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    q1 said, find the co-ords of the stationary point on the curve with equation y = ....
    what was the equation?
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    Does anyone remember?
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    any1?
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    (Original post by manps)
    q1 said, find the co-ords of the stationary point on the curve with equation y = ....
    what was the equation?
    You're Mr Maths, you should know!
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    2x^2 +/- something

    hmm it had stationery point (-3,18) ? or (3,18) ?
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    think stationery points we're (3,-18) so..

    y = 2x^2 + ax + c

    dy/dx = 4x + a

    when dy/dx = 0, x=3

    4x + a = 0
    12 + a = 0
    a=-12

    y=2x^2 - 12x + c

    when x=3,y=-18

    y=2(3)^2 - 12(3) + c
    -18 = 18 - 36 + C
    c=0

    y = 2x^2 - 12x sound familiar?
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    (Original post by Revelation)
    think stationery points we're (3,-18) so..

    y = 2x^2 + ax + c

    dy/dx = 4x + a

    when dy/dx = 0, x=3

    4x + a = 0
    12 + a = 0
    a=-12

    y=2x^2 - 12x + c

    when x=3,y=-18

    y=2(3)^2 - 12(3) + c
    -18 = 18 - 36 + C
    c=0

    y = 2x^2 - 12x sound familiar?
    nah ui can rmbr there being a 24 somewhere along the line of deffrentiating. i think it might hav bin whats on ur mark scheme now, manap. Im pretty sure because i forgot to work out the y co-ordinate. silly mistake.
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    (Original post by Revelation)
    think stationery points we're (3,-18) so..

    y = 2x^2 + ax + c

    dy/dx = 4x + a

    when dy/dx = 0, x=3

    4x + a = 0
    12 + a = 0
    a=-12

    y=2x^2 - 12x + c

    when x=3,y=-18

    y=2(3)^2 - 12(3) + c
    -18 = 18 - 36 + C
    c=0

    y = 2x^2 - 12x sound familiar?
    All I can remember is 12x and a number 2. So that looks right...
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    yes i think that is right. i remember 3, -18 as well.
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    (Original post by manps)
    q1 said, find the co-ords of the stationary point on the curve with equation y = ....
    what was the equation?
    y=12x² - 12x
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    (Original post by Revelation)
    think stationery points we're (3,-18) so..

    y = 2x^2 + ax + c

    dy/dx = 4x + a

    when dy/dx = 0, x=3

    4x + a = 0
    12 + a = 0
    a=-12

    y=2x^2 - 12x + c

    when x=3,y=-18

    y=2(3)^2 - 12(3) + c
    -18 = 18 - 36 + C
    c=0

    y = 2x^2 - 12x sound familiar?
    thats right i think, i'm fairly confident the differential was 4x-12
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    did we have to give the equation in our answer, I thought we just had to find the stationary point cooridinates.????
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    (Original post by boxexpansion)
    did we have to give the equation in our answer, I thought we just had to find the stationary point cooridinates.????
    yes, just the coordinates were needed. ppl here we just trying to figure out the initial question which is:
    y = 2x2 - 12x
 
 
 
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