Stationary Points: Location, Nature, and Curve Sketching

Introduction

A stationary point is any point on a curve where the gradient is zero. These points are fundamental in curve sketching and optimisation problems, both of which appear regularly in 9709 Paper 1. The syllabus requires you to locate stationary points (find their coordinates), determine their nature (maximum or minimum), and use this information when sketching graphs. Points of inflexion are explicitly excluded from this syllabus objective and will not be tested here.

Core Concept

Recall from differentiating powers that $\dfrac{dy}{dx}$ gives the gradient function of a curve $y = f(x)$ . At a stationary point, the curve is momentarily flat — neither increasing nor decreasing — so:

\frac{dy}{dx} = 0

Setting this equal to zero and solving for $x$ locates all stationary points. Substituting each $x$ -value back into $y = f(x)$ gives the full coordinates.

Determining the Nature

Once a stationary point is found, you must decide whether it is a local maximum or a local minimum. There are two reliable methods:

Method 1 — Second Derivative Test

Differentiate $\dfrac{dy}{dx}$ again to obtain $\dfrac{d^2y}{dx^2}$ , then evaluate it at the stationary point.

Value of $\dfrac{d^2y}{dx^2}$ at the point	Nature of stationary point
$\dfrac{d^2y}{dx^2} < 0$	Local maximum
$\dfrac{d^2y}{dx^2} > 0$	Local minimum
$\dfrac{d^2y}{dx^2} = 0$	Test is inconclusive — use Method 2

The reasoning: a negative second derivative means the gradient is decreasing through zero (the curve is concave down, i.e. a peak); a positive second derivative means the gradient is increasing through zero (concave up, i.e. a trough).

Method 2 — First Derivative Sign Change

Examine the sign of $\dfrac{dy}{dx}$ just to the left and right of the stationary point.

Sign of $\dfrac{dy}{dx}$ : left → right	Nature
$+ \to -$	Local maximum
$- \to +$	Local minimum

This method is always valid, including when the second derivative test is inconclusive.

Key Formulae & Definitions

\text{Stationary point: } \frac{dy}{dx} = 0

\text{Second derivative: } \frac{d^2y}{dx^2} = \frac{d}{dx}\!\left(\frac{dy}{dx}\right)

Local maximum at $x = a$ :

\frac{dy}{dx}\bigg|_{x=a} = 0 \quad \text{and} \quad \frac{d^2y}{dx^2}\bigg|_{x=a} < 0

Local minimum at $x = a$ :

\frac{dy}{dx}\bigg|_{x=a} = 0 \quad \text{and} \quad \frac{d^2y}{dx^2}\bigg|_{x=a} > 0

Worked Examples

Example 1 — Polynomial Curve

Find the stationary points of $y = 2x^3 - 9x^2 + 12x - 1$ and determine their nature.

Step 1: Differentiate.

\frac{dy}{dx} = 6x^2 - 18x + 12

Step 2: Set equal to zero and solve.

6x^2 - 18x + 12 = 0

x^2 - 3x + 2 = 0 \quad \text{(divide through by 6)}

(x-1)(x-2) = 0

x = 1 \quad \text{or} \quad x = 2

Step 3: Find the $y$ -coordinates.

x = 1: \quad y = 2(1) - 9(1) + 12(1) - 1 = 4

x = 2: \quad y = 2(8) - 9(4) + 12(2) - 1 = 16 - 36 + 24 - 1 = 3

Stationary points are $(1,\, 4)$ and $(2,\, 3)$ .

Step 4: Find the second derivative.

\frac{d^2y}{dx^2} = 12x - 18

Step 5: Apply the second derivative test.

At $x = 1$ : $\quad \dfrac{d^2y}{dx^2} = 12(1) - 18 = -6 < 0 \quad \Rightarrow \quad \textbf{local maximum}$

At $x = 2$ : $\quad \dfrac{d^2y}{dx^2} = 12(2) - 18 = 6 > 0 \quad \Rightarrow \quad \textbf{local minimum}$

Conclusion: Local maximum at $(1,\,4)$ ; local minimum at $(2,\,3)$ .

