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    Please can someone help, ive tried to do this so many times

    1)find the x coordinate of the point on the curve y= (x^3)/3 - 3x^2 + 2/3 where the tangent is parallel to the tangent drawn at point (1,2)

    thank you
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    (Original post by slothgal)
    Please can someone help, ive tried to do this so many times

    1)find the x coordinate of the point on the curve y= (x^3)/3 - 3x^2 + 2/3 where the tangent is parallel to the tangent drawn at point (1,2)

    thank you
    The point (1,2) isn't on the curve. Is that intentional, or should it be (1,-2) ?
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    yes sorry it should be (1,-2), thanks
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    bump please help
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    (Original post by slothgal)
    bump please help
    unpick the question

    You want 2 parallel lines

    You will need their gradient

    How will you find the gradient of the tangent to the curve at (1, -2)
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    parallel lines have the same gradient, so it has the same gradient as the curve?
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    (Original post by slothgal)
    parallel lines have the same gradient, so it has the same gradient as the curve?
    yes
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    thank you, ill try to work it out now
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    sorry how can i differentiate (x^3)/3
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    (Original post by slothgal)
    sorry how can i differentiate (x^3)/3
    How would you usually differentiate ax^n
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    x^3 is 3x^2
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    (Original post by slothgal)
    x^3 is 3x^2
    x^3 \neq 3x^2

    What you're trying to say is that the first derivative of x^3 with respect to x is 3x^2.
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    (Original post by slothgal)
    x^3 is 3x^2
    yep

    So if you divide that by 3 you will get ... ...
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    X^2! Thank you, im a fool
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    (Original post by slothgal)
    X^2! Thank you, im a fool
    less of a fool if you use the quote facility so that I can see you have responded
 
 
 

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