The gradient of a curve at the point (-2, 0) is $3x^{2} - 4x$ . Find the equation of the curve.

$y = 6x - 4$

$y = 6x^{2} - 4x + 12$

$y = x^{3} - 2x^{2}$

$y = x^{3} - 2x^{2} + 16$

Correct answer is D

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

$0 = (-2^{3}) - 2(-2^{2}) + c$

$0 = -8 - 8 + c \implies c = 16$

$\therefore y = x^{3} - 2x^{2} + 16$

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