Graphs of Trigonometric Functions Practice Questions — ACT

Question 1Question →

Match each of the trigonometric functions listed on the left with the correct description of its amplitude and period listed on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The function

y = 3\sin(2x)

matches the description stating 'Amplitude is 3 and period is

\pi

'. The function

y = 2\cos(3x)

matches the description stating 'Amplitude is 2 and period is

\frac{2\pi}{3}

'. The function

y = 4\sin(\pi x)

matches the description stating 'Amplitude is 4 and period is 2'.

Each trigonometric function of the form

y = a\sin(bx)

or

y = a\cos(bx)

has an amplitude equal to the absolute value of the coefficient of the trigonometric term (

|a|

) and a period equal to

2\pi

divided by the absolute value of the coefficient of the angle variable (

|b|

). Applying these formulas gives the correct properties for each function.

Step-by-Step Solution

1

Identify the general form of the trigonometric functions

The functions are in the form

y = a\sin(bx)

or

y = a\cos(bx)

, where the amplitude is given by the absolute value of the vertical stretch coefficient (

|a|

), and the period is calculated as

\frac{2\pi}{|b|}

.

This establishes the formulas needed to determine the amplitude and period for each equation.

2

Calculate the properties for

y = 3\sin(2x)

The vertical stretch coefficient is 3, so the amplitude is 3. The frequency coefficient is 2, so the period is

\frac{2\pi}{2} = \pi

.

To find the amplitude and period for the first function.

3

Calculate the properties for

y = 2\cos(3x)

The vertical stretch coefficient is 2, so the amplitude is 2. The frequency coefficient is 3, so the period is

\frac{2\pi}{3}

.

To find the amplitude and period for the second function.

4

Calculate the properties for

y = 4\sin(\pi x)

The vertical stretch coefficient is 4, so the amplitude is 4. The frequency coefficient is

\pi

, so the period is

\frac{2\pi}{\pi} = 2

.

To find the amplitude and period for the third function.

Key Concept

Identifying the amplitude and calculating the period of trigonometric functions from their equations

Estimated Time:1m 30s

Question 2Question →

What is the period of the function $f(x) = 4\cos\left(\frac{1}{3}x\right)$ ?

\frac{2\pi}{3}

2\pi

4

6\pi

\frac{4\pi}{3}

Show answer & explanation

Answer: $6\pi$

Answer

The period of the function is

6\pi

.

The standard period of the cosine function

y = \cos(x)

is

2\pi

. For a transformed trigonometric function of the form

y = A\cos(Bx)

, the period is given by the formula

\frac{2\pi}{|B|}

. In the function

f(x) = 4\cos\left(\frac{1}{3}x\right)

, the coefficient of

x

is

B = \frac{1}{3}

. Dividing the standard period

2\pi

by

\frac{1}{3}

yields a period of

2\pi \times 3 = 6\pi

.

Step-by-Step Solution

1

Identify the standard period of the parent cosine function and the coefficient

B

of the variable

x

in the given function.

The parent function is

y = \cos(x)

, which has a standard period of

2\pi

. In the function

f(x) = 4\cos\left(\frac{1}{3}x\right)

, the coefficient of

x

is

B = \frac{1}{3}

.

This sets up the parameters needed for the period formula

\text{Period} = \frac{2\pi}{|B|}

.

2

Substitute the value of

B

into the period formula and simplify.

The period is

\frac{2\pi}{\frac{1}{3}} = 2\pi \times 3 = 6\pi

.

Dividing by a fraction is equivalent to multiplying by its reciprocal, which gives the final horizontal distance for one complete cycle.

Key Concept

Graphs of Trigonometric Functions

Question 3Question →

What is the period of the function $f(x) = \tan(3x)$ ?

\frac{2\pi}{3}

3\pi

\frac{\pi}{3}

\pi

3

Show answer & explanation

Answer: $\frac{\pi}{3}$

Answer

The period of the function is

\frac{\pi}{3}

.

