Combined transformations — order matters — Mathematics

Before this

This leads to

Graphs of sin, cos and tan · coming soon
Exponential functions · coming soon
The chain rule · coming soon

Same two moves, two different graphs

Start with the graph of $y = x^{2}$ . Its lowest point is the origin. Amara and Wei are each asked to do two things to it: stretch it vertically with scale factor 2, and translate it up 3.

Amara stretches first. Every height doubles, so she has $y = 2 x^{2}$ . Then she moves it up 3, and gets $y = 2 x^{2} + 3$ . Its lowest point is $(0, 3)$ .

Wei translates first. He has $y = x^{2} + 3$ . Then he stretches it, so every height of that graph doubles: $y = 2 (x^{2} + 3) = 2 x^{2} + 6$ . Its lowest point is $(0, 6)$ .

They did the same two moves. Neither of them made a mistake. And they drew two different graphs.

The graph of

y = x^{2}

(ink). Stretching first and then moving up 3 gives

y = 2 x^{2} + 3

(teal). Moving up 3 first and then stretching gives

y = 2 x^{2} + 6

(dashed teal). The point

(1, 1)

ends at

(1, 5)

in one order and at

(1, 8)

in the other.

The translations lecture promised that its main sentence would come back for every combined transformation. The reflections lecture promised that order would not always be as harmless as it was for two reflections. This lecture keeps both promises. It answers three questions. When does the order matter? How do you read the order from an equation? And how do you write an equation when the order is given?

Follow one point, one move at a time

The safest way to see what is happening is to follow one point. Take the point $(1, 1)$ on $y = x^{2}$ . Only its height changes in these two moves, so watch the height.

Amara: $1 \times 2 2 + 3 5$ .
Wei: $1 + 3 4 \times 2 8$ .

Now the reason is clear. In Wei's order, the 3 is added before the doubling. So the doubling doubles the 3 as well. That is where the extra 3 comes from: $2 \times (1 + 3) = 2 + 6$ .

Look at Amara's equation, $y = 2 f (x) + 3$ , with $f (x) = x^{2}$ . Read it as a calculation. To find $y$ , you first work out $f (x)$ . Then you multiply by 2. Then you add 3. That is exactly the order of her moves. So for changes outside the function, the order of the moves is the order of the arithmetic.

y = 2 f (x) + 3 : stretch, scale factor 2, then up 3

y = 2 (f (x) + 3) = 2 f (x) + 6 : up 3, then stretch, scale factor 2

The bracket is the whole difference. It tells you that the 3 was added before the stretch.

When order matters, and when it does not

Order does not always matter. Here are three kinds of pairs where it never does.

One horizontal move and one vertical move. A horizontal move changes only the $x$ -coordinate. A vertical move changes only the $y$ -coordinate. They work on different numbers, so neither can affect the other. "Right 2, then up 3" and "up 3, then right 2" both send $(p, q)$ to $(p + 2, q + 3)$ .

Two additions to the same coordinate. Up 3 and then up 5 is up 8 in either order, because $q + 3 + 5 = q + 5 + 3$ .

Two multiplications of the same coordinate. A stretch with scale factor 2 and a stretch with scale factor 3 give a stretch with scale factor 6 in either order. The stretches lecture showed that a reflection is a stretch with scale factor $- 1$ . So a reflection and a stretch in the same direction also give the same graph in either order: $- 1 \times 2 = 2 \times (- 1)$ .

Order matters only when one move adds and the other multiplies, and both change the same coordinate. In other words, a translation together with a stretch or a reflection, in the same direction. Then the multiplication either catches the added amount or it does not. Try every pair in the machine below.

Explore — the order machine

The ink graph is $y = f (x)$ with $f (x) = x^{2} - 2 x$ . Choose a first move and a second move. The solid teal graph does them in your order. The dashed teal graph does them in the other order. When the two graphs lie on top of each other, the order does not matter.

First move:

Then:

Try:

Four pairs of moves applied to $y = f (x)$ , with $f (x) = x^{2} - 2 x$ , in both orders. The point $(1, - 1)$ is the lowest point of the starting graph.

