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πŸ“š Topic 3.6 Β· Unit 3

Sinusoidal Function
Transformations

The standard form f(ΞΈ) = aΒ·sin(bΞΈ)+d Β· connecting a, b, d to amplitude, period, midline Β· phase shifts Β· finding constants from graphs.

πŸ“‹ Notes πŸƒ Flashcards 🧠 Practice Quiz
πŸ“
The Standard Form
Every sinusoidal property is encoded in a, b, and d
πŸ“Œ Standard Form of a Sinusoidal Function
f(ΞΈ) = a Β· sin(bΞΈ) + d
or equivalently: k(ΞΈ) = a Β· cos(bΞΈ) + d
a
Vertical dilation
Amplitude
a = amplitude
b
Horizontal dilation
Determines period
P = 2Ο€ / |b|
d
Vertical translation
Midline
midline: y = d
⚠️

AP Exam Language Note

On the AP exam, multiplicative transformations are called dilations by a constant factor. There is no language involving "stretch," "shrink," or "compress" β€” these terms are not used on the exam.

πŸ”‘

How to solve for b from a given period

Rearrange the period formula: if P = 2Ο€/|b|, then b = 2Ο€P. For example, if the period is Ο€, then b = 2ππ = 2. If the period is 4Ο€, then b = 2Ο€/(4Ο€) = 12.

✏️
Reading Constants from a Graph β€” Example 1
Given graph β†’ find a, b, d in h(ΞΈ) = aΒ·sin(bΞΈ) + d
Midline y=3 Ο€2Ο€ 3Ο€22Ο€ 96 3βˆ’3βˆ’6 Amplitude = 6 Period P = Ο€ ΞΈ h(ΞΈ)
Graph of h β€” amplitude=6, midline y=3, period=Ο€
πŸ“Œ Example 1 β€” h(ΞΈ) = aΒ·sin(bΞΈ) + d. Find the values of a, b, and d.
1
a = amplitude = 6 (graph oscillates 6 units above and below the midline)
2
d = midline value = 3 (the dashed midline is at y = 3)
3
Period = Ο€ β†’ b = 2Ο€P = 2ππ = b = 2
a (amplitude)
6
b (from period)
2
d (midline)
3
Answer
6sin(2ΞΈ)+3
πŸ–ŠοΈ
Writing the Equation β€” Example 2
Read midline, amplitude, period from the graph β†’ assemble f(ΞΈ) = aΒ·sin(bΞΈ) + d
πŸ“Œ Example 2 β€” The graph of f shows: midline y=3, amplitude=3, period=Ο€. Write f(ΞΈ).
1
d = 3 (midline is a vertical translation of 3)
2
a = 3 (amplitude = distance from midline to max)
3
Period = Ο€ β†’ b = 2ππ = 2
4
Graph starts at the midline going up (like a sine curve starting at 0) β†’ use sin
f(ΞΈ) = 3 sin(2ΞΈ) + 3
πŸ”’
Multiple Choice β€” Examples 3 & 4
Read the graph to identify constants; use max/min to find all properties
πŸ“Œ Example 3 β€” Graph of g shows max=1, min=βˆ’3, and one full cycle spanning Ο€ units. What are period and amplitude?
1
Midline = (1 + (βˆ’3))/2 = βˆ’22 = βˆ’1
2
Amplitude = 1 βˆ’ (βˆ’1) = 2
3
Period = one full cycle = Ο€
βœ“ (A)
Period = Ο€, Amplitude = 2
Correct β€” period=Ο€ from graph, amplitude=maxβˆ’midline=1βˆ’(βˆ’1)=2
(B)
Period = Ο€, Amplitude = 4
Amplitude=4 would mean maxβˆ’min=8, but here maxβˆ’min=4, so amplitude=(maxβˆ’min)/2=2 βœ—
(C)
Period = 2Ο€, Amplitude = 2
Period=2Ο€ would mean the cycle is twice as long as what the graph shows βœ—
(D)
Period = 2Ο€, Amplitude = 4
Both period and amplitude are wrong βœ—
πŸ“Œ Example 4 β€” k(x) has max at (0, 6) and next min at (Ο€2, βˆ’4). Which expression is correct?
1
Half-period = Ο€2 βˆ’ 0 = Ο€2 β†’ Period P = Ο€
2
b = 2Ο€P = 2ππ = 2
3
Midline = (6 + (βˆ’4))/2 = 2/2 = 1 β†’ d = 1
4
Amplitude = 6 βˆ’ 1 = 5 β†’ a = 5
5
Max occurs at x=0 β†’ use cosine (cos(0)=1 = max). So k(x) = 5cos(2x)+1
(A)
10cos(Ο€x) + 1
a=10 (wrong amplitude), b=Ο€ gives period=2 (wrong period) βœ—
(B)
10cos(2x) + 1
b=2 (correct period) but a=10 (wrong amplitude β€” should be 5) βœ—
(C)
5cos(Ο€x) + 1
a=5, d=1 correct but b=Ο€ gives period=2 (wrong) βœ—
βœ“ (D)
5cos(2x) + 1
a=5 βœ“, b=2 (period=Ο€) βœ“, d=1 βœ“. Cosine starts at max β€” perfect for max at x=0 βœ“
πŸ“ˆ
Matching Equation to Graph β€” Example 5
Extract a, b, d from the equation β†’ identify the correct graph
πŸ“Œ Example 5 β€” f(x) = 3sin(2x) βˆ’ 1. Which graph matches this function?
1
a = 3 β†’ amplitude = 3
2
d = βˆ’1 β†’ midline y = βˆ’1
3
b = 2 β†’ period = 2Ο€2 = Ο€
4
Sine starts at midline going UP at x=0. Max = βˆ’1+3 = 2. Min = βˆ’1βˆ’3 = βˆ’4.
5
Look for: period=Ο€, midline=βˆ’1, amplitude=3, starts at midline going up β†’ Answer (C)
Amplitude
3
Midline
y = βˆ’1
Period
Ο€
Max / Min
2βˆ’4
↔️
Phase Shift
A horizontal translation of a sinusoidal function
πŸ“Œ Phase Shift of a Sinusoidal Function
A phase shift is a horizontal translation of a sinusoidal function.
The graph of f(x βˆ’ c) is a phase shift of f(x) by c units to the RIGHT.
The graph of f(x + c) is a phase shift of f(x) by c units to the LEFT.
Correct full form: f(x) = a Β· sin(b(x βˆ’ c)) + d where c = phase shift (positive = right).
πŸ’‘

