|f(x)| and f(|x|) are five characters apart on the page and completely different on the graph. The instruction that separates them is which axis you fold about, and it is worth saying out loud every time until it is automatic.
Leave it on the first view with the slider at −2 and ask for both values. The two readouts say 4 and 0, and the bar joining the two markers is four units long. Nobody argues with that.
Then drag the slider right. At x = 2 the bar vanishes and the two markers merge, and from there on they stay merged. Ask why they agree from 2 onwards: because the original was already positive there, so neither fold had anything to do.
Then step through the other three views. The slider drives all of them, so the same x gives the original's value and the transformed value on every graph. The one to dwell on is the last: f(2x + 1) crosses at 0.5, and the dashed original crosses at 2. Ask where 2 went.
| Question | Answer |
|---|---|
| 1. |f(−2)| | |−4| = 4. |
| 2. f(|−2|) | f(2) = 0. |
| 3. 1/f(4) | 1/2 = 0.5. |
| 4. Where f(2x + 1) = 0 | 2x + 1 = 2, so x = 0.5. |
| 5. Roots of f(|x|) | B. Two, at ±2, because the right half is copied over. |
Questions 1 and 2 are the same two numbers, 4 and 0, in the opposite order. A student who swaps them has the arithmetic right and the concept backwards, which is a different diagnosis from one who cannot do either.
1 markThe part of the original kept unchanged.
1 markThe correct fold, reflection or asymptote.
1 markKey features marked: roots, vertices, asymptotes.
These are sketch marks and labels earn them. Tell a class plainly that a rough V with (2, 0) written on it scores more than a beautiful V with nothing written on it, because that is counter-intuitive to anyone who thinks they are being marked on drawing.
| They wrote | What happened |
|---|---|
| 0 on question 1, 4 on question 2 | The error this page exists for. The two moduli swapped. Slider to −2 and read the two labels off the figure. |
| −4 on question 1 | No modulus applied at all. A modulus is never negative, so this is self-checkable. |
| 2 on question 1 | Took |−2| and stopped, giving the modulus of the input when the question wanted it on the output. |
| −2 on question 2 | Gave f(0). Possibly reading the graph's vertex rather than evaluating. |
| 2 on question 3 | Gave f(4) and forgot to invert. Very common, and it means they are reading 1/f(x) as a label rather than an instruction. |
| 0.25 on question 3 | Inverted x rather than f(x). The function is 1/(x − 2), not 1/x. |
| 1 on question 4 | The second error this page exists for. Subtracted the b and stopped. Check it in front of them: f(2(1) + 1) = f(3) = 1, which is not zero. |
| 1.5 on question 4 | Added the 1 instead of subtracting. They are applying the transformation forwards rather than undoing it. |
| 3 on question 4 | Computed 2 + 1. Same direction error, one step shorter. |
| A reflected lower half for f(|x|) | Drew |f(x)| instead. The sketch is wrong across the whole left half and usually earns one of three marks. |
"Why does f(|x|) throw away the left half?" Because the input is |x|, which is never negative, so the function never sees a negative number. Whatever f did on the left is unreachable. Then f(−3) is not consulted at all and f(|−3|) = f(3) is used instead, which is why the left half becomes a copy of the right. Saying "the function never sees a negative input" is the sentence that makes it stick.
"Can |f(x)| have a vertex that is not on the axis?" No, and it is a useful check. Every corner of |f(x)| is at a point where f crossed zero, so every vertex sits on the x-axis. A sketch with a corner floating above the axis is wrong, and that is spottable from across the room.
"What happens to [f(x)]² between −1 and 1?" It gets smaller, which students find surprising. 0.5² = 0.25, so heights inside that band shrink towards the axis and heights outside it grow. That is why the squared graph is flat near its roots and steep further out, and it is worth mentioning because a sketch drawn without it looks wrong in a way students cannot name.
"Where does 1/f(x) cross the axis?" Nowhere. A reciprocal is never zero, so |f(x)| has roots and 1/f(x) has asymptotes in the same places, which is a neat pairing to put on the board. The points that do not move are where f = ±1, and plotting those two first makes the sketch much easier.
| Demonstrate | Store f once, then define the others in terms of it: f2(x)=abs(f1(x)) and f3(x)=f1(abs(x)). Two entries differing only in where the abs sits, and two graphs that share nothing on the left. That typing contrast is the lesson in miniature and takes fifteen seconds. |
|---|---|
| Where they stick | Six functions drawn at once, which is unreadable. Use Hide/Show on the Nspire or deselect with F1 on the Casio, and keep two on screen: the original and one transformation. |
| The check | Zero on f1(2*x+1) returns 0.5. Students who answered 1 will want to argue with the algebra and will not argue with that. |
On the Casio, Abs is under OPTN then NUM, which is not where anyone looks. Worth showing once rather than letting them hunt.
| Step | What |
|---|---|
| 1 | Sketch f(x) = x − 2. Mark the root at 2 and the y-intercept at −2. |
| 2 | Ask for |f(x)| as an instruction, not a formula: what do we do to the picture? |
| 3 | Then f(|x|), and insist it is a different instruction before anyone draws. |
| 4 | The figure, slider at −2. Collect the 4 and the 0. |
| 5 | Count the roots of each: one and two. That settles that they are different functions. |
| 6 | The square: roots stay, nothing negative, the band between −1 and 1 shrinks. |
| 7 | The reciprocal: roots become asymptotes, f = ±1 stays put. |
| 8 | f(2x + 1). Ask where the root goes, collect the 1s, then solve 2x + 1 = 2. |
Do not say "the modulus makes everything positive". It is true of |f(x)| and false of f(|x|), which can be as negative as it likes: f(|0|) = −2 here. The sentence is the reason the two get confused, because it describes one of them and students attach it to both. Say instead which axis you fold about, every time.
Do not teach f(ax + b) as "shift then stretch" without doing the arithmetic. The order is arguable and students can get the right sketch from either description, but the root calculation is not arguable: set the input equal to the old root and solve. Give them that as the reliable method and the transformation language as the picture.