What is wrong rule for the horizontal asymptote?

Use y = 0 only when the numerator degree is strictly smaller. Equal degrees give the ratio of leading coefficients, not 0.

SACE Board SACE 2022-style practice: Calculator-assumed. Let f(x) = 5(x-2)(x+3). Identify the asymptotes and intercepts, and describe the sign of f on each branch.

Vertical asymptotes: the denominator is zero at x = 2 and x = -3. [1 mark] Horizontal asymptote: numerator degree 0, denominator degree 2, so as x → ∞, f(x) → 0; the line y = 0 is a horizontal asymptote. There is no x-intercept (numerator 5 0). The y-intercept is f(0) = 5(-2)(3) = -56. [1 mark] Sign: for x 2 the product (x-2)(x+3) > 0, so f(x) > 0 (above the axis, approaching 0). For -3 < x < 2 the product is negative, so f(x) < 0 on the middle branch, dipping below the axis. [1 mark]

SACE Board SACE 2023-style practice: Calculator-assumed. For f(x) = x^2 + 1x - 1, find all asymptotes and the y-intercept, and state whether the curve has any x-intercepts.

Vertical asymptote: denominator zero at x = 1, numerator there is 2 0, so x = 1 is a vertical asymptote. [1 mark] Oblique asymptote: divide, x^2 + 1x - 1 = x + 1 + 2x - 1. As x → ∞ the remainder term vanishes, so y = x + 1 is an oblique asymptote. [2 marks] Intercepts: y-intercept f(0) = 1-1 = -1; x^2 + 1 = 0 has no real solutions, so no x-intercepts. [1 mark]

SACE Board SACE 2021-style practice: Calculator-assumed. State the horizontal asymptote of g(x) = 2x^2 - 3x^2 + 4 and explain why it has no vertical asymptotes.

The degrees are equal, so the horizontal asymptote is the ratio of leading coefficients y = 21 = 2. [1 mark] The denominator x^2 + 4 is never zero for real x (since x^2 0 gives x^2 + 4 4), so there are no vertical asymptotes. [1 mark]

§-Syllabus dot point

SASpecialist MathematicsSyllabus dot point

How do the zeros of a numerator and denominator dictate the shape of a rational function's graph?

Sketch graphs of rational functions, identifying intercepts, vertical asymptotes, and horizontal or oblique asymptotes.

Sketching rational functions by locating intercepts and vertical asymptotes from the denominator, and finding horizontal or oblique asymptotes from the degrees of numerator and denominator.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

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

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What this dot point is asking
Vertical asymptotes and intercepts
Behaviour far from the origin
Finding an oblique asymptote
Putting the features together into a sketch
A horizontal-asymptote example

What this dot point is asking

You need to sketch graphs of rational functions, identifying their intercepts and their vertical, horizontal and oblique asymptotes.

Vertical asymptotes and intercepts

For $f(x) = \dfrac{P(x)}{Q(x)}$ with $P$ and $Q$ sharing no common factors:

Vertical asymptotes occur at the zeros of $Q(x)$ : the graph shoots to $\pm\infty$ there. Check the sign of $f$ on each side to see whether the branch goes up or down.
$x$ -intercepts occur where $P(x) = 0$ (provided $Q \ne 0$ there).
$y$ -intercept is $f(0)$ , when defined.

If $P$ and $Q$ share a factor, that point is a hole (removable discontinuity), not an asymptote - cancel the common factor first.

Behaviour far from the origin

Finding an oblique asymptote

When the numerator degree is exactly one more than the denominator, polynomial long division writes $f(x) = (\text{linear}) + \dfrac{\text{remainder}}{Q(x)}$ . As $x \to \pm\infty$ the remainder term vanishes, so the line is the oblique asymptote.

Sketching a rational function with an oblique asymptote

Sketch $f(x) = \dfrac{x^{2} + 1}{x - 1}$

Step 1: Find the vertical asymptote and intercepts

A vertical asymptote occurs where the denominator is zero but the numerator is not. The $y$ -intercept is found by evaluating $f(0)$ , and $x$ -intercepts require the numerator to be zero.

Vertical asymptote: denominator zero at $x = 1$ ; numerator there is $2 \ne 0$ , so $x = 1$ is a vertical asymptote.

Intercepts: $y$ -intercept $f(0) = \dfrac{1}{-1} = -1$ . For $x$ -intercepts, $x^2 + 1 = 0$ has no real solutions, so there are no $x$ -intercepts.