The parent function

y = \tan(x)

has a standard period of

\pi

. To find the period of a transformed tangent function of the form

y = \tan(Bx)

, the standard period must be divided by the coefficient of

x

, yielding

\frac{\pi}{|B|}

. Substituting

B = 3

gives

\frac{\pi}{3}

.

Step-by-Step Solution

1

Determine the standard period of the parent function.

The parent function is the tangent function,

y = \tan(x)

, which has a standard period of

\pi

.

The tangent function completes one full cycle of its graph between

-\frac{\pi}{2}

and

\frac{\pi}{2}

.

2

Identify the horizontal compression/stretch coefficient from the given equation.

In

f(x) = \tan(3x)

, the coefficient of

x

is

B = 3

.

This coefficient determines how many cycles occur in a standard interval.

3

Calculate the period using the formula for the tangent function.

Period =

\frac{\pi}{|B|} = \frac{\pi}{3}

.

Dividing the standard period of

\pi

by the absolute value of the coefficient

B

gives the compressed period of the transformed function.

Key Concept

The period of a transformed tangent function

y = \tan(Bx)

is given by

\frac{\pi}{|B|}

.

Question 4Question →

The graph of the function $f(x) = a \cos(b(x - c)) + d$ is shown below for constants $a > 0$ , $b > 0$ , $c \in [0, \pi]$ , and $d$ . The graph has a local maximum at $(\frac{\pi}{3}, 5)$ and the nearest local minimum to its right is at $(\frac{5\pi}{6}, -1)$ . What is the y-intercept of the graph of $f(x)$ ?

\frac{1}{2}

\frac{7}{2}

-\frac{5}{2}

2

-\frac{7}{2}

Show answer & explanation

Answer: $\frac{1}{2}$

Answer

The y-intercept of the graph is

\frac{1}{2}

The correct answer is

\frac{1}{2}

. The amplitude of the function is

a = \frac{5 - (-1)}{2} = 3

, and the midline is

d = \frac{5 + (-1)}{2} = 2

. The distance between the consecutive local maximum and local minimum represents half of a period:

\frac{T}{2} = \frac{5\pi}{6} - \frac{\pi}{3} = \frac{\pi}{2}

, meaning the period is

T = \pi

, so

b = \frac{2\pi}{\pi} = 2

. A local maximum occurs when the argument of the cosine function is

0

, so

2(\frac{\pi}{3} - c) = 0 \implies c = \frac{\pi}{3}

. This gives the equation

f(x) = 3 \cos(2(x - \frac{\pi}{3})) + 2

. Finding the y-intercept requires evaluating the function at

x = 0

:

f(0) = 3 \cos(2(0 - \frac{\pi}{3})) + 2 = 3 \cos(-\frac{2\pi}{3}) + 2 = 3(-\frac{1}{2}) + 2 = \frac{1}{2}

.

Step-by-Step Solution

1

Find the amplitude

a

and midline

d

of the trigonometric function.

a = 3

and

d = 2

The amplitude is half the distance between the maximum and minimum values:

a = \frac{5 - (-1)}{2} = 3

. The midline is the average of these values:

d = \frac{5 + (-1)}{2} = 2

.

2

Determine the period

T

and the frequency coefficient

b

.

T = \pi

and

b = 2

The horizontal distance between a consecutive maximum and minimum is half of the period:

\frac{5\pi}{6} - \frac{\pi}{3} = \frac{\pi}{2}

. Thus, the full period is

T = \pi

. Since

T = \frac{2\pi}{b}

, we find

b = 2

.

3

Determine the horizontal phase shift

c

.

c = \frac{\pi}{3}

A cosine function achieves its maximum when its argument is a multiple of

2\pi

. Since the maximum is at

x = \frac{\pi}{3}

, we set

2(\frac{\pi}{3} - c) = 0

, giving

c = \frac{\pi}{3}

.

