Rewind, Replay, Fast-Forward · Teacher Answers

🔐 Student Activity Unlock Codes (case-insensitive)

Phase 1 → Phase 2

HEIGHT

Phase 2 → Phase 3

SPEED

Phase 3 → Phase 4

REWIND

Phase 4 → Complete

PARENT

🎬 Oral Launch · Read Aloud to Class (about 3 min)

Play any short clip from your phone or YouTube at normal speed, then again at 2× speed. Then say: "Something happened at the 6-second mark of that clip. At 2× speed, when does it happen now? Hold that answer in your head. Today, every graph you draw comes from a video edit like this one, and that one question is going to explain the part of transformations that confuses almost everyone."

Then form visibly random groups, send them to the boards, and have one student per group open the activity from the QR card. The first thing on every board should be a copy of the scenario graph of \(h(t)\); check this before anyone starts Task 1a.

Same Clock, Different Heights

~8 min. Give code HEIGHT when the board shows all four graphs and a rule in coordinate form.

Task 1aReading the footage

\(t\) (s)	2	4	6
Drone A	6	6	10
Drone B	12	12	20
Drone C	3	3	5
Reflection	−6	−6	−10

This is deliberately easy: it builds confidence and gets the marker moving. The only real decision is the sign of the reflection; if a group writes positive values, ask "is the reflection above or below the water?"

Task 1bNaming and sketching

Drone B: \(2h(t)\) · Drone C: \(\tfrac12 h(t)\) · Reflection: \(-h(t)\)

Expected sketch: vertical stretch, scale factor 2

Expected sketch: vertical compression, scale factor ½, and reflection in the horizontal axis

Task 1cThe vertical rule

(a,\ b)\ \text{on}\ y = h(t) \quad\longrightarrow\quad (a,\ kb)\ \text{on}\ y = k\,h(t)

\((a, b) \to (a, kb)\). Invariant points: where \(h(t) = 0\), here \((0, 0)\) and \((8, 0)\).

Points on the horizontal axis never move because \(k \times 0 = 0\) for every \(k\). In the video: when Drone A is on the launch pad, so is every other drone, and so is the reflection.

Push one group further: "Is the reflection a stretch too?" It is the case \(k = -1\). Banking this now makes Phase 3 faster.

Playback Speed

~14 min, the heart of the activity. Give code SPEED only when Task 2d is written convincingly, not just the rule.

Task 2aPredictions

2× speed: peak at \(t = 3\) s, clip ends at \(t = 4\) s.
0.5× speed: peak at \(t = 12\) s, clip ends at \(t = 16\) s.

Almost every student gets this right instantly, and that is the point. Before they watch the clip, say out loud: "You just halved the times. Remember that you did that." You will cash this in at Task 2d.

Task 2b\(F(t) = h(2t)\)

Why the 2 is inside: after \(t\) real seconds at double speed, \(2t\) seconds of the original flight have been shown, so the altitude on screen is the original altitude at time \(2t\), which is \(h(2t)\).

\(t\)	0	1	2	3	4
\(F(t) = h(2t)\)	\(h(0) = 0\)	\(h(2) = 6\)	\(h(4) = 6\)	\(h(6) = 10\)	\(h(8) = 0\)

\(F(5) = h(10)\), which does not exist: the original flight only lasts 8 seconds, so the 2× clip is over after 4 seconds. The domain of \(F\) is \(0 \le t \le 4\).

Expected sketch: horizontal compression, scale factor ½ (towards the vertical axis)

Most common error: plotting \(F(1) = 2h(1)\) or putting points at \(t = 0, 4, 8, 12, 16\). Don't correct it; point at their Task 2a prediction ("you said the peak is at 3 seconds; where is it on your graph?") and walk away. The F(5) question is there to surface domain, a quiet bonus.

Task 2c\(S(t) = h\!\left(\tfrac t2\right)\)

\(t\)	0	4	8	12	16
\(S(t) = h\!\left(\tfrac t2\right)\)	\(h(0) = 0\)	\(h(2) = 6\)	\(h(4) = 6\)	\(h(6) = 10\)	\(h(8) = 0\)

Expected sketch: horizontal stretch, scale factor 2 (away from the vertical axis)

Task 2dWinning the argument

Model reply: "The 2 inside the brackets speeds the clip up, it doesn't make it longer. At 2× speed every moment of the flight happens in half the time: the peak moves from 6 s to 3 s and the clip ends at 4 s instead of 8 s. So the graph is squashed towards the vertical axis by a factor of ½. The heights don't change at all."

(a,\ b)\ \text{on}\ y = h(t) \quad\longrightarrow\quad \left(\tfrac{a}{k},\ b\right)\ \text{on}\ y = h(kt)

\((a, b) \to \left(\dfrac{a}{k},\ b\right)\): horizontal stretch with scale factor \(\dfrac1k\)

Why this matters

Students usually memorise "inside does the opposite" as an arbitrary rule. Here it isn't arbitrary: the input is time, and speeding up time means everything arrives sooner. Any board that can say this in their own words has understood horizontal stretches, not memorised them. Photograph the best reply for the debrief.

