Curve Sketching Strategies

Builds on: aqa-alevel-maths-pure-1 / coordinate-geometry — this lesson is the A2-level deep dive.

This lesson brings together all the tools of coordinate geometry, differentiation, and algebraic analysis to develop a systematic approach to sketching curves. Being able to produce clear, accurate sketches is a vital skill at A-Level — it helps you solve problems, check answers, and communicate mathematical ideas.

Spec Mapping — AQA 7357 Sections C (Coordinate geometry) and G (Differentiation). This lesson is synoptic — it covers the curve-sketching content, linking coordinate-geometry features (intercepts, asymptotes, symmetry) with calculus-based analysis (stationary points, concavity). Refer to the official AQA specification document for exact wording.

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The Systematic Approach

When asked to sketch a curve, work through the following steps:

1. Intercepts — where the curve crosses the axes.
2. Symmetry — is the function even, odd, or periodic?
3. Asymptotes — vertical, horizontal, and oblique.
4. Stationary points — maxima, minima, and points of inflection.
5. Behaviour for large |x| — what happens as x → ±∞?
6. Sign analysis — where is y positive or negative?
7. Put it all together — sketch the curve.

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Step 1: Intercepts

y-intercept

Set x = 0 and find y.

x-intercepts

Set y = 0 and solve for x (these are the roots).

Example 1: Sketch y = x³ − 4x.

y-intercept: x = 0 gives y = 0. Point: (0, 0).

x-intercepts: x³ − 4x = 0
x(x² − 4) = 0
x(x − 2)(x + 2) = 0
x = 0, 2, −2

Points: (−2, 0), (0, 0), (2, 0).

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Step 2: Symmetry

• Even function: f(−x) = f(x). The graph is symmetric about the y-axis. Example: y = x².
• Odd function: f(−x) = −f(x). The graph has rotational symmetry about the origin. Example: y = x³.

Example: For y = x³ − 4x:

f(−x) = (−x)³ − 4(−x) = −x³ + 4x = −(x³ − 4x) = −f(x)

The function is odd, so the graph has rotational symmetry about the origin.

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Step 3: Asymptotes

Vertical Asymptotes

These occur where the denominator of a rational function equals zero (and the numerator does not equal zero at the same point).

Example 2: y = 1/(x − 3). Vertical asymptote at x = 3.

Horizontal Asymptotes

Found by examining the limit as x → ±∞.

For a rational function p(x)/q(x):

• If deg(p) < deg(q): y → 0.
• If deg(p) = deg(q): y → (leading coefficient of p)/(leading coefficient of q).
• If deg(p) > deg(q): no horizontal asymptote (may have an oblique asymptote).

Example 3: y = (2x + 1)/(x − 3).

As x → ∞: y → 2x/x = 2. Horizontal asymptote: y = 2.

Rewriting: y = (2(x − 3) + 7)/(x − 3) = 2 + 7/(x − 3)

This confirms y → 2 as x → ±∞.

Oblique Asymptotes

If deg(p) = deg(q) + 1, perform polynomial long division to find the oblique asymptote.

Example 4: y = (x² + 1)/(x − 1).

Dividing: x² + 1 = (x + 1)(x − 1) + 2
y = x + 1 + 2/(x − 1)

Oblique asymptote: y = x + 1 (as x → ±∞, the fraction 2/(x−1) → 0).

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Step 4: Stationary Points

Find dy/dx, set it equal to zero, and solve. Use the second derivative test or a sign analysis to classify.

Example 5 (continuing Example 1): y = x³ − 4x.

$$\begin{aligned} dy/dx &= 3x^{2} - 4 \\ \text{Set } dy/dx = 0 &: 3x^{2} - 4 = 0 \implies x^{2} = 4/3 \implies x = \pm 2/\sqrt{3} = \pm 2\sqrt{3}/3 \\ d^{2}y/dx^{2} &= 6x \\ \text{At } x = 2\sqrt{3}/3 &: d^{2}y/dx^{2} = 6 \times 2\sqrt{3}/3 = 4\sqrt{3} > 0 \implies \text{minimum} \\ y &= (2\sqrt{3}/3)^{3} - 4(2\sqrt{3}/3) = -16\sqrt{3}/9 \\ \text{At } x = -2\sqrt{3}/3 &: d^{2}y/dx^{2} = -4\sqrt{3} < 0 \implies \text{maximum} \\ y &= 16\sqrt{3}/9 \quad \text{(by odd symmetry)} \end{aligned}$$dy/dxSet dy/dx=0d2y/dx2At x=23​/3yAt x=−23​/3y​=3x2−4:3x2−4=0⟹x2=4/3⟹x=±2/3​=±23​/3=6x:d2y/dx2=6×23​/3=43​>0⟹minimum=(23​/3)3−4(23​/3)=−163​/9:d2y/dx2=−43​<0⟹maximum=163​/9(by odd symmetry)​

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Step 5: Behaviour for Large |x|

Consider what happens as x → +∞ and x → −∞.

