Stationary Points & Curve Sketching

This lesson covers finding stationary points, classifying them, identifying points of inflection, and developing a systematic approach to curve sketching. These are essential skills for A-Level Mathematics and are examined in both pure and applied contexts.

Stationary Points

A stationary point occurs where the gradient of a curve is zero:

dy/dx = 0

At a stationary point, the tangent to the curve is horizontal.

Types of Stationary Point

Local maximum — the curve rises to the point then falls.
Local minimum — the curve falls to the point then rises.
Stationary point of inflection — the curve flattens momentarily but continues in the same direction (either rising–flat–rising or falling–flat–falling).

The First Derivative Test

Examine the sign of dy/dx on either side of the stationary point:

Sign change of dy/dx	Type
+ → 0 → −	Local maximum
− → 0 → +	Local minimum
+ → 0 → +	Rising point of inflection
− → 0 → −	Falling point of inflection

Example 1

Find and classify the stationary points of y = x³ − 3x² − 9x + 5.

dy/dx = 3x² − 6x − 9 = 3(x² − 2x − 3) = 3(x − 3)(x + 1)

Setting dy/dx = 0: x = 3 or x = −1.

At x = −1: y = −1 − 3 + 9 + 5 = 10. Point: (−1, 10).

Check sign of dy/dx:

x = −2: dy/dx = 3(−5)(−1) = 15 > 0
x = 0: dy/dx = 3(−3)(1) = −9 < 0

Sign changes + → − : (−1, 10) is a local maximum.

At x = 3: y = 27 − 27 − 27 + 5 = −22. Point: (3, −22).

Check sign of dy/dx:

x = 2: dy/dx = 3(−1)(3) = −9 < 0
x = 4: dy/dx = 3(1)(5) = 15 > 0

Sign changes − → + : (3, −22) is a local minimum.

The Second Derivative Test

An alternative (and often quicker) approach:

If d²y/dx² > 0 at the stationary point → local minimum.
If d²y/dx² < 0 at the stationary point → local maximum.
If d²y/dx² = 0 → test is inconclusive (use the first derivative test).

Example 2 (Continuing Example 1)

d²y/dx² = 6x − 6

At x = −1: d²y/dx² = −6 − 6 = −12 < 0 → maximum ✓ At x = 3: d²y/dx² = 18 − 6 = 12 > 0 → minimum ✓

Points of Inflection

A point of inflection is where the curve changes concavity — from concave up (curving upwards, d²y/dx² > 0) to concave down (curving downwards, d²y/dx² < 0), or vice versa.

At a point of inflection:

d²y/dx² = 0

But d²y/dx² = 0 is necessary, not sufficient. You must also check that d²y/dx² changes sign either side of the point.

A stationary point of inflection is a point of inflection where dy/dx = 0 as well.

Example 3

Find the points of inflection of y = x⁴ − 6x² + 8x + 1.

dy/dx = 4x³ − 12x + 8
d²y/dx² = 12x² − 12 = 12(x² − 1) = 12(x − 1)(x + 1)

Set d²y/dx² = 0: x = 1 or x = −1.

Check sign change:

At x = 1:

x = 0.5: d²y/dx² = 12(0.25 − 1) = −9 < 0
x = 1.5: d²y/dx² = 12(2.25 − 1) = 15 > 0

Sign changes − → + : point of inflection at x = 1 ✓

At x = −1:

x = −1.5: d²y/dx² = 12(2.25 − 1) = 15 > 0
x = −0.5: d²y/dx² = 12(0.25 − 1) = −9 < 0

Sign changes + → − : point of inflection at x = −1 ✓

Curve Sketching Strategy

To sketch a curve y = f(x), follow this systematic approach:

Step 1: Find where the curve crosses the axes

y-intercept: Set x = 0, find y = f(0).
x-intercepts: Set y = 0, solve f(x) = 0.

Step 2: Find stationary points

Set dy/dx = 0 and solve. Classify each using the second derivative test (or first derivative test).

Step 3: Determine behaviour as x → ±∞

For polynomials, the leading term dominates. E.g. if y = 2x³ − ..., then y → +∞ as x → +∞ and y → −∞ as x → −∞.

Step 4: Check for symmetry

If f(−x) = f(x), the curve is symmetric about the y-axis (even function).
If f(−x) = −f(x), the curve has rotational symmetry about the origin (odd function).

Step 5: Find any asymptotes (for rational functions)

Vertical asymptotes: where the denominator equals zero.
Horizontal asymptotes: the value of y as x → ±∞.

Step 6: Plot key points and draw

Mark intercepts, stationary points, asymptotes, and connect them with a smooth curve respecting the gradient information.

Example 4: Full Curve Sketch

Sketch y = x³ − 3x.