4

Evaluate the function at

x = 0

to find the y-intercept.

f(0) = \frac{1}{2}

Substitute the parameters into the function:

f(x) = 3 \cos(2(x - \frac{\pi}{3})) + 2

. Substituting

x = 0

gives

f(0) = 3 \cos(-\frac{2\pi}{3}) + 2 = 3(-\frac{1}{2}) + 2 = \frac{1}{2}

.

Key Concept

Determining the equation of a transformed trigonometric function from key features (maximum and minimum points) and evaluating it.

Estimated Time:3m 0s

Question 5Question →

For the trigonometric equations on the left, which description on the right correctly matches the graphical features of each equation?

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct matches are: the equation

y = -3\sin\left(2x - \frac{\pi}{3}\right) + 1

matches the description with a period of

\pi

, a range of

[-2, 4]

, and a positive

y

-intercept; the equation

y = 2\cos\left(\frac{1}{2}x + \frac{\pi}{4}\right) - 1

matches the description with a period of

4\pi

, a range of

[-3, 1]

, and a phase shift of

\frac{\pi}{2}

units to the left; the equation

y = -\tan\left(3x - \frac{\pi}{2}\right) + 2

matches the description with a period of

\frac{\pi}{3}

that is undefined at the

y

-axis; and the equation

y = 3\sin\left(\frac{2}{3}x + \pi\right) - 2

matches the description with a period of

3\pi

, a range of

[-5, 1]

, and a

y

-intercept of

(0, -2)

.

Each equation is correctly paired with the unique graphical properties determined by calculating its amplitude, period, range, phase shift, and

y

-intercept. The first equation features a period of

\pi

, a range of

[-2, 4]

, and a positive

y

-intercept. The second equation has a period of

4\pi

, a range of

[-3, 1]

, and a phase shift of

\frac{\pi}{2}

units left. The third equation has a period of

\frac{\pi}{3}

and is undefined at the

y

-axis since

x=0

creates a vertical asymptote. The fourth equation has a period of

3\pi

, a range of

[-5, 1]

, and a

y

-intercept of

(0, -2)

.

Step-by-Step Solution

1

Determine the period, range, and y-intercept for the function

y = -3\sin\left(2x - \frac{\pi}{3}\right) + 1

.

The period is

\pi

, range is

[-2, 4]

, and

y

-intercept is

1 + \frac{3\sqrt{3}}{2} > 0

.

The period is

\frac{2\pi}{2} = \pi

, the range is

[1-3, 1+3] = [-2, 4]

, and evaluating at

x=0

yields

y = -3\sin(-\pi/3) + 1 > 0

.

2

Determine the period, range, and phase shift for the function

y = 2\cos\left(\frac{1}{2}x + \frac{\pi}{4}\right) - 1

.

The period is

4\pi

, range is

[-3, 1]

, and phase shift is

\frac{\pi}{2}

units left.

The period is

\frac{2\pi}{1/2} = 4\pi

, the range is

[-1-2, -1+2] = [-3, 1]

, and factoring the argument yields

\frac{1}{2}(x + \frac{\pi}{2})

, giving a shift of

\frac{\pi}{2}

units left.

3

Determine the period and domain restriction for the function

y = -\tan\left(3x - \frac{\pi}{2}\right) + 2

.

The period is

\frac{\pi}{3}

and the function is undefined at

x = 0

(the

y

-axis).

For tangent, the period is

\frac{\pi}{3}

. At

x=0

, the argument is

-\frac{\pi}{2}

, where tangent is undefined, meaning the function has a vertical asymptote at the

y

-axis.

4

Determine the period, range, and y-intercept for the function

y = 3\sin\left(\frac{2}{3}x + \pi\right) - 2

.

The period is

3\pi

, range is

[-5, 1]

, and

y

-intercept is

(0, -2)

.

The period is

\frac{2\pi}{2/3} = 3\pi

, the range is

[-2-3, -2+3] = [-5, 1]

, and at

x=0

,

y = 3\sin(\pi) - 2 = -2

.