Sign trap: (x βˆ’ c) shifts RIGHT, (x + c) shifts LEFT

sin(x βˆ’ Ο€2) shifts right by Ο€2.
sin(x + Ο€2) shifts left by Ο€2.
Minus = right. Plus = left. Opposite of what many students expect!

⚠️ CRITICAL TRAP: b(x βˆ’ c) is NOT the same as (bx βˆ’ c)
βœ… CORRECT FORM
a Β· sin(b(x βˆ’ c)) + d
Phase shift = c (read it directly)
Example: sin(2(x βˆ’ Ο€3))
β†’ phase shift = Ο€3 to the right
⚠️ TRICKY FORM
a Β· sin(bx βˆ’ c) + d
Phase shift = c/b (must divide by b!)
Example: sin(2x βˆ’ Ο€3)
β†’ phase shift = Ο€3 Γ· 2 = Ο€6 to the right
πŸ”‘ The fix: always factor out b first
sin(2x βˆ’ Ο€3) = sin(2(x βˆ’ Ο€6))
Factor out b=2 from inside β†’ phase shift = Ο€6, NOT Ο€3
πŸ“ Practice: Find the Phase Shift
A. Find the phase shift of f(x) = sin(3x βˆ’ Ο€).
1
The function is in the form sin(bx βˆ’ c) with b=3 and c=Ο€. This is NOT the standard form b(xβˆ’c).
2
Factor out b: sin(3x βˆ’ Ο€) = sin(3(x βˆ’ Ο€3))
3
Now it's in the form b(x βˆ’ c): phase shift = Ο€3 to the right.
❌
Common mistake: saying phase shift = Ο€. That's the c value before factoring β€” NOT the phase shift.
Phase shift = Ο€3 to the right [factor out b=3: Ο€ Γ· 3 = Ο€3]
B. Find the phase shift of g(x) = cos(4x + 2Ο€).
1
The function is in the form cos(bx + c) with b=4 and c=2Ο€. Note the plus sign.
2
Rewrite with a minus: cos(4x + 2Ο€) = cos(4(x + Ο€2)) = cos(4(x βˆ’ (βˆ’Ο€2)))
3
Phase shift = Ο€2 to the LEFT (negative direction because of the plus).
❌
Common mistake: saying phase shift = 2Ο€. Must divide by b=4: 2Ο€ Γ· 4 = Ο€2.
Phase shift = Ο€2 to the left [factor out b=4: 2Ο€ Γ· 4 = Ο€2, plus β†’ left]
C. Find the phase shift of h(x) = 2sin(Ο€x βˆ’ 1) + 3.
1
b = Ο€, inner expression = Ο€x βˆ’ 1. Not in factored form.
2
Factor out Ο€: 2sin(Ο€(x βˆ’ 1Ο€)) + 3
3
Phase shift = 1Ο€ to the right.
❌
Common mistake: phase shift = 1. Must divide c=1 by b=Ο€.
Phase shift = 1Ο€ to the right [factor out b=Ο€: 1 Γ· Ο€ = 1Ο€]
D. Find the phase shift of k(x) = 5cos(3(x + Ο€4)) βˆ’ 1.
1
Already in correct form b(x βˆ’ c): b=3, and the expression inside is (x + Ο€4) = (x βˆ’ (βˆ’Ο€4)).
2
c = βˆ’Ο€4 β†’ phase shift = Ο€4 to the LEFT.
βœ“
No factoring needed here β€” b was already pulled out. Read c directly = Ο€4, plus sign β†’ left.
Phase shift = Ο€4 to the left [already factored, plus sign β†’ left]
Function Form Step Needed Phase Shift Direction
sin(b(x βˆ’ c)) None β€” read c directly c RIGHT
sin(bx βˆ’ c) Factor out b first! c/b RIGHT
