State the vertical asymptote of y = 3x + 4. [1 mark]

State the horizontal asymptote of y = 2x + 1x - 3. [1 mark]

Northern IrelandFurther MathsSyllabus dot point

How do you sketch rational functions and find their stationary points and asymptotes?

Curve sketching of rational functions, finding vertical and horizontal asymptotes, oblique asymptotes, the range of values a rational function can take, and locating stationary points.

A CCEA AS Further Maths answer on sketching rational functions, finding vertical, horizontal and oblique asymptotes, determining the set of values a rational function can take, and locating stationary points to complete the sketch.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-16

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

Have a quick question? Jump to the Q&A page

Jump to a section

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

What this dot point is asking

CCEA wants you to sketch rational functions: find vertical asymptotes (where the denominator is zero), horizontal or oblique asymptotes (the behaviour for large $|x|$ ), axis intercepts, the set of values $y$ can take, and any stationary points. A clean, labelled sketch showing all of these features is the goal.

The answer

Vertical asymptotes

Horizontal and oblique asymptotes

The range of values

Stationary points

Finding the range of a rational function

Find the set of values that $y = \dfrac{x}{x^2 + 1}$ can take.

step Rearrange into a quadratic in x

Multiply out: $y(x^2 + 1) = x$ , so $yx^2 - x + y = 0$ .

step Apply the discriminant condition

For real $x$ , the discriminant $b^2 - 4ac \geq 0$ with $a = y$ , $b = -1$ , $c = y$ :

(-1)^2 - 4(y)(y) \geq 0 \;\Rightarrow\; 1 - 4y^2 \geq 0 \;\Rightarrow\; y^2 \leq \tfrac{1}{4}.

step State the range

-\tfrac{1}{2} \leq y \leq \tfrac{1}{2}.

The function never goes outside this band, and reaches the ends at its turning points

x = \pm 1

Examples in context

Example 1. Reaction rate against concentration. Many saturating processes follow $v = \dfrac{V_{\max}\,s}{K + s}$ , whose graph rises and levels off at the horizontal asymptote $v = V_{\max}$ . Reading the asymptote tells a scientist the maximum rate before any data plateau is reached.

Example 2. Designing a band-limited response. An engineer who needs a quantity to stay within fixed bounds uses the discriminant method to confirm a transfer function such as $\frac{x}{x^2 + 1}$ never leaves $[-\frac{1}{2}, \frac{1}{2}]$ . The range calculation is a design guarantee, not just an exam exercise.

Try this

Q1. State the vertical asymptote of $y = \dfrac{3}{x + 4}$ . [1 mark]

Cue. $x + 4 = 0$ , so $x = -4$ .

Q2. State the horizontal asymptote of $y = \dfrac{2x + 1}{x - 3}$ . [1 mark]

Cue. Equal degrees, ratio of leading coefficients $\frac{2}{1}$ , so $y = 2$ .

Q3. For $y = \dfrac{x^2 + 3}{x}$ , by dividing, state the oblique asymptote. [2 marks]

Cue. $y = x + \frac{3}{x}$ , so the oblique asymptote is $y = x$ .

Exam-style practice questions

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

CCEA AS 20206 marksA curve has equation

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

. Find the equations of the asymptotes and the coordinates of the points where the curve crosses the axes.

Show worked answer →

Vertical asymptote where the denominator is zero: $x - 2 = 0$ , so $x = 2$ .

Horizontal asymptote: divide top and bottom by $x$ , $y = \dfrac{1 + 1/x}{1 - 2/x} \to \dfrac{1}{1} = 1$ as $x \to \pm\infty$ , so $y = 1$ .

Crosses the $x$ -axis when $y = 0$ : $x + 1 = 0$ , so $x = -1$ , point $(-1, 0)$ .

Crosses the $y$ -axis when $x = 0$ : $y = \dfrac{1}{-2} = -\dfrac{1}{2}$ , point $\left(0, -\dfrac{1}{2}\right)$ .

Markers reward the vertical asymptote from the denominator, the horizontal asymptote from the behaviour at infinity, and both intercepts.

CCEA AS 20197 marksA curve has equation

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

. Show that the curve has an oblique asymptote

y = x + 1

, and find the coordinates of its stationary points.

Show worked answer →

Divide: $\dfrac{x^2}{x - 1} = x + 1 + \dfrac{1}{x - 1}$ (since $x^2 = (x - 1)(x + 1) + 1$ ). As $x \to \pm\infty$ the remainder $\frac{1}{x-1} \to 0$ , so $y \to x + 1$ : the oblique asymptote is $y = x + 1$ .

Differentiate using the quotient rule on $y = \dfrac{x^2}{x - 1}$ :

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

Stationary points where the numerator is zero: $x = 0$ giving $y = 0$ , and $x = 2$ giving $y = \dfrac{4}{1} = 4$ .

So the stationary points are $(0, 0)$ and $(2, 4)$ . Markers reward the division to the oblique asymptote, the quotient-rule derivative, and both stationary points.

Related dot points

Sources & how we know this

CCEA GCE Further Mathematics specification — CCEA (2018)