Key Concept

Identifying graphs of trigonometric functions from their equations by determining amplitude, period, phase shift, midline, range, and asymptotes.

Question 6Question →

The graph of the function $f(x) = A \sin(Bx + C) + D$ has a minimum point at $\left(\frac{\pi}{4}, -2\right)$ and its consecutive maximum point at $\left(\frac{5\pi}{4}, 8\right)$ , where $A > 0$ and $B > 0$ . What is the value of $f\left(\frac{7\pi}{4}\right)$ ?

-2

0

3

5

8

Show answer & explanation

Answer: 3

Answer

3

The midline of a sinusoidal curve is the average of its maximum and minimum values, which is

\frac{8 + (-2)}{2} = 3

. The horizontal distance between consecutive minimum and maximum points is half of a period:

\frac{5\pi}{4} - \frac{\pi}{4} = \pi

. Thus, the full period is

2\pi

, and the next minimum occurs at

\frac{5\pi}{4} + \pi = \frac{9\pi}{4}

. The value

x = \frac{7\pi}{4}

lies halfway between the maximum at

\frac{5\pi}{4}

and the minimum at

\frac{9\pi}{4}

. Midway between a maximum and a minimum, the function crosses its midline, so

f\left(\frac{7\pi}{4}\right) = 3

.

Step-by-Step Solution

1

Determine the midline (vertical shift

D

) of the trigonometric function.

The midline is

D = \frac{\text{Maximum} + \text{Minimum}}{2} = \frac{8 + (-2)}{2} = 3

.

The midline lies exactly halfway between the vertical peak and trough.

2

Find the horizontal distance for a half-period and determine the full period.

Half-period =

\frac{5\pi}{4} - \frac{\pi}{4} = \pi

, so the full period

T = 2\pi

.

The horizontal distance between consecutive minimum and maximum points equals half of one full period.

3

Identify the behavior of the graph at

x = \frac{7\pi}{4}

.

Since

x = \frac{7\pi}{4}

is halfway between the maximum at

x = \frac{5\pi}{4}

and the next minimum at

x = \frac{9\pi}{4}

, the function value equals the midline height

D = 3

.

A sinusoidal wave crosses its midline exactly midway between a peak and a trough.

Key Concept

Key features of transformed trigonometric functions (amplitude, midline, period, and symmetry)

Estimated Time:2m 0s

Question 7Question →

Match each transformed trigonometric function listed on the left with the correct description of its key graphical features listed on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The trigonometric functions correctly match their graphical features as follows:

f(x)

matches the description with period

\pi

and midline

y = 1

;

g(x)

matches the description with period

4\pi

and phase shift

2\pi

left;

h(x)

matches the description with period

\pi/3

and y-intercept

(0, 3)

;

k(x)

matches the description with period

2

and midline

y = -3

.

Each trigonometric equation is mapped to its unique set of graphical properties by evaluating its period, midline, phase shift, and specific points like y-intercepts or extrema using standard trigonometric transformation formulas.

Step-by-Step Solution

1

Analyze

f(x) = -3\cos\left(2x - \frac{\pi}{2}\right) + 1

.

Factor out the coefficient of

x

:

f(x) = -3\cos\left(2\left(x - \frac{\pi}{4}\right)\right) + 1

. The period is

\frac{2\pi}{B} = \frac{2\pi}{2} = \pi

. The midline is

y = D = 1

. The maximum value occurs when the cosine term equals

-1

(due to the

-3

coefficient):

-3(-1) + 1 = 4

, which happens at

2x - \frac{\pi}{2} = \pi \implies x = \frac{3\pi}{4}

.

Identify period, midline, phase shift, and extrema from standard form

y = A\cos(B(x-C)) + D

.

2

Analyze

g(x) = 2\sin\left(\frac{1}{2}x + \pi\right) - 1

.

