What is wrong side near a vertical asymptote?

Always test the sign just left and just right of the asymptote; the branches can go opposite ways.

WASpecialist MathematicsSyllabus dot point

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

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

WACE Specialist Unit 3 rational functions: vertical asymptotes from denominator zeros, horizontal and oblique asymptotes from degree comparison, intercepts, holes from common factors, and sign analysis for sketching, with a worked example.

Generated by Claude Opus 4.77 min answerUpdated 2026-05-29

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

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What this dot point is asking
Vertical asymptotes and holes
End behaviour: horizontal and oblique asymptotes
Intercepts and sign analysis
Assembling the sketch

What this dot point is asking

SCSA wants a structured sketch of a rational function: intercepts, every asymptote, behaviour near each asymptote, and stationary points where relevant.

Vertical asymptotes and holes

Factor numerator and denominator. A zero of the denominator that is not cancelled by the numerator gives a vertical asymptote, where the curve shoots to $\pm\infty$ . A factor common to both gives a hole (removable discontinuity) at that $x$ -value, not an asymptote. Near a vertical asymptote, check the sign of $y$ on each side to decide whether the branch goes up or down.

End behaviour: horizontal and oblique asymptotes

Compare degrees of numerator and denominator:

For an oblique asymptote, divide $P$ by $Q$ ; the quotient (a linear expression) is the asymptote and the remainder term vanishes as $x \to \pm\infty$ .

Intercepts and sign analysis

The $y$ -intercept is the value at $x = 0$ (if defined). The $x$ -intercepts are the zeros of the numerator that are not also zeros of the denominator. A sign table across the intercepts and vertical asymptotes shows which regions sit above or below the $x$ -axis, fixing the shape of each branch.

Assembling the sketch

Draw the asymptotes as dashed lines first, mark the intercepts, then connect each branch consistently with the sign analysis and end behaviour. Stationary points can be located with calculus if the question requires them.

Worked example

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

Numerator factors as $(x - 1)(x + 1)$ ; denominator zero at $x = 2$ is not cancelled, so $x = 2$ is a vertical asymptote.

Degrees: $\deg P = 2 = \deg Q + 1$ , so there is an oblique asymptote. Divide:

\frac{x^2 - 1}{x - 2} = x + 2 + \frac{3}{x - 2}.

So the oblique asymptote is $y = x + 2$ , and the curve sits just above it for large positive $x$ and just below for large negative $x$ .

Intercepts: $x$ -intercepts at $x = 1$ and $x = -1$ ; $y$ -intercept at $x = 0$ gives $y = \tfrac{-1}{-2} = \tfrac{1}{2}$ . A sign check shows the right branch rises through the intercepts toward the asymptote $y = x + 2$ , and the left branch approaches it from below.