Analysis only. The guidance names the order explicitly, and gives sketching y = 3x² + 2 from y = x² as the example, so both halves of this page are the examined content.
Ask for 3x² + 2 and 3(x² + 2) at x = 1 before anything is on the screen. Most of a class gets both right, as arithmetic, without noticing they are two different curves. Then put the figure up and the two answers they just produced are the two dots.
Move the slider and the gap stays at 4. That constancy is what makes it a real difference rather than a coincidence at one point, and it is worth saying out loud: a gap that never closes cannot be a rounding.
View 2 is the sign, with y = x² dashed underneath. The point marked at (−3, 36) is the one that does the work: if a student thinks the vertex is at −3, the curve is 36 above them there.
| Question | Answer |
|---|---|
| 1. 3x² + 2 at x = 1 | 3 + 2 = 5. |
| 2. 3(x² + 2) at x = 1 | 3 × 3 = 9. |
| 3. Vertex of (x − 3)² | x = 3. |
| 4. −f(2) for f(x) = x² | −4. |
| 5. Why the orders differ | B. Translating first means the stretch acts on the 2 as well. |
Questions 1 and 2 are the page in two lines. Set them together, before any theory, and the 4 arrives from their own arithmetic. The vertices, (0, 2) and (0, 6), are worth adding on the board afterwards because they make the same point at a single glance.
1 markEach transformation named, with its direction and factor.
1 markThe order, where more than one is involved.
1 markThe resulting equation, or a sketch with one named point's image marked.
The vocabulary is marked. A translation needs a vector or a direction and a distance; a stretch needs a direction and a factor. "It moves up and gets thinner" describes the picture and earns nothing, and students are often surprised by that, so it is worth marking one set strictly early.
| They wrote | What happened |
|---|---|
| 9 on question 1 | The other order. Worth asking which part the 3 is multiplying. |
| 3 | 3x² with the 2 dropped. Arithmetic rather than method. |
| 11 | 3(1 + 2) + 2: the 2 used twice. They have merged the two orders into one expression. |
| 25 | (3 + 2)². The squaring applied last rather than to the x. |
| 7 on question 2 | 3 + 2 + 2. The 3 has been added rather than multiplied through the bracket. |
| −3 on question 3 | The sign error this page exists for. Send them to the figure's (−3, 36). |
| 0 on question 3 | The vertex of x² before the translation. They have not applied it. |
| 9 on question 3 | The curve's VALUE at x = 0, which is a y not an x. |
| 4 on question 4 | f(−2) rather than −f(2). The two reflections confused, and for an even function one of them does nothing, which hides the error. |
| −2 | Negated the input instead of the output. Apply f first. |
"Why does the minus inside move it the other way?" Because to get the same output you now have to feed in a bigger x. Pick a point: y = x² is 0 at x = 0, and y = (x − 3)² is 0 at x = 3, so the whole picture has slid right. Tracking one point is more convincing than any rule, and it generalises to every inside transformation.
"Which order should I use if the question does not say?" Read the equation. The outermost operation was applied last, so work from the outside in. If a question gives you a list of transformations instead, apply them in the order written. The ambiguity students worry about is almost never there once they look at which expression they have been handed.
"Does the order always matter?" No, and saying so is more honest than implying it always does. Two translations commute, and so do two vertical stretches. It matters when one transformation acts on the thing another has just produced, which is exactly the stretch-and-translate case. A class that knows when it matters is faster than one that checks every time.
"Is f(2x + 1) on the syllabus?" Not at Standard Level; the guidance says so explicitly. It is AHL 2.16. If a strong student asks, the honest answer is that it is a horizontal stretch and a horizontal translation in one bracket, and that the order inside it is precisely what makes it harder.
| Demonstrate | Enter the base function as Y1 or f1, then write both orders IN TERMS OF IT: 3*f1(x)+2 and 3*(f1(x)+2). The brackets on the screen are then the order, which is a better display of the idea than two expanded quadratics that look unrelated. Trace between the curves at a fixed x and the 4 reads off. |
|---|---|
| Where they stick | Referring to one function from inside another. On the Casio the Y symbols come from VARS → GRAPH, and typing the letter Y with a 1 after it gives Y times 1, which graphs a straight line and confuses everybody. On the Nspire, f1(x) has to include the (x). Both take one demonstration and then work for ever. |
| The check | One point, tracked. Decide where (1, 1) on y = x² should end up under the transformation you are describing, then check the machine agrees. If your description sends it to (1, 5) and the graph has it at (1, 9), the order is wrong and you know immediately. |
Analysis Paper 1 has no calculator and sketching a transformed graph from a given one is Paper 1 work. Use the machine for the order comparison, then sketch by hand with a tracked point.
| Step | What |
|---|---|
| 1 | Questions 1 and 2 cold, as arithmetic. Collect 5 and 9. |
| 2 | Point out that those were two curves, not two sums. Figure view 1. |
| 3 | Slide the slider. The gap stays at 4. |
| 4 | Expand 3(x² + 2) and watch the 2 become a 6. |
| 5 | The outside and inside rule, as one sentence, then the table of six. |
| 6 | (x − 3)²: ask where the vertex is. Collect the −3. |
| 7 | View 2, and the point at (−3, 36). |
| 8 | Both reflections on x², with the even-function trap named. |
Do not say "inside the bracket does the opposite". It is a memory trick with no content, and students apply it to the outside transformations too. Say that inside the function the x is being changed before f sees it, so to get the old output you need a new input, and then track one point.
Do not say "the order never matters as long as you do them all". It is false and this page is the counterexample. If you want a simple rule, the true one is that the outermost operation in the final expression was applied last, which is short enough to say and always works.