State the vertical asymptote of y = 1x + 2. [1 mark]

SQA AH style: asymptotes-style practice: Find the vertical and non-vertical asymptotes of y = x^2x - 1.

Vertical asymptote where the denominator is zero: x - 1 = 0, so x = 1 (1 mark). Divide to find the non-vertical asymptote: x^2x - 1 = x + 1 + 1x - 1 (2 marks). As x → ∞ the remainder 1x - 1 → 0, so the oblique asymptote is y = x + 1 (1 mark). Markers reward the vertical asymptote, the algebraic division, and identifying the slant asymptote from the polynomial part.

SQA AH style: optimisation-style practice: A closed cylinder has volume 1000\,cm^3. Find the radius that minimises its surface area.

Volume π r^2 h = 1000, so h = 1000π r^2 (1 mark). Surface area S = 2π r^2 + 2π r h = 2π r^2 + 2000r (1 mark). dSdr = 4π r - 2000r^2; set to zero: 4π r^3 = 2000, so r^3 = 500π and r = (500π)^1/3 ≈ 5.42\,cm (2 marks). d^2Sdr^2 = 4π + 4000r^3 > 0, confirming a minimum (1 mark). Markers reward the constraint, the surface-area function, the derivative set to zero, and the second-derivative check.

ScotlandMathsSyllabus dot point

How do you analyse a function's features to sketch its curve and apply rates of change?

Examine the properties of functions, including domain, asymptotes, symmetry and stationary points, to sketch the graph of a rational or other function, and apply differentiation to rates of change and optimisation problems.

A focused answer to the SQA Advanced Higher Mathematics functions and graph sketching content, covering domain and symmetry, vertical and non-vertical asymptotes of rational functions, stationary points and their nature, the systematic sketching of a curve, and applying differentiation to rates of change and optimisation.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-14

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Properties of functions
Asymptotes of rational functions
Stationary points and sketching
Rates of change and optimisation
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What this dot point is asking

The SQA wants you to analyse a function's features, domain, symmetry, asymptotes and stationary points, well enough to sketch its graph, and to apply differentiation to rates of change and optimisation. This pulls the differentiation toolkit into a single, structured task.

Properties of functions

Before sketching, read off the structural features. The domain excludes any input that breaks the function, such as a value making a denominator zero or a square root negative. Symmetry saves work: a function is even if $f(-x) = f(x)$ (symmetric in the $y$ -axis) and odd if $f(-x) = -f(x)$ (symmetric under a half-turn about the origin).

Asymptotes of rational functions

A rational function can run off to infinity in two ways. A vertical asymptote occurs at any $x$ where the denominator is zero but the numerator is not. A non-vertical (horizontal or oblique) asymptote describes the behaviour as $x \to \pm\infty$ , found by algebraic division.

Finding asymptotes of a rational function

Find the asymptotes of $y = \dfrac{2x + 1}{x - 3}$ .

Step 1: Find the vertical asymptote

A vertical asymptote occurs where the denominator is zero and the numerator is not. Set $x - 3 = 0$ , giving $x = 3$ . The numerator at $x = 3$ is $7 \ne 0$ , confirming a genuine asymptote rather than a hole:

x = 3\ \text{is a vertical asymptote.}

Step 2: Divide to find the behaviour as $x \to \pm\infty$

Polynomial division reveals the non-vertical asymptote. Divide the numerator by the denominator:

\frac{2x + 1}{x - 3} = 2 + \frac{7}{x - 3}.

Step 3: Read the horizontal asymptote

As $x \to \pm\infty$ the remainder term $\dfrac{7}{x - 3} \to 0$ , so the curve approaches $y = 2$ :

y = 2\ \text{is the horizontal asymptote.}

Final answer: vertical asymptote $x = 3$ ; horizontal asymptote $y = 2$ .

When the numerator degree is exactly one more than the denominator, division leaves a linear part, giving an oblique (slant) asymptote instead of a horizontal one.

Stationary points and sketching

Stationary points occur where $\dfrac{dy}{dx} = 0$ ; classify each as a maximum, minimum or point of inflection using the second derivative or a nature table. With the domain, intercepts, asymptotes and stationary points in hand, a clean sketch follows.

Locating and classifying stationary points

Locate and classify the stationary points of $y = x^3 - 3x^2$ .

Step 1: Differentiate and solve for $x$

Stationary points occur where the gradient is zero. Factorise the derivative to find both values of $x$ at once:

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

Step 2: Classify each stationary point using the second derivative

The sign of $\dfrac{d^2y}{dx^2}$ tells us whether each point is a maximum (negative) or minimum (positive):

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

At $x = 0$ : $6(0) - 6 = -6 < 0$ , so this is a local maximum.
At $x = 2$ : $6(2) - 6 = 6 > 0$ , so this is a local minimum.

Step 3: State the coordinates

Evaluate $y$ at each stationary point:

y(0) = 0, \quad y(2) = 8 - 12 = -4.

Final answer: $(0,\,0)$ is a local maximum and $(2,\,-4)$ is a local minimum.

Rates of change and optimisation

A rate-of-change problem links two changing quantities through a derivative; an optimisation problem maximises or minimises a quantity subject to a constraint. The method is the same: express the target as a function of one variable using the constraint, differentiate, set the derivative to zero, and confirm the nature with the second derivative. A connected-rates problem additionally uses the chain rule to link two rates, for example $\dfrac{dV}{dt} = \dfrac{dV}{dr}\dfrac{dr}{dt}$ , when a volume and a radius both change with time. In every case the structure is the same: find the relationship, differentiate, and substitute the known values at the instant in question.

Try this

Q1. State the vertical asymptote of $y = \dfrac{1}{x + 2}$ . [1 mark]

Cue. $x = -2$ , where the denominator is zero.

Q2. Is $f(x) = x^3$ even or odd? [1 mark]

Cue. $f(-x) = -x^3 = -f(x)$ , so it is odd.

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AH style: asymptotes4 marksFind the vertical and non-vertical asymptotes of

y = \dfrac{x^2}{x - 1}

Show worked answer →

Vertical asymptote where the denominator is zero: $x - 1 = 0$ , so $x = 1$ (1 mark).

Divide to find the non-vertical asymptote: $\dfrac{x^2}{x - 1} = x + 1 + \dfrac{1}{x - 1}$ (2 marks).

As $x \to \pm\infty$ the remainder $\dfrac{1}{x - 1} \to 0$ , so the oblique asymptote is $y = x + 1$ (1 mark). Markers reward the vertical asymptote, the algebraic division, and identifying the slant asymptote from the polynomial part.

AH style: optimisation5 marksA closed cylinder has volume

1000\,\text{cm}^3

. Find the radius that minimises its surface area.

Show worked answer →

Volume $\pi r^2 h = 1000$ , so $h = \dfrac{1000}{\pi r^2}$ (1 mark).

Surface area $S = 2\pi r^2 + 2\pi r h = 2\pi r^2 + \dfrac{2000}{r}$ (1 mark).

$\dfrac{dS}{dr} = 4\pi r - \dfrac{2000}{r^2}$ ; set to zero: $4\pi r^3 = 2000$ , so $r^3 = \dfrac{500}{\pi}$ and $r = \left(\dfrac{500}{\pi}\right)^{1/3} \approx 5.42\,\text{cm}$ (2 marks).

$\dfrac{d^2S}{dr^2} = 4\pi + \dfrac{4000}{r^3} > 0$ , confirming a minimum (1 mark). Markers reward the constraint, the surface-area function, the derivative set to zero, and the second-derivative check.

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Sources & how we know this

SQA Advanced Higher Mathematics Course Specification — SQA (2019)