The Student Room Group

Differentiation

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+cy=mx +c where mm is your gradient, your equation is y+7x=21    y=7x+21y + 7x = -21 \iff y = -7x + 21 are you sure the gradient is +7?
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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