DWG NO. 4.6 — Lesson 6 of 6

Amplitude, Period, Phase Shift, and Vertical Shift

Unit 4: Trigonometric Functions and the Unit Circle · ~25 min

Objective Identify the amplitude, period, phase shift, and vertical shift of y = a sin(b(x − h)) + k and use them to graph the transformed wave.

A sinusoidal function in the form y = a·sin(b(x − h)) + k is just y = sin x run through the same transformation toolkit from 1.4 — a stretch, a horizontal compression, and two shifts. The letters just get new names suited to a wave: a controls how tall it is, b controls how fast it repeats, h slides it sideways, and k slides it up or down.

Amplitude = |a| — half the distance from the wave's highest point to its lowest point (vertical stretch).
Period = \(\frac{2\pi}{|b|}\) — the length of one complete cycle (horizontal stretch/compression).
Phase shift = h — the horizontal shift; the wave that would start at x = 0 instead starts at x = h.
Vertical shift = k — moves the entire wave, and its midline, up or down; the new midline is y = k.

y = sin x (dashed) vs. y = 2 sin(x − \(\frac{\pi}{2}\)) + 1 (solid) — amplitude 2, phase shift right \(\frac{\pi}{2}\), midline y = 1

Worked Example 1 · Reading the parameters

ProblemIdentify the amplitude, period, phase shift, and vertical shift of y = 3 sin(2(x + \(\frac{\pi}{4}\))) − 1.

1Match to the form y = a sin(b(x − h)) + k: a = 3, b = 2, and since it's (x + \(\frac{\pi}{4}\)) = (x − (−\(\frac{\pi}{4}\))), h = −\(\frac{\pi}{4}\). k = −1.

2Amplitude = |3| = 3. Period = \(\frac{2\pi}{2}\) = π. Phase shift = −\(\frac{\pi}{4}\) (left). Vertical shift = −1 (down).

Amplitude 3, period π, phase shift \(\frac{\pi}{4}\) left, vertical shift 1 down

Worked Example 2 · Writing an equation from a description

ProblemWrite a cosine equation with amplitude 4, period 4π, no phase shift, and midline y = 2.

1a = 4 directly. For the period, solve \(\frac{2\pi}{b}\) = 4π, so b = 2π ÷ 4π = \(\frac{1}{2.}\)

2No phase shift means h = 0; midline y = 2 means k = 2.

y = 4 cos(\(\frac{1}{2}\) x) + 2

Worked Example 3 · Key points of a transformed wave

ProblemFind the maximum and minimum values, and the x-value of the first maximum on x ≥ 0, for y = 5 sin(x − π) + 3.

1Amplitude 5, midline y = 3, so max = 3 + 5 = 8 and min = 3 − 5 = −2.

2Parent sine hits its first max at x = \(\frac{\pi}{2}\); shifting right by π moves that to x = \(\frac{\pi}{2}\) + π = \(\frac{3\pi}{2}\).

max = 8, min = −2, first maximum at x = \(\frac{3\pi}{2}\)

Guided practice

1.State the amplitude and period of y = −2 sin(3x).
Show answer

Amplitude = |−2| = 2. Period = \(\frac{2\pi}{3,}\) so period = \(\frac{2\pi}{3}\).
2.State the phase shift and vertical shift of y = cos(x − \(\frac{\pi}{3}\)) + 4.
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Phase shift \(\frac{\pi}{3}\) right. Vertical shift 4 up.
3.Write a sine equation with amplitude 1, period 2π, phase shift \(\frac{\pi}{2}\) right, and no vertical shift.
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a=1, b=1 (since \(\frac{2\pi}{b=2\pi}\)), h=\(\frac{\pi}{2}\), k=0: y = sin(x − \(\frac{\pi}{2}\)).
4.Find the maximum and minimum values of y = 3 cos(x) − 2.
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Midline y = −2, amplitude 3: max = −2 + 3 = 1, min = −2 − 3 = −5.
5.Explain why increasing b makes the period shorter, in terms of how fast x needs to change.
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Since period = \(\frac{2\pi}{|b|,}\) a larger b divides 2π into a smaller number — the input b(x−h) reaches a full 2π cycle with a smaller change in x, so the wave completes each repetition faster (b acts as a horizontal compression).

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