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    At which points on the curve y = x^3 -5x^2 is the anger parallel to the line y+7x =-21.

    Gradient of line is 7
    Therefore 3x^2 -10x -7

    How do you factorise this?
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    (Original post by Custardcream000)
    At which points on the curve y = x^3 -5x^2 is the anger parallel to the line y+7x =-21.

    Gradient of line is 7
    Therefore 3x^2 -10x -7

    How do you factorise this?
    Equation of a straight line y=mx +c where m is your gradient, your equation is y + 7x = -21 \iff y = -7x + 21 are you sure the gradient is +7?
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    (Original post by Custardcream000)
    At which points on the curve y = x^3 -5x^2 is the anger parallel to the line y+7x =-21.

    Gradient of line is 7
    Therefore 3x^2 -10x -7

    How do you factorise this?
    that's not correct
    the equation of a straight line is usually in the form y=mx+c where m is the gradient, so y=-7x-21, the gradient is -7
    so the quadratic you want to solve is actually 3x^2-10x+7=0 which is easy enough to solve.
 
 
 
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