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
MP
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