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Functions and sketching graphs (WACE Year 12 Mathematics Specialist Unit 3)

Syllabus dot point

“Sketch rational functions, reciprocal and modulus graphs, and use transformations and asymptotic behaviour”

8 min read

Quick answer

Sketch rational, reciprocal and modulus functions by finding intercepts, vertical asymptotes (where the denominator is zero), horizontal or oblique asymptotes (from end behaviour), and any turning points, then join the pieces respecting sign and asymptotic approach.

Jump to a section
  1. What this dot point is asking
  2. Rational functions and asymptotes
  3. The reciprocal of a graph
  4. Modulus functions
  5. A systematic sketching checklist
  6. Sign tables make the branches unambiguous
  7. Transformations of standard graphs
  8. Exam-style questions

What this dot point is asking

SCSA expects you to analyse and sketch graphs of rational functions y=P(x)Q(x)y = \tfrac{P(x)}{Q(x)}, graphs of y=1f(x)y = \tfrac{1}{f(x)} from the graph of y=f(x)y = f(x), and modulus graphs y=∣f(x)∣y = |f(x)| and y=f(∣x∣)y = f(|x|), using asymptotes, intercepts and transformations rather than plotting points.

Rational functions and asymptotes

A vertical asymptote occurs where Q(x)=0Q(x) = 0 but P(x)≠0P(x) \neq 0. Near it the graph shoots to ±∞\pm\infty; the sign on each side is found from a sign table.

For end behaviour, compare degrees of PP and QQ:

  • deg⁡P<deg⁡Q\deg P < \deg Q: horizontal asymptote y=0y = 0.
  • deg⁡P=deg⁡Q\deg P = \deg Q: horizontal asymptote y=leading coeff Pleading coeff Qy = \tfrac{\text{leading coeff } P}{\text{leading coeff } Q}.
  • deg⁡P=deg⁡Q+1\deg P = \deg Q + 1: an oblique (slant) asymptote, found by polynomial division.
Crossing an asymptote

A curve can cross a horizontal or oblique asymptote (the asymptote describes only end behaviour), but it can never cross a vertical asymptote.

The reciprocal of a graph

To sketch y=1f(x)y = \tfrac{1}{f(x)} from y=f(x)y = f(x):

  • Zeros of ff become vertical asymptotes of 1f\tfrac{1}{f}.
  • Where f→±∞f \to \pm\infty, 1f→0\tfrac{1}{f} \to 0.
  • Maxima of ff (above the axis) become minima of 1f\tfrac{1}{f} and vice versa.
  • Points where f=±1f = \pm 1 are fixed, since 1±1=±1\tfrac{1}{\pm 1} = \pm 1.
  • The sign of 1f\tfrac{1}{f} matches the sign of ff.

Modulus functions

For y=∣f(x)∣y = |f(x)|, reflect any part of y=f(x)y = f(x) that lies below the xx-axis up across the axis. For y=f(∣x∣)y = f(|x|), take the part of y=f(x)y = f(x) for x≥0x \ge 0 and reflect it in the yy-axis to give an even function.

A systematic sketching checklist

SCSA rewards a methodical approach, not point-plotting. Work through the same list every time:

  1. Domain. State any values excluded by a zero denominator or even root.
  2. Intercepts. Set y=0y = 0 for xx-intercepts and x=0x = 0 for the yy-intercept.
  3. Vertical asymptotes. Solve Q(x)=0Q(x) = 0 where P(x)≠0P(x) \neq 0, then a sign table on each side.
  4. End behaviour. Compare the degrees of numerator and denominator for a horizontal or oblique asymptote.
  5. Turning points. Differentiate and solve dydx=0\tfrac{dy}{dx} = 0 where the question needs them.
  6. Join the pieces respecting the sign of the function between consecutive features.

Sign tables make the branches unambiguous

The single most reliable tool for a rational sketch is the sign table. List every xx-value where the numerator or denominator is zero, in order; these split the number line into intervals. On each interval the function keeps a constant sign, found by testing one value or counting the negative factors. The sign table tells you whether each branch approaches a vertical asymptote from +∞+\infty or −∞-\infty and on which side of the xx-axis each branch lives, which is exactly the information a hand sketch needs.

Transformations of standard graphs

Many Specialist graphs are transformations of a known parent curve. The graph of y=a f(b(x−h))+ky = a\,f(b(x - h)) + k takes y=f(x)y = f(x), stretches it by factor aa vertically and 1b\tfrac{1}{b} horizontally, then translates it by hh right and kk up. Recognising a rational function as a translated reciprocal, for instance y=1x−2+3y = \dfrac{1}{x - 2} + 3 as the curve y=1xy = \dfrac{1}{x} shifted right 22 and up 33, gives the asymptotes x=2x = 2 and y=3y = 3 instantly without further work.

Worked example

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

Step 1: Find the intercepts.

