The point at which the slope of the tangent line is -3

F(x) = 6x^2 + 3x - 6

The derivative, and thus the slope of the T. L. to the graph, is f '(x) = 12x + 3.  Set this result = to -3 and solve for x:

12x + 3 = -3  =>  12x = - 6 and x = -1/2
                                                                       6 + 6 - 24
If x = -1/2, y = f(-1/2) = 6(1/4) - 3(-1/2) - 6 =  = -6
                                                                             4

The slope of the T. L. to the graph will be -3 at the point (-1/2, -6). 

Where's the rest of the

Ibelieve the answer is c