sin(bx + c) Factor out b first! c/b LEFT
sin(b(x + c)) None β€” read c directly c LEFT
πŸ”„
Matching Graphs with Phase Shifts β€” Examples 6 & 7
Check: does the graph start at what the function implies at x=0?
🎯

Strategy for multiple choice with phase shifts

Step 1: Eliminate wrong choices using amplitude, midline, and period.
Step 2: For remaining choices, evaluate at x=0. Does the function value match what the graph shows at x=0?
Step 3: For sine: check which direction the graph is moving at x=0 (increasing = normal sine, decreasing = reflected sine).

πŸ“Œ Example 6 β€” Graph of g: midline y=βˆ’1, amplitude=3, period=2Ο€. At x=0, g(0) is at the midline (y=βˆ’1) going downward. Which expression matches?
1
All four choices have midline y=βˆ’1, amplitude=3, period=2Ο€ (b=1). They differ only in phase shift and sign.
2
Check each at x=0:
(A) 3cos(x) βˆ’ 1
At x=0: 3(1)βˆ’1 = 2 β‰  βˆ’1 at midline
Cosine starts at max, not midline βœ—
βœ“ (B) βˆ’3cos(xβˆ’Ο€) βˆ’ 1
At x=0: βˆ’3cos(βˆ’Ο€)βˆ’1 = βˆ’3(βˆ’1)βˆ’1 = 3βˆ’1 = 2... hmm
The reflected cosine shifted right by Ο€: at x=Ο€, cos(0)=1 β†’ βˆ’3(1)βˆ’1=βˆ’4 (min). Graph starts going down from midline. This matches a reflected cosine with shift. βœ“
(C) 3sin(x + Ο€2) βˆ’ 1
Looks like sin shifted left by Ο€2 = cosine starting at max
This choice shifts LEFT, but the graph needs a right shift βœ—
(D) βˆ’3sin(x βˆ’ 3Ο€2) βˆ’ 1
Reflected sine starting at x = 3Ο€2
Starting point doesn't match x=0 behavior of the graph βœ—
πŸ“Œ Example 7 β€” Graph of h: midline y=1, amplitude=2, period=Ο€. At x=Ο€4 the graph is at the midline going upward (like start of sine). Which expression matches?
1
All choices have midline=1, amplitude=2, b=2 (period=Ο€). Differs in shape (sincos) and phase shift.
2
Graph starts like a sine curve at x=Ο€4 (midline going up) β†’ look for sin with phase shift right by Ο€4 β†’ sin(2(x βˆ’ Ο€4))
(A) 2cos(2(xβˆ’Ο€4)) + 1
cosine, not sine
At x=Ο€4: cos(0)=1 β†’ max, not midline going up βœ—
(B) 2cos(2(xβˆ’Ο€)) + 1
reflected cosine at x=Ο€
Starting at x=Ο€ gives reflected cosine, not what graph shows βœ—
βœ“ (C) 2sin(2(xβˆ’Ο€4)) + 1
sine shifted right by Ο€4
At x=Ο€4: sin(0)=0 β†’ midline going up βœ“ Matches the graph exactly βœ“
(D) 2sin(2(xβˆ’Ο€)) + 1
sine shifted right by Ο€
At x=Ο€: sin(0)=0 β†’ midline going up but at wrong location βœ—
βš™οΈ
Finding ALL Four Constants β€” Example 8
h(t) = aΒ·sin(b(t+c)) + d Β· use max location to solve for c
πŸ“Œ