Rewrite as

g(x) = 2\sin\left(\frac{1}{2}(x - (-2\pi))\right) - 1

. The period is

\frac{2\pi}{1/2} = 4\pi

, and the phase shift is

2\pi

units to the left. The minimum value is

2(-1) - 1 = -3

, which occurs when

\frac{1}{2}x + \pi = \frac{3\pi}{2} \implies x = \pi

.

Determine horizontal shift, period, and minimum location.

3

Analyze

h(x) = -\tan\left(3x + \frac{3\pi}{4}\right) + 2

.

The period for tangent is

\frac{\pi}{B} = \frac{\pi}{3}

. Consecutive vertical asymptotes occur every period

\frac{\pi}{3}

. Evaluating at

x = 0

gives

h(0) = -\tan\left(\frac{3\pi}{4}\right) + 2 = -(-1) + 2 = 3

, giving a y-intercept of

(0, 3)

.

Apply tangent period formula

\frac{\pi}{|B|}

and evaluate y-intercept.

4

Analyze

k(x) = 4\cos\left(\pi x - \frac{\pi}{2}\right) - 3

.

Rewrite as

k(x) = 4\cos\left(\pi\left(x - \frac{1}{2}\right)\right) - 3

. The period is

\frac{2\pi}{\pi} = 2

. The phase shift is

\frac{1}{2}

unit to the right, and the midline is

y = -3

.

Extract parameters from cosine function with

\pi

in argument.

Key Concept

Graphical transformations of trigonometric functions (amplitude, period

T = \frac{2\pi}{|B|}

or

\frac{\pi}{|B|}

, phase shift

C

, and midline

D

).

Estimated Time:2m 0s

Question 8Question →

The vertical displacement, $d(t)$ in centimeters, of a particle executing simple harmonic motion is modeled by the trigonometric function $d(t) = A \sin(B(t - C)) + D$ , where $A > 0$ , $B > 0$ , and $C$ represents the smallest non-negative phase shift in seconds. The graph of $d(t)$ completes one full cycle every $\frac{2\pi}{3}$ seconds, has a maximum value of $7\text{ cm}$ at $t = \frac{5\pi}{18}\text{ seconds}$ , and has a minimum value of $-3\text{ cm}$ . What is the value of $C$ ?

\frac{\pi}{18}

\frac{\pi}{9}

\frac{2\pi}{9}

\frac{\pi}{3}

\frac{5\pi}{18}

Show answer & explanation

Answer: $\frac{\pi}{9}$

Answer

\frac{\pi}{9}

To find the phase shift

C

, first determine

B

using the period formula

\text{Period} = \frac{2\pi}{B}

. Since the period is

\frac{2\pi}{3}

,

B = 3

. The function achieves a maximum when its sine argument equals

\frac{\pi}{2} + 2k\pi

. Setting

3\left(\frac{5\pi}{18} - C\right) = \frac{\pi}{2}

yields

\frac{5\pi}{6} - 3C = \frac{\pi}{2}

, which simplifies to

3C = \frac{\pi}{3}

, giving

C = \frac{\pi}{9}

.

Step-by-Step Solution

1

Determine the value of

B

from the period of the function.

B = 3

The standard period for a sine function is

2\pi

. Given that the period is

\frac{2\pi}{3}

, we set

\frac{2\pi}{B} = \frac{2\pi}{3}

, which yields

B = 3

.

2

Determine the maximum value equation for the parent sine function.

Argument equals

\frac{\pi}{2}

The standard sine function

\sin(\theta)

achieves its first positive maximum at

\theta = \frac{\pi}{2}

. Thus, for

d(t)

, the maximum occurs when

B(t - C) = \frac{\pi}{2}

.

3

Substitute

B = 3

and

t = \frac{5\pi}{18}

into the argument equation and solve for

C

.

