−8)^2}+64$$

  42. perl
    • one year ago
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    \[ \Large { -x^2 + 16x \\ = -x^2 + 16x \\ = -(x^2 - 16x ) \\ = -(x^2 - 16 x + 0) \\ = -(x^2 - 16x + 64 - 64) \\ = - (x^2 -16x +64) - (-64) \\ = - \color{red}{(x^2 -16x +64)} + 64 \\ = - \color{red}{(x-8)^2} +64 }\]

  43. anonymous
    • one year ago
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    ?, but x squared = x^2 and 8 squared equals 64, so it would be x^2 - 64... and why do u only subtact the -64

  44. anonymous
    • one year ago
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    \[-(x^2 - 16x + 64) (+64)..(why single out)( -64)\]

  45. anonymous
    • one year ago
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    and... is there a method to turn this\[-(x^2 - 16x + 64)\] into the one u had said, if so ive prob forgot it

  46. anonymous
    • one year ago
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    @perl

  47. perl
    • one year ago
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    The goal here is to find the 'vertex form' of the quadratic function $$ \Large y = a(x-h)^2 + k $$

  48. perl
    • one year ago
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    The vertex form tells us the max/min is at (h,k)

  49. anonymous
    • one year ago
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    this is what they did... The sum of two numbers is 16. Let x and y be the two numbers. x + y = 16 -----------> (1) y = 16 - x P (x) is a product function. P (x) = xy is maximum. P (x) = x(16 - x) P (x) = 16x - x2 P (x) = -x2 + 16x = -(x2 - 16x + 64 - 64) P (x) = -(x - 8)2 + 64 The values of a is negative. So, the function is maximum. The vertex of the function is (8, 64) and the maximum value is 64.

  50. anonymous
    • one year ago
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    and i dont understand how this = -(x2 - 16x + 64 - 64) turned ito this P (x) = -(x - 8)2 + 64

  51. perl
    • one year ago
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    did you follow the steps above

  52. anonymous
    • one year ago
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    why do u multiply the negative by the -64 but not the other numbers, why are the other units in perenthesis and -64 isnt?

  53. anonymous
    • one year ago
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    hello?

  54. anonymous
    • one year ago
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    @perl

  55. perl
    • one year ago
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    I will label the steps \[ \Large { \\ 1) ~~-x^2 + 16x \\ 2) ~~-(x^2 - 16x ) \\ 3) ~~~-(x^2 - 16 x + 0) \\ 4)~~ -(x^2 - 16x + 64 - 64) \\ 5)~~ - (x^2 -16x +64) - (-64) \\ 6)~~ - \color{red}{(x^2 -16x +64)} + 64 \\ 7)~~ - \color{red}{(x-8)^2} +64 }\]

  56. perl
    • one year ago
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    which step is unclear , let me know :)

  57. anonymous
    • one year ago
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    ive told u 3 times what step is unclear

  58. anonymous
    • one year ago
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    why do u multiply the negative by the -64 but not the other numbers, why are the other units in perenthesis and -64 isnt?

  59. perl
    • one year ago
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    in order to complete the square

  60. perl
    • one year ago
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    We can wait to multiply by -64 if you like.

  61. anonymous
    • one year ago
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    o,o...what part dont u understand? why arent u answering my question....? i will figure it out by myself

  62. perl
    • one year ago
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    Here is alternative steps. \[ \Large { -x^2 + 16x \\ = -(x^2 - 16x ) \\ = -(x^2 - 16 x + 0) \\ = -(x^2 - 16x + 64 - 64) \\ = -( (x^2 - 16x + 64) - 64) \\ = - ((x-8)^2 -64) \\ = -(x-8)^2-(-64) \\ = -(x-8)^2+ 64 }\]

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