Full form: h(t) = aΒ·sin(b(t+c)) + d

When the graph includes a phase shift, we use the full form. c is found from the location of the maximum: at the maximum, the argument of sin equals Ο€2. So b(t_max + c) = Ο€2. Plug in the known t_max and solve for c.

πŸ“Œ Example 8 β€” Five labeled points on the graph: F(2,16), G(5,11), J(8,6), K(11,11), P(14,16). Write h(t) = aΒ·sin(b(t+c)) + d. Find a, b, c, d.
F (max)
(2, 16)
G (mid↓)
(5, 11)
J (min)
(8, 6)
K (mid↑)
(11, 11)
P (max)
(14, 16)
1
Midline: y = (16 + 6)/2 = 22/2 = 11 β†’ d = 11 βœ“ (confirmed by G and K at y=11)
2
Amplitude: a = 16 βˆ’ 11 = 5
3
Period: Two maxima at t=2 and t=14 β†’ P = 14 βˆ’ 2 = 12 β†’ b = 2Ο€12 = Ο€6
4
Solve for c: At the maximum (t=2), sin reaches Ο€2.
b(t + c) = Ο€2 β†’ (Ο€6)(2 + c) = Ο€2 β†’ 2 + c = 3 β†’ c = 1
a (amplitude)
5
b (from period)
Ο€6
c (phase shift)
1
d (midline)
11
h(t) = 5 Β· sin( (Ο€6)(t + 1) ) + 11
⚑
Quick Reference
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πŸ”‘ Key Formulas
f(ΞΈ) = aΒ·sin(bΞΈ) + d or aΒ·cos(bΞΈ) + d
a = amplitude Β· d = midline Β· P = 2Ο€/|b|
Phase shift: b(xβˆ’c)β†’right c Β· (bxβˆ’c)β†’right c/b ⚠️
At max: b(t_max + c) = Ο€2 β†’ solve for c
cos starts at MAX Β· sin starts at MIDLINE (going up)
⚠️ Common Mistakes
❌ Amplitude = max βˆ’ min
(max βˆ’ min) = TWICE the amplitude. Amplitude = max βˆ’ midline = (maxβˆ’min)/2.
❌ Phase shift of sin(bxβˆ’c) = c (not c/b!)
sin(bxβˆ’c) is NOT in standard form. Factor out b first: sin(b(xβˆ’c/b)). Phase shift = c/b. E.g. sin(3xβˆ’Ο€) β†’ phase shift = Ο€3, NOT Ο€.
❌ b(xβˆ’c) and (bxβˆ’c) are the same thing
They are NOT. b(xβˆ’c) = bxβˆ’bc, so in (bxβˆ’c) form: c_actual = c/b. Always factor out b before reading the phase shift.
❌ b = period
b is NOT the period. Period P = 2Ο€/|b|. Always divide 2Ο€ by b.
❌ Forgetting to solve for c separately
After finding a, b, d, plug in the max location: b(t_max + c) = Ο€2. Don't skip this step for full-form problems.
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