Throughout this Topic, translation distances are positive, and horizontal and vertical stretch factors are positive unless a reflection is stated separately.

1. Why Horizontal and Vertical Transformations Behave Differently

Let $y=g(x)$ be the original graph, and let $y=G(x)$ be the transformed graph.

If $g(p)=q$, then $(p,q)$ lies on the original graph. The table follows this point through each transformation and derives the corresponding function rule.

The key distinction is:

$$\boxed{ \text{Horizontal transformations change the input of }g, \qquad \text{vertical transformations change its output.} }$$

Do not treat these as rules to memorise.
The important idea is how the new point on the graph determines the new function. Once that is understood, the same reasoning applies even when the current function is already transformed.

2. Transforming a Function That Has Already Been Transformed

The same reasoning applies when the current graph is already a transformed version of $f$.

Suppose the current graph is $y=g(x)$, where, for example, $g(x)=A f(Bx+C)+D$.

Let $y=G(x)$ be the graph obtained after applying one more transformation.

The important point is that the new transformation acts on the current function $g$ as a whole.

$$\boxed{ f(x) \longrightarrow g(x) \longrightarrow G(x) }$$

We do not return to the original $f(x)$ when applying the next transformation.

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Horizontal Transformations of the Current Function

Suppose $g(p)=q$.

Then $(p,q)$ is a point on the current graph.

If that current graph is moved horizontally, the same reasoning as before applies.

Suppose the current graph is $g(x)=A f(Bx+C)+D$.

Then the horizontal transformations can be written directly in terms of the original function $f$.

For example, $g(x)=f(2x)$.

Moving the current graph right by $a$ gives $G(x)=g(x-a) =f\bigl(2(x-a)\bigr) =f(2x-2a)$.

Hence $f(2x) \xrightarrow{\text{right by }a} f(2x-2a),$ not $f(2x-a)$.

The latter would move the graph right by only $a/2$.

3. Composing Several Transformations

When several transformations are applied in succession, each one acts on the current function produced by the previous step.

Write

$$g_0(x)=f(x),$$

then update the function one step at a time:

$$\boxed{ g_0 \longrightarrow g_1 \longrightarrow g_2 \longrightarrow \cdots }$$

Following a Transformation Chain

Starting with

$$g_0(x)=f(x),$$

apply, in order:

1. translate $2$ units to the right;
2. reflect in the $y$-axis;
3. stretch parallel to the $x$-axis by scale factor $3$;
4. translate $4$ units upwards.

Hence $G(x)=f\left(-\frac{x}{3}-2\right)+4.$

The whole chain is

$$f(x) \longrightarrow f(x-2) \longrightarrow f(-x-2) \longrightarrow f\left(-\frac{x}{3}-2\right) \longrightarrow f\left(-\frac{x}{3}-2\right)+4.$$

Each arrow acts on the expression immediately before it.

Order Matters

The same transformations applied in a different order can produce a different graph.

Translate right by $2$, then stretch horizontally by factor $3$:

$$f(x) \xrightarrow{\text{right }2} f(x-2) \xrightarrow{\text{horizontal stretch }3} f\left(\frac{x}{3}-2\right) = f\left(\frac{x-6}{3}\right).$$

Reverse the order:

$$f(x) \xrightarrow{\text{horizontal stretch }3} f\left(\frac{x}{3}\right) \xrightarrow{\text{right }2} f\left(\frac{x-2}{3}\right).$$

In general, $f\left(\frac{x-6}{3}\right) \ne f\left(\frac{x-2}{3}\right)$.

Therefore $\text{the order of transformations can change the final graph}.$

When the order is changed, do not try to rearrange the transformations mentally. Apply each sequence to the current function and compare the resulting expressions.