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Simple Quadratic Factor question

  1. Jan 6, 2006 #1
    okay i need to solve the quadratic y = -x^2 + 5x + 6 and need to find the coordinates of the vertex of the curve

    by factorising

    y = -(x^2 - 5x - 5)
    y = -(x + 1)(x - 6)

    so when y = 0, x = -1 and x = 6 (that parts simple)
    .........................................................................
    At the vertex x = (-1 + 6) / 2 = 5/2 (as the curve is symetrical)

    so as y = -x^2 + 5x + 6

    y = -5/2^2 + 5(-5/2) + 6
    y = 25/4 + 50/4 + 24/4
    y = 99/4

    (this is the answer i got which doesn't look right, and answer book gives a different answer)

    can u tell me where i went wrong? thanx
     
  2. jcsd
  3. Jan 6, 2006 #2
    Sorry, this sites changed a bit since i last came on, i've posted it in the wrong section
     
  4. Jan 6, 2006 #3

    mathman

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    First line should read -(5/2)^2+5(5/2)+6
    Second line should read -25/4+50/4+24/4
    Result y=49/4
    I suggest you carefully check your arithmetic before posting.
     
  5. Jan 6, 2006 #4
    thanx i got that answer too, wasn't sure which way to work it out, the negative sign threw me a bit, the book is wrong then as i thought, they have 29/4 for their answer!

    thanx for ur help
     
  6. Jan 21, 2006 #5
    mathman is right & here's why:
    -5/2*5/2=-25/4 because -1*1=-1
    5 * 5/2 = 25/2 = 50/4
    6 = 24/4
    50 + 24 - 25 = 50 - 1 = 49/4
    There's your answer
     
  7. Jan 22, 2006 #6

    HallsofIvy

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    You can also do this by completing the square:
    y= -x2+ 5x+ 6= -(x2- 5x)+ 6
    = -(x2- 5x+ 25/4- 25/4)+ 6
    = -(x- 5/2)2+ 25/4+ 6
    = -(x- 5/2)2+ 25/4+ 24/4
    = -(x- 5/2)2+ 49/4

    The vertex is at (5/2, 49/4).
     
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