Setting the numerator to zero gives the xx-intercepts, and setting x=0x = 0 gives the yy-intercept. These points anchor the sketch on the axes.

xx-intercepts where x2−1=0x^2 - 1 = 0, so x=±1x = \pm 1. yy-intercept: x=0x = 0 gives y=−1−2=12y = \tfrac{-1}{-2} = \tfrac{1}{2}.

Step 2: Find the vertical asymptote.

A vertical asymptote occurs where the denominator is zero and the numerator is non-zero. The curve shoots to ±∞\pm\infty on each side of this line.

x−2=0x - 2 = 0 gives x=2x = 2 (numerator non-zero there).

Step 3: Find the oblique asymptote by polynomial division.

When the degree of the numerator exceeds the degree of the denominator by exactly one, dividing gives a linear quotient that describes the slant the curve approaches as x→±∞x \to \pm\infty.

Degree of numerator exceeds denominator by one, so divide:

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

As x→±∞x \to \pm\infty the remainder vanishes, so the oblique asymptote is y=x+2y = x + 2.

Step 4: Determine the behaviour near the vertical asymptote and assemble the sketch.

Examining the sign of 3x−2\tfrac{3}{x-2} on each side of x=2x = 2 shows how each branch approaches the asymptote. The remainder 3x−2\tfrac{3}{x-2} is positive for x>2x > 2 and negative for x<2x < 2.

As x→2+x \to 2^+, 3x−2→+∞\tfrac{3}{x-2} \to +\infty, so y→+∞y \to +\infty; as x→2−x \to 2^-, y→−∞y \to -\infty.

Putting it together: a branch for x<2x < 2 passing through (±1,0)(\pm 1, 0) and (0,12)(0, \tfrac12), dropping to −∞-\infty near x=2x = 2, and a branch for x>2x > 2 rising from +∞+\infty and settling onto y=x+2y = x + 2.

Common traps
Cancelling a common factor silently
If a factor cancels, you get a hole (removable discontinuity), not an asymptote. State the excluded xx-value with an open circle.
Assuming the curve never meets a horizontal asymptote
It often does in the middle of the domain; only vertical asymptotes are truly forbidden.
Reflecting the wrong axis for modulus
∣f(x)∣|f(x)| reflects below-axis parts upward; f(∣x∣)f(|x|) reflects the right half leftward. They are different graphs.

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 2022-style7 marks
Calculator-assumed. Consider f(x)=2x2−8x2−1f(x) = \dfrac{2x^2 - 8}{x^2 - 1}. (a) State the equations of all asymptotes. (b) Find the coordinates of the axis intercepts. (c) Sketch the graph.
Show worked answer →

A full rational-function analysis.

(a) Vertical asymptotes where x2−1=0x^2 - 1 = 0, so x=1x = 1 and x=−1x = -1 (numerator non-zero there). Degrees are equal, so the horizontal asymptote is y=21=2y = \dfrac{2}{1} = 2.

(b) xx-intercepts where 2x2−8=02x^2 - 8 = 0, so x2=4x^2 = 4, giving (2,0)(2, 0) and (−2,0)(-2, 0). yy-intercept at x=0x = 0: y=−8−1=8y = \dfrac{-8}{-1} = 8, so (0,8)(0, 8).

(c) Three branches. The left branch for x<−1x < -1 comes down from y=2y = 2, passes through (−2,0)(-2, 0) and rises to +∞+\infty as x→−1−x \to -1^-. The middle branch for −1<x<1-1 < x < 1 passes through (0,8)(0, 8) as a local maximum and drops to −∞-\infty at both ends. The right branch for x>1x > 1 rises from −∞-\infty, crosses (2,0)(2, 0) and settles onto y=2y = 2.

Markers reward both vertical asymptotes, the horizontal asymptote y=2y = 2 from equal degrees, all intercepts, and a sketch with correct branch behaviour.

WACE 2024-style4 marks
Calculator-free. The graph of y=f(x)y = f(x) has a single xx-intercept at x=3x = 3 and a local maximum at (1,2)(1, 2). Describe the key features of the graph of y=1f(x)y = \dfrac{1}{f(x)}.
Show worked answer →

A reciprocal-graph reasoning question.

The zero of ff at x=3x = 3 becomes a vertical asymptote of 1f\dfrac{1}{f}. The local maximum of ff at (1,2)(1, 2), which is above the axis, becomes a local minimum of 1f\dfrac{1}{f} at (1,12)\left(1, \dfrac{1}{2}\right).

Where f→±∞f \to \pm\infty, 1f→0\dfrac{1}{f} \to 0, so 1f\dfrac{1}{f} has the xx-axis as a horizontal asymptote in those directions. The sign of 1f\dfrac{1}{f} matches the sign of ff on every interval. Markers reward the asymptote at x=3x = 3, the max-to-min swap at (1,12)\left(1, \dfrac{1}{2}\right), and the sign and end-behaviour statements.

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