Reciprocal/composite transformations | Year 12

01 / LEARN

Understand the idea.

Reciprocate ordinates

For $g (x) = 1 / f (x)$ , keep each allowed input and reciprocate its nonzero output. Signs are preserved. Large output magnitudes become small, while outputs close to zero create large reciprocal magnitudes. Ordinates $1$ and $- 1$ stay fixed.

Every zero of $f$ is excluded from the reciprocal domain. When $f$ approaches zero there, investigate the one-sided reciprocal behaviour to identify a vertical asymptote. Use the formula and domain rather than drawing through the excluded input.

Use exact transformations

For $g (x) = \frac{a}{x - h} + k$ with $a \neq 0$ , the vertical and horizontal asymptotes are $x = h$ and $y = k$ . Their intersection is the centre of symmetry; the sign of $a$ determines the branch orientation.

Where $f$ is differentiable and nonzero, the reciprocal derivative is $g^{'} = - f^{'} / f^{2}$ . Its denominator is positive, so its sign is opposite to that of $f^{'}$ . Separate derivative intervals at poles as well as stationary inputs.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

The reciprocal of a quadratic

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

Graph of y=1/(x squared minus 1), with three separate branches, dashed vertical asymptotes at x=-1 and x=1, and the marked point (0,-1). Coordinate ranges are x from -6 to 6 and y from -8 to 8.

Worked solution

1
Find the domain and vertical asymptotes.
$g (x) = \frac{1}{x^{2} - 1} = \frac{1}{(x - 1) (x + 1)}, x \neq - 1, 1.$
The numerator is nonzero at both denominator zeros, so $x = - 1$ and $x = 1$ are vertical asymptotes.
2
Determine the intercepts.
$g (0) = - 1, \frac{1}{x^{2} - 1} \neq 0.$
The vertical intercept is $(0, - 1)$ and the nonzero numerator prevents horizontal intercepts.
3
Differentiate and solve for a stationary input.
$g^{'} (x) = - \frac{2 x}{{(x^{2} - 1)}^{2}} = 0 \Rightarrow x = 0.$
Only the numerator can be zero on the domain.
4
Classify the stationary point.
$g^{'} (x) > 0 for - 1 < x < 0, g^{'} (x) < 0 for 0 < x < 1.$
The central branch increases then decreases, so $(0, - 1)$ is a local maximum, not a global maximum across the poles.

Answer

Vertical asymptotes $x = \pm 1$ ; y-intercept and local maximum $(0, - 1)$ ; no x-intercepts.

EXAMPLE 02

A translated reciprocal

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

Worked solution

1
Read the translated asymptotes.
$g (x) = \frac{2}{x - 1} + 3 : x = 1, y = 3.$
The denominator vanishes at one, and the reciprocal term tends to zero as $∣ x ∣$ grows.
2
Solve for the horizontal intercept.
$\begin{aligned} 0 & = \frac{2}{x - 1} + 3, \\ 0 & = 2 + 3 (x - 1), \\ 3 x - 1 & = 0 \Rightarrow x = \frac{1}{3} . \end{aligned}$
Multiplication by $x - 1$ is valid at this allowed solution, giving $(\frac{1}{3}, 0)$ .
3
Find the vertical intercept.
$g (0) = \frac{2}{- 1} + 3 = 1.$
The graph meets the vertical axis at $(0, 1)$ .
4
State the centre of symmetry.
$(h, k) = (1, 3) .$
In shifted coordinates $u = x - 1$ and $v = y - 3$ , the equation $v = 2 / u$ is unchanged by $(u, v) \mapsto (- u, - v)$ .

Answer

$x = 1$ , $y = 3$ ; x-intercept $1 / 3$ , y-intercept $1$ ; centre $(1, 3)$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

What happens to an ordinate of -5 after reciprocation?

Show answer & reasoning

$- 1 / 5$ .

The reciprocal of $- 5$ is $\frac{1}{- 5} = - \frac{1}{5}$ . The result stays negative because the denominator is negative.

Keep practising

6 worksheet questions · 19 marks

Open worksheet Worked solutions

Read the questions online

For $f (x) = x^{2} - 1$ , identify vertical asymptotes, intercepts and the stationary point of $\frac{1}{f (x)}$ .
4 marks
Find the range and asymptote of $y = \frac{1}{x^{2} + 1}$ .
3 marks
Which points on a graph $y = f (x)$ remain fixed when changing to $y = \frac{1}{f (x)}$ ?
3 marks
For $g (x) = \frac{2}{x - 1} + 3$ , state its asymptotes, intercepts and centre of symmetry.
4 marks
Find the domain of $y = \frac{1}{\sqrt{x}}$ and explain the endpoint.
3 marks
If $f (a) = - 4$ , find the corresponding ordinate on $y = \frac{1}{f (x)}$ . 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 $\frac{1}{f (x)}$ .
4 marks
$g (x) = \frac{1}{x^{2} - 1}$ .
$g^{'} (x) = - \frac{2 x}{{(x^{2} - 1)}^{2}}$ , 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 $y = \frac{1}{x^{2} + 1}$ .
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 $y = \frac{1}{f (x)}$ ?
3 marks
Solve $y = \frac{1}{y}$ , with $y$ nonzero.
This gives $y^{2} = 1$ .
Points where $f (x) = 1$ or $f (x) = - 1$ .
For $g (x) = \frac{2}{x - 1} + 3$ , 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 $\frac{1}{3}$ , y-intercept $1$ ; centre $(1, 3)$ .
Find the domain of $y = \frac{1}{\sqrt{x}}$ 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 $y = \frac{1}{f (x)}$ . Does its sign change?
2 marks
Reciprocation preserves the sign of a nonzero real number.
$- \frac{1}{4}$ ; 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 ↗

In this lesson

Understand the idea.

Reciprocate ordinates

Use exact transformations

See the method in action.

The reciprocal of a quadratic

Find the domain and vertical asymptotes.

Determine the intercepts.

Differentiate and solve for a stationary input.

Classify the stationary point.

A translated reciprocal

Read the translated asymptotes.

Solve for the horizontal intercept.

Find the vertical intercept.

State the centre of symmetry.

Try it yourself.

Check your understanding

Keep practising