Intercepts:

y-intercept: (0, 0)
x-intercepts: x(x² − 3) = 0, so x = 0, x = ±√3

Stationary points:

dy/dx = 3x² − 3 = 0
x = ±1

At x = 1: y = 1 − 3 = −2. Point: (1, −2). At x = −1: y = −1 + 3 = 2. Point: (−1, 2).

d²y/dx² = 6x

At x = 1: d²y/dx² = 6 > 0 → minimum. At x = −1: d²y/dx² = −6 < 0 → maximum.

End behaviour: As x → +∞, y → +∞. As x → −∞, y → −∞.

Symmetry: f(−x) = −f(x), so the curve has rotational symmetry about the origin (odd function).

The curve passes through (−√3, 0), rises to the maximum at (−1, 2), falls through (0, 0), continues falling to the minimum at (1, −2), then rises through (√3, 0) to infinity.

Asymptotic Behaviour

For rational functions like y = (2x + 1)/(x − 3):

Vertical asymptote: x = 3 (denominator = 0).

Horizontal asymptote: As x → ±∞, y → 2x/x = 2. So y = 2 is a horizontal asymptote.

Behaviour near asymptotes:

As x → 3⁺: denominator → 0⁺, so y → +∞.
As x → 3⁻: denominator → 0⁻, so y → −∞.

Example 5: Sketching a Rational Function

Sketch y = 1/(x² − 4) = 1/((x − 2)(x + 2)).

Vertical asymptotes: x = 2, x = −2.

Horizontal asymptote: As x → ±∞, y → 0.

y-intercept: y = 1/(0 − 4) = −1/4. Point: (0, −1/4).

x-intercepts: 1/(x² − 4) = 0 has no solution — the curve never crosses the x-axis.

Stationary points:

dy/dx = −2x/(x² − 4)² = 0
x = 0

At x = 0: y = −1/4. This is a maximum for the section between the asymptotes (check the second derivative or signs of dy/dx).

Summary

Stationary points occur where dy/dx = 0.
Classify using the second derivative test (d²y/dx² > 0 → min, d²y/dx² < 0 → max) or the first derivative test.
Points of inflection occur where d²y/dx² = 0 and d²y/dx² changes sign.
To sketch a curve: find intercepts, stationary points, asymptotes, end behaviour, and symmetry.
For rational functions, identify vertical and horizontal asymptotes by examining where the denominator is zero and the limit as x → ±∞.

Exam Tip: In curve sketching questions, the examiner expects to see labelled coordinates of all key points (intercepts, turning points), correct shape, and asymptotes clearly marked with dashed lines. Even a rough sketch should demonstrate that you understand the essential features. Do not join points with straight lines — draw smooth curves.

A-Level Deep Dive: Stationary Points and Curve Sketching

Spec mapping

AQA 7357 specification, Section G — Differentiation covers differentiation to find gradients, tangents and normals, maxima and minima and stationary points, points of inflection (refer to the official specification document for exact wording). This appears in Papers 1 and 2 (Pure Mathematics) and feeds Paper 3 mechanics through optimisation of force/displacement models. The classification machinery ( $\frac{d^2y}{dx^2}$ test, sign-change test, concavity) is examined alongside Section F (graphs and transformations). The AQA formula booklet does not list the stationary-point classification rule — it must be memorised and applied with the correct logical flow.

Worked example with full mark scheme

Question (8 marks): Sketch the curve $y = \dfrac{x^2 - 4}{x - 1}$ for $x \in \mathbb{R}$ , $x \neq 1$ , indicating clearly all intercepts, asymptotes, stationary points and the behaviour as $x \to \pm \infty$ .

Solution with mark scheme:

Step 1 — intercepts.

For $y = 0$ : $x^2 - 4 = 0 \implies x = \pm 2$ . The curve crosses the $x$ -axis at $(-2, 0)$ and $(2, 0)$ . For $x = 0$ : $y = \dfrac{-4}{-1} = 4$ . The $y$ -intercept is $(0, 4)$ .

M1 A1 — both intercepts identified.

Step 2 — vertical asymptote.

The denominator vanishes at $x = 1$ , while the numerator gives $1 - 4 = -3 \neq 0$ , so $x = 1$ is a genuine vertical asymptote (not a removable hole). As $x \to 1^-$ , numerator $\to -3$ , denominator $\to 0^-$ , so $y \to +\infty$ . As $x \to 1^+$ , denominator $\to 0^+$ , so $y \to -\infty$ .

M1 — asymptote identified with sign analysis.

Step 3 — oblique asymptote via polynomial division.

$\dfrac{x^2 - 4}{x - 1} = x + 1 - \dfrac{3}{x - 1}$

So as $x \to \pm \infty$ , $y \to x + 1$ . The line $y = x + 1$ is an oblique asymptote.