Stretch vertically (scale factor 2) and up 2. Stretch first: $y = 2 f (x) + 2$ , and $(1, - 1)$ moves to $(1, 0)$ . Up first: $y = 2 f (x) + 4$ , and it moves to $(1, 2)$ . Different graphs.
Stretch horizontally (scale factor 2) and right 2. Stretch first: $y = f (\frac{1}{2} x - 1)$ , and $(1, - 1)$ moves to $(4, - 1)$ . Right first: $y = f (\frac{1}{2} x - 2)$ , and it moves to $(6, - 1)$ . Different graphs.
Reflect in the $x$ -axis and up 2. Reflect first: $y = - f (x) + 2$ , and $(1, - 1)$ moves to $(1, 3)$ . Up first: $y = - f (x) - 2$ , and it moves to $(1, - 1)$ . Different graphs.
Up 2 and right 2. Either order: $y = f (x - 2) + 2$ , and $(1, - 1)$ moves to $(3, 1)$ . The same graph.

What should you have seen? Every pair with one horizontal and one vertical move gave a single graph. So did two stretches, or a stretch and a reflection in the same direction. Only a translation paired with a stretch or a reflection in the same direction split into two graphs.

Stop and think

A graph is reflected in the $x$ -axis and translated up 3. Does the order matter? If it does, write both equations.

Show the answer

Yes. Both moves change the $y$ -coordinate, and one multiplies (by $- 1$ ) while the other adds. Reflect first: every height $q$ becomes $- q$ , then $- q + 3$ . So $y = - f (x) + 3$ . Translate first: $q$ becomes $q + 3$ , then $- (q + 3) = - q - 3$ . So $y = - f (x) - 3$ . Many students expect "a reflection is just a flip, so it cannot depend on order". But a reflection in the $x$ -axis multiplies every height by $- 1$ , and multiplying after adding catches the added amount.

Inside the bracket, the order runs backwards

Now the hard part of the lecture. Spend time here. What happens when both changes are inside the bracket? Before you read on, make a prediction.

Predict — where does the lowest point go?

The graph of $y = x^{2}$ has its lowest point at $(0, 0)$ . Where is the lowest point of $y = (2 x + 6)^{2}$ ? Choose before you look.

The figure shows $y = x^{2}$ , then $y = (x + 6)^{2}$ , then $y = (2 x + 6)^{2}$ . The lowest point moves from $(0, 0)$ to $(- 6, 0)$ , and then to $(- 3, 0)$ .

The lowest point of $y = (2 x + 6)^{2}$ is $(- 3, 0)$ , because the bracket is zero when $2 x + 6 = 0$ .

Here is the reason, built from the sentence the translations lecture gave you. A change to the input moves the graph the opposite way, because the input must be changed back before the function sees the number it needs.

Look at what $y = f (2 x + 6)$ does to an input $x$ before $f$ sees it. It multiplies by 2. Then it adds 6. Suppose the old graph had a point $(p, q)$ , so $f (p) = q$ . Which new input $x$ makes the function see $p$ ? You must undo the two steps. You undo the last step first, as you do when you take off your shoes before your socks:

p - 6 p - 6 \div 2 \frac{p - 6}{2}

So the point $(p, q)$ moves to $(\frac{p - 6}{2}, q)$ . Read the arrows as moves of the graph. First every $x$ -coordinate decreases by 6: a translation 6 to the left. Then every $x$ -coordinate is divided by 2: a horizontal stretch with scale factor $\frac{1}{2}$ . For the lowest point, $0 \to - 6 \to - 3$ . That matches the figure.

Look at what just happened. The calculation inside the bracket was "multiply, then add". The moves of the graph were "subtract, then divide". Each step was undone, so each move is the opposite of what is written. And the steps were undone from the last one backwards, so the order is reversed as well. This is the same rule that the inverse functions lecture used to undo a function: undo each step with its opposite, in the reverse order.

There is a second correct description. Factorise the inside first: $2 x + 6 = 2 (x + 3)$ . Now the calculation is "add 3, then multiply by 2". Undo it backwards: divide by 2, then subtract 3. So the graph is stretched horizontally with scale factor $\frac{1}{2}$ first, and then translated 3 to the left. Check the lowest point: $0 \to 0 \to - 3$ . The same answer, from a different pair of moves.

