State the asymptotes of y = 5x - 2. [2 marks]

Determine whether f(x) = x^3 - x is even, odd or neither. [2 marks]

SingaporeMathsSyllabus dot point

What are the key features of a curve, and how do asymptotes and symmetry guide a sketch?

Identify and use the key features of a curve - intercepts, turning points, asymptotes, symmetry and behaviour at infinity - to produce and interpret graph sketches

A focused answer to the H2 Mathematics outcome on curve features. Vertical, horizontal and oblique asymptotes, symmetry, behaviour at infinity, and how these features combine to determine a sketch.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

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

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to recognise and use the structural features of a curve - axis intercepts, turning points, asymptotes, symmetry and behaviour at large $|x|$ - both to draw a sketch and to read information off a given graph. These features are the vocabulary of all curve sketching.

The answer

The three kinds of asymptote

Vertical asymptote: a line $x = a$ that the curve approaches as $y \to \pm\infty$ , occurring where a denominator is zero but the numerator is not.
Horizontal asymptote: a line $y = b$ that the curve approaches as $x \to \pm\infty$ , found by examining the limit of the function at infinity.
Oblique (slant) asymptote: a line $y = mx + c$ approached at infinity, arising when the degree of the numerator is one more than the denominator.

Behaviour at infinity

To find horizontal or oblique behaviour, divide by the highest power of $x$ in the denominator, or carry out polynomial division. Whatever the remainder fraction tends to zero leaves the dominant part, which is the asymptote. Noting whether the curve approaches from above or below sharpens the sketch.

Symmetry

A function is even if $\mathrm{f}(-x) = \mathrm{f}(x)$ ; its graph is symmetric about the $y$ -axis.
A function is odd if $\mathrm{f}(-x) = -\mathrm{f}(x)$ ; its graph has rotational symmetry of order two about the origin.

Spotting symmetry halves the sketching work.

Turning points and intercepts

Turning points come from $\dfrac{dy}{dx} = 0$ ; intercepts come from setting $x = 0$ and $y = 0$ . Together with the asymptotes they pin down where the curve must go.

Worked example

Find all the asymptotes and the symmetry of $y = \dfrac{x}{x^2 - 1}$ , and describe the curve.

Step 1: Vertical asymptotes

$x^2 - 1 = (x - 1)(x + 1) = 0$ at $x = 1$ and $x = -1$ ; the numerator is nonzero there, so both are vertical asymptotes.

Step 2: Behaviour at infinity

Divide by $x^2$ : $y = \dfrac{1/x}{1 - 1/x^2}$ . As $x \to \pm\infty$ this tends to $\dfrac{0}{1} = 0$ , so $y = 0$ is a horizontal asymptote.

Step 3: Symmetry

Replace $x$ by $-x$ : $\dfrac{-x}{x^2 - 1} = -\dfrac{x}{x^2 - 1}$ , so the function is odd and the graph has rotational symmetry about the origin.

Step 4: Intercept and shape

The only intercept is the origin (numerator zero at $x = 0$ ). The curve passes through $(0,0)$ , has three branches divided by $x = -1$ and $x = 1$ , and the odd symmetry means the left half is the right half rotated $180^\circ$ about the origin.

Common traps

Calling a removable factor a vertical asymptote: If a factor cancels between numerator and denominator, that $x$ -value is a hole, not an asymptote.
Assuming a horizontal asymptote exists: When the numerator degree exceeds the denominator degree by one, the asymptote is oblique, not horizontal; by more than one there is no linear asymptote.
Confusing even and odd: Even means symmetry in the $y$ -axis ( $\mathrm{f}(-x) = \mathrm{f}(x)$ ); odd means rotational symmetry about the origin ( $\mathrm{f}(-x) = -\mathrm{f}(x)$ ).
Believing the curve cannot cross a horizontal asymptote: It may cross a horizontal or oblique asymptote at finite $x$ ; it only cannot cross a vertical one.
Ignoring the side of approach: Stating the asymptote without noting whether the curve nears it from above or below loses the finer marks.

Examples in context

Example 1. A logistic-style limit. A response curve $y = \dfrac{100x}{x + 5}$ has horizontal asymptote $y = 100$ , telling you the response saturates at $100$ units however large the input. Reading the asymptote answers the practical question before any plotting.

Example 2. Using symmetry to save work. Because $y = \dfrac{x}{x^2 + 1}$ is odd, you sketch it for $x \geq 0$ (a hump peaking where $\frac{dy}{dx} = 0$ , then decaying to the asymptote $y = 0$ ) and rotate it about the origin to complete the picture.

Try this

Q1. State the asymptotes of $y = \dfrac{5}{x - 2}$ . [2 marks]

Cue. Vertical $x = 2$ ; horizontal $y = 0$ .

Q2. Determine whether $\mathrm{f}(x) = x^3 - x$ is even, odd or neither. [2 marks]

Cue. $\mathrm{f}(-x) = -x^3 + x = -\mathrm{f}(x)$ , so odd.

Q3. Explain how to detect an oblique asymptote and find its equation. [2 marks]

Cue. Numerator degree one more than denominator; polynomial division gives a linear quotient, which is the asymptote.

Exam-style practice questions

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

Original4 marksState the equations of all asymptotes of the curve

y = \dfrac{1}{x^2 - 9}

and describe its symmetry.

Show worked answer →

The denominator $x^2 - 9 = (x - 3)(x + 3)$ is zero at $x = 3$ and $x = -3$ , giving vertical asymptotes $x = 3$ and $x = -3$ .

As $x \to \pm\infty$ , $\dfrac{1}{x^2 - 9} \to 0$ , so $y = 0$ is a horizontal asymptote.

Replacing $x$ by $-x$ leaves the function unchanged, so the curve is symmetric about the $y$ -axis (it is an even function).

Markers reward both vertical asymptotes, the horizontal asymptote, and identifying the $y$ -axis symmetry from the even structure.

Original4 marksExplain how to determine the behaviour of

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

x \to \pm\infty

, and state the resulting asymptote.

Show worked answer →

Divide numerator and denominator by $x^2$ : $y = \dfrac{2}{1 + \frac{1}{x^2}}$ .

As $x \to \pm\infty$ , $\dfrac{1}{x^2} \to 0$ , so $y \to \dfrac{2}{1} = 2$ .

Hence $y = 2$ is a horizontal asymptote, approached from below since $\dfrac{1}{x^2} > 0$ makes the denominator slightly larger than $1$ , giving $y$ slightly less than $2$ .

Markers reward dividing through by the highest power, taking the limit, and stating the asymptote with the side of approach.