The gradient of a curve at the point (-2, 0) is \(3x^{2} – 4x\). Find the equation of the curve.
- \(y = 6x - 4\)
- \(y = 6x^{2} - 4x + 12\)
- \(y = x^{3} - 2x^{2}\)
-
\(y = x^{3} - 2x^{2} + 16\)
The correct answer is: D
Explanation
The gradient of a curve is gotten by differentiating the equation of the curve. Therefore, given the gradient, integrate to get the equation of the curve back.
\(\frac{\mathrm d y}{\mathrm d x} = 3x^{2} - 4x\)
\(y = \int {(3x^{2} - 4x)} \mathrm {d} x = \frac{3x^{2+1}}{2+1} - \frac{4x^{1+1}}{1+1} + c\)
= \(x^{3} - 2x^{2} + c \)
To find c (the constant of integration), when x = -2, y = 0
\(0 = (-2^{3}) - 2(-2^{2}) + c\)
\(0 = -8 - 8 + c \implies c = 16\)
\(\therefore y = x^{3} - 2x^{2} + 16\)