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    Hey I really need help with this because I can never get it right

    Express x2 + 6x + 5 in the form (x+a)2 + b
    Write down the co-ordinates of the minimum point for the graph x2 + 6x + 5

    I got (x+3)2 + 14 but apparently b should be -4 so why the negative square of 3 rather than the positive?
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    (Original post by Dowel)
    Hey I really need help with this because I can never get it right

    Express x2 + 6x + 5 in the form (x+a)2 + b
    Write down the co-ordinates of the minimum point for the graph x2 + 6x + 5

    I got (x+3)2 + 14 but apparently b should be -4 so why the negative square of 3 rather than the positive?
    You did this:

    (X+3)^2+9 + 5

    But if you expand this out do you get the original expression?



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    x^2+6x+5 goes to (x+3)^2-(3)^2+5, which is (x+3)^2-4. a=3, b=5. Differentiate to get 2x +6, as it is a minimum set the differential to equal 0 thus the gradient at this point on the curve =-3. Then sub -3 into the original equation to get -4 so the minimum co-ordinate is (-3,-4). When completing the square remember it is (x+a/2)^2-(a/2)^2+c, not that we only square the value the negative applies to the value after it is squared (brackets first-BODMAS). Hope that helps all the best for tomorrow. ; )
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    (Original post by kingaaran)
    You did this:

    (X+3)^2+9 + 5

    But if you expand this out do you get the original expression?



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    so you get x^2 + 6x + 23 with my one and you get the right answer with -9 rather than plus 9, so are you just putting the number that fits the completed square to get the original equation?
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    (Original post by Dowel)
    so you get x^2 + 6x + 23 with my one and you get the right answer with -9 rather than plus 9, so are you just putting the number that fits the completed square to get the original equation?
    Yeah. Completing the square is all about taking a quadratic and writing it in a different way, without changing it. You see when you have your (x+3)^2, you get x^2+6x and then a +9. This is not what you had in your initial expression, so you must take it away
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    (Original post by kingaaran)
    Yeah. Completing the square is all about taking a quadratic and writing it in a different way, without changing it. You see when you have your (x+3)^2, you get x^2+6x and then a +9. This is not what you had in your initial expression, so you must take it away
    Thank you so much
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    (Original post by Maz A)
    x^2+6x+5 goes to (x+3)^2-(3)^2+5, which is (x+3)^2-4. a=3, b=5. Differentiate to get 2x +6, as it is a minimum set the differential to equal 0 thus the gradient at this point on the curve =-3. Then sub -3 into the original equation to get -4 so the minimum co-ordinate is (-3,-4). When completing the square remember it is (x+a/2)^2-(a/2)^2+c, not that we only square the value the negative applies to the value after it is squared (brackets first-BODMAS). Hope that helps all the best for tomorrow. ; )
    Thank you for that as well , is it a coincidence that -3,-4 is so similar to (x+3)^2 -4?
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    (Original post by kingaaran)
    Yeah. Completing the square is all about taking a quadratic and writing it in a different way, without changing it. You see when you have your (x+3)^2, you get x^2+6x and then a +9. This is not what you had in your initial expression, so you must take it away
    When you complete the square and get the answer in this format e.g. : (x+3)^2 - 4,
    how do you know what the coordinates for the minimum and maximum points are?
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    (Original post by SANTR)
    When you complete the square and get the answer in this format e.g. : (x+3)^2 - 4,
    how do you know what the coordinates for the minimum and maximum points are?
    Say you have the function y=(x+3)^2 - 4.

    The minimum value of the x-coordinate will be whichever value of x gives us the lowest value of y.

    (x+3)^2 can never be negative and so the smallest value it can take is 0. This will happen when x+3=0, which will be when x = -3.

    To find the y coordinate, we can then just substitute that in: y=(-3+3)^2 -4 = -4. Hence the minimum point is (-3,-4).
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    (Original post by kingaaran)
    Say you have the function y=(x+3)^2 - 4.

    The minimum value of the x-coordinate will be whichever value of x gives us the lowest value of y.

    (x+3)^2 can never be negative and so the smallest value it can take is 0. This will happen when x+3=0, which will be when x = -3.

    To find the y coordinate, we can then just substitute that in: y=(-3+3)^2 -4 = -4. Hence the minimum point is (-3,-4).
    Also, how would you work out the maximum point?
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    (Original post by Dowel)
    Hey I really need help with this because I can never get it right

    Express x2 + 6x + 5 in the form (x+a)2 + b
    Write down the co-ordinates of the minimum point for the graph x2 + 6x + 5

    I got (x+3)2 + 14 but apparently b should be -4 so why the negative square of 3 rather than the positive?

    Dude, the answer is in the form (x+b)2 + c
    As a rule of thumb, you subtract b2 to give you the correct c value.

    E.g x2 + 6x + 5

    Step 1: (x+3)2 -(32) +5

    Step 2: (x+3)2 -9 +5 (Notice how I subtract the square of b)

    Step 3: (x+3)2 -4 (Simplify c for last step)

    KEEP PRACTICING!
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    (Original post by SANTR)
    Also, how would you work out the maximum point?
    Similar method, you look at how to maximise the function, i.e. y = 4-(x+3)^2. You want (x+3)^2 to be as small as possible, its smallest value is 0 and this happened when x+3=0, x=-3. You then sub this in to find y
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    (Original post by SANTR)
    Also, how would you work out the maximum point?
    The equation of the graph is a positive parable (x^2). Therefore it doesn't have a maximum only a minimum (curve is at bottom of graph), therefore you can get one value for the co-ordinate of the minimum. You would use the same principle if it was a negative quadratic set the differential to equal 0 then work out the co-ordinate for the turning point. It would be the same with a cubic graph but you would get two values both for a maximum/minimum which by drawing will help to work out which is which.
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    This is quite useful to explain completing the square, especially if the coefficient of x^2 isn't 1

    http://www.mathcentre.ac.uk/resource...re2-2009-1.pdf
 
 
 
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