Find the gradient to the normal of the curve \(y = x^{3} – x^{2}\) at the point where x = 2.
-
\(\frac{-1}{8}\)
- \(\frac{1}{8}\)
- \(\frac{-1}{24}\)
- \(1\)
The correct answer is: A
Explanation
Given : \(y = x^{3} - x^{2}\)
\(\frac{\mathrm d y}{\mathrm d x} = 3x^{2} - 2x\)
\(\therefore \text{The gradient of the tangent at point (x = 2)} = 3(2^{2}) - 2(2) \)
= \(12 - 4 = 8\)
Recall, the tangent and the normal are perpendicular to each other and the product of the gradients of perpendicular lines = -1.
\(\implies \text{the gradient of the normal} = \frac{-1}{8}\)