WAEC Further Mathematics 2015 Paper

Concept:For a quadratic expression with a positive coefficient of $x^2$, the turning point is a minimum.This minimum occurs where the derivative $\frac{dy}{dx} = 0$.Explanation:Given $y = x^{2} + 6x - 12$.Differentiate with respect to $x$:$\frac{dy}{dx} = 2x + 6$.Set $\frac{dy}{dx} = 0$ to find the critical point:$2x + 6 = 0 \Rightarrow x = -3$.Since the coefficient of $x^2$ is positive, $x = -3$ gives the minimum.Substitute $x = -3$ into $y$:$y = (-3)^{2} + 6(-3) - 12 = 9 - 18 - 12 = -21$.Therefore, the minimum value of $y$ is $-21$.Answer:Minimum value $= -21$Correct option: A.