What is the equation of the normal line to the curve y = 3x^3 - 6x^2 at the point (1, 4)?

First we must find the gradient of the curve at the point (1, 4) - we do this by finding the derivative dy/dx, as we do to find any gradient. Differentiating the equation of the curve gives us that dy/dx = 9x^2 - 12x. We then substitute in the value of x, the x-coordinate 1, so that we obtain dy/dx = 9(1)^2 – 12(1) = -3. Given this, we can find the gradient of the normal by using the equation m2 = - 1/m1, so that the gradient of the curve at this point, m1, is the negative reciprocal of the gradient of the normal line, m2. Using this equation we find that m2 = -1/-3 = 1/3, and so we then find the equation of the normal line to the curve at (1, 4) by using the equation y - y1 = m(x - x1), which gives us the equation of a line given a point and a gradient. Using this equation we obtain y - 4 = 1/3 (x - 1), and so simplifying this gives us y = 1/3 (x - 11).