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How to find MINIMUM and MAXIMUM values when completing the square?? Watch

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    I have my maths exam tomorrow and seem to be very stuck on this part.
    Someone post a site where it explains it well.
    I know how to complete the square but am unsure about how to find the min. and max. values??

    Thanks
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    look for the Range. thats the value.
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    (Original post by Intellect.x)
    I have my maths exam tomorrow and seem to be very stuck on this part.
    Someone post a site where it explains it well.
    I know how to complete the square but am unsure about how to find the min. and max. values??

    Thanks
    Haven't done this for a while, but it is just saying the positive and negative answer. When you work out the answer, it will either be positive or negative, so that's your answer.
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    To complete the square you'll need a positive or negative number to cancel the number out and leave X.

    Say if it's like X-2, then +2 will cancel out the -2 and leave X.

    IDK if that's what you wanted or not?
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    (Original post by Intellect.x)
    I have my maths exam tomorrow and seem to be very stuck on this part.
    Someone post a site where it explains it well.
    I know how to complete the square but am unsure about how to find the min. and max. values??
    Thanks
    You will most probably have to complete the square but only occasionally do they ask you to take it any further.

    y = x^2 - 6x + 17
    complete the square
    y = (x - 3)^2 - 9 + 17
    y = (x - 3)^2 + 8

    the minimum value of this is when x - 3 = 0, i.e. x=3 since any other negative or positive value of x creates a positive number when squared.
    The minimum value when x=3 is obviously 8.
    Whenever you complete the square you can immediately 'see' the coordinates of the minimum point.

    Never seen a GCSE question where it was -x^2 which would mean that you were looking at a maximum.

    You can also use complete the square to 'solve' when it doesn't factorise
    y = x^2 - 4x + 1
    y = (x - 2)^2 - 3

    (x - 2)^2 - 3 = 0
    (x - 2)^2 = 3
    square root both sides
    x - 2 = + or - sqrt 3
    x = 2 + sqrt 3 or 2 - sqrt 3
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    Are you a-level/gcse? If you're doing a-level second order differentiate the curve----> if you get a positive value it is a minimum point. If it is a negative value it is a maximum point.
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    (Original post by 130ss)
    look for the Range. thats the value.
    This is a very interesting point of view. Your academic and intellectual knowledge is clearly above average. However, I cannot help but think your explanation is lacking in depth and detail. Considering the context and contextual analysis of the topic, perhaps you could take the time to write a slightly more focused and critical response. The use of the noun 'range', could suggest that you are trying to put the reader into a state of insecurity and confusion, by the lack of a definite end point. I hope you take my constructive criticism beneficially and it helps you with your further studies. Thank you for your time, kind regards, LBBJ
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    once you have

    y = ( x + a )2 + b

    the minimum ( vertex ) is at ( -a, b )

    so the y values are ≥ b

    if it is the other way up then

    y = -( x + a )2 + b

    the maximum ( vertex ) is at ( -a, b )

    so the y values are ≤ b
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    (Original post by LBBJ)
    This is a very interesting point of view. Your academic and intellectual knowledge is clearly above average. However, I cannot help but think your explanation is lacking in depth and detail. Considering the context and contextual analysis of the topic, perhaps you could take the time to write a slightly more focused and critical response. The use of the noun 'range', could suggest that you are trying to put the reader into a state of insecurity and confusion, by the lack of a definite end point. I hope you take my constructive criticism beneficially and it helps you with your further studies. Thank you for your time, kind regards, LBBJ
    (Original post by the bear)
    once you have

    y = ( x + a )2 + b

    the minimum ( vertex ) is at ( -a, b )

    so the y values are ≥ b

    if it is the other way up then

    y = -( x + a )2 + b

    the maximum ( vertex ) is at ( -a, b )

    so the y values are ≤ b
    Helpful, but you're both about 6 years late.
 
 
 
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