Quick questions on Stationary points and their nature explained: O-Level A-Maths

A stationary point is where the tangent is horizontal, so the gradient is zero:

The fastest classification uses the second derivative $\dfrac{d^2y}{dx^2}$ at each stationary point:

The most valued application is optimisation, and the setup is where marks are won or lost. The routine is: write the quantity to be optimised as a function of one variable (using a constraint to eliminate any second variable), differentiate, set the derivative to zero to find the stationary point, then confirm it is the maximum or minimum required using the second derivative. For a box of fixed volume, you would express the surface area in terms of one dimension using the volume constraint, then minimise. The skill is reducing the problem to a single-variable function before differentiating, because the calculus only starts once the quantity is written in terms of one variable.

Positive second derivative is a minimum (valley), negative is a maximum (hill); reversing them is common.

Find the stationary point of $y = x^2 - 4x + 1$. [3 marks]

State the nature of a stationary point where $\dfrac{d^2y}{dx^2} = 5$. [1 mark]

Find and classify the stationary points of $y = x^3 - 12x$. [4 marks]