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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 4.79 min answer

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  1. What this dot point is asking
  2. Vertical asymptotes and intercepts
  3. Behaviour far from the origin
  4. Finding an oblique asymptote
  5. 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)=P(x)Q(x)f(x) = \dfrac{P(x)}{Q(x)} with PP and QQ sharing no common factors:

  • Vertical asymptotes occur at the zeros of Q(x)Q(x): the graph shoots to ±\pm\infty there. Check the sign of ff on each side to see whether the branch goes up or down.
  • xx-intercepts occur where P(x)=0P(x) = 0 (provided Q0Q \ne 0 there).
  • yy-intercept is f(0)f(0), when defined.

If PP and QQ 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)=(linear)+remainderQ(x)f(x) = (\text{linear}) + \dfrac{\text{remainder}}{Q(x)}. As x±x \to \pm\infty the remainder term vanishes, so the line is the oblique asymptote.

A horizontal-asymptote example

For g(x)=2x23x2+4g(x) = \dfrac{2x^2 - 3}{x^2 + 4}, the degrees are equal, so the horizontal asymptote is the ratio of leading coefficients y=21=2y = \dfrac{2}{1} = 2. The denominator x2+4x^2 + 4 has no real zeros, so there are no vertical asymptotes; the yy-intercept is 34-\tfrac{3}{4} and the xx-intercepts solve 2x23=02x^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.

2017 SACE Stage 23 marksLet f(x) = 5/((x-2)(x+3)). Draw the graph of y = f(x).
Show worked answer →

Find the key features before sketching.

Vertical asymptotes: the denominator is zero at x = 2 and x = -3, so there are vertical asymptotes at x = 2 and x = -3. [1 mark]

Horizontal asymptote: the numerator has degree 0 and the denominator degree 2, so as x -> plus or minus infinity, f(x) -> 0. The line y = 0 (the x-axis) is a horizontal asymptote. There is no x-intercept (the numerator 5 is never 0). The y-intercept is f(0) = 5/((-2)(3)) = -5/6. [1 mark]

Sign and shape: for x < -3 and x > 2 the product (x-2)(x+3) is positive, so f(x) > 0 (curve above the axis, approaching 0). For -3 < x < 2 the product is negative, so f(x) < 0; this middle branch dips below the axis with a maximum (least negative) value near x = -1/2. The three branches together, hugging the asymptotes x = -3, x = 2 and y = 0, give the sketch. [1 mark]