Find the asymptotes of y = 3x - 1x + 2. [2 marks]

SingaporeMathsSyllabus dot point

How do we sketch a rational function by finding its intercepts, asymptotes and turning points?

Sketch graphs of rational functions of the form a linear over linear and a quadratic over linear, finding intercepts, asymptotes, stationary points and the regions where the curve lies

A focused answer to the H2 Mathematics outcome on sketching rational functions. Finding intercepts, vertical and oblique asymptotes, stationary points, and assembling a correct sketch of linear-over-linear and quadratic-over-linear curves.

Generated by Claude Opus 4.810 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 sketch rational functions, principally of the form $\dfrac{ax + b}{cx + d}$ (linear over linear) and $\dfrac{ax^2 + bx + c}{dx + e}$ (quadratic over linear), by systematically finding intercepts, asymptotes and stationary points and then assembling a clear sketch with the correct branches.

The answer

The systematic checklist

Every rational-function sketch follows the same routine:

Axis intercepts. Set $x = 0$ for the $y$ -intercept; set the numerator to zero for $x$ -intercepts.
Vertical asymptotes. Where the denominator is zero (and the numerator is not).
Horizontal or oblique asymptote. Divide the numerator by the denominator and look at the behaviour as $x \to \pm\infty$ .
Stationary points. Differentiate and set the derivative to zero (often needed for the quadratic-over-linear case).
Assemble. Plot the features and draw smooth branches approaching the asymptotes.

Linear over linear

For $\dfrac{ax + b}{cx + d}$ , rewrite by division as a constant plus a fraction: this is a translated rectangular hyperbola. The horizontal asymptote is $y = \dfrac{a}{c}$ and the vertical asymptote is $x = -\dfrac{d}{c}$ . There are no turning points.

Quadratic over linear

For $\dfrac{ax^2 + bx + c}{dx + e}$ , polynomial division gives a linear term plus a remainder fraction, so the curve has an oblique (slant) asymptote. These curves usually have a maximum and a minimum found by differentiation, and the two stationary values are separated by the vertical asymptote.

Worked example

Sketch the curve $y = \dfrac{x^2 - x - 2}{x - 1}$ , showing intercepts, asymptotes and stationary points.

Step 1: Axis intercepts

$x$ -intercepts: numerator $x^2 - x - 2 = (x - 2)(x + 1) = 0$ gives $x = 2$ and $x = -1$ . $y$ -intercept: $x = 0$ gives $y = \dfrac{-2}{-1} = 2$ .

Step 2: Vertical asymptote

Denominator zero at $x = 1$ (numerator is $-2 \neq 0$ there), so $x = 1$ is a vertical asymptote.

Step 3: Oblique asymptote by division

\frac{x^2 - x - 2}{x - 1} = x + \frac{-2}{x - 1} = x - \frac{2}{x - 1}

As $x \to \pm\infty$ the fraction vanishes, so $y = x$ is the oblique asymptote.

Step 4: Stationary points

$\dfrac{dy}{dx} = 1 + \dfrac{2}{(x - 1)^2}$ . This is always positive, so there are no stationary points and the curve is strictly increasing on each branch.

Step 5: Assemble

Two branches separated by $x = 1$ , each approaching the line $y = x$ at infinity, passing through $(-1, 0)$ , $(0, 2)$ and $(2, 0)$ , rising throughout.

Examples in context

Example 1. A concentration model. A drug concentration $C(t) = \dfrac{4t}{t + 2}$ for $t \geq 0$ is linear over linear with horizontal asymptote $C = 4$ . The curve rises from the origin toward $4$ , modelling a concentration that saturates, never quite reaching the limiting value.

Example 2. Reading the slant asymptote. For $y = \dfrac{x^2 + 3}{x}$ , division gives $y = x + \dfrac{3}{x}$ , so the curve hugs the line $y = x$ far from the origin. This tells you the long-run trend at a glance before any detailed plotting.

Try this

Q1. Find the asymptotes of $y = \dfrac{3x - 1}{x + 2}$ . [2 marks]

Cue. Vertical $x = -2$ ; horizontal $y = 3$ (ratio of leading coefficients).

Q2. The curve $y = \dfrac{x^2 + 4}{x}$ has stationary points. Find them and their nature. [3 marks]

Cue. $y = x + \dfrac{4}{x}$ , $\dfrac{dy}{dx} = 1 - \dfrac{4}{x^2} = 0$ gives $x = \pm 2$ ; $(2, 4)$ minimum, $(-2, -4)$ maximum.

Q3. Explain how to find an oblique asymptote of a rational function. [2 marks]

Cue. Carry out polynomial division; the polynomial (linear) part is the asymptote as the remainder fraction tends to zero.

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.

Original5 marksThe curve

C

has equation

y = \dfrac{2x + 1}{x - 3}

. Find the equations of the asymptotes, the axis intercepts, and hence sketch

C

Show worked answer →

Vertical asymptote where the denominator is zero: $x = 3$ .

For the horizontal asymptote, divide: $y = \dfrac{2x + 1}{x - 3} = 2 + \dfrac{7}{x - 3}$ (since $2(x - 3) = 2x - 6$ and $2x + 1 - (2x - 6) = 7$ ). As $x \to \pm\infty$ , $y \to 2$ , so $y = 2$ is a horizontal asymptote.

Intercepts: when $x = 0$ , $y = \dfrac{1}{-3} = -\tfrac{1}{3}$ ; when $y = 0$ , $2x + 1 = 0$ , so $x = -\tfrac{1}{2}$ .

The sketch is a rectangular hyperbola: one branch in the region $x > 3$ approaching $y = 2$ from above and rising near $x = 3^+$ , the other for $x < 3$ passing through the intercepts and approaching $y = 2$ from below.

Markers reward both asymptotes, both intercepts, and a correctly shaped two-branch sketch.

Original6 marksThe curve has equation

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

. Find the asymptotes and the coordinates of the stationary points, and sketch the curve.

Show worked answer →

Write $y = x + \dfrac{1}{x}$ . Vertical asymptote $x = 0$ ; oblique asymptote $y = x$ (since $\dfrac{1}{x} \to 0$ ).

Differentiate: $\dfrac{dy}{dx} = 1 - \dfrac{1}{x^2}$ . Set to zero: $x^2 = 1$ , so $x = \pm 1$ .

At $x = 1$ , $y = 2$ (minimum); at $x = -1$ , $y = -2$ (maximum). There are no axis intercepts since $x^2 + 1 > 0$ .

The sketch has two branches: for $x > 0$ a curve with minimum $(1, 2)$ sitting above the line $y = x$ ; for $x < 0$ a curve with maximum $(-1, -2)$ below the line.

Markers reward the split form, both asymptotes, both stationary points with their nature, and a correct sketch.