That is it. That is the whole idea. Outside the bracket, the moves happen in the order of the arithmetic. Inside the bracket, each step is undone, so the moves are the opposites, in the reverse order. And a horizontal move never interferes with a vertical one, so the inside and the outside can be dealt with separately.

Combined transformations

For $y = a f (b x + c) + d$ , every point $(p, q)$ on $y = f (x)$ moves to

$(\frac{p - c}{b}, a q + d)$

Vertical moves, in the order of the arithmetic: stretch with scale factor $a$ , then translate by $d$ .

Horizontal moves, undoing the inside in reverse: translate by $- c$ , then stretch with scale factor $\frac{1}{b}$ . Or, with $b x + c = b (x + \frac{c}{b})$ : stretch with scale factor $\frac{1}{b}$ , then translate by $- \frac{c}{b}$ .

Worked examples

Worked Example 1 — Reading four moves from an equation

Describe the transformations that map $y = f (x)$ to $y = 3 f (2 x - 4) + 5$ . The point $(2, 1)$ lies on $y = f (x)$ . Find its image.

Outside, the arithmetic is "multiply by 3, then add 5". So: a vertical stretch with scale factor 3, then a translation up 5.

Inside, the arithmetic is "multiply by 2, then subtract 4". Undo it backwards: add 4, then divide by 2. So: a translation 4 to the right, then a horizontal stretch with scale factor $\frac{1}{2}$ .

Or factorise: $2 x - 4 = 2 (x - 2)$ . The arithmetic is now "subtract 2, then multiply by 2". Undo it backwards: divide by 2, then add 2. So: a horizontal stretch with scale factor $\frac{1}{2}$ , then a translation 2 to the right. Both descriptions are correct. "Translate right 2, then stretch" is not. It gives $f (2 x - 2)$ , a different graph.

The image of $(2, 1)$ : $x$ : $2 \to 6 \to 3$ . $y$ : $1 \to 3 \to 8$ . So the image is $(3, 8)$ .

Check by substitution. At $x = 3$ , the inside is $2 \times 3 - 4 = 2$ , and $f (2) = 1$ . So $y = 3 \times 1 + 5 = 8$ . ✓

Worked Example 2 — Following a point through three moves

The point $P (4, 6)$ lies on $y = f (x)$ . Find its image on $y = - f (x + 2) - 1$ , and on $y = - (f (x + 2) - 1)$ .

For the first curve, the inside is "add 2", so the graph moves 2 to the left: $x$ goes from 4 to 2. The outside is "multiply by $- 1$ , then subtract 1": the height goes $6 \to - 6 \to - 7$ . The image is $(2, - 7)$ .

For the second curve, the inside is the same, so $x$ is again 2. But the outside is now "subtract 1, then multiply by $- 1$ ": $6 \to 5 \to - 5$ . The image is $(2, - 5)$ .

Check: the second equation is $y = - f (x + 2) + 1$ . At $x = 2$ , $f (4) = 6$ , so $y = - 6 + 1 = - 5$ . ✓ One bracket moved the point 2 units.

Worked Example 3 — Writing the equation from a description

Write the equation of the graph after each sequence of moves on $y = f (x)$ . (a) Translate 3 to the left, then stretch vertically with scale factor 4. (b) Stretch horizontally with scale factor $\frac{1}{2}$ , then translate 3 to the left. (c) Translate 3 to the left, then stretch horizontally with scale factor $\frac{1}{2}$ .

Build the equation one layer at a time. Each new move acts on the whole of what you have so far. A horizontal move replaces every $x$ by something. A vertical move acts on the whole output.

(a) Left 3: $f (x + 3)$ . Stretch vertically: multiply the whole output by 4. $y = 4 f (x + 3)$ . One move is horizontal and the other vertical, so the order does not matter here.

(b) Stretch: $f (2 x)$ . Then left 3: replace every $x$ by $(x + 3)$ . $y = f (2 (x + 3)) = f (2 x + 6)$ .

