WAEC Further Mathematics 2021 Paper

Concept:The gradient of a curve at any point equals the value of the derivative $\frac{dy}{dx}$ at that point.We use this derivative to solve for the $x$-coordinate, then substitute to find the $y$-coordinate.Explanation:The given curve is $y = 3x^2 + 11x + 7$.Differentiate with respect to $x$: $\frac{dy}{dx} = 6x + 11$.At point $P(x,y)$, the gradient is given as $-1$.So, set $6x + 11 = -1$.Subtract 11 from both sides: $6x = -12$.Divide by 6: $x = -2$.Now substitute $x = -2$ into the original equation to find $y$.$y = 3(-2)^2 + 11(-2) + 7$$y = 3(4) - 22 + 7$$y = 12 - 22 + 7 = -3$Therefore, the coordinates of $P$ are $(-2, -3)$.Answer:The coordinates of $P$ are $(-2, -3)$.This corresponds to option B.