f x x x3 43 2= - -(a) Find the gradient of the graph of ( )fy x= where x 1= .

To find the gradient of the graph of the function f(x) = -x^3 + 43x - 2 at the point where x = 1, we need to find the derivative of the function first.

The derivative of a function gives us the slope of the tangent line at any point on the graph of the function.

The derivative of f(x) = -x^3 + 43x - 2 is found using the power rule for differentiation, which states that the derivative of x^n is n*x^(n-1).

So, the derivative of -x^3 is -3x^2, the derivative of 43x is 43, and the derivative of -2 is 0.

Therefore, the derivative of f(x) = -x^3 + 43x - 2 is f'(x) = -3x^2 + 43.

To find the gradient of the graph at the point where x = 1, we substitute x = 1 into the derivative:

f'(1) = -3(1)^2 + 43 = -3 + 43 = 40.

So, the gradient of the graph of the function at the point where x = 1 is 40.