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Minimum Point

  1. Oct 17, 2005 #1
    I'm a bit stuck on this question:
    the curve C has the equation y=x2 + 2x + 4
    a) Express x2 +2x + 4 in the form a(x+b)2 + c and hence the coordinates of the minimum point C.
    This is what i've done:
    x2 +2x +4 = a(x+b)2+c
    a(x+b)(x-b) + c = x2+2x+4
    a(x2 + xb+ xb +b2) + c = x2+2x+4
    ax2 + 2abx + ab2 + c = x2 +2x+4
    Therefore: a= 1, 2ab= 2, ab= 1, b= 1, ab2 + c= 4
    1+c = 4
    c= 3
    (x=0) y= 4
    y= 1(x+1)2+ 2
    ..and that's all i've got so far. Please let me know if it's all wrong and how do i go about getting the minimum point from here.
    Thanks!
     
  2. jcsd
  3. Oct 17, 2005 #2

    hotvette

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    You are almost there. Take a look at your final equation ([itex]y= 1(x+1)^2+ 2[/itex]). What value of x will make y the smallest it can possibly be?
     
    Last edited: Oct 17, 2005
  4. Oct 17, 2005 #3

    HallsofIvy

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    I have a small problem with that. If y= (x+1)2+ 2 then
    y= x2+ 2x+ 1+ 2= x2+ 2x+ 3 which doesn't appear to be what you started with!
     
  5. Oct 17, 2005 #4
    sorry, that would be me unable to read my own scribbled notes! it's meant to be a 3.
     
  6. Oct 17, 2005 #5
    Standard (and easier way) of completing the square toget a(x+b)2 + c is

    1x2 +2x +4 = 1 (x2 +2 x + 1) + 3 =
    (x + 1)2 + 3.

    Graphing (especially by hand) would also give a good idea of the minimum.

    Looking into an algebra or precalculus textbook would work too.
     
    Last edited: Oct 17, 2005
  7. Oct 18, 2005 #6

    VietDao29

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    Since you have:
    A2 ≥ 0.
    So (x + 1)2 ≥ 0.
    Adding 3 to both sides gives:
    (x + 1)2 + 3 ≥ 3.
    So what's the smallest value y can have, what x makes y smallest?
    Viet Dao,
     
  8. Oct 18, 2005 #7
    right, so..the minimum point is (-1,3)?
     
  9. Oct 18, 2005 #8

    HallsofIvy

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    Yes, that's correct. If x=-1, y= 02+ 3= 3. If x is any number other than -1, x+ 1 is not 0 so (x+1)2 is positive and (x+1)2+ 3 is greater than 3.
     
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