Example 2 — Involving the Chain Rule

A curve has equation $y = (3x - 1)^4 - 8x$ . Find any stationary points and determine their nature.

Step 1: Differentiate using the chain rule.

\frac{dy}{dx} = 4(3x-1)^3 \cdot 3 - 8 = 12(3x-1)^3 - 8

Step 2: Set equal to zero.

12(3x-1)^3 = 8

(3x-1)^3 = \frac{2}{3}

3x - 1 = \left(\frac{2}{3}\right)^{\!\frac{1}{3}}

x = \frac{1}{3}\left[1 + \left(\frac{2}{3}\right)^{\!\frac{1}{3}}\right]

Evaluating numerically: $\left(\tfrac{2}{3}\right)^{1/3} \approx 0.874$ , so $x \approx \dfrac{1+0.874}{3} \approx 0.625$ .

Step 3: Find $y$ .

y = (3(0.625)-1)^4 - 8(0.625) = (0.874)^4 - 5 \approx 0.584 - 5 = -4.416

Stationary point approximately at $(0.625,\; -4.416)$ .

Step 4: Find the second derivative.

\frac{d^2y}{dx^2} = 12 \cdot 3(3x-1)^2 \cdot 3 = 108(3x-1)^2

Step 5: Apply the test.

At the stationary point, $(3x-1)^2 = \left(\tfrac{2}{3}\right)^{2/3} > 0$ , so:

\frac{d^2y}{dx^2} = 108\left(\frac{2}{3}\right)^{\!\frac{2}{3}} > 0 \quad \Rightarrow \quad \textbf{local minimum}

Common Mistakes & Examiner Pitfalls

Forgetting the $y$ -coordinate. Solving $\dfrac{dy}{dx} = 0$ gives only the $x$ -value. Examiners award a mark specifically for the full coordinate — always substitute back.
Stopping at location. The question almost always says "find and determine the nature of"; omitting the nature test loses marks.
Misreading the second derivative sign. Remember: $\dfrac{d^2y}{dx^2} < 0$ is a maximum (many students instinctively reverse this). Think of a hill: the curve curves downwards, so the rate of change of gradient is negative.
Arithmetic slips when factorising $\dfrac{dy}{dx} = 0$ . Always check your roots by substitution before proceeding.
Using the second derivative test when it gives zero. If $\dfrac{d^2y}{dx^2} = 0$ at the stationary point, the test tells you nothing — switch immediately to the sign-change method.
Sketching without stationary point information. When sketching, mark the stationary points with their coordinates, indicate whether each is a max or min, and show the general shape of the curve in each region. Intercepts with the axes should also be shown where possible.

Practice Questions

Q1. Find the coordinates of the stationary point of $y = x^2 - 6x + 11$ and state its nature.

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\frac{dy}{dx} = 2x - 6 = 0 \implies x = 3

y = 9 - 18 + 11 = 2

\frac{d^2y}{dx^2} = 2 > 0 \implies \textbf{minimum at } (3,\,2)

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Q2. A curve is defined by $y = 2x^3 + 3x^2 - 36x + 5$ . Find the coordinates of both stationary points and determine their nature.

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\frac{dy}{dx} = 6x^2 + 6x - 36 = 6(x^2 + x - 6) = 6(x+3)(x-2) = 0

x = -3 \quad \text{or} \quad x = 2

At $x = -3$ : $y = 2(-27) + 3(9) - 36(-3) + 5 = -54 + 27 + 108 + 5 = 86$

At $x = 2$ : $y = 2(8) + 3(4) - 36(2) + 5 = 16 + 12 - 72 + 5 = -39$

\frac{d^2y}{dx^2} = 12x + 6

At $x = -3$ : $12(-3)+6 = -30 < 0 \implies \textbf{local maximum at }(-3,\,86)$

At $x = 2$ : $12(2)+6 = 30 > 0 \implies \textbf{local minimum at }(2,\,-39)$

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Q3. The curve $y = x + \dfrac{4}{x}$ has a stationary point for $x > 0$ . Find it and determine its nature.

