Apply translations, reflections and dilations to the graph of a function and identify the resulting equation

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.
• IMATH_53 -intercepts (zeros) shift horizontally and rescale by horizontal factors.
• IMATH_54 -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.

Worked examples

A composite transformation

Start with $y = x^2$. Find the equation after stretching vertically by $3$, reflecting in the $x$-axis, shifting right by $2$, and shifting up by $4$.

Vertical: $y = -3 x^2$ (vertical stretch and reflection together, since both are outside).

Then horizontal shift: $y = -3 (x - 2)^2$.

Then vertical shift: $y = -3 (x - 2)^2 + 4$.

Vertex at $(2, 4)$, opening downward.

Inside and outside dilations

Sketch $y = 2 \sin(3 x)$ starting from $y = \sin x$.

Inside: $\sin(3 x)$ compresses horizontally by factor $\frac{1}{3}$. Period changes from $2 \pi$ to $\frac{2 \pi}{3}$.

Outside: factor $2$ doubles the amplitude. Maximum $2$, minimum $-2$.

Working backwards

The graph of $y = 4 - (x + 1)^2$ is the parabola $y = x^2$ reflected in the $x$-axis (giving $y = -x^2$), shifted left by $1$ (giving $y = -(x + 1)^2$), and shifted up by $4$. Vertex at $(-1, 4)$, opens downward.

Tracking a single point

The point $(0, 0)$ on $y = \sin x$ becomes which point on $y = 5 \sin(2(x - \pi / 4)) + 3$?

$x$: solve $2(x - \pi / 4) = 0$, so $x = \pi / 4$.

$y$: $5 \sin 0 + 3 = 3$.

New point: $(\pi / 4, 3)$.

Common traps

Sign flip on horizontal translations. $y = f(x - h)$ moves the graph right by $h$, not left. Drawing one specific point through the transformation is a fast check.

Applying inside operations in the wrong order. $f(2x - 4)$ is not the same as $f(2(x - 4))$. Factor first: $f(2x - 4) = f(2(x - 2))$, so the horizontal compression by $\frac{1}{2}$ is followed by a shift right by $2$, not $4$.

Treating a horizontal dilation as a vertical effect. $y = f(2 x)$ rescales the $x$-axis, not the $y$-axis. The $y$-values of any single point are unchanged.

Confusing $-f(x)$ with $f(-x)$. The first reflects in the $x$-axis (flip top to bottom), the second in the $y$-axis (flip left to right). For an even function they look the same; for an odd function $f(-x) = -f(x)$, so they coincide there too.

Missing the impact on the domain. Reflecting or dilating horizontally changes the natural domain of a function that has a restricted domain (such as $\sqrt{x}$, $\ln x$, or $\arccos x$). Track the new domain along with the new equation.

In one sentence

The general transformed function $y = a f(b(x - h)) + k$ takes the base graph $y = f(x)$, dilates and translates inside the function to act on $x$ and outside the function to act on $y$, with horizontal operations running in the opposite order to their natural reading.