Investigation

## Solution

Choose two of these transformations and apply them in turn starting with the function $f(x)$. Sketch the resulting graph after you have applied one transformation and then the other. Does it matter which order you apply the two transformations?

Does the order matter for all pairs, some pairs or none of the pairs?

Translation by $\begin{pmatrix}-2\\0\end{pmatrix}$

Translation by $\begin{pmatrix}0\\-4\end{pmatrix}$

Stretch by factor $3$ parallel to $x$

Stretch by factor $\frac{1}{2}$ parallel to $y$

Would your answer be different if we had chosen a different set of transformations? Or a different starting function?

Transformations (1) and (2)

Applying (1) then (2) translates the graph $2$ units to the left and then $4$ units down. Applying (2) then (1) translates it down then left. The result is the same; the order does not matter.

One of the key points on the graph is the local maximum at $(0,4)$. Applying the transformations in either order takes this point to $(-2,0)$ as shown dotted in the diagram.

Algebraically, we replace $x$ with $x+2$, and replace $y$ with $y+4$, giving us $y+4=f(x+2) \quad\implies\quad y=f(x+2)-4.$

Transformations (1) and (3)

Applying (1) then (3) translates the graph $2$ units to the left and then stretches it horizontally. The local maximum at $(0,4)$ moves to $(-2,4)$ which is then stretched away from the $y$-axis to $(-6,4)$.

Applying (3) to the point $(0,4)$ does not move it as it is on the $y$-axis. Transformation (1) then moves it to $(-2,4)$. This time, the order has made a difference to where the local maximum ends up and we get two different graphs, as shown here in blue and green.

What is it that makes the order matter this time whereas it didn’t matter for transformations (1) and (2)?

How are the two resulting graphs related? What is the transformation from one to the other?

Algebraically, transformation (1) replaces $x$ with $x+2$ while (3) replaces $x$ with $\frac{x}{3}$. So the different orders look like this. \begin{align*} y&=f(x) \quad\stackrel{(1)}{\longrightarrow}\quad y=f(x+2) \quad\stackrel{(3)}{\longrightarrow}\quad y=f\left(\frac{x}{3}+2\right) \\ y&=f(x) \quad\stackrel{(3)}{\longrightarrow}\quad y=f\left(\frac{x}{3}\right) \quad\stackrel{(1)}{\longrightarrow}\quad y=f\left(\frac{x+2}{3}\right) \end{align*}

Transformations (1) and (4)

The order does not matter. Algebraically we have $y=\frac{1}{2}\,f(x+2)$.

Transformations (2) and (3)

The order does not matter. Algebraically we have $y=f\left(\frac{x}{3}\right)-4$.

Transformations (2) and (4)

The order does matter. Algebraically, transformation (2) replaces $y$ with $y+4$ while (4) replaces $y$ with $2y$. So the different orders look like this. \begin{align*} y&=f(x) \quad\stackrel{(2)}{\longrightarrow}\quad y+4=f(x) \quad\stackrel{(4)}{\longrightarrow}\quad 2y+4=f(x) \implies y=\frac{1}{2}\big(f(x)-4\big) \\ y&=f(x) \quad\stackrel{(4)}{\longrightarrow}\quad 2y=f(x) \quad\stackrel{(2)}{\longrightarrow}\quad 2(y+4)=f(x) \implies y=\frac{1}{2}\,f(x)-4 \end{align*}

As before, how are the two resulting graphs related?

Transformations (3) and (4)

The order does not matter. Algebraically we have $y=\frac{1}{2}\,f\left(\frac{x}{3}\right)$.

How would a translation by $\begin{pmatrix}3\\-2\end{pmatrix}$ behave when combined with other transformations?

What if we combined three of the original transformations? Would the order always matter?