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Rational functions and asymptotes (WACE Year 12 Mathematics Specialist Unit 3)

Syllabus dot point

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

8 min read

Quick answer

For a rational function y=P(x)Q(x)y = \tfrac{P(x)}{Q(x)}: vertical asymptotes occur where Q=0Q = 0 but P≠0P \neq 0; the end behaviour gives a horizontal asymptote if deg⁡P≤deg⁡Q\deg P \le \deg Q or an oblique (slant) asymptote if deg⁡P=deg⁡Q+1\deg P = \deg Q + 1. Add intercepts and a sign analysis and the graph follows.

Jump to a section
  1. What this dot point is asking
  2. Vertical asymptotes and holes
  3. End behaviour: horizontal and oblique asymptotes
  4. Reading degree differences quickly
  5. Intercepts and sign analysis
  6. Crossing a horizontal or oblique asymptote
  7. Behaviour near a vertical asymptote
  8. Assembling the sketch
  9. Exam-style questions

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 xx-value, not an asymptote. Near a vertical asymptote, check the sign of yy on each side to decide whether the branch goes up or down.

End behaviour: horizontal and oblique asymptotes

Compare degrees of numerator and denominator:

Degree comparison for end behaviour

If deg⁡P<deg⁡Q\deg P < \deg Q the horizontal asymptote is y=0y = 0. If deg⁡P=deg⁡Q\deg P = \deg Q it is y=leading coeff Pleading coeff Qy = \tfrac{\text{leading coeff }P}{\text{leading coeff }Q}. If deg⁡P=deg⁡Q+1\deg P = \deg Q + 1 there is an oblique asymptote found by polynomial division.

For an oblique asymptote, divide PP by QQ; the quotient (a linear expression) is the asymptote and the remainder term vanishes as x→±∞x \to \pm\infty.

Reading degree differences quickly

The degree comparison can be made mechanical. Divide the numerator and denominator by the highest power of xx in the denominator and let x→±∞x \to \pm\infty. If the numerator degree is smaller, every surviving term has a positive power of xx in the denominator and the quotient tends to 00, giving the asymptote y=0y = 0. If the degrees are equal, only the ratio of leading coefficients survives, giving a non-zero horizontal asymptote. If the numerator degree is exactly one greater, the polynomial division leaves a linear quotient plus a remainder that vanishes, giving the oblique asymptote. A numerator two or more degrees higher has no linear asymptote and the curve grows without bound, which is beyond the standard SCSA cases but worth recognising.

Intercepts and sign analysis

The yy-intercept is the value at x=0x = 0 (if defined). The xx-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 xx-axis, fixing the shape of each branch.

Crossing a horizontal or oblique asymptote

A frequent misconception is that a curve can never touch its asymptote. This is true only for vertical asymptotes. A horizontal or oblique asymptote describes end behaviour as x→±∞x \to \pm\infty, so the curve is free to cross it in the middle of the domain. To find where y=P(x)Q(x)y = \dfrac{P(x)}{Q(x)} meets its horizontal asymptote y=cy = c, solve P(x)Q(x)=c\dfrac{P(x)}{Q(x)} = c; a solution is a genuine crossing point that should be marked on the sketch.

Behaviour near a vertical asymptote

The sign of the function immediately on each side of a vertical asymptote determines whether the branch rises to +∞+\infty or falls to −∞-\infty. The quickest method is to substitute a value just above and just below the asymptote into the factored form and read the sign, or to use a full sign table. An odd-power factor in the denominator gives opposite signs on the two sides (the branches go opposite ways), while an even-power factor gives the same sign on both sides (both branches go the same way).

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.

Sketch y=x2−1x−2y = \dfrac{x^2 - 1}{x - 2}

We need to find all asymptotes, intercepts, and the sign of each branch to produce the sketch.

