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# Differentiation watch

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

Therefore 3x^2 -10x -7

How do you factorise this?
2. (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.

Therefore 3x^2 -10x -7

How do you factorise this?
3. (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.

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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Updated: November 11, 2015
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