WAEC Further Mathematics 2015 Paper

Concept:The gradient function is the derivative of the curve's equation.So, integrate the gradient and use the given point to find the constant of integration.Explanation:Let the equation of the curve be $y$.The gradient is given as $\frac{dy}{dx} = 3x^2 - 4x$.Integrate both sides with respect to $x$.$y = \int (3x^2 - 4x) \, dx$$y = \frac{3x^{2+1}}{2+1} - \frac{4x^{1+1}}{1+1} + c$$y = x^3 - 2x^2 + c$The curve passes through the point $(-2, 0)$.Substitute $x = -2$ and $y = 0$ into the equation.$0 = (-2)^3 - 2(-2)^2 + c$$0 = -8 - 8 + c$$0 = -16 + c$$c = 16$Substitute $c = 16$ back into the integrated equation.$y = x^3 - 2x^2 + 16$Answer:The equation of the curve is $y = x^3 - 2x^2 + 16$.Therefore, the correct option is D.