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Year 11 Maths - Methods (Unit 1 and Unit 2) Transformations

Transformations of Graphs with Matrices

20 practice questions 0 video lessons Theory + worked examples

Apply matrix transformations to graphs for Victorian Year 11 Mathematical Methods (VCAA) — using matrices and translations to move and reshape a curve.

You will learn to represent dilations, reflections and translations with matrices, apply them to a function, and find the equation of the image curve, connecting transformations to the graphs you already know.

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Theory

In Year 11 Mathematical Methods (Unit 2), a transformation of the plane can be written as a matrix equation \(\begin{bmatrix}x'\\y'\end{bmatrix}=\mathbf{M}\begin{bmatrix}x\\y\end{bmatrix}+\begin{bmatrix}h\\k\end{bmatrix}\). This page shows how to find the image of a point by multiplying, and the image equation of a graph by making \(x\) and \(y\) the subjects and substituting — for reflections, dilations, translations, and combinations of these.

A linear transformation of the plane multiplies the position vector \(\begin{bmatrix}x\\y\end{bmatrix}\) by a \(2\times 2\) matrix \(\mathbf{M}\). A following translation adds a column vector \(\begin{bmatrix}h\\k\end{bmatrix}\). Together they map each point \((x,y)\) to its image \((x',y')\).

A reflection flips the plane across an axis, a dilation stretches it away from an axis by a factor, and a translation slides it without turning or stretching. Each is captured by a standard matrix (and, for a translation, a vector).

To transform a point, substitute its coordinates and multiply. To transform a whole graph \(y=f(x)\), you must go the other way: write \(x'\) and \(y'\) in terms of \(x\) and \(y\), make \(x\) and \(y\) the subjects, then substitute into \(y=f(x)\).

Points multiply forward; equations substitute backward. For a point, plug in and multiply. For a graph, rearrange the mapping so old coordinates are written in terms of new ones, then substitute.
Reflection in the x-axisThe parabola y equals x squared and its reflection y equals minus x squared, mirror images across the x-axis. x y y=x^2 y=-x^2
Reflection in the \(x\)-axis, matrix \(\begin{bmatrix}1&0\\0&-1\end{bmatrix}\): each \(y\)-coordinate changes sign.
Dilation of factor 2 from the x-axisThe parabola y equals x squared and the taller image y equals 2x squared, stretched away from the x-axis. x y y=x^2 y=2x^2
Dilation of factor \(2\) from the \(x\)-axis, matrix \(\begin{bmatrix}1&0\\0&2\end{bmatrix}\): each \(y\)-coordinate doubles.

The general matrix form of a transformation is:

\[\begin{bmatrix}x'\\y'\end{bmatrix}=\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}+\begin{bmatrix}h\\k\end{bmatrix}\]
[x′y′]=M[xy]+[hk]

The standard \(2\times 2\) matrices (with dilation factor \(k\)) are:

TransformationMatrixEffect on \((x,y)\)
Reflection in the \(x\)-axis\(\begin{bmatrix}1&0\\0&-1\end{bmatrix}\)\((x,\,-y)\)
Reflection in the \(y\)-axis\(\begin{bmatrix}-1&0\\0&1\end{bmatrix}\)\((-x,\,y)\)
Dilation factor \(k\) from the \(y\)-axis\(\begin{bmatrix}k&0\\0&1\end{bmatrix}\)\((kx,\,y)\)
Dilation factor \(k\) from the \(x\)-axis\(\begin{bmatrix}1&0\\0&k\end{bmatrix}\)\((x,\,ky)\)
Translation by \(\begin{bmatrix}h\\k\end{bmatrix}\)add \(\begin{bmatrix}h\\k\end{bmatrix}\)\((x+h,\,y+k)\)
Image of a graph: from \(x'=\dots\) and \(y'=\dots\), make \(x\) and \(y\) the subjects, substitute into \(y=f(x)\), then drop the dashes to write the image in \(x\) and \(y\).

