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How do transformations and structure determine the shape of a graph?

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

WACE Specialist Unit 3 functions and graphs: rational functions, vertical and horizontal asymptotes, the reciprocal of a graph, modulus functions, and curve sketching from intercepts, asymptotes and turning points.

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  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

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:

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

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, 1f0\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 x0x \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=af(b(xh))+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=1x2+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.

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 20227 marksCalculator-assumed. Consider f(x)=2x28x21f(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 x21=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 2x28=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=81=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 x1x \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 20244 marksCalculator-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, 1f0\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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