Check (b) and (c) with a point $(4, q)$ . In (b), the stretch sends 4 to 2, and the translation sends 2 to $- 1$ . The formula gives $\frac{4 - 6}{2} = - 1$ . ✓ In (c), $4 \to 1 \to 0.5$ , and the formula gives $\frac{4 - 3}{2} = 0.5$ . ✓ So (b) and (c) are different graphs.

Worked Example 4 — Sketching a quadratic from its moves

Describe $y = - 2 (x - 1)^{2} + 8$ as transformations of $y = x^{2}$ . Find its turning point and its roots.

With $f (x) = x^{2}$ , the equation is $y = - 2 f (x - 1) + 8$ . Inside: a translation 1 to the right. Outside, in the order of the arithmetic: multiply by $- 2$ (a stretch with scale factor 2 and a reflection in the $x$ -axis, in either order), then add 8 (up 8).

The turning point $(0, 0)$ goes to $(1, 0)$ , then $(1, 0)$ , then $(1, 8)$ . It is now the highest point, because of the reflection.

Roots: $- 2 (x - 1)^{2} + 8 = 0$ , so $(x - 1)^{2} = 4$ , and $x - 1 = \pm 2$ . The roots are $x = - 1$ and $x = 3$ .

Check: at $x = 3$ , $- 2 \times 4 + 8 = 0$ . ✓ The roots are equally spaced on either side of $x = 1$ , as they must be for a parabola. ✓

Three classic mistakes

Trap 1 — Reading the inside from left to right

$y = f (2 x - 4)$ is not "stretch, then translate 4 to the right". That order gives $f (2 (x - 4)) = f (2 x - 8)$ . Inside the bracket the moves come in reverse: translate right 4, then stretch. Or factorise first: stretch, then translate right 2. If you are unsure, follow one point.

Trap 2 — Losing the bracket

"Translate up 3, then stretch vertically with scale factor 2" is $y = 2 (f (x) + 3) = 2 f (x) + 6$ , not $2 f (x) + 3$ . When a stretch comes after a translation, it acts on the whole output so far, so it needs a bracket around it.

Trap 3 — Translating by the wrong amount after factorising

In $f (2 x + 6)$ , the 6 is a translation only if you do it first. After you factorise to $f (2 (x + 3))$ , the translation is 3, and it comes after the stretch. Mixing the two descriptions, "stretch, then left 6", gives $f (2 x + 12)$ , a different graph.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

The point $(3, - 2)$ lies on $y = f (x)$ . Find its image on (a) $y = 2 f (x) - 5$ , (b) $y = 2 (f (x) - 5)$ , (c) $y = - f (x) + 4$ .

Hint 1
All three changes are outside the bracket, so only the height changes. The $x$ -coordinate stays 3.

Hint 2
Follow the arithmetic in order, starting from the height $- 2$ .

Full answer
(a) $- 2 \to - 4 \to - 9$ . The image is $(3, - 9)$ . (b) $- 2 \to - 7 \to - 14$ . The image is $(3, - 14)$ . (c) $- 2 \to 2 \to 6$ . The image is $(3, 6)$ . Check (b) by expanding: $2 f (x) - 10$ at $x = 3$ gives $- 4 - 10 = - 14$ . ✓ (a) and (b) use the same two moves in different orders, and differ by 5.
Describe $y = f (3 x + 6)$ as two moves in two different ways. The point $(3, 0)$ lies on $y = f (x)$ . Find its image.

Hint 1
The inside does "multiply by 3, then add 6". Undo it from the last step backwards.

Hint 2
For the second description, factorise: $3 x + 6 = 3 (x + 2)$ .

