Find the gradient to the normal of the curve \(y = x^{3} - x^{2}\) at the point where x = 2.

Assessment: WAEC SSCE - Further Mathematics - 2016 (Objective) Subject: Further Mathematics

Question 1 Report

Find the gradient to the normal of the curve \(y = x^{3} - x^{2}\) at the point where x = 2.

Answer Details
To find the gradient to the normal, we need to find the gradient of the tangent and then take the negative reciprocal of that value. To find the gradient of the tangent, we need to differentiate the equation of the curve. \[\frac{dy}{dx} = 3x^{2} - 2x\] At the point where x = 2, the gradient of the tangent is: \[\frac{dy}{dx}\Bigr|_{x=2} = 3(2)^{2} - 2(2) = 8\] So the gradient to the normal is the negative reciprocal of 8: \[\frac{-1}{8}\] Therefore, the answer is (a) \(\frac{-1}{8}\).

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