A Notation for Transformations Revision Notes for VCE SSCE Mathematical Methods

A Notation for Transformations

Understanding transformation notation

Transformations of the plane can be thought of as functions that take a point and move it to a new location. Just as we use function notation for single-variable functions, we can use similar notation for transformations that work with two variables.

A transformation $T$ maps points from $\mathbb{R}^2$ (the coordinate plane) to $\mathbb{R}^2$ . We write this as:

T : \mathbb{R}^2 \to \mathbb{R}^2

T(x, y) = (ax + h, by + k)

where $a \neq 0$ and $b \neq 0$

This notation tells us that $T$ takes a point $(x, y)$ and transforms it to a new point $(ax + h, by + k)$ .

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The maximal domain and range of transformations of this type is $\mathbb{R}^2$ , meaning they work on all points in the coordinate plane. This is important because it tells us that these transformations can be applied to any point without restriction.

Basic transformations of the plane

Several fundamental transformations form the building blocks for more complex movements. These include reflections, dilations, and translations. Understanding these basic operations is essential for working with more complex transformations.

In the table above, the point $(x', y')$ represents the image of the original point $(x, y)$ after the transformation has been applied.

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Understanding the General Form

All of these basic transformations, and combinations of them, can be expressed in the general form:

$(x, y) \to (ax + h, by + k)$

where $a \neq 0$ and $b \neq 0$

This powerful representation allows us to describe any combination of reflections, dilations, and translations using just four parameters: $a$ , $b$ , $h$ , and $k$ .

Linear transformations

Linear transformations are a special category of transformations. They include dilations, reflections in the $x$ -axis, reflections in the $y$ -axis, and combinations of these operations.

A transformation is defined as linear when it has the form:

T : \mathbb{R}^2 \to \mathbb{R}^2

T(x, y) = (ax + by, cx + dy)

Notice that linear transformations don't include constant terms in their expressions. This is an important distinction.

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Critical Distinction for Linear Transformations

Transformations of the form $T(x, y) = (ax + h, by + k)$ are only linear when $h = k = 0$ .

If either $h$ or $k$ is non-zero, the transformation includes a translation component and is therefore not linear.

Reflection in the line $y = x$

Another example of a linear transformation is reflection in the line $y = x$ . This transformation swaps the $x$ and $y$ coordinates.

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Connection to Inverse Functions

This transformation is particularly important because it relates to inverse functions. When you reflect the graph of $y = f(x)$ in the line $y = x$ , you obtain the graph of the inverse function $x = f(y)$ .

This geometric relationship provides a visual way to understand why inverse functions "swap" the roles of $x$ and $y$ .

Worked example: Evaluating transformations

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Worked Example: Evaluating a Transformation

Let $T : \mathbb{R}^2 \to \mathbb{R}^2$ , $T(x, y) = (3x + 2, 4y - 6)$

Part a: Evaluate $T(4, 5)$ and $T(-1, 2)$

Solution:

For $T(4, 5)$ :

T(4, 5) = (3 \times 4 + 2, 4 \times 5 - 6)

T(4, 5) = (14, 14)

For $T(-1, 2)$ :

T(-1, 2) = (3 \times (-1) + 2, 4 \times 2 - 6)

T(-1, 2) = (-1, 2)

Notice that the point $(-1, 2)$ is mapped to itself under this transformation - it's a fixed point!

Part b: Find the equation of the image of the graph $y = f(x)$ , where $f(x) = 2^x$ , under this transformation.

Solution:

We have $T(x, y) = (3x + 2, 4y - 6)$ and let $(x', y') = (3x + 2, 4y - 6)$

From the transformation:

x' = 3x + 2

Solving for $x$ :

x = \frac{x' - 2}{3}

y' = 4y - 6

Solving for $y$ :

y = \frac{y' + 6}{4}

Since the original graph is $y = 2^x$ , we substitute:

\frac{y' + 6}{4} = 2^{\frac{x' - 2}{3}}

Simplifying to standard form (dropping the primes):

y = 4 \times 2^{\frac{x - 2}{3}} - 6

Worked example: Finding transformation parameters

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Worked Example: Determining Unknown Parameters

Let $T : \mathbb{R}^2 \to \mathbb{R}^2$ , $T(x, y) = (ax + h, by + k)$ . Given that $T(1, 6) = (-7, -3)$ and $T(-2, 3) = (11, 12)$ , determine the values of $a$ , $b$ , $h$ and $k$ .

Solution:

From $T(1, 6) = (-7, -3)$ :

a + h = -7 \quad \text{... (1)}

6b + k = -3 \quad \text{... (2)}

From $T(-2, 3) = (11, 12)$ :

-2a + h = 11 \quad \text{... (3)}

3b + k = 12 \quad \text{... (4)}

Subtracting equation (3) from equation (1):

3a = -18

a = -6

Substituting into equation (1):

-6 + h = -7

h = -1

Subtracting equation (4) from equation (2):

3b = -15

b = -5

Substituting into equation (2):

6(-5) + k = -3

-30 + k = -3

k = 27

Therefore: a = -6, b = -5, h = -1, k = 27

Composition of transformations

Composition of transformations combines multiple transformations in sequence. This works similarly to function composition, where we apply one transformation and then apply another to the result.

The composition of transformations $T$ and $S$ can be written as $T \circ S$ , which means "apply $S$ first, then apply $T$ ".

