How to find coordinates from the gradient?

I know how to do the differentiation but I don't know what to do next.





Eg.





Given the curve y= x^3 - 3x^2 - 9x + 11, find dy/dx. Hence obtain


a) the x-coordinates of the points where the gradient is 15.


b) the coordinates of the points where the gradient is zero.How to find coordinates from the gradient?
dy/dx = 3x^2 - 6x - 9





a) Gradient = 15 ==%26gt; dy/dx = 15. And, we have :


3x^2 - 6x - 9 = 15


3x^2 - 6x - 24 = 0


x^2 - 2x - 8 = 0


(x-4)(x+2) = 0


x=4 or x=-2


Hence, the x-coordinates of the points where the gradient is 15 are


x=4 and x=-2 (answer)





b) Gradient = 0 ==%26gt; dy/dx = 0


we have 3x^2 - 6x - 9 = 0


x^2 - 2x - 3 = 0


(x-3)(x+1) = 0


x=3 or x=-1


when x=3, y = (3)^3 - 3(3)^2 - 9(3) + 11 = -16


when x=-1, y = (-1)^3 - 3(-1)^2 - 9(-1) + 11 = 16


Hence, the coordinates of the points where gradient is zero are


(3, -16) and (-1, 16) (answer)





Note :


Please take note that ';x-coordinates'; and ';coordinates'; are different, as follows :


x-coordinates refers to the x-values


and coordinates refers to both the x-value and y-value


For example, let us refer to the point Q (2, 15).


Then, the x-coordinate of Q is 2


and, the coordinate of Q is (2, 15)How to find coordinates from the gradient?
There is a wording problem here: I think the teacher is using ';gradient'; and ';derivative'; as if they are interchangeable, when in fact they are slightly different (the gradient is a vector quantity, which isn't relevant for a problem like this).





Take the derivative, set it equal to the specified value, and solve for x. You may or may not need to use the quadratic formula. That's all!
y= x^3 - 3x^2 - 9x + 11





dy/dx = 3x^2 - 6x - 9





15 = 3(x)^2 - 6x -9


0 = 3(x)^2 - 6x - 24


0 = x^2 -2x - 8


0 = (x-4)(x+2) x= {-2, 4}


slope is 15 at x= {-2, 4}





0 = 3x^2 - 6x -9


0 = x^2 - 2x -3


0 = (x -3)(x+1) x = {-1,3}


slope is 0 at x= {-2, 4}
there is a problem there and it has to be sorted out