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WASpecialist MathematicsSyllabus dot point

Given the graph of y equals f of x, how do we sketch the graph of its reciprocal one over f of x?

Sketch the reciprocal of a function, relating zeros to asymptotes and turning points to turning points

WACE Specialist Unit 3 reciprocal graphs: how zeros of f become vertical asymptotes of one over f, where the reciprocal is large or small, sign preservation, fixed points at plus and minus one, and turning point behaviour, with a worked example.

Generated by Claude Opus 4.87 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Zeros become asymptotes
Large becomes small, small becomes large
A worked structural reading
Sign and fixed points
Turning points swap type
Why the turning-point swap happens
Asymptotes of f become zeros of the reciprocal
Sketching routine

What this dot point is asking

SCSA wants you to transform a known graph into the graph of its reciprocal using these structural rules, without computing a new formula point by point.

Zeros become asymptotes

Wherever $f(x) = 0$ , the reciprocal $\tfrac{1}{f(x)}$ is undefined and blows up, so each zero of $f$ gives a vertical asymptote of the reciprocal. The sign of $f$ just either side of the zero tells you whether the reciprocal branch goes to $+\infty$ or $-\infty$ .

Large becomes small, small becomes large

A worked structural reading

Take a cubic $f$ with zeros at $x = -1, 1, 3$ and the usual cubic shape. The reciprocal $\dfrac{1}{f}$ then has three vertical asymptotes, one at each of $x = -1, 1, 3$ . Between consecutive zeros $f$ changes sign, so the reciprocal also changes sign across each asymptote. Wherever $f$ has a local maximum or minimum, the reciprocal has a stationary point of the opposite type (provided $f$ is non-zero there), and far out where the cubic grows large the reciprocal hugs the $x$ -axis. Reading the reciprocal off the original graph this way, feature by feature, is faster and less error-prone than substituting numbers, and it is exactly the reasoning SCSA expects in a sketch.

Sign and fixed points

The reciprocal has the same sign as $f$ , since $\tfrac{1}{f}$ is positive exactly when $f$ is positive. The graphs cross where $f(x) = \tfrac{1}{f(x)}$ , that is where $f(x) = \pm 1$ ; these points are shared by both curves.

Turning points swap type

Where $f$ has a local maximum (and is positive there), $\tfrac{1}{f}$ has a local minimum at the same $x$ , because dividing one into a peak gives a trough. Likewise a positive local minimum of $f$ becomes a local maximum of the reciprocal. Where $f$ is negative the roles invert accordingly.

Why the turning-point swap happens

The swap of maxima and minima follows from calculus, which is worth knowing even though the sketch is done by reasoning. If $g(x) = \dfrac{1}{f(x)}$ then $g'(x) = -\dfrac{f'(x)}{[f(x)]^2}$ . The denominator $[f(x)]^2$ is always positive, so $g'(x) = 0$ exactly where $f'(x) = 0$ : the reciprocal has its stationary points at the same $x$ -values as $f$ . The sign of $g'$ is the opposite of the sign of $f'$ , so the curve that was rising is now falling and vice versa, which is precisely why a maximum of a positive $f$ becomes a minimum of $\dfrac{1}{f}$ .

Asymptotes of f become zeros of the reciprocal

The relationship runs both ways. Just as a zero of $f$ becomes a vertical asymptote of $\dfrac{1}{f}$ , a vertical asymptote of $f$ becomes a point where $\dfrac{1}{f} \to 0$ . So if $f$ itself is a rational function with a vertical asymptote at $x = a$ , the reciprocal approaches the $x$ -axis there. Tracking both directions of this duality lets you sketch the reciprocal of quite complicated graphs purely structurally.

Sketching routine

Mark the zeros of $f$ as asymptotes of the reciprocal, mark the $f = \pm 1$ crossing points, send the reciprocal toward $y = 0$ where $f$ is large, and turn the peaks into troughs. Then join smoothly.

