WAEC Mathematics 2012 Paper

Concept:$y$ is negative where the parabola lies below the x-axis; the gradient of $y$ is its derivative $\frac{dy}{dx}$.Explanation:First find the roots by putting $y=0$:$$2x^{2}-5x-3=0$$Factorise:$$(2x+1)(x-3)=0$$This gives $x=-\frac{1}{2}$ or $x=3$.The coefficient of $x^2$ is positive, so the parabola opens upwards.Hence the curve is below the x-axis between the roots:$$-\frac{1}{2} < x < 3$$For the gradient, differentiate the function:$$\frac{dy}{dx}=4x-5$$At $x=4$:$$\frac{dy}{dx}=4(4)-5=11$$On the given graph, this slope is read as approximately $11.1$.Answer:$y$ is negative for $-\frac{1}{2} < x < 3$; the gradient at $x=4$ is closest to $11.1$, which is Option A.