C = \frac{\pi}{9}

Substitute the given values:

3\left(\frac{5\pi}{18} - C\right) = \frac{\pi}{2} \implies \frac{5\pi}{6} - 3C = \frac{\pi}{2}

. Subtract

\frac{5\pi}{6}

from both sides to get

-3C = \frac{3\pi}{6} - \frac{5\pi}{6} = -\frac{2\pi}{6} = -\frac{\pi}{3}

. Dividing by

-3

gives

C = \frac{\pi}{9}

.

Key Concept

Phase shift and parameter identification from graphs of transformed sine functions

Question 9Question →

In the standard $(x,y)$ coordinate plane, what is the distance along the $x$ -axis between any two consecutive maximum points on the graph of the function $f(x) = 5 \cos\left(3x - \frac{\pi}{2}\right) + 4$ ?

\frac{\pi}{3}

\frac{\pi}{2}

\frac{2\pi}{3}

2\pi

6\pi

Show answer & explanation

Answer: $\frac{2\pi}{3}$

Answer

The distance between any two consecutive maximum points is

\frac{2\pi}{3}

.

The distance along the

x

-axis between consecutive peaks of a cosine wave represents one full period. For the function

f(x) = 5 \cos\left(3x - \frac{\pi}{2}\right) + 4

, the coefficient of

x

is 3. The period formula for cosine is

\frac{2\pi}{|B|}

, which evaluates to

\frac{2\pi}{3}

.

Step-by-Step Solution

1

Relate consecutive maximum points to the function's period

The distance along the

x

-axis between consecutive maximum points of a periodic trigonometric function equals one period length,

T

.

Cosine graphs repeat their peak values once per complete wave cycle.

2

Identify the coefficient

B

of the variable

x

In

f(x) = 5 \cos\left(3x - \frac{\pi}{2}\right) + 4

, the coefficient of

x

is

B = 3

.

The standard transformation model is

y = A \cos(Bx - C) + D

.

3

Calculate the period using

T = \frac{2\pi}{|B|}

T = \frac{2\pi}{3}

.

Dividing the base cosine period of

2\pi

by

|B| = 3

gives the period of the transformed function.

Key Concept

The period of a transformed cosine function

y = A \cos(Bx - C) + D

is

\frac{2\pi}{|B|}

, which measures the horizontal distance between consecutive peak values.

Question 10Question →

What is the period of the trigonometric function $f(x) = 3 \cos\left(\frac{\pi}{5}x + \frac{\pi}{2}\right) - 4$ ?

10

5

2\pi

\frac{\pi}{5}

\frac{2}{5}

Show answer & explanation

Answer: $10$

Answer

10

The period of a function of the form

f(x) = A \cos(Bx + C) + D

is given by

T = \frac{2\pi}{|B|}

. Substituting

B = \frac{\pi}{5}

yields

T = \frac{2\pi}{\frac{\pi}{5}} = 2\pi \cdot \frac{5}{\pi} = 10

.

Step-by-Step Solution

1

Identify the coefficient of

x

(

B

) in the given trigonometric function.

B = \frac{\pi}{5}

The standard form for a transformed cosine function is

f(x) = A \cos(Bx + C) + D

, where

B

controls the horizontal stretch or compression.

2

Apply the period formula for cosine,

T = \frac{2\pi}{|B|}

.

T = \frac{2\pi}{\frac{\pi}{5}}

The fundamental period of

\cos(x)

is

2\pi

, which is scaled inversely by

|B|

.

3

Simplify the fraction by multiplying by the reciprocal.

T = 2\pi \cdot \frac{5}{\pi} = 10

Dividing by a fraction is equivalent to multiplying by its reciprocal, and the factor of

\pi

cancels out.

