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y = 12 x - 16 The gradient of the tangent at any particular point is given by the derivative. f(x) = x^3 :. f'(x) = 3x^2 At (2,8), x=2 => f'(2) = (3)2^2 = 12 So the tangent passes through (2,8) and has gradient m=12 So, Using y - y_1 = m(x - x_1 ) , the tangent equation is y - 8 = 12( x - 2 ) :. y - 8 = 12 x - 24 :. y = 12 x - 16.