Reading All Four Parameters

For $y = 3\sin(2(x - \frac{\pi}{6})) - 1$, find the amplitude, period, phase shift, midline and range.

Match the general form: Compare with $y = A\sin(B(x - C)) + D$ = $A = 3,\ B = 2,\ C = \frac{\pi}{6},\ D = -1$

Amplitude and period: Amplitude $= |3| = 3$, period $= \frac{2\pi}{2} = \pi$ = Amplitude $3$, period $\pi$

Phase shift and midline: $C = \frac{\pi}{6} > 0$, so the shift is to the right; $D = -1$ = Shift $\frac{\pi}{6}$ right, midline $y = -1$

Range: $[D - |A|,\ D + |A|] = [-1 - 3,\ -1 + 3]$ = $[-4, 2]$

Answer: Amplitude $3$, period $\pi$, phase shift $\frac{\pi}{6}$ to the right, midline $y = -1$, range $[-4, 2]$.

Factor Out B Before Reading the Shift

Find the phase shift and the period of $y = \sin(3x - \pi)$.

Factor out $B = 3$: $3x - \pi = 3\left(x - \frac{\pi}{3}\right)$ = $y = \sin(3(x - \frac{\pi}{3}))$

Read the phase shift: $C = \frac{\pi}{3} > 0$ = $\frac{\pi}{3}$ to the right

Find the period: $\frac{2\pi}{B} = \frac{2\pi}{3}$ = Period $\frac{2\pi}{3}$

Check where the cycle starts: The cycle starts where the bracket is $0$: $3x - \pi = 0$ = $x = \frac{\pi}{3}$, it matches

Answer: The phase shift is $\frac{\pi}{3}$ to the right (not $\pi$) and the period is $\frac{2\pi}{3}$.

Graphing with the Five Key Points

Graph one period of $y = 2\sin(2(x - \frac{\pi}{4})) + 1$.

Read the parameters: Compare with $y = A\sin(B(x - C)) + D$ = $A = 2,\ B = 2,\ C = \frac{\pi}{4},\ D = 1$

Period and step: Period $= \frac{2\pi}{2} = \pi$, step $= \frac{\pi}{4}$ = Step $\frac{\pi}{4}$

The five x-values: Start at $x = C = \frac{\pi}{4}$ and add $\frac{\pi}{4}$ four times = $\frac{\pi}{4},\ \frac{\pi}{2},\ \frac{3\pi}{4},\ \pi,\ \frac{5\pi}{4}$

The five y-values (sine pattern): Midline $1$, max $1 + 2 = 3$, midline $1$, min $1 - 2 = -1$, midline $1$ = $1,\ 3,\ 1,\ -1,\ 1$

Plot and connect: Draw a smooth wave through the five points and repeat it to both sides = Range $[-1, 3]$

Answer: One cycle runs from $x = \frac{\pi}{4}$ to $x = \frac{5\pi}{4}$ through $(\frac{\pi}{4}, 1)$, $(\frac{\pi}{2}, 3)$, $(\frac{3\pi}{4}, 1)$, $(\pi, -1)$ and $(\frac{5\pi}{4}, 1)$. The range is $[-1, 3]$.

A Reflected Cosine

Graph one period of $y = -3\cos(\frac{1}{2}(x + \pi)) + 2$.

Read the parameters: $x + \pi = x - (-\pi)$ = $A = -3,\ B = \frac{1}{2},\ C = -\pi,\ D = 2$

Period and step: Period $= \frac{2\pi}{1/2} = 4\pi$, step $= \frac{4\pi}{4} = \pi$ = Step $\pi$

The five x-values: Start at $x = -\pi$ and add $\pi$ four times = $-\pi,\ 0,\ \pi,\ 2\pi,\ 3\pi$

The five y-values: Cosine starts at a max, but $A < 0$ flips it: min, midline, max, midline, min with min $2 - 3$ and max $2 + 3$ = $-1,\ 2,\ 5,\ 2,\ -1$

Domain and range: $[D - |A|,\ D + |A|] = [2 - 3,\ 2 + 3]$ = Domain all real numbers, range $[-1, 5]$

Answer: Key points $(-\pi, -1)$, $(0, 2)$, $(\pi, 5)$, $(2\pi, 2)$, $(3\pi, -1)$. The range is $[-1, 5]$.

The Order of Transformations

Describe, in the right order, how $y = \sin(x)$ becomes $y = 2\sin(3x + \frac{\pi}{2}) - 1$, and find the first maximum to the right of the start of the cycle.

