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    ok, my teacher was going on about how to do this by putting the equation in complete the square form and then using that 2 find out the vertex of the graph but i've kinda 4gotten how 2 do this and need some help. i'm not sure if this will come up or not but it'll probz be the last question. so how do u find out the vertex of a quadratic graph is basically wat i need an answer to. thanks
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    (Original post by urban_flavaz)
    ok, my teacher was going on about how to do this by putting the equation in complete the square form and then using that 2 find out the vertex of the graph but i've kinda 4gotten how 2 do this and need some help. i'm not sure if this will come up or not but it'll probz be the last question. so how do u find out the vertex of a quadratic graph is basically wat i need an answer to. thanks
    by vertex wat do u mean?
    using complete the square to give the poistive and neg points of the graph?
    Its jsut lyk factorisin..what dont u get completing the square?
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    If you could give us an example of some sorts it would be better.
    Vertex could mean the points where the line intercepts the y axis??
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    (Original post by jamal1425)
    If you could give us an example of some sorts it would be better.
    Vertex could mean the points where the line intercepts the y axis??
    say for eg u have a question saying:

    "what are the minimum/max points on the graph 2x^2 + 8x - 5
    you would do (x+4)(x+4) which gives you x^2 + 8x + 16 but we want -5 so we would take off 21 thus giving:
    (x+4)^2 -21
    this then makes
    x= +/- the square root of -21 -4
    althought i just thought the question up you get the method: as im sure you cant get soln's for minus roots! LOL but do u understand i am not so gud at maths so sorry if its all rong!
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    (Original post by gemma.....)
    say for eg u have a question saying:

    "what are the minimum/max points on the graph 2x^2 + 8x - 5
    you would do (x+4)(x+4) which gives you x^2 + 8x + 16 but we want -5 so we would take off 21 thus giving:
    (x+4)^2 -21
    this then makes
    x= +/- the square root of -21 -4
    althought i just thought the question up you get the method: as im sure you cant get soln's for minus roots! LOL but do u understand i am not so gud at maths so sorry if its all rong!
    Yes I understand. Completing the square is straight forward, but I think you can get solutions for minus roots?!?! Im not sure. But thats the answer to 'urbanflavaz' question, gemmas response.
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    (Original post by urban_flavaz)
    ok, my teacher was going on about how to do this by putting the equation in complete the square form and then using that 2 find out the vertex of the graph but i've kinda 4gotten how 2 do this and need some help. i'm not sure if this will come up or not but it'll probz be the last question. so how do u find out the vertex of a quadratic graph is basically wat i need an answer to. thanks
    Complete the square of a quadratic.

    Using:

    y = f(x)

    When: y = f(x + a) , graph shifted a places to the left.

    When: y = f(x) + b , graph shifted b places up.

    So: An example:

    y = x² + 4x - 20 =0

    y = (x + 2)² - 4 - 20 = 0

    y = (x + 2)² - 24 = 0

    Therefore:

    The graph is y = x² to the left two places and down 24 places.

    If you take the vertex of the parlabra (sp) to be (0,0) initially then it is (-2, -24) now.

    I hope that helps.
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    Minus Square Roots are Imaginary Numbers.
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    Can anyone confirm / deconfirm what I have said above?
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    You complete the square to find the minimum point of a quadratic. So you will be finding (a) the minimum value for ax² + bx + c and (b) the value of x which gives this minimum value. If you draw a graph of y = ax² + bx + x you will therefore be finding the minimum value for y and the corresponding value for x which gives this. So you end up with two coordinates.

    Completing the square gives:

    ax² + bx + c
    a(x² + (b/a)x + (c/a))
    a[ (x + (b/2a))² - (b²/4a²) + (c/a) ]
    a[ (x + (b/2a))² + (4ac - b²)/4a² ]
    a[ (x + (b/2a))² ] + (4ac - b²)/4a

    Now put the part inside the brackets equal to 0:

    x + (b/2a) = 0
    x = (-b/2a)

    To get the value of x which given a min. value for (ax² + bx + c), then substitute this back into the equation to find what it gives:

    a(-b/2a)² + b(-b/2a) + c
    a(b²/4a²) - (b²/2a) + c
    ab²/4a² - 2ab²/4a² + 4a²c/4a²
    (ab² - 2ab² + 4a²c)/4a²
    (4a²c - ab²)/4a²
    (4ac - b²)/4a

    Recognise this from anywhere? Have a look up if you don't - it's the part outside the square brackets. So we can conclude that by completing the square and equating the part inside the brackers containing x to zero, and finding a value for x (which gives a minimum), we don't need to substitute the whole thing back into the equation - just look at the part outside and that's your minimum point.

    This might seem complicated for GCSE, but try it with numbers following it through - the a, b and c work just as well (but look more complicated), and mean that I can prove it for all values of these constants, and not just 4x² - 2x + 1.
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    (Original post by mik1a)
    You complete the square to find the minimum point of a quadratic. So you will be finding (a) the minimum value for ax² + bx + c and (b) the value of x which gives this minimum value. If you draw a graph of y = ax² + bx + x you will therefore be finding the minimum value for y and the corresponding value for x which gives this. So you end up with two coordinates.

    Completing the square gives:

    ax² + bx + c
    a(x² + (b/a)x + (c/a))
    a[ (x + (b/2a))² - (b²/4a²) + (c/a) ]
    a[ (x + (b/2a))² + (4ac - b²)/4a² ]
    a[ (x + (b/2a))² ] + (4ac - b²)/4a

    Now put the part inside the brackets equal to 0:

    x + (b/2a) = 0
    x = (-b/2a)

    To get the value of x which given a min. value for (ax² + bx + c), then substitute this back into the equation to find what it gives:

    a(-b/2a)² + b(-b/2a) + c
    a(b²/4a²) - (b²/2a) + c
    ab²/4a² - 2ab²/4a² + 4a²c/4a²
    (ab² - 2ab² + 4a²c)/4a²
    (4a²c - ab²)/4a²
    (4ac - b²)/4a

    Recognise this from anywhere? Have a look up if you don't - it's the part outside the square brackets. So we can conclude that by completing the square and equating the part inside the brackers containing x to zero, and finding a value for x (which gives a minimum), we don't need to substitute the whole thing back into the equation - just look at the part outside and that's your minimum point.

    This might seem complicated for GCSE, but try it with numbers following it through - the a, b and c work just as well (but look more complicated), and mean that I can prove it for all values of these constants, and not just 4x² - 2x + 1.
    I thought that was quite neat.. didn't expect it to work so well!
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    (Original post by mik1a)
    I thought that was quite neat.. didn't expect it to work so well!
    that was very good - u r a genius, give me ur brain for the maths exam on tuesday i dont grasp maths! :rolleyes:
 
 
 
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