For polynomials, the leading term dominates:

• y = x³ − 4x: as x → ∞, y → ∞; as x → −∞, y → −∞.
• y = −2x⁴ + x: as x → ±∞, y → −∞.

For rational functions, compare with the asymptotes found in Step 3.

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Step 6: Sign Analysis

Determine where y > 0 and y < 0. This helps ensure your sketch is on the correct side of the x-axis between roots.

Example: For y = x(x − 2)(x + 2):

Interval	Sign of x	Sign of (x−2)	Sign of (x+2)	Sign of y
x < −2	−	−	−	−
−2 < x < 0	−	−	+	+
0 < x < 2	+	−	+	−
x > 2	+	+	+	+

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Sketching Rational Functions — Full Example

Example 6: Sketch y = (x + 1)/(x² − 4).

Step 1 — Factor the denominator. $x^2 - 4 = (x-2)(x+2)$x2−4=(x−2)(x+2).

Step 2 — Intercepts.

• y-intercept: $x = 0 \Rightarrow y = 1/(-4) = -1/4$x=0⇒y=1/(−4)=−1/4.
• x-intercept: $x+1 = 0 \Rightarrow x = -1$x+1=0⇒x=−1; the curve passes through $(-1, 0)$(−1,0).

Step 3 — Asymptotes.

• Vertical: $x = 2$x=2 and $x = -2$x=−2.
• Horizontal: $\deg(\text{num}) < \deg(\text{den})$deg(num)<deg(den), so $y \to 0$y→0 as $x \to \pm\infty$x→±∞.

Step 4 — Stationary points. By the quotient rule, $\frac{dy}{dx} = \frac{(x^2-4)(1) - (x+1)(2x)}{(x^2-4)^2} = \frac{-x^2 - 2x - 4}{(x^2-4)^2}.$dxdy​=(x2−4)2(x2−4)(1)−(x+1)(2x)​=(x2−4)2−x2−2x−4​.

Setting the numerator equal to zero gives $x^2 + 2x + 4 = 0$x2+2x+4=0, whose discriminant is $4 - 16 = -12 < 0$4−16=−12<0. No stationary points.

Step 5 — Behaviour near the asymptotes.

Approach	Sign of denominator	Sign of numerator	Limit
$x \to 2^+$x→2+	small positive	$\to 3$→3	$y \to +\infty$y→+∞
$x \to 2^-$x→2−	small negative	$\to 3$→3	$y \to -\infty$y→−∞
$x \to -2^+$x→−2+	small negative	$\to -1$→−1	$y \to +\infty$y→+∞
$x \to -2^-$x→−2−	small positive	$\to -1$→−1	$y \to -\infty$y→−∞

For large $|x|$∣x∣, $y \sim x/x^2 = 1/x$y∼x/x2=1/x, so $y \to 0^+$y→0+ as $x \to +\infty$x→+∞ and $y \to 0^-$y→0− as $x \to -\infty$x→−∞.

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Sketching Curves Given Parametrically

For parametric curves, use the table-of-values approach from the parametric equations lesson, combined with:

• dy/dx = (dy/dt)/(dx/dt) for stationary points and tangent directions.
• Axis crossings found by setting x = 0 or y = 0 in the parametric equations.

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Transformation of Curves — Quick Reference

If you know the graph of y = f(x):

Transformation	Effect
y = f(x) + a	Translate up by a
y = f(x − a)	Translate right by a
y = af(x)	Vertical stretch, scale factor a
y = f(ax)	Horizontal stretch, scale factor 1/a
y = −f(x)	Reflect in x-axis
y = f(−x)	Reflect in y-axis

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Specimen Question (Modelled on the AQA Paper Format)

Note: this question is constructed to model AQA Paper 1/2/3 style; it is not a reproduction of any published past paper.

Worked Examination-Style Problem

Problem: Sketch the curve y = (x² − 1)/(x² + 1), showing clearly any asymptotes, intercepts, and stationary points.

