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    So say I had this:
    x^2 + 8x +3
    (x + 4)^2 -16 +3
    (x + 4)^2 -13
    Therefore, the minimum point is (-4,-13)

    What if the coefficient was a minus:
    -3x^2 + 5x + 5

    What would be the minimum points then?
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    = -3(x-5/6)^2 + 85/12
    Minimum point = (5/6, 85/12)
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    (Original post by Neevan)
    = -3(x-5/6)^2 + 85/12
    Minimum point = (5/6, 85/12)

    Thank you so much!!! So regardless of the minus, the x coordinate of the minimum point will be positive and not timesed by 3? It's basically never affected by the coefficient right...oh and the y as well for that matter?
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    (Original post by Nieveee)
    So say I had this:
    x^2 + 8x +3
    (x + 4)^2 -16 +3
    (x + 4)^2 -13
    Therefore, the minimum point is (-4,-13)

    What if the coefficient was a minus:
    -3x^2 + 5x + 5

    What would be the minimum points then?
    It won't be a minimum though will it?

    What shape is a quadratic with a negative x^2 coefficient?
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    (Original post by Muttley79)
    It won't be a minimum though will it?

    What shape is a quadratic with a negative x^2 coefficient?
    Ahhh, it'd be a maximum right, considering it's the other way round?
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    (Original post by Nieveee)
    Ahhh, it'd be a maximum right, considering it's the other way round?
    Indeed
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    sorry about that
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    (Original post by Muttley79)
    Indeed
    Oh, so would I be right in saying the maximum of this is, for example:

    -2(x-2)^2 + 18
    = (2,18)

    and this:
    4(x-3)^2 + 23
    = (3, 23) is the minimum point?
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    (Original post by Neevan)
    sorry about that
    No problem!
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    (Original post by Nieveee)
    Oh, so would I be right in saying the maximum of this is, for example:

    -2(x-2)^2 + 18
    = (2,18)

    and this:
    4(x-3)^2 + 23
    = (3, 23) is the minimum point?
    Yes - remember you can always check by differentiating and putting dy/dx =0 if you have time.
 
 
 
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