Combining Functions
Investigation

Order! Order!
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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?

Cubic shaped graph crossing the x-axis at -1 and touching it at 2, and crossing the y-axis at 4

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.

The transformed graph touches the x-axis at 2 and crosses at 4, and crosses the y-axis at -4

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.

The result of the transformation by (1) followed by (3) is a curve which crosses the x-axis, at -9 and touches it at the origin and has a local maximum at (-6,4); The result of transformation by (3) followed by (1) is a curve which crosses the x-axis at -5 and touches it at 4 and has a local maximum at (-2,4)

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?

The green and blue graphs are related by a horizontal translation through a distance of \(4\). This is the difference between a translation of \(2\), and a translation of \(2\) that is then scaled up by a factor of \(3\).

The order matters because both transformations are acting horizontally.

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 transformed graph crosses the x-axis at -3 and touches it at the origin and has a local maximum at (-2,2)

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

Transformations (2) and (3)

The transformed graph touches the x-axis at the origin and crosses it at 9, and has a local minimum at (6,-4)

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

Transformations (2) and (4)

The result of the transformation by (2) followed by (4) crosses the y-axis at -2 and has the local minimum at (2,-4); The result of transformation by (4) followed by (2) touches the x-axis at the origin and has the local minimum at (2,-2)
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 transformed graph has the local maximum where it crosses the y axis at 2 and the local minimum where it touches the x axis at 6

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

Of our four transformations, (1) and (3) are in the \(x\) direction while (2) and (4) are in the \(y\) direction. The order matters whenever we combine a stretch and a translation in the same direction.

In each of the cases (1)(3) and (2)(4) above, the resulting graphs are related by a translation. Can you explain the magnitude of the translation in each case?

The function used in this example has the shape of a cubic. The same conclusions would be true of any other function, but notice that with simple functions such as \(f(x)=x^2\) you cannot distinguish visually between a horizontal and a vertical stretch.

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?

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Last updated 25-Apr-17
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