Rewind and Remix

~9 min. Give code REWIND when all three mystery clips have a function and a description.

Task 3a\(R(t) = h(-t)\)

\(t\)	−8	−6	−4	−2	0
\(R(t) = h(-t)\)	\(h(8) = 0\)	\(h(6) = 10\)	\(h(4) = 6\)	\(h(2) = 6\)	\(h(0) = 0\)

Expected sketch: reflection in the vertical axis

Mirror: the vertical axis (\(t = 0\)). In Phase 1 the mirror was the horizontal axis.

Connect the two reflections explicitly: minus outside flips heights (reflection in the horizontal axis); minus inside flips time (reflection in the vertical axis). It is the same "inside versus outside" story as Phase 2, with \(k = -1\).

Task 3bThree mystery clips

Clip A: 2× speed and twice as high, \(y = 2h(2t)\)

Clip B: reflected in the lake and half as high, \(y = -\tfrac12 h(t)\)

Clip C: rewind at 2× speed, \(y = h(-2t)\)

Clip C is the hardest: groups usually see "backwards" and miss "faster" (or the reverse). Ask: "How long does Clip C last?" (4 s, so it's at double speed.) Accept any correct equivalent such as \(h(2 \cdot (-t))\). All values verified computationally.

Back to the Parents

~8 min. Final phase; give code PARENT when 4a to 4c are done. 4d is for fast groups and the debrief.

Task 4a\(4x^2\) versus \((2x)^2\)

\(x\)	−2	−1	0	1	2
\(4f(x) = 4x^2\)	16	4	0	4	16
\(f(2x) = (2x)^2\)	16	4	0	4	16

\(y = 4f(x)\) is a vertical stretch with scale factor 4; \(y = f(2x)\) is a horizontal stretch with scale factor ½. They produce exactly the same curve, because \((2x)^2 = 2^2x^2 = 4x^2\).

Expected sketch: both transformations land on the same parabola

Task 4b\(2\sqrt x\) versus \(\sqrt{4x}\)

\(x\)	0	1	4	9
\(2g(x) = 2\sqrt x\)	0	2	4	6
\(g(4x) = \sqrt{4x}\)	0	2	4	6

\sqrt{4x} = \sqrt4\,\sqrt x = 2\sqrt x \qquad (x \ge 0)

Expected sketch: vertical stretch factor 2 is the same as horizontal stretch factor ¼

Task 4cFinding \(k\)

\(9x^2 = (kx)^2\): \(k^2 = 9\), so \(k = 3\) (or \(k = -3\))
\(3\sqrt x = \sqrt{kx}\): \(\sqrt k = 3\), so \(k = 9\)
\(5|x| = |kx|\): \(|k| = 5\), so \(k = 5\) (or \(k = -5\))

If a group finds \(k = -3\), celebrate it: \(y = (-3x)^2\) is a horizontal compression and a reflection in the vertical axis, and the parabola is symmetric so nothing visibly changes. For \(\sqrt{kx}\), \(k = -9\) fails because the domain flips to \(x \le 0\).

Task 4dChallenge: faking Drone B

No. Changing the playback speed only moves points left and right; every height on the graph stays the same, so the drone never goes above 10 m. Drone B reaches 20 m. Formally: \(h(kt)\) has a maximum of 10 for every \(k\), but \(2h(t)\) has a maximum of 20.

The trick works for \(x^2\), \(\sqrt x\) and \(|x|\) because scaling the input of these functions scales the output by a fixed factor everywhere: \((kx)^2 = k^2 x^2\). The drone's flight has no such property: for example \(h(2 \times 3) = h(6) = 10\), but \(2h(3) = 2 \times 6 = 12\).

Watch for this: a group testing only \(t = 1\) will find \(h(2) = 6 = 2h(1)\) and conclude it works. That happens because the first piece of the flight is a straight line through the origin (\(h(t) = 3t\)), which is itself a "parent-like" function. Lovely debrief moment if it comes up.

Big idea to land

For most functions, a vertical stretch and a horizontal stretch are genuinely different edits. For special functions like \(x^2\) and \(\sqrt x\) they can coincide, and that is a property of the function, not a rule about transformations.

📋 Consolidation · select 2–3 boards to debrief (about 5 min)

Task 2a: "You all halved the times instantly. Why?" Then show the reveal video below.
Task 2d: read out the most convincing reply. Land the phrase: inside the brackets changes the input (time), so it does the opposite of what it looks like.
Phase 3: minus outside flips heights, minus inside flips time.
Task 4a: play the parent reveal. Same curve, two different stories.

Debrief reveal 1: key points moving for \(2h(t)\), \(h(2t)\), \(h\!\left(\tfrac t2\right)\) and \(h(-t)\). Show after Phase 2 or at the end.

Debrief reveal 2: \(4x^2\) and \((2x)^2\) land on the same curve. Show after Phase 4.