A1 — oblique asymptote stated correctly.

Step 4 — stationary points.

Differentiating using the quotient rule:

$\dfrac{dy}{dx} = \dfrac{2x(x - 1) - (x^2 - 4)(1)}{(x - 1)^2} = \dfrac{2x^2 - 2x - x^2 + 4}{(x - 1)^2} = \dfrac{x^2 - 2x + 4}{(x - 1)^2}$

Setting $\dfrac{dy}{dx} = 0$ requires $x^2 - 2x + 4 = 0$ . The discriminant is $4 - 16 = -12 < 0$ , so there are no real stationary points.

M1 A1 — derivative correct, conclusion that no stationary points exist.

Step 5 — sketch.

The curve has two branches separated by $x = 1$ . Left branch: passes through $(-2, 0)$ and $(0, 4)$ , rises toward $+\infty$ as $x \to 1^-$ , follows $y = x + 1$ as $x \to -\infty$ . Right branch: comes from $-\infty$ at $x = 1^+$ , passes through $(2, 0)$ , follows $y = x + 1$ as $x \to +\infty$ .

B1 — sketch shows correct branch behaviour and asymptote alignment.

Total: 8 marks (M3 A3 B1 plus structure mark).

Specimen question modelled on the AQA 7357 Paper 1 format

Question (6 marks): A curve has equation $y = x^3 - 6x^2 + 9x + 2$ .

(a) Find the coordinates of the stationary points and classify each. (4)

(b) Determine the range of values of $x$ for which the curve is concave up. (2)

Mark scheme decomposition by AO:

(a)

M1 (AO1.1a) — $\dfrac{dy}{dx} = 3x^2 - 12x + 9 = 3(x - 1)(x - 3)$ , so stationary at $x = 1, 3$ .
A1 (AO1.1b) — $y(1) = 6$ , $y(3) = 2$ , giving $(1, 6)$ and $(3, 2)$ .
M1 (AO1.1b) — $\dfrac{d^2y}{dx^2} = 6x - 12$ . At $x = 1$ : $-6 < 0$ (max). At $x = 3$ : $6 > 0$ (min).
A1 (AO2.5) — both classifications stated with reference to the sign of the second derivative.

(b)

M1 (AO1.1b) — concave up requires $\dfrac{d^2y}{dx^2} > 0$ , i.e. $6x - 12 > 0$ .
A1 (AO2.5) — $x > 2$ .

Total: 6 marks split AO1 = 4, AO2 = 2.

Synoptic links

Connects to:

Section F — Graphs and transformations: stationary points of $y = f(x)$ map predictably under transformations. $y = f(x) + c$ shifts stationary $y$ -values by $c$ but preserves $x$ -coordinates; $y = f(x - a)$ shifts $x$ -coordinates by $+a$ ; $y = -f(x)$ swaps maxima and minima. Recognising this saves re-differentiating transformed curves.
Section G — First derivative (gradients, tangents, normals): the tangent at a stationary point is horizontal ( $y = k$ ), and the normal is vertical ( $x = h$ ). Curve-sketching questions often pair "find stationary points" with "find equation of normal at a non-stationary point" — same derivative, two uses.
Section G — Second derivative and concavity: $\dfrac{d^2y}{dx^2} > 0$ marks regions of concave-up behaviour (curve lies above its tangents); $< 0$ marks concave-down. Points of inflection occur where concavity changes, which requires $\dfrac{d^2y}{dx^2} = 0$ and a sign change — not just $\dfrac{d^2y}{dx^2} = 0$ alone.
Mechanics (Paper 3): velocity-time graphs are curves whose stationary points correspond to instantaneous zero acceleration, and whose concavity indicates whether acceleration is increasing or decreasing (jerk). A particle's displacement curve has stationary points exactly when velocity is zero — i.e. at turning points of motion.
Modelling and optimisation: real-world maxima and minima problems (largest volume, shortest distance, minimum cost) reduce to finding stationary points of a function on a closed interval, then comparing critical values to endpoint values. The sketching skill is the diagnostic — without a sketch, candidates often miss boundary maxima.

Mark-scheme literacy

Curve-sketching questions on 7357 split AO marks more evenly than algebraic procedural topics:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	40–55%	Differentiating correctly, solving $\dfrac{dy}{dx} = 0$ , computing $\dfrac{d^2y}{dx^2}$ , identifying intercepts and asymptotes
AO2 (reasoning / interpretation)	30–40%	Classifying stationary points with justification, distinguishing genuine asymptotes from removable holes, articulating end behaviour
AO3 (problem-solving)	10–20%	Choosing an efficient method (sign-change vs second derivative), integrating multiple features into a coherent sketch, handling parameters or piecewise definitions

Stationary Points & Curve Sketching

Stationary Points & Curve Sketching