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    Find the two points on the curve y=2x^3 - 5x^2 +9x -1 whos gradient is 13

    I have no idea how to do this I diffrntiated to get 6x^2 -10x +9 but dunno what to do then
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    dy/dx = 13....solve it.
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    6x^2-10x+9=13
    Solve to find 2 x values
    Sub both x's into original equation to find corresponding y values.
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    (Original post by jit987)
    6x^2-10x+9=13
    Solve to find 2 x values
    Sub both x's into original equation to find corresponding y values.
    Ah right thanks
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    How would i go about this would i factorise it im really tired and confused now Worked it out i was being stupid sorry
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    (Original post by hazbaz)
    How would i go about this would i factorise it im really tired and confused now Worked it out i was being stupid sorry
    Bring the thirteen over to the other side such that:
    6x^2-10x+9=13

    6x^2-10x-4=0

    Then use the quadratic formula if you find that easier.
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    Or you could divide by 2:
    3x^2-5x-2=0 And see if that factorises.
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    Ye i ddi it thank you . For this one A normal to the curve y=x^2 has a gradient of 2 find where it meers the curve. How would i go about this i have only just started these so ima bit stuck
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    (Original post by hazbaz)
    Ye i ddi it thank you . For this one A normal to the curve y=x^2 has a gradient of 2 find where it meers the curve. How would i go about this i have only just started these so ima bit stuck
    if the gradient of the normal is 2, the gradient of the tangent would be -1/2
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    Ah right ye so then i find y using that which = - 1/4
 
 
 
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