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.