For the curve, the gradient at $(x, y)$ is $\frac{dy}{dx} = 2\sqrt{x + 3} - x$. It has a stationary point at $(a, 14)$, with $a$ a positive constant.
(a)[3]
Find $a$.
(b)[3]
Determine what type of stationary point it is.
(c)[4]
Find the curve's equation.
Worked solution & mark scheme
This 10-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Sets $\frac{dy}{dx}=0$, leading to $2\sqrt{a+3}-a=0$” …