Full answer
Undoing "×3, then +6" backwards gives "−6, then ÷3": translate 6 to the left, then stretch horizontally with scale factor $\frac{1}{3}$ . With $3 (x + 2)$ , the inside is "+2, then ×3", undone as "÷3, then −2": stretch with scale factor $\frac{1}{3}$ , then translate 2 to the left. The image of $(3, 0)$ : $3 \to - 3 \to - 1$ , or $3 \to 1 \to - 1$ . So it is $(- 1, 0)$ . Check: at $x = - 1$ , the inside is $3 (- 1) + 6 = 3$ , and $f (3) = 0$ . ✓
Let $f (x) = x$ . Write the equation of the graph after each pair of moves. (a) Reflect in the $y$ -axis, then translate 2 to the right. (b) Translate 2 to the right, then reflect in the $y$ -axis. Give the domain of each, and the image of the starting point $(0, 0)$ .

Hint 1
Build each equation in layers. A horizontal move replaces every $x$ in what you have so far.

Hint 2
For (a): $f (- x) = - x$ , then replace $x$ by $(x - 2)$ .

Full answer
(a) $- x$ , then $- (x - 2) = 2 - x$ . The domain is $x \leq 2$ . The point $(0, 0)$ goes to $(0, 0)$ , then $(2, 0)$ . (b) $x - 2$ , then $- x - 2$ . The domain is $x \leq - 2$ . The point goes to $(2, 0)$ , then $(- 2, 0)$ . Check (b): at $x = - 2$ , $2 - 2 = 0$ . ✓ Both moves change $x$ , and one adds while the other multiplies by $- 1$ , so the order matters.
A weather station records the temperature during one day. The graph $y = g (t)$ gives the temperature in °C at $t$ hours after midnight. (a) To change °C to °F, you multiply by 1.8 and then add 32. Write the equation of the same graph in °F. (b) Explain why "add 32, then multiply by 1.8" gives the wrong graph, and by how many degrees it is wrong. (c) The station now measures time in hours after 6 a.m. Write the equation of the graph in °F with this new time $x$ . Check your answer at 9 a.m., when the temperature was 20 °C.

Hint 1
(a) and (b) are changes to the output. (c) is a change to the input: 9 a.m. is $x = 3$ , but it was $t = 9$ .

Hint 2
For (c), the function $g$ needs the time after midnight. Given $x$ , that time is $x + 6$ .

Full answer
(a) $y = 1.8 g (t) + 32$ : a vertical stretch with scale factor 1.8, then a translation up 32. (b) $1.8 (g (t) + 32) = 1.8 g (t) + 57.6$ . The stretch also multiplies the 32, so every value is 25.6 degrees too high. (c) The function needs $t = x + 6$ , so $y = 1.8 g (x + 6) + 32$ . This is a translation 6 to the left: the graph starts 6 hours later, so every time is 6 smaller. Check: at 9 a.m., $x = 3$ , and $g (3 + 6) = g (9) = 20$ . So $y = 1.8 \times 20 + 32 = 68$ °F. ✓ The wrong order would give $1.8 \times 52 = 93.6$ °F.

The bigger idea

Order matters far beyond graphs. Putting on your socks and then your shoes is not the same as shoes then socks. Turning a book over and then rotating it ends differently from rotating it first. In mathematics, two operations that give the same result in either order are said to commute. Adding and multiplying the same number do not commute, and that single fact explains this whole lecture.

The backwards order inside the bracket is the same rule you met with inverse functions. To undo a chain of steps, you undo the last one first. You will meet it again in calculus, where the chain rule works through a composite function from the outside in. And you will meet it with matrices, where the order of two transformations of the plane matters in the same way.

Key concepts in this lesson

The order of two transformations matters only when both change the same coordinate and one adds while the other multiplies, such as a translation and a stretch in the same direction.
Changes outside the function happen in the order of the arithmetic, so $y = a f (x) + d$ is a stretch with scale factor $a$ followed by a translation by $d$ .
Changes inside the bracket are undone before the function sees the number, so they happen as opposite moves in the reverse order: $f (b x + c)$ is a translation by $- c$ followed by a stretch with scale factor $\frac{1}{b}$ .
Factorising the inside, $b x + c = b (x + \frac{c}{b})$ , gives a second correct description: the stretch first, then a translation by $- \frac{c}{b}$ .
For $y = a f (b x + c) + d$ , every point $(p, q)$ moves to $(\frac{p - c}{b}, a q + d)$ , and following one point is the safest check of any description.