Sketch

y = \dfrac{1}{f(x)}

where

f(x) = x^2 - 4

We apply the structural rules for reciprocal graphs: identify the key features of $f$ and translate each one into the corresponding feature of the reciprocal.

Step 1: Identify the key features of $f$

The parabola $f(x) = x^2 - 4$ has zeros at $x = \pm 2$ and a minimum value of $-4$ at $x = 0$ . Knowing these features is the foundation for reading off the reciprocal without further calculation.

Step 2: Locate the vertical asymptotes and sign of the reciprocal

Each zero of $f$ becomes a vertical asymptote of the reciprocal. Between $x = -2$ and $x = 2$ the parabola is negative, so the reciprocal is also negative in this region, with a turning point at $x = 0$ where $f$ reaches its minimum of $-4$ . Because the minimum of a negative $f$ is the least negative value, it becomes the maximum of this negative branch: $\tfrac{1}{-4} = -\tfrac{1}{4}$ .

Step 3: Describe the outer branches and end behaviour

For $|x| > 2$ the parabola is positive and grows without bound, so the reciprocal is positive and tends to $y = 0^+$ as $x \to \pm\infty$ . The $x$ -axis is a horizontal asymptote of the reciprocal.

Step 4: Find where $f$ and the reciprocal meet

The two curves share the same $y$ -value wherever $f(x) = \pm 1$ , that is wherever $x^2 - 4 = \pm 1$ , giving $x^2 = 3$ or $x^2 = 5$ .

Final answer: Vertical asymptotes at $x = \pm 2$ ; the middle branch is negative with a maximum at $\bigl(0, -\tfrac{1}{4}\bigr)$ ; the outer branches are positive and approach $y = 0$ ; intersection points at $x = \pm\sqrt{3}$ and $x = \pm\sqrt{5}$ .

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 20236 marksCalculator-assumed. The graph of

y = f(x)

is a straight line through

(0, -2)

and

(4, 0)

. (a) Find the rule for

f(x)

. (b) Sketch

y = \dfrac{1}{f(x)}

, showing the asymptote, the

f = \pm 1

crossing points and the end behaviour.

Show worked answer →

A reciprocal of a linear function.

(a) Slope $= \dfrac{0 - (-2)}{4 - 0} = \dfrac{1}{2}$ , so $f(x) = \dfrac{1}{2}x - 2$ .

(b) Zero of $f$ at $x = 4$ becomes a vertical asymptote of $\dfrac{1}{f}$ . As $x \to \pm\infty$ , $\dfrac{1}{f} \to 0$ , so $y = 0$ is a horizontal asymptote. Crossing points where $f = \pm 1$ : $\dfrac{1}{2}x - 2 = 1 \Rightarrow x = 6$ , and $\dfrac{1}{2}x - 2 = -1 \Rightarrow x = 2$ . The reciprocal is negative for $x < 4$ (where $f < 0$ ) and positive for $x > 4$ .

Markers reward the rule, the asymptote at $x = 4$ , the crossing points $(6, 1)$ and $(2, -1)$ , and the correct sign on each side.

WACE 20214 marksCalculator-free. The function

f

has a local minimum at

(2, 3)

and is positive everywhere. State, with reasons, the nature and coordinates of the corresponding feature of

y = \dfrac{1}{f(x)}

x = 2

Show worked answer →

A turning-point swap question.

Since $f$ is positive everywhere, $\dfrac{1}{f}$ is positive everywhere and has no asymptotes. At a local minimum of $f$ where $f > 0$ , the reciprocal has a local maximum, because making the denominator smallest makes the fraction largest.

So $y = \dfrac{1}{f(x)}$ has a local maximum at $\left(2, \dfrac{1}{3}\right)$ . Markers reward identifying the swap from minimum to maximum, the reason (smallest positive denominator gives largest value), and the coordinates $\left(2, \dfrac{1}{3}\right)$ .