Transformations of functions (VCE Maths Methods Unit 1)

What this dot point is asking

VCAA wants you to identify and apply translations, dilations and reflections to function graphs, working with the standard transformed form $y = a f(b(x - h)) + k$.

The general transformed form

Vertical dilation (factor $a$)

$y = af(x)$ stretches the graph vertically by factor $|a|$. $y$-values multiply by $a$. The $x$-axis is fixed.

Horizontal dilation (factor $1/b$)

$y = f(bx)$ compresses horizontally by factor $b$ (or equivalently dilates by factor $1/b$). $x$-values divide by $b$. The $y$-axis is fixed.

The reciprocal feature trips students: $y = f(2x)$ is a horizontal compression, not a stretch.

Translations

$y = f(x - h)$ shifts the graph $h$ units to the right (if $h > 0$). Counterintuitively, the bracket has the opposite sign to the direction of shift.

$y = f(x) + k$ shifts the graph $k$ units up (if $k > 0$). Straightforward.

Reflections

In the $x$-axis: $y = -f(x)$. Achieved with $a = -1$.

In the $y$-axis: $y = f(-x)$. Achieved with $b = -1$.

In the line $y = x$: swap $x$ and $y$ (inverse function, next dot point).

Order of transformations

The order matters when combining. A safe order is:

1. Apply horizontal dilation/reflection (inside the bracket, with $b$).
2. Apply horizontal translation ($-h$ shift).
3. Apply vertical dilation/reflection (with $a$).
4. Apply vertical translation ($+k$).

Mixing the order can produce different results.

Worked example

Starting from $y = \sqrt{x}$, sketch $y = -2\sqrt{x - 3} + 1$.

$a = -2$, $b = 1$, $h = 3$, $k = 1$.

• Domain: $x \ge 3$ (the $\sqrt{}$ requires non-negative argument).
• Range: $y \le 1$ (the $-2$ flips down from $y = 1$).
• Starting point of the graph (where the original $(0, 0)$ goes): $(3, 1)$.
• Reflection makes the graph descend to the right; dilation by $2$ makes it descend twice as steeply.

Common traps

Direction of horizontal translation

$(x - 3)$ shifts right, $(x + 3)$ shifts left.

Mixing up horizontal and vertical dilation factors

$y = 2f(x)$ doubles $y$-values; $y = f(2x)$ halves $x$-values.

Forgetting that reflection is a special case of dilation

$a = -1$ is a reflection in the $x$-axis with no scaling.

Order matters

Apply transformations consistently; the standard order is inside-out within the function and then outside-in.

In one sentence

For $y = af(b(x - h)) + k$: $a$ dilates/reflects vertically, $b$ dilates/reflects horizontally (with reciprocal scale factor $1/|b|$), $h$ shifts horizontally (right if positive, opposite sign in the bracket), and $k$ shifts vertically, with the order inside-out within the function and then outside-in.

Examples in context

Example 1. Repositioning a parabolic arch. A bridge arch is modelled by $y = -x^2$ centred at the origin. To raise the apex to a point $3$ m right and $4$ m up, apply $y = -(x - 3)^2 + 4$ ($h = 3$, $k = 4$). The apex sits at $(3, 4)$. If the arch must also be twice as tall, use $a = 2$: $y = -2(x - 3)^2 + 4$, giving apex $(3, 4)$ but steeper sides.

Example 2. Stretching a tide curve. A tide is modelled by $y = f(x)$ over a $24$-hour cycle. To model a location where the same pattern repeats every $12$ hours and reaches twice the height, apply $y = 2f(2x)$: the $b = 2$ compresses the period to half (a $12$-hour cycle) and $a = 2$ doubles the amplitude. Note the horizontal factor is the reciprocal: $f(2x)$ halves the period, it does not double it.

Try this

Q1. Describe the transformations of $y = x^2$ giving $y = (x - 4)^2 - 2$, and state the vertex. [2 marks]

• Cue. Translate $4$ right and $2$ down; vertex $(4, -2)$.

Q2. The graph of $y = \sqrt{x}$ is transformed to $y = 3\sqrt{x + 2}$. State the domain, and the image of the point $(0,0)$. [3 marks]

• Cue. Domain $x \ge -2$; the origin maps to $(-2, 0)$ (translate $2$ left, then vertical dilation fixes $y = 0$).

Q3. Starting from $y = f(x)$, write the rule after a reflection in the $x$-axis, then a translation $5$ up. [2 marks]

• Cue. $y = -f(x) + 5$.