Step 1: Find vertical asymptotes and check for holes

Factor the numerator to identify any cancelled factors. The numerator is (x−1)(x+1)(x - 1)(x + 1), which shares no factor with the denominator (x−2)(x - 2), so x=2x = 2 is a genuine vertical asymptote, not a hole.

Step 2: Find the oblique asymptote by polynomial division

The numerator degree (2) is exactly one more than the denominator degree (1), so an oblique asymptote exists. Perform polynomial division:

x2−1x−2=x+2+3x−2.\frac{x^2 - 1}{x - 2} = x + 2 + \frac{3}{x - 2}.

The oblique asymptote is y=x+2y = x + 2. The remainder term 3x−2\tfrac{3}{x-2} is positive for x>2x > 2 and negative for x<2x < 2, so the curve lies above the asymptote to the right and below it to the left.

Step 3: Find the intercepts

Setting x=0x = 0 gives the yy-intercept: y=−1−2=12y = \tfrac{-1}{-2} = \tfrac{1}{2}. The xx-intercepts come from the numerator zeros: x=1x = 1 and x=−1x = -1.

Step 4: Perform a sign analysis and sketch

A sign check in each region shows that the right branch rises through the intercepts and follows the oblique asymptote from above. The left branch approaches the asymptote from below.

Final answer: Vertical asymptote x=2x = 2, oblique asymptote y=x+2y = x + 2, xx-intercepts at x=±1x = \pm 1, yy-intercept 12\tfrac{1}{2}.

Common traps
Calling a hole an asymptote
A cancelled factor produces a hole, not a vertical asymptote.
Missing the oblique asymptote
When the numerator degree is one more than the denominator, you must divide; there is no horizontal asymptote.
Wrong side near a vertical asymptote
Always test the sign just left and just right of the asymptote; the branches can go opposite ways.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2021-style8 marks
Calculator-assumed. Let f(x)=x2−4x+3x−1f(x) = \dfrac{x^2 - 4x + 3}{x - 1}. (a) Show that ff has a removable discontinuity, stating its location. (b) Simplify f(x)f(x) for x≠1x \neq 1. (c) State all asymptotes and intercepts and sketch the graph.
Show worked answer →

A hole-versus-asymptote question.

(a) Factor the numerator: x2−4x+3=(x−1)(x−3)x^2 - 4x + 3 = (x - 1)(x - 3). The factor (x−1)(x - 1) cancels with the denominator, so x=1x = 1 is a removable discontinuity (a hole), not an asymptote.

(b) For x≠1x \neq 1, f(x)=(x−1)(x−3)x−1=x−3f(x) = \dfrac{(x - 1)(x - 3)}{x - 1} = x - 3.

(c) So the graph is the line y=x−3y = x - 3 with an open circle (hole) at x=1x = 1, where y=1−3=−2y = 1 - 3 = -2, i.e. the point (1,−2)(1, -2) is excluded. There are no asymptotes. xx-intercept at x=3x = 3, yy-intercept at (0,−3)(0, -3).

Markers reward the factor cancellation, identifying the hole at (1,−2)(1, -2), the simplified line, and a sketch with the open circle.

WACE 2024-style5 marks
Calculator-free. State the equations of the asymptotes of y=3x+5x−2y = \dfrac{3x + 5}{x - 2} and describe the behaviour of the graph as x→2x \to 2 from each side.
Show worked answer →

A standard asymptote-and-behaviour question.

Vertical asymptote where x−2=0x - 2 = 0, so x=2x = 2 (numerator non-zero there). Degrees are equal, so the horizontal asymptote is y=31=3y = \dfrac{3}{1} = 3.

Near x=2x = 2: for x→2+x \to 2^+ the denominator is a small positive number and the numerator is 3(2)+5=11>03(2) + 5 = 11 > 0, so y→+∞y \to +\infty. For x→2−x \to 2^- the denominator is a small negative number, so y→−∞y \to -\infty.

Markers reward both asymptotes and the correct +∞+\infty and −∞-\infty behaviour determined from the sign of the denominator on each side.

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