Reciprocal and Absolute Value Functions Revision Notes for HSC SSCE Mathematics Advanced

Reciprocal and Absolute Value Functions

Introduction

In this note, you will learn how to transform reciprocal and absolute value functions using translations, reflections, and dilations. These transformations change the position, shape, and orientation of the graphs. You will learn to find equations of transformed functions, determine key features like asymptotes and vertices, identify domain and range, and sketch graphs of transformed functions.

Transformations of reciprocal functions

The general form

The reciprocal function $f(x) = \frac{1}{x}$ can be transformed using translations, reflections, and dilations. When transformations are applied, the general form becomes:

$g(x) = \frac{l}{x - a} + b$

This equation contains three important parameters that control the transformation:

infoNote

Understanding each parameter is crucial for predicting how the graph will change. The parameter $l$ affects the shape and orientation, while $a$ and $b$ control the position of the graph.

Parameter $l$ (dilation factor):

When $|l| > 1$ , the graph stretches vertically (becomes steeper)
When $0 < |l| < 1$ , the graph compresses vertically (becomes flatter)
When $l < 0$ , the graph reflects over the x-axis

Parameter $a$ (horizontal translation):

When $a > 0$ , the graph shifts right by $a$ units
When $a < 0$ , the graph shifts left by $|a|$ units

Parameter $b$ (vertical translation):

When $b > 0$ , the graph shifts up by $b$ units
When $b < 0$ , the graph shifts down by $|b|$ units

Domain, range and asymptotes

When a reciprocal function is transformed, its domain and range change according to the translation parameters. The function also has two asymptotes that shift with the transformations.

Domain: The domain excludes the value where the denominator equals zero. This occurs at $x = a$ . The domain is written as $x \neq a$ , or using interval notation: $(-\infty, a) \cup (a, \infty)$ .

Range: The range excludes the value of the horizontal asymptote. This is $y \neq b$ , or using interval notation: $(-\infty, b) \cup (b, \infty)$ .

Vertical asymptote: This is a vertical line that the graph approaches but never touches. It occurs at x = a.

Horizontal asymptote: This is a horizontal line that the graph approaches but never touches. It occurs at y = b.

chatImportant

Remember the pattern: For reciprocal functions, both the domain and range are restricted. The vertical asymptote is at $x = a$ (affecting domain), and the horizontal asymptote is at $y = b$ (affecting range). The graph has two separate curved sections that never cross these asymptotes.

Worked example: reciprocal function transformation

lightbulbExample

Worked Example: Transforming a Reciprocal Function

Question: Transform $f(x) = \frac{1}{x}$ by translating right by 2 units and up by 3 units, then reflecting over the $x$ -axis.

Part a) Determine the equation of the transformed function

To find the transformed function, we apply the transformations step by step using the general form $g(x) = \frac{l}{x - a} + b$ with $a = 2$ , $b = 3$ , and $l = -1$ (negative because of reflection).

Start with the original function:

$f(x) = \frac{1}{x}$

Translate right by 2 and up by 3:

$g(x) = \frac{1}{x - 2} + 3$

Reflect over the $x$ -axis (multiply by $-1$ ):

$g(x) = -\left(\frac{1}{x - 2} + 3\right)$

Distribute the negative sign:

$g(x) = -\frac{1}{x - 2} - 3$

The transformed function is g(x) = -1/(x - 2) - 3.

Part b) Determine the asymptotes

The asymptotes are found using the parameters $a$ and $b$ from the equation.

The vertical asymptote is at $x = a$ :

$x = 2$

The horizontal asymptote is at $y = b$ :

$y = -3$

Part c) Determine the domain and range

The domain excludes the vertical asymptote where $x = a$ :

Domain: x ≠ 2, or $(-\infty, 2) \cup (2, \infty)$

The range excludes the horizontal asymptote where $y = b$ :

Range: y ≠ -3, or $(-\infty, -3) \cup (-3, \infty)$

Part d) Sketch the graph, labelling the asymptotes, a key point and the y-intercept

To sketch the graph, we need to find a key point and the $y$ -intercept.

Find a key point by substituting $x = 3$ :

g(x) = -\frac{1}{x - 2} - 3 g(3) = -\frac{1}{3 - 2} - 3 g(3) = -\frac{1}{1} - 3 g(3) = -1 - 3 g(3) = -4

The point at $x = 3$ is $(3, -4)$ .

Find the $y$ -intercept by substituting $x = 0$ :

g(0) = -\frac{1}{0 - 2} - 3 g(0) = -\frac{1}{-2} - 3 g(0) = \frac{1}{2} - 3 g(0) = -\frac{5}{2} \text{ or } -2.5

The $y$ -intercept is at $(0, -2.5)$ .

The graph shows the transformed function $g(x) = -\frac{1}{x - 2} - 3$ with:

Vertical asymptote at $x = 2$ (shown as a vertical dashed line)
Horizontal asymptote at $y = -3$ (shown as a horizontal dashed line)
Key points at $(0, -2.5)$ and $(3, -4)$

Notice how the asymptotes have shifted from their original positions at $x = 0$ and $y = 0$ to the new positions at $x = 2$ and $y = -3$ . This confirms that the transformations were applied correctly.

