y = 2cos(3(x − 4)) + 1 has period 2.094, not 6.283. The 2π answer is three times too long, and the diagram shows it: three complete cycles sit in the space one was supposed to take.
The amplitude and the principal axis do not move as b changes. Only the horizontal squashing does, which is why b is the one people misread.
Every one of the four does one job, and three of them are easy:
| Letter | What it is | Read it off |
|---|---|---|
| a | Amplitude: the distance from the principal axis to a peak. | (max − min) ÷ 2 |
| d | Principal axis: the line the curve oscillates about. | (max + min) ÷ 2 |
| b | Horizontal squash, which sets the period = 2π/b. | 2π ÷ period |
| c | Horizontal shift, by −c. | where a peak has moved to |
For y = 2cos(3(x − 4)) + 1: a = 2, b = 3, c = −4, d = 1. So the amplitude is 2, the principal axis is y = 1, the maximum is d + a = 3, the minimum is d − a = −1, and the period is 2π/3 = 2.094.
The maximum is d + a and not a. Writing 2 for the maximum of this curve is a common single-mark loss, and the principal axis is what you have forgotten when you do it.
The period is 2π/b, and 2π is only right when b is 1. That is the whole content of the b column and it is the most expensive line on the page.
Here is why, without memorising it. A cosine completes one cycle as its input goes through 2π. The input here is 3(x − 4), so the input runs three times as fast as x does, and x only needs to cover a third of 2π for the input to cover all of it. A bigger b means a shorter period, which is the opposite of what the word "bigger" suggests, and that is why it gets written the wrong way up.
Move the slider. At b = 3 the period is 2.094 and the 2π span beside it covers three complete cycles. At b = 0.5 the period is 12.566 and the 2π span covers only half a cycle. The two agree in exactly one place, b = 1, and every question will give you a b that is not 1.
Those same four numbers are much harder to read when the bracket has been multiplied out, and exam questions give it to you both ways.
2cos(3(x − 4)) + 1 and 2cos(3x − 12) + 1 are the same function. They agree at every x, and you can check it at one: both give 3 at x = 4 and both give −0.98 at x = 5.
But the second form invites you to read the shift as 12, and it is 4. The 12 is 3 × 4, so:
factorise the b out before you read anything horizontal
3x − 12 = 3(x − 4), so the shift is 4 to the right. Reading 12 puts your curve 8 units from where it belongs, which is nearly four whole periods away. Nothing about the shape of your sketch will look wrong; it will simply be in the wrong place.
The vertical numbers do not need this treatment. a and d are outside the function and behave exactly as written, which is the general rule from 2.11: outside means vertical and does what it says, inside means horizontal and needs unpicking.
Where are the peaks? A cosine peaks when its input is zero, so 3(x − 4) = 0 gives the peak at x = 4, which is the shift read straight off. The others follow one period apart: 1.906, 4, 6.094. The minima sit halfway between, at 2.953 and 5.047, where the curve reaches −1.
That is the fastest way to sketch one of these: find one peak from the shift, step by the period, and draw the principal axis first.
y = 3 sin 2x. There is no c and no d, which makes the two remaining numbers easier to see.
Its first maximum is at x = π/4, where it reaches 3: a quarter of the way through the period, as a sine always is.
The machine will draw these perfectly and tell you nothing, unless you set the window from the period you calculated. Doing it that way round makes the graph a check on your arithmetic rather than a replacement for it.
When you may use it. Analysis Paper 1 is non-calculator, and reading a, b, c and d off a given graph is classic Paper 1 work. Paper 2 is where a real-life context appears, such as a tide or a temperature, and there the machine finds the maximum and the crossing times.
The mark people lose. Giving 2π as the period because the function is a cosine. It is the commonest single error in the sub-topic and it survives a sketch, because a sketch with the wrong period still looks like a cosine. The habit: write "period = 2π/b" before you write a number, every time, even when b is 1. The line costs four seconds and it is also frequently a method mark.
Questions 1 to 3 are about y = 2cos(3(x − 4)) + 1.
1. Write down the period, to 3 decimal places.
2. Write down the maximum value.
3. The same curve can be written 2cos(3x − 12) + 1. How far has it been shifted to the right?
4. Write down the period of y = 3 sin 2x, to 3 decimal places.
5. Why does a bigger b give a shorter period?
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 down.
1 markThe shift, with the b factorised out first.
On a question that gives you a graph and asks for the equation, all four are separate marks and they can be earned in any order. Get the principal axis first: it makes the amplitude a subtraction rather than a guess.
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