Image of a point

  1. Substitute the coordinates into \(\begin{bmatrix}x'\\y'\end{bmatrix}=\mathbf{M}\begin{bmatrix}x\\y\end{bmatrix}\left(+\begin{bmatrix}h\\k\end{bmatrix}\right)\).
  2. Multiply the matrix by the column, row by row, then add any translation vector.
  3. Read off \(x'\) and \(y'\) as the image coordinates.

Image of a graph \(y=f(x)\)

  1. Write the mapping: expand the matrix equation to get \(x'\) and \(y'\) in terms of \(x\) and \(y\).
  2. Make \(x,y\) the subjects: rearrange to write \(x\) and \(y\) in terms of \(x'\) and \(y'\).
  3. Substitute these into \(y=f(x)\) and simplify; then replace \(x'\to x\), \(y'\to y\) for the image equation.
Example 1 — Image of a point
The point \((5,2)\) is transformed by \(\begin{bmatrix}x'\\y'\end{bmatrix}=\begin{bmatrix}1&0\\0&-1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}\). State the coordinates of the image.
Solution

Substitute the point and multiply:

\[\begin{bmatrix}x'\\y'\end{bmatrix}=\begin{bmatrix}1&0\\0&-1\end{bmatrix}\begin{bmatrix}5\\2\end{bmatrix}\]

Row by row:

\(x'\)\(=\)\((1)(5)+(0)(2)\)
\(=\)\(5\)
\(y'\)\(=\)\((0)(5)+(-1)(2)\)
\(=\)\(-2\)

This matrix reflects points in the \(x\)-axis, so the \(x\)-coordinate is unchanged and the \(y\)-coordinate changes sign.

The image is \((5,\,-2)\).

A point reflected in the x-axisThe point P at five two and its image P dash at five minus two, reflected across the x-axis. x y P(5,2) P'(5,-2)
(5,-2)
Example 2 — Image equation (dilation)
The graph of \(y=x^2\) is dilated by a factor of \(2\) from the \(y\)-axis (parallel to the \(x\)-axis), described by \(\begin{bmatrix}x'\\y'\end{bmatrix}=\begin{bmatrix}2&0\\0&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}\). Find the image equation.
Solution

Write the mapping from the matrix:

\(x'\)\(=\)\(2x\)
\(y'\)\(=\)\(y\)

Make \(x\) and \(y\) the subjects:

\(x\)\(=\)\(\dfrac{x'}{2}\)
\(y\)\(=\)\(y'\)

Substitute into \(y=x^2\):

\(y'\)\(=\)\(\left(\dfrac{x'}{2}\right)^2\)
\(=\)\(\dfrac{(x')^2}{4}\)

Dropping the dashes gives the image in \(x\) and \(y\).

The image equation is \(y=\dfrac{1}{4}x^2\).

Dilation of factor 2 from the y-axisThe parabola y equals x squared and the wider image y equals one quarter x squared. x y y=x^2 y=¼x^2
y=14x2
Example 3 — Image equation (translation)
The graph of \(y=x^2\) is translated by \(\begin{bmatrix}x'\\y'\end{bmatrix}=\begin{bmatrix}x\\y\end{bmatrix}+\begin{bmatrix}-2\\3\end{bmatrix}\). Find the image equation.
Solution

Write the mapping:

\(x'\)\(=\)\(x-2\)
\(y'\)\(=\)\(y+3\)

Make \(x\) and \(y\) the subjects:

\(x\)\(=\)\(x'+2\)
\(y\)\(=\)\(y'-3\)

Substitute into \(y=x^2\):

\(y'-3\)\(=\)\((x'+2)^2\)
\(y'\)\(=\)\((x'+2)^2+3\)

Dropping the dashes gives the image equation.

The image equation is \(y=(x+2)^2+3\).