Transformations of absolute value functions

The general form

The absolute value function $f(x) = |x|$ creates a V-shaped graph. When transformations are applied, the general form becomes:

$g(x) = l|x - a| + b$

This equation uses the same three parameters as the reciprocal function, but they affect the graph differently:

Parameter $l$ (dilation factor):

When $|l| > 1$ , the V-shape becomes narrower (steeper)
When $0 < |l| < 1$ , the V-shape becomes wider (less steep)
When $l < 0$ , the V-shape flips upside down (reflects over the $x$ -axis)

Parameter $a$ (horizontal translation):

When $a > 0$ , the graph shifts right by $a$ units
When $a < 0$ , the graph shifts left by $|a|$ units

Parameter $b$ (vertical translation):

When $b > 0$ , the graph shifts up by $b$ units
When $b < 0$ , the graph shifts down by $|b|$ units

Vertex, domain and range

Unlike reciprocal functions, absolute value functions do not have asymptotes. Instead, they have a vertex where the V-shape meets.

Vertex: The vertex is the turning point of the V-shape. It is located at (a, b). This is the lowest point when $l > 0$ , or the highest point when $l < 0$ .

chatImportant

Be careful with the vertex location! In the equation $g(x) = l|x - a| + b$ , the vertex is at $(a, b)$ , NOT at $(-a, b)$ . This is a common mistake to avoid.

Domain: The domain of an absolute value function is all real numbers: $(-\infty, \infty)$ .

Range: The range depends on the sign of $l$ :

If $l > 0$ , the V-shape opens upward and the range is $[b, \infty)$
If $l < 0$ , the V-shape opens downward and the range is $(-\infty, b]$

infoNote

The key difference between reciprocal and absolute value functions is that absolute value functions have no restrictions on their domain, while reciprocal functions exclude one value. Additionally, absolute value functions have a clear vertex, while reciprocal functions have asymptotes instead.

Worked example: absolute value function transformation

lightbulbExample

Worked Example: Transforming an Absolute Value Function

Question: Transform $f(x) = |x|$ by dilating vertically by a factor of 2, translating left by 1 unit, and down by 2 units.

Part a) Determine the equation of the transformed function

To find the transformed function, we use the general form $g(x) = l|x - a| + b$ with $l = 2$ , $a = -1$ , and $b = -2$ .

Start with the original function:

$f(x) = |x|$

Apply dilation with $l = 2$ :

$g(x) = 2|x|$

Apply translations with $a = -1$ and $b = -2$ :

$g(x) = 2|x + 1| - 2$

The transformed function is g(x) = 2|x + 1| - 2.

Part b) Determine the vertex

The vertex is located at $(a, b)$ from the equation $g(x) = l|x - a| + b$ .

For $g(x) = 2|x - (-1)| - 2$ , we have $a = -1$ and $b = -2$ .

The vertex is at (-1, -2).

Part c) Determine the range

The range depends on the value of $l$ and $b$ .

Since $l = 2 > 0$ , the V-shape opens upward, and the range starts at the vertex and extends to infinity:

Range: [[-2, ∞)]

Part d) Determine the x-intercepts and sketch the graph

To find the $x$ -intercepts, set $g(x) = 0$ and solve for $x$ .

Start with the transformed function:

$g(x) = 2|x + 1| - 2$

Set equal to zero:

$0 = 2|x + 1| - 2$

Add 2 to both sides:

$2 = 2|x + 1|$

Divide both sides by 2:

$1 = |x + 1|$

Solve the absolute value equation. This gives two cases:

Case 1: $x + 1 = 1$

$x = 0$

Case 2: $x + 1 = -1$

$x = -2$

The $x$ -intercepts are at (0, 0) and (-2, 0).

The graph shows the transformed function $g(x) = 2|x + 1| - 2$ with:

Vertex at $(-1, -2)$
$x$ -intercepts at $(0, 0)$ and $(-2, 0)$
A steeper V-shape due to the dilation factor of 2

The vertex has moved from the origin to $(-1, -2)$ , and the graph is narrower than the parent function because it has been stretched vertically by a factor of 2.

Exam tips

When working with transformed functions, follow these helpful strategies:

infoNote

For reciprocal functions:

Always identify the asymptotes first using $x = a$ and $y = b$
Remember that the domain and range both have restrictions
Plot at least two points on each branch of the hyperbola
The graph has two separate curved sections that never touch the asymptotes

For absolute value functions:

Find the vertex first using $(a, b)$
The $x$ -intercepts require solving an absolute value equation (two cases)
The graph forms a V-shape with the vertex at the tip
Check whether $l$ is positive (opens up) or negative (opens down)

General transformation tips:

Work step by step through each transformation
Write down the parameter values $l$ , $a$ , and $b$ before starting
Sketch a rough graph to visualize the transformation
Double-check your asymptotes, vertex, domain and range

Remember!

bookmarkSummary

Key Points to Remember:

Reciprocal functions have the form $g(x) = \frac{l}{x - a} + b$ , with vertical asymptote at $x = a$ and horizontal asymptote at $y = b$ . The domain is $x \neq a$ and range is $y \neq b$ .
Absolute value functions have the form $g(x) = l|x - a| + b$ , with vertex at $(a, b)$ . The domain is all real numbers and the range is $[b, \infty)$ if $l > 0$ , or $(-\infty, b]$ if $l < 0$ .
The parameter $l$ controls dilation and reflection. When $|l| > 1$ , the graph stretches; when $0 < |l| < 1$ , it compresses; when $l < 0$ , it reflects over the $x$ -axis.
The parameters $a$ and $b$ control translations. The value $a$ shifts the graph horizontally (right if positive, left if negative), and $b$ shifts it vertically (up if positive, down if negative).
To sketch transformed graphs, identify key features first (asymptotes for reciprocal functions, vertex for absolute value functions), then find intercepts and additional points to complete the sketch.

Reciprocal and Absolute Value Functions (HSC SSCE Mathematics Advanced): Revision Notes