Let f (x) = ax ^ 3 + BX + C (a is not equal to 0) be an odd function, the tangent of its image at point (1, f (1)) is perpendicular to the straight line x-6y-7 = 0, and the minimum value of derivative f '(x) is - 12 Ask for: 1) The value of a and B I want to ask what is the explanation for the lower tangent being perpendicular to the straight line x-6y-7 = 0? How can we calculate the slope of x-6y-7 = 0 to be 1 / 6? How can we get f '(1) = 3A + B = - 6? I don't quite understand these questions,

Let f (x) = ax ^ 3 + BX + C (a is not equal to 0) be an odd function, the tangent of its image at point (1, f (1)) is perpendicular to the straight line x-6y-7 = 0, and the minimum value of derivative f '(x) is - 12 Ask for: 1) The value of a and B I want to ask what is the explanation for the lower tangent being perpendicular to the straight line x-6y-7 = 0? How can we calculate the slope of x-6y-7 = 0 to be 1 / 6? How can we get f '(1) = 3A + B = - 6? I don't quite understand these questions,

If we change x-6y-7 = 0 to y = 1 / 6x-7 / 6 (like y = KX + b), the slope is naturally 1 / 6L
Secondly, the two lines are perpendicular to each other. If the slope of the line exists, the product of the two slopes is - 1, so f '(1) = 3A + B = - 6