Solution:

Intercepts:
y-intercept: x = 0 gives y = −1/1 = −1. Point: (0, −1).
x-intercepts: x² − 1 = 0 ⟹ x = ±1. Points: (−1, 0), (1, 0).

Symmetry: f(−x) = ((−x)² − 1)/((−x)² + 1) = (x² − 1)/(x² + 1) = f(x). Even function — symmetric about y-axis.

Asymptotes:
No vertical asymptotes (denominator x² + 1 > 0 for all x).
Horizontal: as x → ±∞, y → x²/x² = 1. Asymptote: y = 1.

Note: y = 1 − 2/(x² + 1), confirming y → 1 and y < 1 for all finite x.

Stationary points:
dy/dx = [2x(x²+1) − (x²−1)(2x)] / (x²+1)²
      = [2x³ + 2x − 2x³ + 2x] / (x²+1)²
      = 4x / (x²+1)²

dy/dx = 0 ⟹ x = 0.

At x = 0: y = −1.
d²y/dx² at x = 0: Since dy/dx = 4x/(x²+1)², applying the quotient rule or noting the sign change: for x < 0, dy/dx < 0; for x > 0, dy/dx > 0. So x = 0 is a minimum.

Minimum point: (0, −1).

Behaviour: y increases from −1, passes through (1, 0), and approaches 1 from below.

The curve is a smooth valley shape, symmetric about the y-axis, with minimum at (0, −1), crossing the x-axis at ±1, and asymptotically approaching y = 1 from below.

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Synoptic Links

This topic connects to:

• Coordinate geometry foundations (aqa-alevel-maths-pure-1 / coordinate-geometry) — intercepts, axes, and the underlying sketch-anatomy framework.
• Differentiation techniques (aqa-alevel-maths-pure-2 / differentiation-techniques) — first- and second-derivative tests for stationary points and concavity.
• Algebra and functions (aqa-alevel-maths-pure-1 / algebra-and-functions) — domain, range, asymptote and transformation reasoning behind a sketch.

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Summary

• Systematic approach: intercepts → symmetry → asymptotes → stationary points → end behaviour → sign analysis → sketch.
• Vertical asymptotes: denominator = 0, numerator ≠ 0.
• Horizontal asymptotes: compare degrees of numerator and denominator.
• Oblique asymptotes: polynomial division when deg(num) = deg(denom) + 1.
• Stationary points: dy/dx = 0; classify using second derivative or sign change.
• Always mark key features on your sketch: label intercepts, asymptotes, and stationary points.

Exam Tip: When sketching curves in the exam, label all key features clearly — coordinates of intercepts and stationary points, equations of asymptotes. Draw smooth curves (not jagged lines). Make sure your curve approaches asymptotes correctly and does not cross a vertical asymptote. If the question says "sketch", a rough but accurate picture with labels is expected — you do not need to plot every point precisely.

A-Level Deep Dive: Curve Sketching Strategies

Spec mapping

AQA 7357 specification, Pure Mathematics, Section C — Coordinate Geometry covers curves defined by simple equations including polynomials, the modulus of a linear function, $y = a/x$y=a/x and $y = a/x^2$y=a/x2 (including their vertical and horizontal asymptotes), exponential and logarithmic functions, and parametrically defined curves (refer to the official specification document for exact wording). Curve-sketching is examined in both Paper 1 and Paper 2 and is the synoptic spine that connects Section E (differentiation, stationary points and points of inflection), Section F (integration, area under a curve), Section G (numerical methods, location of roots) and the transformations strand within Section B (algebra and functions). The AQA formula booklet does not list standard curve shapes — they must be memorised.

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Worked example with full mark scheme

Question (8 marks):

Sketch the curve with equation $y = \dfrac{x^2 - 4}{x^2 - 1}$y=x2−1x2−4​, showing clearly all intercepts with the axes, all asymptotes, and the coordinates of any stationary points. (8)

Solution with mark scheme:

Step 1 — factorise and identify zeros and poles.

$y = \dfrac{(x - 2)(x + 2)}{(x - 1)(x + 1)}$y=(x−1)(x+1)(x−2)(x+2)​

The numerator is zero at $x = \pm 2$x=±2, so the $x$x-intercepts are $(2, 0)$(2,0) and $(-2, 0)$(−2,0). The denominator is zero at $x = \pm 1$x=±1 — and the numerator is non-zero there — so $x = 1$x=1 and $x = -1$x=−1 are vertical asymptotes (no holes).

M1 — factorise correctly. A1 — both intercepts and both asymptote locations identified.