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Worked Example: Composing Transformations

Let $T_1(x, y) = (x + 3, 2y)$ and $T_2(x, y) = (3x + 2, y - 3)$ . Find the rule for:

Part a: $T_2(T_1(x, y))$

Solution:

T_2(T_1(x, y)) = T_2(x + 3, 2y)

= (3(x + 3) + 2, 2y - 3)

= (3x + 11, 2y - 3)

Part b: $T_1(T_2(x, y))$

Solution:

T_1(T_2(x, y)) = T_1(3x + 2, y - 3)

= (3x + 2 + 3, 2(y - 3))

= (3x + 5, 2y - 6)

Notice that T₂(T₁(x, y)) ≠ T₁(T₂(x, y)), demonstrating that composition is not commutative in this case.

Important properties of composition

Several key properties apply to composition of transformations:

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Key Properties of Composition

Two transformations $T_1$ and $T_2$ are equal if $T_1(x, y) = T_2(x, y)$ for all $(x, y) \in \mathbb{R}^2$
The composition can be written using circle notation: $T \circ S$
In general, composition is not commutative. That is, $T(S(x, y)) \neq S(T(x, y))$ , or $T \circ S \neq S \circ T$
Some families of transformations do commute. For example, if $S$ and $T$ are both translations, then $T(S(x, y)) = S(T(x, y))$

Inverses of transformations

An inverse transformation "undoes" the effect of the original transformation, returning points to their original positions.

A transformation $T$ is one-to-one if $T(x_1, y_1) = T(x_2, y_2)$ implies $x_1 = x_2$ and $y_1 = y_2$ . All transformations of the form we're studying are one-to-one.

For a transformation $T$ , we define its inverse $T^{-1}$ by:

T^{-1}(x', y') = (x, y) \text{ if } T(x, y) = (x', y')

The inverse transformation $T^{-1}$ is also one-to-one, and $T$ is the inverse of $T^{-1}$ .

Key properties of inverse transformations:

T \circ T^{-1}(x, y) = (x, y) \text{ for all } (x, y) \in \mathbb{R}^2

T^{-1} \circ T(x, y) = (x, y) \text{ for all } (x, y) \in \mathbb{R}^2

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Worked Example: Finding an Inverse Transformation

Find the inverse of the transformation $T : \mathbb{R}^2 \to \mathbb{R}^2$ , $T(x, y) = (3x - 2, -5y + 3)$

Solution:

We know that $T \circ T^{-1}(x, y) = (x, y)$

Let $T^{-1}(x, y) = (w, z)$

Then:

T(T^{-1}(x, y)) = (x, y)

T(w, z) = (x, y)

(3w - 2, -5z + 3) = (x, y)

From this, we get two equations:

3w - 2 = x

-5z + 3 = y

Solving for $w$ :

3w = x + 2

w = \frac{x + 2}{3}

Solving for $z$ :

-5z = y - 3

z = \frac{3 - y}{5}

Therefore:

T^{-1}(x, y) = \left(\frac{x + 2}{3}, \frac{3 - y}{5}\right)

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You can verify this is correct by checking that $T \circ T^{-1}(x, y) = T^{-1} \circ T(x, y) = (x, y)$ . This verification step confirms that the inverse truly "undoes" the original transformation.

Transformations on subsets of $\mathbb{R}^2$

When working with functions that have restricted domains (not all of $\mathbb{R}$ ), we need to consider how transformations affect both the domain and range.

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Worked Example: Transformations with Restricted Domains

Consider the function $f : [0, 3] \to \mathbb{R}$ , $f(x) = -x^2 + 2x$ .

Part a: Find the range of $f$

Solution:

The graph of $y = f(x)$ has a local maximum at $(1, 1)$ .

The endpoints have coordinates $(0, 0)$ and $(3, -3)$ .

Therefore the range is -3, 1

Part b: Find the image of $f$ under the transformation with rule $T(x, y) = (2x, -2y + 3)$ . State the domain and range of this image.

Solution:

Let $T(x, y) = (x', y')$

Therefore, $x' = 2x$ and $y' = -2y + 3$

Solving for $x$ and $y$ :

x = \frac{x'}{2}

y = \frac{y' - 3}{-2}

The original equation $y = -x^2 + 2x$ becomes:

\frac{y' - 3}{-2} = -\left(\frac{x'}{2}\right)^2 + 2\left(\frac{x'}{2}\right)

Simplifying (and dropping the primes):

y = \frac{x^2}{2} - 2x + 3

The domain is calculated from the original domain $[0, 3]$ :

[2 \times 0, 2 \times 3] = [0, 6]

The turning point, which is now a local minimum, has coordinates:

(2 \times 1, -2 \times 1 + 3) = (2, 1)

The range can be calculated from the domain and equation of the image as 1, 9.

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Alternative Method for Finding Range

Alternatively, you can consider the transformation of the original range $[-3, 1]$ . Note that there is a reflection in the x-axis (the coefficient of $y$ is negative), so care must be taken with the endpoints of the range when using this method.

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Key Points to Remember:

Transformations are written as $T : \mathbb{R}^2 \to \mathbb{R}^2$ , $T(x, y) = (ax + h, by + k)$ where $a \neq 0$ and $b \neq 0$
Linear transformations have the form $T(x, y) = (ax + by, cx + dy)$ with no constant terms
To find the image of a function under a transformation, express $x$ and $y$ in terms of $x'$ and $y'$ , then substitute into the original equation
Composition of transformations is generally not commutative: $T \circ S \neq S \circ T$ in most cases
The inverse transformation $T^{-1}$ satisfies $T \circ T^{-1}(x, y) = T^{-1} \circ T(x, y) = (x, y)$

A Notation for Transformations (VCE SSCE Mathematical Methods): Revision Notes