Find the gradient of the tangent and the normal to the curve f(x)= 4x^3 - 7x - 10 at the point (2, 8)

y = 4x³ - 7x -10The gradient of the function at any point can be found using its derivative:dy/dx = 12x² - 7The gradient of the function, m₁, at (2,8) is equal to the gradient of the tangent at that point:m₁ = 12(2)² - 7 = 48 - 7 = 41Since the tangent and normal to a given point are perpendicular, their respective gradients form the equation below:m₁m₂ = -1, where m₂ is the gradient of the normal=> m₂ = -1/m₁ = -1/41