Graph transformations: Translations, reflections and dilations for HSC Maths Advanced functions

What this dot point is asking

NESA wants you to take the graph of a base function $y = f(x)$ and produce the graph or the equation that results from any combination of translations, reflections and dilations. You must know which transformations act inside the function (on $x$) and which act outside (on $y$), and the order in which combined transformations apply.

The answer

Vertical transformations (act on $y$, outside the function)

Starting from $y = f(x)$:

• Vertical translation by $k$: $y = f(x) + k$ shifts the graph up by $k$ (down if $k < 0$).
• Vertical dilation by $a$: $y = a f(x)$ stretches vertically by factor $|a|$. If $a < 0$, also reflects in the $x$-axis.
• Reflection in the $x$-axis: $y = -f(x)$.

These do what they say: a point $(x_0, y_0)$ on $y = f(x)$ becomes $(x_0, a y_0 + k)$ on $y = a f(x) + k$.

Horizontal transformations (act on $x$, inside the function)

Starting from $y = f(x)$:

• Horizontal translation by $h$: $y = f(x - h)$ shifts the graph right by $h$ (left if $h < 0$). Note the sign flip: $x - h$ means right by $h$.
• Horizontal dilation by $b$: $y = f(b x)$ compresses horizontally by factor $b$ (stretches if $0 < b < 1$). The sign flip applies here too: dividing by $b$ inside means stretching by $b$.
• Reflection in the $y$-axis: $y = f(-x)$.

A point $(x_0, y_0)$ on $y = f(x)$ becomes $\left(\frac{x_0 + h}{1}, y_0\right) = \left(x_0 + h, y_0\right)$ after a shift, and $\left(\frac{x_0}{b}, y_0\right)$ after a horizontal dilation.

The general form

The most general single-variable transformation is

The graph is obtained from $y = f(x)$ by:

1. (Inside) horizontal dilation by factor $\frac{1}{b}$, then translation right by $h$.
2. (Outside) vertical dilation by factor $a$, then translation up by $k$.

A point $(x_0, y_0)$ on $y = f(x)$ maps to $\left( \frac{x_0}{b} + h, a y_0 + k \right)$.

Order of operations

Inside the function: apply the dilation $b$ first, then the translation $h$. Outside: apply the dilation $a$ first, then the translation $k$. Mixing the order changes the answer.

A useful mnemonic: inside transformations act in the opposite of the natural reading order on the algebra; outside transformations act in the natural order.

Effect on key features

Translations move features without changing them. Dilations rescale distances. Reflections flip orientation.

• Asymptotes shift with translations and rescale with dilations.
• $x$-intercepts (zeros) shift horizontally and rescale by horizontal factors.
• $y$-intercept changes under any horizontal transformation that moves $x = 0$.
• The maximum value of $|a f(x) + k|$ on the same domain changes by $|a|$ in spread and $+ k$ in centre.

Build a transformed graph one step at a time

The reliable way to sketch a combined transformation is to apply one operation at a time and redraw, rather than trying to jump to the final picture. Below, the curve $y = -3(x - 2)^2 + 4$ is built from $y = x^2$ in the exact order set by the general form: dilations and reflections first, then translations, with outside acting on $y$ and inside acting on $x$. Each stage keeps the previous curve as a dashed ghost so you can see what moved. (This is the same worked Step 1 below, drawn out.)

Stage 1, start from the base curve. Draw $y = x^2$: a parabola with its vertex at $(0, 0)$, opening upward and symmetric about the $y$-axis. Every other curve in this sequence is this one in disguise.

Stage 2, apply the outside dilation and reflection. The factor $-3$ acts outside, so it does what it says: multiply every height by $3$ (a vertical stretch) and flip top-to-bottom (because of the minus). The result $y = -3x^2$ opens downward and is three times as steep. The vertex stays at $(0, 0)$, because scaling and reflecting a height of $0$ leaves it at $0$.

Stage 3, apply the inside translation. Replacing $x$ with $x - 2$ acts inside the function, so it does the opposite of what it looks like: it shifts the graph right by $2$, not left. The whole parabola slides across so the vertex now sits at $(2, 0)$. Heights are unchanged; only the horizontal position moves.

Stage 4, apply the outside translation. Adding $4$ acts outside, so again it does what it says: lift the whole graph up by $4$. The vertex rises from $(2, 0)$ to $(2, 4)$. The finished curve $y = -3(x - 2)^2 + 4$ is a downward parabola with vertex $(2, 4)$, and now it crosses the $x$-axis (because the maximum is above the axis), which the earlier stages did not.

If you ever apply the translation before the dilation, you get the wrong answer: shifting first and then stretching scales the shift as well. Doing the inside dilation before the inside translation, and the outside dilation before the outside translation, is what keeps the vertex landing at $(2, 4)$.

How exam questions ask about transformations

The wording changes but the task is one of a few shapes. Learn to match the phrasing to the move:

• "The graph of $y = f(x)$ has a [feature] at $(p, q)$. Find the corresponding point on $y = a f(b(x - h)) + k$." Track that one point: its image is $\left(\frac{p}{b} + h, \, a q + k\right)$. This is the single most common 2 to 3 mark question (see the first HSC-style question above).
• "Describe the sequence of transformations that maps $y = f(x)$ to [equation]." Read the equation outside-in or use the general form: name each of $a$, $b$, $h$, $k$ and state stretch/reflect/shift with direction and factor.
• "Sketch $y = a f(b(x - h)) + k$, showing key features." Mark the transformed intercepts, asymptotes, and turning points or vertex; do not just translate the shape vaguely. Markers look for moved features at the right coordinates.
• "Write the equation of the graph obtained by [a list of transformations]." Apply them in order to the algebra, keeping inside and outside operations separate.
• "On the same axes, sketch $y = f(x)$ and $y = f(-x)$ (or $-f(x)$, or $f(x) + c$)." A reflection or single shift: state which axis the reflection is in, or which way the shift goes, and draw both curves.
• A horizontal factor written un-factored, like $f(2x - 4)$. Factor the inside first to $f(2(x - 2))$ before reading off the shift, or you will shift by the wrong amount.