Step 2: Perform polynomial division to find the oblique asymptote

Because the numerator degree exceeds the denominator degree by exactly one, there is an oblique (slant) asymptote. Dividing $x^2+1$ by $x-1$ expresses the function as a linear part plus a remainder that vanishes as $x\to\pm\infty$ .

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

As $x \to \pm\infty$ the term $\dfrac{2}{x-1} \to 0$ , so the oblique asymptote is $y = x + 1$ .

Step 3: Determine behaviour near the asymptotes and sketch

Examining the sign of the remainder term tells us which side of the oblique asymptote each branch lies on. Checking the sign of $f$ on each side of the vertical asymptote tells us the direction of each branch there.

For large positive $x$ the $\dfrac{2}{x-1}$ term is positive, so the curve lies just above $y = x + 1$ ; for large negative $x$ it lies just below. Near $x = 1$ , the function $\to +\infty$ from the right and $\to -\infty$ from the left. Sketch two branches hugging the asymptotes $x = 1$ and $y = x + 1$ .

Putting the features together into a sketch

A reliable sketching routine for any rational function is: factorise numerator and denominator; mark vertical asymptotes (denominator zeros) and intercepts (numerator zeros and $f(0)$ ); determine the end behaviour from the degree comparison; then test the sign of $f$ on each interval the vertical asymptotes create. The sign test tells you which way each branch heads near an asymptote, and the end behaviour tells you how each branch settles far out. Joining these constraints smoothly, never crossing a vertical asymptote, produces an accurate graph. The order matters: features first, curve last.

A horizontal-asymptote example

For $g(x) = \dfrac{2x^2 - 3}{x^2 + 4}$ , the degrees are equal, so the horizontal asymptote is the ratio of leading coefficients $y = \dfrac{2}{1} = 2$ . The denominator $x^2 + 4$ has no real zeros, so there are no vertical asymptotes; the $y$ -intercept is $-\tfrac{3}{4}$ and the $x$ -intercepts solve $2x^2 - 3 = 0$ .

Exam-style practice questions

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

SACE 20223 marksCalculator-assumed. Let

f(x) = \dfrac{5}{(x-2)(x+3)}

. Identify the asymptotes and intercepts, and describe the sign of

f

on each branch.

Show worked answer →

Vertical asymptotes: the denominator is zero at $x = 2$ and $x = -3$ . [1 mark]

Horizontal asymptote: numerator degree $0$ , denominator degree $2$ , so as $x \to \pm\infty$ , $f(x) \to 0$ ; the line $y = 0$ is a horizontal asymptote. There is no $x$ -intercept (numerator $5 \ne 0$ ). The $y$ -intercept is $f(0) = \dfrac{5}{(-2)(3)} = -\dfrac{5}{6}$ . [1 mark]

Sign: for $x < -3$ and $x > 2$ the product $(x-2)(x+3) > 0$ , so $f(x) > 0$ (above the axis, approaching $0$ ). For $-3 < x < 2$ the product is negative, so $f(x) < 0$ on the middle branch, dipping below the axis. [1 mark]

SACE 20234 marksCalculator-assumed. For

f(x) = \dfrac{x^2 + 1}{x - 1}

, find all asymptotes and the

y

-intercept, and state whether the curve has any

x

-intercepts.

Show worked answer →

Vertical asymptote: denominator zero at $x = 1$ , numerator there is $2 \ne 0$ , so $x = 1$ is a vertical asymptote. [1 mark]

Oblique asymptote: divide, $\dfrac{x^2 + 1}{x - 1} = x + 1 + \dfrac{2}{x - 1}$ . As $x \to \pm\infty$ the remainder term vanishes, so $y = x + 1$ is an oblique asymptote. [2 marks]

Intercepts: $y$ -intercept $f(0) = \dfrac{1}{-1} = -1$ ; $x^2 + 1 = 0$ has no real solutions, so no $x$ -intercepts. [1 mark]

SACE 20212 marksCalculator-assumed. State the horizontal asymptote of

g(x) = \dfrac{2x^2 - 3}{x^2 + 4}

and explain why it has no vertical asymptotes.

Show worked answer →

The degrees are equal, so the horizontal asymptote is the ratio of leading coefficients $y = \dfrac{2}{1} = 2$ . [1 mark]

The denominator $x^2 + 4$ is never zero for real $x$ (since $x^2 \ge 0$ gives $x^2 + 4 \ge 4$ ), so there are no vertical asymptotes. [1 mark]