IB Maths AI HL Further Functions & Graphs Paper 2 Amplitude, period, phase shift ~8 min read

Sinusoidal Functions & Graphs

Both y = a sin(b(x − c)) + d and y = a cos(b(x − c)) + d draw waves with the same four parameters: amplitude, period, principal axis and phase shift. Identify those four and the sketch follows.

📘 What you need to know

Standard form: y = a sin(b(x − c)) + d or y = a cos(b(x − c)) + d.
Amplitude = |a|: the vertical distance from the principal axis to a maximum.
Period = 360°/b (degrees) or 2π/b (radians): the horizontal length of one full cycle.
Principal axis: y = d, the horizontal line midway between max and min.
Max = a + d (when a > 0), min = −a + d; for negative a the two swap.
Phase shift = c: horizontal translation of the standard sine or cosine wave. y-intercept: substitute x = 0 directly.
Set the GDC to degrees or radians to match the question before plotting.

The standard sinusoidal form

Every sine or cosine question on AI HL can be read off the standard form. The four parameters do four distinct jobs: a stretches the wave vertically (amplitude), b compresses it horizontally (period), c slides it horizontally (phase shift), and d lifts the whole curve up to its principal axis. Sine starts on the principal axis and rises; cosine starts at the maximum — otherwise their geometry is identical.

Amplitude, period and principal axis

Read a, b, d straight from the equation. The amplitude is |a|. The period is 360°/b (or 2π/b in radians) — smaller b gives a longer wave, larger b a shorter one. The principal axis is the horizontal line y = d midway between the maximum y = a + d and the minimum y = −a + d (using positive a by convention).

Sine starts on the principal axis and rises; cosine starts at the maximum. Both share amplitude, period and principal axis.

Sinusoidal function at a glance y = a sin(b(x − c)) + d or y = a cos(b(x − c)) + d amplitude |a| · period 360°b or 2πb · principal axis y = d · phase shift c

Phase shift, y-intercept and sketching

The phase shift c tells you how far the standard wave has been pushed sideways: positive c shifts right, negative c shifts left. The y-intercept is whatever pops out when you substitute x = 0 — for a sine that’s −a sin(bc) + d; for a cosine, a cos(bc) + d. In practice the easiest approach on a calculator paper is to type the function into the GDC and let the analyse-graph menu hand back the maxima, minima, intercepts and zeros.

Check the angle mode before pressing any keys: if the equation uses degrees (e.g. 30(t − 3)^°) but the GDC is in radians, everything reads wrong.

🧭 Recipe — sketching a sinusoidal function

Read off the four parameters: a, b, c, d.
Amplitude |a|, period 360°/b (or 2π/b), principal axis y = d.
Max = a + d, min = −a + d (swap if a is negative).
y-intercept: substitute x = 0 into the equation.
Plot the principal axis, mark max and min lines lightly, then draw the wave starting from the appropriate point (axis for sine, max for cosine) shifted by c.

Worked examples

WE 1

Amplitude, period, principal axis

For y = 4 sin(3x^°) + 2, state the amplitude, period and equation of the principal axis.

read a, b, d a = 4, b = 3, d = 2 amplitude |a| amplitude = 4 period 360°/b period = 360°/3 = 120° principal axis y = d amplitude 4 · period 120° · axis y = 2

WE 2

Maximum and minimum values

Find the maximum and minimum values of y = 5 cos(x/2) − 1.

read a, d a = 5, d = −1 max = a + d (cos = 1) max = 5 + (−1) = 4 min = −a + d (cos = −1) min = −5 + (−1) = −6 max 4 · min −6

WE 3

y-intercept of a phase-shifted sine

Find the y-intercept of y = 3 sin(2(x^° − 30^°)) + 1, to 2 d.p.

substitute x = 0 y = 3 sin(2(−30°)) + 1 = 3 sin(−60°) + 1 = 3(−√3/2) + 1 = 1 − 3√3/2 ≈ −1.60 (0, −1.60) use the GDC in degree mode to evaluate directly.

WE 4

Negative amplitude coefficient

For y = −2 cos(4x^°) + 3, state the amplitude, period, principal axis, maximum and minimum.

read a, b, d a = −2, b = 4, d = 3 amplitude = |a| = 2 period = 360°/4 = 90° principal axis y = 3 max when −2 cos = +2 i.e. cos = −1 max = 2 + 3 = 5 min when cos = 1: −2 + 3 = 1 amplitude 2 · period 90° · axis y = 3 · max 5 · min 1 negative a flips which value of cos gives the max.

WE 5

Sketch over a domain

For y = 2 sin(x^°) + 3 over 0° ≤ x ≤ 360°, state the key features and the coordinates of the maxima, minima and axis crossings.

parameters a = 2, b = 1, d = 3; amp 2, period 360°, axis y = 3 y-intercept y(0) = 2(0) + 3 = 3 ⇒ (0°, 3) max at x = 90°: y = 2 + 3 = 5 axis crossing at x = 180°: y = 3 min at x = 270°: y = −2 + 3 = 1 end of cycle at x = 360°: y = 3 (0, 3), (90, 5), (180, 3), (270, 1), (360, 3)

WE 6

Applied: tide depth

The depth of water in a harbour after t hours is D(t) = 4 sin(30(t − 3)^°) + 8 for 0 ≤ t ≤ 24. (a) State the maximum and minimum depths. (b) State the period. (c) Find the first time at which D = 10.

(a) max = a + d, min = −a + d max = 4 + 8 = 12 m min = −4 + 8 = 4 m (b) period = 360°/b = 360°/30 = 12 hours (c) set D = 10 4 sin(30(t − 3)°) = 2 ⇒ sin(30(t − 3)°) = 0.5 30(t − 3) = 30° ⇒ t − 3 = 1 ⇒ t = 4 max 12 m · min 4 m · period 12 h · D = 10 first at t = 4 h

💡 Top tips

Amplitude is positive: take |a|, never just a.
Period and frequency are reciprocals: b ↑ means period ↓.
Principal axis is the constant d; the curve oscillates symmetrically about it.
For applied questions (tides, day-length, temperature), max = peak, min = trough, principal axis = average.
Confirm angle mode (degrees vs radians) before any GDC work.

⚠ Common mistakes

Using a instead of |a| as amplitude when a is negative.
Swapping period and frequency — period is 360°/b, not b/360°.
Reading c with the wrong sign: b(x − c) has the shift as written; b(x + c) shifts the other way.
Calculator in the wrong angle mode: a question in degrees evaluated in radians gives nonsense.
Forgetting to swap max and min when a is negative.

Chapter complete — you now have the full toolkit of polynomial, exponential and trigonometric graphs for AI HL Paper 1 & 2 sketches.

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