Reciprocal/composite transformations | Year 12

01 / LEARN

The idea, step by step

Reciprocate ordinates

For y=1/f(x), nonzero ordinates become their reciprocals. Their signs stay the same; large magnitudes become small, and values close to zero produce large magnitudes. Points with ordinate 1 or -1 remain fixed.

Zeros of the original function are excluded inputs for its reciprocal. A genuine zero generally creates a vertical asymptote; check the complete formula and domain instead of guessing from a picture.

Use exact transformations

In a/(x-h)+k, the asymptotes are x=h and y=k, and their intersection is the centre of symmetry. The sign of a determines branch orientation.

For a differentiable function $f$ , the reciprocal derivative is $- f^{'} (x) / {[f (x)]}^{2}$ wherever $f (x) \neq 0$ . Use this derivative to locate stationary behaviour. Do not draw a graph through a pole merely to connect plotted points.

A common trap

Changing the sign when taking a reciprocal.

Instead: The reciprocal of a negative number remains negative.

02 / WORKED EXAMPLES

See the thinking

Example 1

The reciprocal of a quadratic

For $f (x) = x^{2} - 1$ , identify vertical asymptotes, intercepts and the stationary point of $1 / f (x)$ .

$g (x) = 1 / (x^{2} - 1)$ .
$g^{'} (x) = - 2 x / {(x^{2} - 1)}^{2}$ , positive before 0 and negative after 0 within the central branch.

Vertical asymptotes

x = \pm 1

; y-intercept and local maximum

(0, - 1)

; no x-intercepts.

Example 2

A translated reciprocal

For $g (x) = 2 / (x - 1) + 3$ , state its asymptotes, intercepts and centre of symmetry.

For y=0, solve $2 + 3 (x - 1) = 0$ .
Substitute x=0 for the y-intercept.

x = 1

y = 3

; x-intercept

1 / 3

, y-intercept

1

; centre

(1, 3)

Check your understanding

What happens to an ordinate of -5 after reciprocation?

Show answer

$- 1 / 5$ .

Only magnitude changes, not sign.

03 / WORKSHEET

Your turn

6 questions · 19 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

For $f (x) = x^{2} - 1$ , identify vertical asymptotes, intercepts and the stationary point of $1 / f (x)$ .
4 marks
Find the range and asymptote of $y = 1 / (x^{2} + 1)$ .
3 marks
Which points on a graph $y = f (x)$ remain fixed when changing to $y = 1 / f (x)$ ?
3 marks
For $g (x) = 2 / (x - 1) + 3$ , state its asymptotes, intercepts and centre of symmetry.
4 marks
Find the domain of $y = 1 / \sqrt{x}$ and explain the endpoint.
3 marks
If $f (a) = - 4$ , find the corresponding ordinate on $y = 1 / f (x)$ . Does its sign change?
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

For $f (x) = x^{2} - 1$ , identify vertical asymptotes, intercepts and the stationary point of $1 / f (x)$ .
4 marks
$g (x) = 1 / (x^{2} - 1)$ .
$g^{'} (x) = - 2 x / {(x^{2} - 1)}^{2}$ , positive before 0 and negative after 0 within the central branch.
Vertical asymptotes $x = \pm 1$ ; y-intercept and local maximum $(0, - 1)$ ; no x-intercepts.
Find the range and asymptote of $y = 1 / (x^{2} + 1)$ .
3 marks
The denominator is at least 1 and grows without bound.
The maximum 1 occurs at x=0; zero is approached but never reached.
Range $(0, 1]$ ; horizontal asymptote $y = 0$ .
Which points on a graph $y = f (x)$ remain fixed when changing to $y = 1 / f (x)$ ?
3 marks
Solve $y = 1 / y$ , with y nonzero.
This gives y squared equal to 1.
Points where $f (x) = 1$ or $f (x) = - 1$ .
For $g (x) = 2 / (x - 1) + 3$ , state its asymptotes, intercepts and centre of symmetry.
4 marks
For y=0, solve $2 + 3 (x - 1) = 0$ .
Substitute x=0 for the y-intercept.
$x = 1$ , $y = 3$ ; x-intercept $1 / 3$ , y-intercept $1$ ; centre $(1, 3)$ .
Find the domain of $y = 1 / \sqrt{x}$ and explain the endpoint.
3 marks
The square root needs x nonnegative.
The reciprocal also excludes x=0.
$x > 0$ .
If $f (a) = - 4$ , find the corresponding ordinate on $y = 1 / f (x)$ . Does its sign change?
2 marks
Reciprocation preserves the sign of a nonzero real number.
$- 1 / 4$ ; the sign stays negative.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist functions and sketching graphs. Calculus-based analysis assumes concurrent Methods differentiation; real-polynomial division detail is taught in the separate Real Polynomials chapter.

Official curriculum source ↗

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