United StatesPrecalculusSyllabus dot point

How does each transformation of a sinusoid change its graph, and how do you combine them?

Topic 3.6 Sinusoidal Function Transformations: describe how changing each parameter transforms a sinusoid, and combine vertical and horizontal stretches, reflections and shifts.

A focused answer to AP Precalculus Topic 3.6, covering how each of the four sinusoidal parameters transforms the graph, how vertical and horizontal changes combine, and how to read a transformed sinusoid back into its equation.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Quick answer

Each parameter of $y = a\sin(b(x - c)) + d$ performs a specific transformation of the parent sine wave. Multiplying by $a$ stretches the graph vertically by $|a|$ (changing the amplitude) and, if $a$ is negative, reflects it over the midline. The factor $b$ stretches or compresses horizontally, setting the period to $\frac{2\pi}{|b|}$ ; a larger $b$ squeezes the wave. Subtracting $c$ inside the argument shifts the graph right by $c$ (the phase shift). Adding $d$ shifts it up by $d$ , moving the midline to $y = d$ . Vertical transformations act on the output and horizontal transformations act on the input, so they can be applied independently. Combining them, you scale and reflect first, then shift, to land on the final graph.

Jump to a section

What this topic is asking
The four transformations
Why the horizontal shift is c, not bc
Reading a graph back to an equation
Try this

What this topic is asking

The College Board (Topic 3.6) wants you to describe how each transformation changes a sinusoid and to combine them. The four parameters of $y = a\sin(b(x - c)) + d$ map to a vertical stretch or reflection ( $a$ ), a horizontal stretch or compression that sets the period ( $b$ ), a horizontal shift ( $c$ ), and a vertical shift ( $d$ ). You must apply these singly and together, and read a transformed graph back into an equation.

The four transformations

Vertical changes ( $a$ , $d$ ) act on the outputs; horizontal changes ( $b$ , $c$ ) act on the inputs. Because they touch different variables, they do not interfere, which is why you can read each one off the graph separately.

Why the horizontal shift is c, not bc

A subtlety: the form is written $b(x - c)$ , with the period factor multiplying $(x - c)$ as a unit. Written this way, $c$ is exactly the rightward shift. If instead you expand to $\sin(bx - bc)$ , the constant inside is $bc$ , not $c$ , which is why factoring the $b$ out first is the safe habit. Always identify $c$ from the factored form to avoid scaling the shift by mistake.

Combining transformations

Describe the graph of $y = -2\sin\left(3\left(x - \frac{\pi}{6}\right)\right) + 1$ relative to $y = \sin x$ , and state its amplitude, period, phase shift and midline.

step 1 Read the vertical stretch and reflection

$a = -2$ : the amplitude is $|-2| = 2$ , and the negative sign reflects the wave over its midline.

step 2 Read the period

$b = 3$ , so the period is $\frac{2\pi}{3}$ : the wave is compressed to one third of the parent period.

step 3 Read the phase shift

The factored argument $3\left(x - \frac{\pi}{6}\right)$ shows $c = \frac{\pi}{6}$ : the graph shifts right by $\frac{\pi}{6}$ .

step 4 Read the vertical shift and combine

$d = 1$ , so the midline is $y = 1$ . Altogether: amplitude $2$ , period $\frac{2\pi}{3}$ , shifted right $\frac{\pi}{6}$ and up $1$ , and flipped vertically. The maximum is $1 + 2 = 3$ and the minimum is $1 - 2 = -1$ .

Reading a graph back to an equation

To recover an equation from a transformed graph, read the amplitude and midline from the extremes (giving $a$ and $d$ ), the period from peak-to-peak distance (giving $b = \frac{2\pi}{\text{period}}$ ), and the phase from a known landmark (a maximum for cosine, an upward midline crossing for sine). This is the reverse of the worked routine and is exactly what the modelling questions in Topic 3.7 ask.

A point worth stating once is that order matters only between scaling and shifting, not within them. Apply the vertical stretch and reflection before the vertical shift, and set the period before applying the phase shift; otherwise the shifts get scaled. Following scale-then-shift on each axis keeps every transformation landing where the equation says it should.

Try this

Q1. How does the graph of $y = \cos x$ change to become $y = \cos x - 4$ ? [1 point]

Cue. It shifts down $4$ units; the midline moves to $y = -4$ , with amplitude and period unchanged.

Q2. What transformation does the negative sign in $y = -\sin x$ produce? [1 point]

Cue. A reflection over the midline (the $x$ -axis here), flipping peaks to troughs.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)1 marksSection I, Part A (multiple choice, no calculator). The graph of

y = \sin x

is stretched vertically by a factor of

2

, then shifted up

3

. Which equation results? (A)

y = 2\sin x + 3

(B)

y = \sin(2x) + 3

(C)

y = 2\sin(x + 3)

(D)

y = \sin x + 6

Show worked answer →

The correct answer is (A), $y = 2\sin x + 3$ .

A vertical stretch by $2$ multiplies the output by $2$ , giving $2\sin x$ . Shifting up $3$ adds $3$ to the output, giving $2\sin x + 3$ . Choice (B) changes the period, not the amplitude, and (C) shifts horizontally, not vertically.

AP 2024 (style)4 marksSection II (free response, no calculator). Starting from

y = \cos x

, a graph is reflected over its midline, has its period halved, and is shifted right by

\frac{\pi}{4}

. (a) State the value of

a

and of

b

for the new function. (b) Write the transformed equation in the form

y = a\cos(b(x - c)) + d

with

d = 0

Show worked answer →

A 4-point question on combining sinusoidal transformations.

(a) Parameters (2 points): reflecting over the midline makes $a$ negative, so $a = -1$ . Halving the period from $2\pi$ to $\pi$ requires $b = \frac{2\pi}{\pi} = 2$ .
(b) Equation (2 points): a right shift of $\frac{\pi}{4}$ gives $c = \frac{\pi}{4}$ , so $y = -\cos\left(2\left(x - \frac{\pi}{4}\right)\right)$ . The horizontal shift is applied inside the argument as $(x - c)$ , after the period change is set by $b$ .

Related dot points

Sources & how we know this

AP Precalculus Course and Exam Description — College Board (2023)