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Rewrite: $y = x + 4x^{-1}$

\frac{dy}{dx} = 1 - \frac{4}{x^2} = 0 \implies x^2 = 4 \implies x = 2 \; (\text{since } x > 0)

y = 2 + \frac{4}{2} = 4

\frac{d^2y}{dx^2} = \frac{8}{x^3}; \quad \text{at } x=2: \frac{8}{8} = 1 > 0 \implies \textbf{minimum at }(2,\,4)

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Q4. A curve has equation $y = 3x^4 - 16x^3 + 24x^2 + 5$ . Find all stationary points and determine their nature.

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\frac{dy}{dx} = 12x^3 - 48x^2 + 48x = 12x(x^2 - 4x + 4) = 12x(x-2)^2 = 0

x = 0 \quad \text{or} \quad x = 2

At $x = 0$ : $y = 5$ . At $x = 2$ : $y = 3(16) - 16(8) + 24(4) + 5 = 48 - 128 + 96 + 5 = 21$

\frac{d^2y}{dx^2} = 36x^2 - 96x + 48

At $x = 0$ : $48 > 0 \implies \textbf{local minimum at }(0,\,5)$

At $x = 2$ : $36(4) - 96(2) + 48 = 144 - 192 + 48 = 0$ — second derivative test inconclusive.

Use sign change for $x = 2$ : $\frac{dy}{dx} = 12x(x-2)^2$ . Since $(x-2)^2 \geq 0$ always and $12x > 0$ for $x > 0$ , the sign of $\frac{dy}{dx}$ is positive on both sides of $x = 2$ . There is no sign change, so $(2, 21)$ is neither a maximum nor a minimum (it is a stationary point of inflexion — but note this type of identification is not required by the syllabus; recognising the second derivative is inconclusive and the sign test shows no change is sufficient).

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Q5. Sketch the curve $y = x^3 - 3x^2$ , marking the coordinates of all stationary points and any intercepts with the axes.

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Stationary points:

\frac{dy}{dx} = 3x^2 - 6x = 3x(x-2) = 0 \implies x = 0 \text{ or } x = 2

At $x=0$ : $y = 0$ . At $x=2$ : $y = 8 - 12 = -4$ .

\frac{d^2y}{dx^2} = 6x - 6

At $x = 0$ : $-6 < 0 \implies$ local maximum at $(0,\,0)$ At $x = 2$ : $6 > 0 \implies$ local minimum at $(2,\,-4)$

Axis intercepts: $y=0$ : $x^3 - 3x^2 = x^2(x-3) = 0 \implies x = 0$ (repeated) and $x = 3$ .

Sketch shows: curve rising from bottom-left, reaching a local max at $(0,0)$ , falling to a local min at $(2,-4)$ , then rising through $(3,0)$ to the top-right.

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Connections

Prerequisite subtopics you should be confident with:

Differentiating Powers of $x$ — essential for forming $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$ from polynomial and rational expressions.
The Chain Rule — needed whenever the function involves a composite expression such as $(ax+b)^n$ , as seen in Example 2 above.

Likely next subtopics:

Increasing and Decreasing Functions — uses the sign of $\dfrac{dy}{dx}$ on intervals, extending the sign-change ideas developed here.
Applications of Differentiation (Optimisation) — real-world maximum/minimum problems (e.g. maximising volume, minimising cost) rely entirely on the stationary point techniques in this note.
Curve Sketching (further) — combining stationary point analysis with knowledge of asymptotes and transformations for more complete sketches.

Introduction

Core Concept

Determining the Nature

Key Formulae & Definitions

Worked Examples

Example 1 — Polynomial Curve

Example 2 — Involving the Chain Rule

Common Mistakes & Examiner Pitfalls

Practice Questions

Connections

Figures

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