Translation by the vector (-2, 3)The parabola y equals x squared with vertex at the origin and its image y equals x plus two all squared plus three with vertex at minus two three. x y y=x^2 image
y=(x+2)²+3
Example 4 — Combined transformation
The graph of \(y=x^2\) is reflected in the \(x\)-axis, then dilated by a factor of \(2\) from the \(x\)-axis, then translated \(5\) units up, combining as \(\begin{bmatrix}x'\\y'\end{bmatrix}=\begin{bmatrix}1&0\\0&-2\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}+\begin{bmatrix}0\\5\end{bmatrix}\). Find the image equation.
Solution

Write the mapping (matrix, then add the vector):

\(x'\)\(=\)\(x\)
\(y'\)\(=\)\(-2y+5\)

Make \(x\) and \(y\) the subjects:

\(x\)\(=\)\(x'\)
\(y\)\(=\)\(\dfrac{y'-5}{-2}\)

Substitute into \(y=x^2\), then clear the fraction:

\(\dfrac{y'-5}{-2}\)\(=\)\((x')^2\)
\(y'-5\)\(=\)\(-2(x')^2\)
\(y'\)\(=\)\(-2(x')^2+5\)

Dropping the dashes gives the image equation.

The image equation is \(y=5-2x^2\).

Combined transformation of y=x squaredThe parabola y equals x squared opening upward and the image y equals five minus two x squared opening downward with vertex at zero five. x y y=x^2 y=5-2x^2
y=5-2x2

Common pitfalls

Substituting the wrong way for a graph. For a graph you cannot plug \(x'=2x\) straight into \(y=f(x)\). First make \(x\) and \(y\) the subjects, then substitute — the equation transforms backward.
Swapping the two dilations. "Factor \(k\) from the \(x\)-axis\" scales \(y\), matrix \(\begin{bmatrix}1&0\\0&k\end{bmatrix}\); \"from the \(y\)-axis\" scales \(x\), matrix \(\begin{bmatrix}k&0\\0&1\end{bmatrix}\).
Adding the translation before multiplying. In \(\mathbf{M}\begin{bmatrix}x\\y\end{bmatrix}+\begin{bmatrix}h\\k\end{bmatrix}\) the matrix acts first, then the vector is added.
Sign flips when rearranging. From \(x'=x-2\) the subject is \(x=x'+2\); moving a term across the equals sign changes its sign.

Frequently asked questions

How do you find the image of a point using a matrix?

Substitute the coordinates into \(\begin{bmatrix}x'\\y'\end{bmatrix}=\mathbf{M}\begin{bmatrix}x\\y\end{bmatrix}\), multiply row by row, then add any translation vector.

How do you find the equation of the image of a graph?

Write \(x'\) and \(y'\) in terms of \(x\) and \(y\), make \(x\) and \(y\) the subjects, substitute into \(y=f(x)\), then drop the dashes.

What matrix reflects a graph in an axis?

Reflection in the \(x\)-axis is \(\begin{bmatrix}1&0\\0&-1\end{bmatrix}\) (\(y\) changes sign); in the \(y\)-axis it is \(\begin{bmatrix}-1&0\\0&1\end{bmatrix}\) (\(x\) changes sign).

What matrix represents a dilation?

A dilation of factor \(k\) from the \(x\)-axis is \(\begin{bmatrix}1&0\\0&k\end{bmatrix}\); from the \(y\)-axis it is \(\begin{bmatrix}k&0\\0&1\end{bmatrix}\).

How do you translate a graph with matrices?

Add a column vector \(\begin{bmatrix}h\\k\end{bmatrix}\): the point \((x,y)\) maps to \((x+h,\,y+k)\).

Why do you invert the mapping for an equation but not for a point?

A point is a specific input, so you multiply it forward. A graph is the set of all points on \(y=f(x)\), so you must express the old coordinates in terms of the new ones before substituting.

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