Sketch rational functions, identifying vertical, horizontal and oblique asymptotes.

Sketching rational functions using intercepts, vertical asymptotes, horizontal and oblique asymptotes via division, and sign analysis, with worked examples for TCE Mathematics Specialised Unit 3.

Generated by Claude Opus 4.79 min answerUpdated 2026-05-30

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What this dot point is asking

This dot point pulls together everything in the sketching strand. A rational function combines zeros, asymptotes and sign behaviour, and a clean sketch comes from working through a fixed checklist rather than plotting random points.

The checklist for a rational function

Domain and vertical asymptotes. Set $Q(x) = 0$ . Each zero of the denominator that is not cancelled by the numerator gives a vertical asymptote.
Intercepts. The $y$ intercept is the value at $x = 0$ . The $x$ intercepts are where $P(x) = 0$ (and $Q(x) \ne 0$ ).
End behaviour. Compare degrees of $P$ and $Q$ to find horizontal or oblique asymptotes.
Sign analysis. Use the zeros and asymptotes to split the number line into intervals and test the sign in each.

End behaviour rules

Finding an oblique asymptote

When the numerator degree is one more than the denominator, divide $P(x)$ by $Q(x)$ . The quotient is a linear expression $y = mx + c$ , which is the oblique asymptote, and the remainder term vanishes as $x \to \pm\infty$ .

Sketching a rational function with a horizontal asymptote

Worked example

Sketch $y = \dfrac{2x}{x - 1}$ .

Vertical asymptote: denominator zero at $x = 1$ . Intercepts: at $x = 0$ , $y = 0$ , so the graph passes through the origin, which is both the $x$ and $y$ intercept. End behaviour: degrees are equal (both $1$ ), so the horizontal asymptote is $y = \tfrac{2}{1} = 2$ .

Sign analysis around $x = 1$ : just left of $1$ the denominator is small negative and the numerator near $2$ , so $y \to -\infty$ ; just right of $1$ , $y \to +\infty$ . For large positive $x$ , $y$ approaches $2$ from above; for large negative $x$ , it approaches $2$ from below. The curve has two branches separated by the asymptote $x = 1$ , with horizontal asymptote $y = 2$ .

Sketching a rational function with an oblique asymptote

Worked example

Sketch $y = \dfrac{x^2 + 1}{x}$ .

Vertical asymptote at $x = 0$ . There are no $x$ intercepts because $x^2 + 1 > 0$ always. The numerator degree ( $2$ ) is one more than the denominator degree ( $1$ ), so divide:

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

The oblique asymptote is $y = x$ , and the leftover term $\tfrac{1}{x}$ vanishes as $x \to \pm\infty$ . For $x > 0$ the curve lies above the line $y = x$ , with a minimum where the derivative is zero; differentiating $x + \tfrac1x$ gives $1 - \tfrac{1}{x^2} = 0$ , so $x = 1$ , $y = 2$ . By the odd symmetry of $x + \tfrac1x$ , the third quadrant branch has a maximum at $(-1, -2)$ . The graph approaches $y = x$ at the ends and the vertical asymptote near $x = 0$ .

Sign and shape

After placing intercepts and asymptotes, a sign table over the intervals between them tells you which side of each asymptote the curve sits on. This is usually enough to sketch without finding every turning point, though setting the derivative to zero pins down any local maxima or minima precisely.

Why this matters

Rational functions appear throughout the course, from partial fractions in integration to modelling problems with limiting behaviour. A systematic checklist, intercepts then asymptotes then sign, produces a correct sketch quickly and avoids the guesswork of point plotting.

Exam-style practice questions

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

2022 TASC7 marksConsider f(x) = (x^2 - 5x + 7)/(x - 2). (a) Determine the locations of any zeros and critical points of f(x). (b) Sketch the graph of y = f(x).

Show worked answer →

(a) Zeros: set the numerator to zero. x^2 - 5x + 7 = 0 has discriminant 25 - 28 = -3 < 0, so there are no real zeros. Vertical asymptote at x = 2 (denominator zero). Divide to find the oblique asymptote: x^2 - 5x + 7 = (x - 2)(x - 3) + 1, so f(x) = x - 3 + 1/(x - 2) and the oblique asymptote is y = x - 3. For critical points, f'(x) = 1 - 1/(x - 2)^2 = 0 gives (x - 2)^2 = 1, so x = 1 or x = 3. f(1) = -3 (local maximum) and f(3) = 1 (local minimum). (4 marks)

(b) Sketch the two branches separated by the vertical asymptote x = 2, approaching the oblique asymptote y = x - 3 at both ends. The left branch passes through the local maximum (1, -3); the right branch passes through the local minimum (3, 1). No x-intercepts; y-intercept at f(0) = 7/(-2) = -3.5. (3 marks)

2023 TASC7 marksConsider f(x) = (x^2 + 3)/(x - 1). (a) Determine the intercepts and critical points of f(x). (b) Sketch the graph of y = f(x).

Show worked answer →

(a) y-intercept: f(0) = 3/(-1) = -3. x-intercepts: x^2 + 3 = 0 has no real solution, so there are none. Vertical asymptote x = 1. Polynomial division: x^2 + 3 = (x - 1)(x + 1) + 4, so f(x) = x + 1 + 4/(x - 1) with oblique asymptote y = x + 1. Critical points: f'(x) = 1 - 4/(x - 1)^2 = 0 gives (x - 1)^2 = 4, so x = 3 or x = -1. f(3) = 6 (local minimum on the right branch) and f(-1) = -2 (local maximum on the left branch). (5 marks)

(b) Draw the vertical asymptote x = 1 and oblique asymptote y = x + 1. The left branch rises to the local maximum (-1, -2) and passes through the y-intercept (0, -3); the right branch dips to the local minimum (3, 6). (2 marks)

2021 TASC9 marksConsider f(x) = (x^2 - 4x + 20)/(x - 2). (a) Determine the intercepts and critical points of f(x). (b) Find the behaviour of f(x) as x tends to infinity. (c) Sketch the graph of y = f(x).

Show worked answer →

(a) y-intercept: f(0) = 20/(-2) = -10. x-intercepts: x^2 - 4x + 20 = 0 has discriminant 16 - 80 < 0, so none. Vertical asymptote x = 2. Since x^2 - 4x + 20 = (x - 2)^2 + 16, f(x) = (x - 2) + 16/(x - 2). f'(x) = 1 - 16/(x - 2)^2 = 0 gives (x - 2)^2 = 16, so x = 6 or x = -2. f(6) = 8 (local minimum) and f(-2) = -8 (local maximum). (4 marks)

(b) As x tends to plus or minus infinity, 16/(x - 2) tends to 0, so f(x) approaches the oblique asymptote y = x - 2. (1 mark)

(c) Sketch the vertical asymptote x = 2 and oblique asymptote y = x - 2. The left branch peaks at the local maximum (-2, -8); the right branch bottoms at the local minimum (6, 8). (4 marks)