Work out the equation of the normal to the curve y = x^3 + 2x^2 - 5 at the point where x = -2. [5 marks]

Firstly, we need to differentiate the function of the curve y = x^3 + 2x^2 - 5. Using d/dx (x^n) = nx^(n - 1), we get dy/dx = 3x^2 + 4x. To find the gradient of this curve, we then need to evaluate dy/dx at x = -2 -> dy/dx = 3(-2)^2 + 4(-2) = 4. Since we know that a normal is perpendicular to the curve, the gradient of the normal is -1/gradient of the curve -> gradient of normal = -1/4. We then need to find the y coordinate of the point where x = -2. We do this by substituting x = -2 into y = x^3 + 2x^2 - 5 -> y = (-2)^3 + 2(-2)^2 - 5 = -5. Next, we use the equation y - y1 = m(x - x1) with m = -1/4 and (x1, y1) = (-2, -5). This gives us y - - 5 = -1/4(x - - 2) -> y + 5 = (-1/4)x - 1/2 -> y = (-1/4)x - 11/2. Therefore, the equation of the normal to the curve y = x^3 + 2x^2 - 5 at the point where x = -2 is y = (-1/4)x - 11/2.