Factor out $B = 3$: $3x + \frac{\pi}{2} = 3\left(x + \frac{\pi}{6}\right)$ = $C = -\frac{\pi}{6}$

Horizontal changes (inside): First compress by the factor $3$ (period $\frac{2\pi}{3}$), then shift = $\frac{\pi}{6}$ to the left

Vertical changes (outside): First stretch by $2$ (amplitude $2$), then shift = $1$ down, midline $y = -1$

First maximum: Start $x = -\frac{\pi}{6}$, step $= \frac{2\pi/3}{4} = \frac{\pi}{6}$, so max at $-\frac{\pi}{6} + \frac{\pi}{6}$ with $y = -1 + 2$ = $(0, 1)$

Check in the original equation: $2\sin(3 \cdot 0 + \frac{\pi}{2}) - 1 = 2 \cdot 1 - 1$ = $1$, correct

Answer: Compress horizontally by $3$, shift $\frac{\pi}{6}$ left, stretch vertically by $2$, shift $1$ down. The cycle starts at $x = -\frac{\pi}{6}$ and the first maximum is $(0, 1)$.

Mistake: Reading the phase shift of $y = \sin(2x - \pi)$ as $\pi$

Why: The shift is the number subtracted from $x$ itself, not from $2x$. $B$ multiplies the whole bracket, so it also squeezes the shift.

Correct: Factor out $B$ first: $\sin(2(x - \frac{\pi}{2}))$, so the shift is $\frac{\pi}{2}$ to the right.

Mistake: Shifting $y = \cos(x + \frac{\pi}{3})$ to the right

Why: The general form has $x - C$. Here $x + \frac{\pi}{3} = x - (-\frac{\pi}{3})$, so $C = -\frac{\pi}{3}$ is negative.

Correct: A plus sign inside the bracket means a shift to the left: $\frac{\pi}{3}$ to the left.

Mistake: Starting the key points at $x = 0$ or stepping by a whole period

Why: After a phase shift the cycle no longer starts at $0$, and five points must split ONE period into four equal parts.

Correct: Start at $x = C$ and step by $\frac{\text{period}}{4}$. For period $\pi$ the step is $\frac{\pi}{4}$.

Mistake: Keeping the usual max/min order when $A < 0$

Why: A negative $A$ reflects the wave across its midline, so every maximum becomes a minimum and the other way round.

Correct: For $y = -2\sin(x)$ the pattern is midline, min, midline, max, midline. The amplitude is still $2$.

Ferris Wheel Height

The height of a seat on a turning Ferris wheel is a transformed cosine: the center of the wheel gives the midline and the radius gives the amplitude.

A wheel with radius 20 m whose center is 22 m above the ground turns once every 30 minutes. A seat that starts at the bottom has height $h(t) = -20\cos(\frac{\pi}{15}t) + 22$ meters after $t$ minutes, and the period is $\frac{2\pi}{\pi/15} = 30$.

Tides in a Harbor

Harbor masters predict the water depth with a sine model whose period is about 12 hours, so ships know when they can enter.

The depth $d(t) = 1.5\sin(\frac{\pi}{6}(t - 3)) + 4$ meters, $t$ hours after midnight, has midline 4 m, amplitude 1.5 m and period $\frac{2\pi}{\pi/6} = 12$ hours. The depth swings between 2.5 m and 5.5 m.

Hours of Daylight

The number of daylight hours over the year follows a sine wave with a period of 365 days. The midline is 12 hours, the day length at the equinoxes.

In a city in northern Europe, $L(t) = 4\sin(\frac{2\pi}{365}(t - 80)) + 12$ hours, where $t$ is the day of the year. Day 80 (around March 21) is the spring equinox, and the day length swings between 8 and 16 hours.

Q: Why is the phase shift of $\sin(2x - \pi)$ not $\pi$?

A: Because $B = 2$ multiplies the whole bracket. Factor it out: $\sin(2x - \pi) = \sin(2(x - \frac{\pi}{2}))$. The cycle starts where the bracket is $0$, at $x = \frac{\pi}{2}$, so the shift is $\frac{\pi}{2}$.

Q: Does the order of transformations really matter?

A: Yes, for a shift and a stretch in the same direction. Shifting $\sin(x)$ right by $\frac{\pi}{2}$ and THEN compressing by $2$ gives $\sin(2x - \frac{\pi}{2}) = \sin(2(x - \frac{\pi}{4}))$, a shift of only $\frac{\pi}{4}$. Horizontal and vertical changes do not affect each other, so you may do the two groups in either order.

Q: Do the key points work the same way for cosine?

A: Yes. Start at $x = C$ and step by $\frac{\text{period}}{4}$ as for sine. Only the pattern of y-values changes: cosine goes max, midline, min, midline, max (or the reverse if $A < 0$).

Q: What if $B$ is negative?

A: Rewrite it first. Sine is odd, $\sin(-\theta) = -\sin(\theta)$, so $\sin(-2x) = -\sin(2x)$. Cosine is even, $\cos(-\theta) = \cos(\theta)$, so $\cos(-2x) = \cos(2x)$. Then $B > 0$ and the method works as usual.