Step 2 — y-intercept.

Set $x = 0$x=0: $y = \dfrac{-4}{-1} = 4$y=−1−4​=4. So $(0, 4)$(0,4) lies on the curve.

B1 — $y$y-intercept.

Step 3 — horizontal asymptote (end behaviour).

Both numerator and denominator are degree 2, so divide leading coefficients: $y \to 1$y→1 as $x \to \pm\infty$x→±∞. The horizontal asymptote is $y = 1$y=1.

M1 — recognise leading-coefficient ratio gives the horizontal asymptote.

Step 4 — stationary points.

Use the quotient rule with $u = x^2 - 4$u=x2−4, $v = x^2 - 1$v=x2−1:

$\dfrac{dy}{dx} = \dfrac{2x(x^2 - 1) - 2x(x^2 - 4)}{(x^2 - 1)^2} = \dfrac{2x[(x^2 - 1) - (x^2 - 4)]}{(x^2 - 1)^2} = \dfrac{6x}{(x^2 - 1)^2}$dxdy​=(x2−1)22x(x2−1)−2x(x2−4)​=(x2−1)22x[(x2−1)−(x2−4)]​=(x2−1)26x​

Setting $\dfrac{dy}{dx} = 0$dxdy​=0 gives $x = 0$x=0, so the only stationary point is $(0, 4)$(0,4) — coinciding with the $y$y-intercept.

M1 — apply quotient rule and simplify. A1 — stationary point at $(0, 4)$(0,4).

Step 5 — sign analysis and shape.

Sign of $\dfrac{dy}{dx} = \dfrac{6x}{(x^2 - 1)^2}$dxdy​=(x2−1)26x​: denominator is positive (away from asymptotes), so the gradient has the sign of $x$x. Negative for $x < 0$x<0, positive for $x > 0$x>0, so on the central branch (between $x = -1$x=−1 and $x = 1$x=1) the stationary point at $(0, 4)$(0,4) is a local maximum on that branch. On the outer branches $|x| > 1$∣x∣>1, $y$y approaches $1$1 from above as $x \to \pm \infty$x→±∞, falls through the $x$x-axis at $\pm 2$±2, and tends to $+\infty$+∞ as $x \to \pm 1$x→±1 from outside.

Inside $|x| < 1$∣x∣<1, the curve passes through the maximum at $(0, 4)$(0,4) and tends to $-\infty$−∞ as $x \to \pm 1$x→±1 from inside (since the numerator is negative — close to $-4$−4 — and the denominator approaches zero through negative values).

A1 — correct shape with three branches, asymptotic behaviour shown both sides of each vertical asymptote, central turning point identified as a maximum on the central branch.

Total: 8 marks (M3 A3 B1 + 1 shape mark).

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Specimen question modelled on the AQA 7357 Paper 1 format

Question (6 marks): The curve $C$C has equation $y = x e^{-x}$y=xe−x for $x \in \mathbb{R}$x∈R.

(a) Find the coordinates of the stationary point of $C$C and determine its nature. (4)

(b) Describe the behaviour of $y$y as $x \to +\infty$x→+∞ and as $x \to -\infty$x→−∞, and hence sketch $C$C. (2)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — product rule: $\dfrac{dy}{dx} = e^{-x} - x e^{-x} = (1 - x)e^{-x}$dxdy​=e−x−xe−x=(1−x)e−x.
• A1 (AO1.1b) — set to zero, $e^{-x} > 0$e−x>0, so $x = 1$x=1, $y = e^{-1}$y=e−1.
• M1 (AO2.4) — second derivative or sign test. $\dfrac{d^2 y}{dx^2} = -e^{-x} - (1 - x)e^{-x} = (x - 2)e^{-x}$dx2d2y​=−e−x−(1−x)e−x=(x−2)e−x, which is negative at $x = 1$x=1.
• A1 (AO2.4) — local maximum at $(1, 1/e)$(1,1/e).

(b)

• B1 (AO1.2) — as $x \to +\infty$x→+∞, $y \to 0^+$y→0+ (exponential decay dominates linear growth); as $x \to -\infty$x→−∞, $y \to -\infty$y→−∞.
• B1 (AO2.5) — sketch with curve passing through origin, rising to maximum at $(1, 1/e)$(1,1/e), decaying to zero from above, and dropping steeply on the left.

Total: 6 marks split AO1 = 3, AO2 = 3. End-behaviour marks are awarded for the direction of approach (from above / below), not just the limiting value.

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Synoptic links

Connects to: