That sentence is counter-intuitive enough that students write the formula upside down, and it costs a mark on nearly every graph question in the topic. The figure makes it a count of cycles rather than a formula to trust.
Start at b = 1 and walk the slider up slowly. At b = 1 the two spans sit exactly on top of each other and the verdict reads "exactly right". That is the moment to say: this is the only b for which the answer 2π is correct, and no exam question will give you this b.
Then keep going to 3 and ask how many complete cycles the dashed 2π span now covers. Three. Not "the period is wrong by a factor" but "the thing you called one cycle is three". Counting beats dividing here.
Then go down to b = 0.5, where the period is 12.566 and the dashed span runs off the end of the window. Students who have only seen b > 1 think the formula is "divide to make it smaller"; the half shows them it is a division either way.
Note what does not move throughout: the principal axis at y = 1 and the lines at 3 and −1 stay put, because a and d are outside the function. Point at that explicitly, because it is the fastest route into the inside-and-outside rule.
| Question | Answer |
|---|---|
| 1. The period | 2π/3 = 2.094. |
| 2. The maximum | d + a = 1 + 2 = 3. |
| 3. The shift | 4 to the right, because 3x − 12 = 3(x − 4). |
| 4. The period of 3 sin 2x | 2π/2 = π = 3.142. |
| 5. Why a bigger b is shorter | B. The input runs b times as fast, so x covers a bth of the distance. |
Question 4 is deliberately a different b on a friendlier function. A student who writes 2.094 again has pattern-matched the previous answer rather than used the formula, and that is a useful thing to catch before an exam does.
1 markAmplitude, from (max − min)/2 or read off a.
1 markPrincipal axis, from (max + min)/2.
1 markPeriod as 2π/b, with the formula written.
1 markThe shift, with the b factorised out first.
On "find the equation from this graph" all four are independent, so a student who gets the period wrong still banks the other three. Teach the order: principal axis, then amplitude, then period, then shift. Each one makes the next easier and the shift is hardest, so it goes last.
| They wrote | What happened |
|---|---|
| 6.283 for the period | The error this page exists for. They have read "cosine" and recalled 2π. Send them to the cycle count rather than the formula. |
| 18.85 | 2π × 3, multiplying where they should divide. This is the formula remembered upside down, and it is the one the slider fixes: a bigger b visibly squashes. |
| 3 for the period | Gave b itself. Usually a reading error rather than a misconception, but check they know what the period means. |
| 1.047 | π/3, which is HALF the period. They have counted peak to trough. Worth naming, because on a sine it is a very natural half to count. |
| 2 for the maximum | Gave the amplitude. They have forgotten the principal axis, which is why it is the first thing to find and not the third. |
| 4 for the maximum | max − min, which is twice the amplitude. Two errors compounding; ask them to draw the three horizontal lines. |
| 12 for the shift | The second error this page exists for. They read 3x − 12 without factorising. It puts the curve 8 units out, which is nearly four periods, and the sketch still looks like a cosine. |
| −4 for the shift | Right size, wrong direction. Inside the function a minus moves it the positive way, which is the 2.11 rule and worth linking back to. |
| 2.094 on question 4 | Copied the previous answer. The b changed and they did not look. |
"Why is it 2π and not 360?" Because Analysis assumes radians unless a degree symbol appears, and the guide says so plainly. In degrees the period of this curve is 120, and the formula is 360/b. Both are right and mixing them is not, so pick radians and stay there unless the question uses degrees.
"Does the shift change the period?" No, and it is worth demonstrating rather than asserting: the slider changes b and the peaks move and bunch up; a change to c would slide every peak without changing the gap. The two are independent, which is why they are separate marks.
"Which is the amplitude if a is negative?" The amplitude is |a|, because a distance cannot be negative. A negative a reflects the curve in the principal axis, so a cosine with a = −2 starts at its minimum instead of its maximum. Students usually spot the reflection and then write −2 for the amplitude, which loses the mark.
"Can I just read the period off the graph?" Yes, and on a "find the equation" question that is exactly the method: measure peak to peak, then b = 2π/period. It is the same formula rearranged, and students find that direction easier, so give them both forms and let them pick.
| Demonstrate | Graph it in the DEFAULT window first and let them see the mess: 9.55 cycles crammed across the screen and no period readable. Then set Xmin 0 and Xmax 8 from the period they calculated. The point is that the window comes from the arithmetic, not the other way round, and that is a transferable habit. |
|---|---|
| Where they stick | Degrees mode. In degrees this curve has a period of 120, so a window of 0 to 8 shows an almost flat line and students conclude the equation is wrong. Set the angle unit first and say why. |
| The check | Two consecutive maxima, then subtract: 4 and 6.094 give 2.094. That difference IS the period, measured rather than recalled, and it settles the 2π argument without an appeal to authority. |
The maximum finder needs bounds either side of one peak. With nine and a half cycles on screen students put the bounds around three peaks and get whichever one the machine prefers, which looks like a calculator fault and is a window fault.
| Step | What |
|---|---|
| 1 | Plain sin x and cos x. Period 2π, amplitude 1, and where each starts. |
| 2 | a and d first, because they are outside and behave: amplitude and principal axis, max = d + a. |
| 3 | Now b, with the figure. Start at 1, walk up to 3, count the cycles. |
| 4 | Period = 2π/b, derived from the cycle count rather than given. |
| 5 | Down to b = 0.5, so the division works both ways. |
| 6 | c, as the position of a peak: 3(x − 4) peaks at 4. |
| 7 | The multiplied-out form 3x − 12, and the factorising habit. Collect the 12s. |
| 8 | Sketch 3 sin 2x from scratch: axis, amplitude, period, one peak. |
Do not say "b stretches the graph". It squashes when b > 1, and "stretch by a factor of 1/b" is the accurate phrasing. Saying stretch is how students end up multiplying 2π by 3. If you want one sentence: b tells you how many cycles fit where one used to.
Do not let the multiplied-out form go unmentioned until an exam. If every example in class is given as a(b(x + c)) + d, factorised and tidy, the first 3x − 12 they meet will be in a paper. Put one multiplied-out example in the first lesson, before the habit of reading c straight off has set.