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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.

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  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

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:

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.

Exam-style practice questions

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

WACE 20218 marksCalculator-assumed. Let f(x)=x24x+3x1f(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 x1x \neq 1. (c) State all asymptotes and intercepts and sketch the graph.
Show worked answer →

A hole-versus-asymptote question.

(a) Factor the numerator: x24x+3=(x1)(x3)x^2 - 4x + 3 = (x - 1)(x - 3). The factor (x1)(x - 1) cancels with the denominator, so x=1x = 1 is a removable discontinuity (a hole), not an asymptote.

(b) For x1x \neq 1, f(x)=(x1)(x3)x1=x3f(x) = \dfrac{(x - 1)(x - 3)}{x - 1} = x - 3.

(c) So the graph is the line y=x3y = x - 3 with an open circle (hole) at x=1x = 1, where y=13=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 20245 marksCalculator-free. State the equations of the asymptotes of y=3x+5x2y = \dfrac{3x + 5}{x - 2} and describe the behaviour of the graph as x2x \to 2 from each side.
Show worked answer →

A standard asymptote-and-behaviour question.

Vertical asymptote where x2=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 x2+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 x2x \to 2^- the denominator is a small negative number, so yy \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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