Key Concept

Period of Transformed Cosine Functions

Question 11Question →

Match each transformed trigonometric function on the left with its set of defining graphical properties on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings are:

f(x) = 4\sin(3x) + 2

matches with 'Amplitude of 4, period of 2π/3, and a midline at y = 2';

g(x) = -2\cos(2x - π)

matches with 'Amplitude of 2, period of π, and a phase shift of π/2 units to the right';

h(x) = 3\tan(x/2) - 1

matches with 'Period of 2π, vertical shift of 1 unit down, with vertical asymptotes at x = π + 2kπ'; and

k(x) = \cos(x + π/3) - 4

matches with 'Midline at y = -4, amplitude of 1, and a phase shift of π/3 units to the left'.

Each trigonometric function is matched according to its standard parameter transformation rules:

y = A\sin(B(x - C)) + D

or

y = A\cos(B(x - C)) + D

, where

|A|

is amplitude, period is

\frac{2\pi}{|B|}

for sine/cosine and

\frac{\pi}{|B|}

for tangent,

C

is horizontal phase shift, and

y = D

is the midline.

Step-by-Step Solution

1

Analyze

f(x) = 4\sin(3x) + 2

using the standard form

y = A\sin(B(x - C)) + D

Amplitude

= |A| = 4

, period

= \frac{2\pi}{B} = \frac{2\pi}{3}

, midline

= D \implies y = 2

.

Direct extraction of parameters for sine graphs.

2

Factor out

B = 2

from

g(x) = -2\cos(2x - \pi)

g(x) = -2\cos\left(2\left(x - \frac{\pi}{2}\right)\right)

, so amplitude

= |-2| = 2

, period

= \frac{2\pi}{2} = \pi

, phase shift

= \frac{\pi}{2}

to the right.

Factoring

B

is necessary to correctly identify the horizontal phase shift.

3

Analyze tangent function parameters for

h(x) = 3\tan\left(\frac{1}{2}x\right) - 1

Period

= \frac{\pi}{B} = \frac{\pi}{1/2} = 2\pi

, shifted down

1

unit (

y = -1

). Asymptotes occur when

\frac{1}{2}x = \frac{\pi}{2} + k\pi \implies x = \pi + 2k\pi

.

Tangent period uses

\frac{\pi}{B}

instead of

\frac{2\pi}{B}

, and asymptotes occur where tangent arguments equal odd multiples of

\frac{\pi}{2}

.

4

Analyze transformation parameters for

k(x) = \cos\left(x + \frac{\pi}{3}\right) - 4

Amplitude

= 1

, midline

= y = -4

, phase shift

= \frac{\pi}{3}

to the left.

Addition inside the function argument

(x + C)

corresponds to a horizontal shift to the left.

Key Concept

Identifying amplitude, period, midline, phase shift, and asymptotes from transformed trigonometric equations

Question 12Question →

In the standard $(x,y)$ coordinate plane, a cosine function of the form $y = a \cos(b(x - c)) + d$ , where $a > 0$ and $b > 0$ , has a local maximum at $(2, 9)$ and the very next local minimum at $(6, 1)$ . What is the value of $b$ ?

\frac{\pi}{8}

\frac{\pi}{4}

\frac{\pi}{2}

\pi

2\pi

Show answer & explanation

Answer: $\frac{\pi}{4}$

Answer

\frac{\pi}{4}

The distance along the

x

-axis from a maximum to the consecutive minimum is half of one full period of the cosine wave. Here, that distance is

6 - 2 = 4

, which means the full period is

2 \times 4 = 8

. Using the relationship

\text{Period} = \frac{2\pi}{b}

, we solve for

b

to get

b = \frac{2\pi}{8} = \frac{\pi}{4}

.

Step-by-Step Solution

1

Determine the horizontal distance between the consecutive maximum and minimum points.

The horizontal distance is

6 - 2 = 4

units.

The

x

-coordinates of the maximum and minimum points are

2

and

6

, respectively.

2

Calculate the period of the cosine function.

\text{Period} = 2 \times 4 = 8

units.

The horizontal distance between a peak and the immediately following trough of a cosine wave represents exactly half of one full period.

3

Solve for the coefficient

b

using the period formula

b = \frac{2\pi}{\text{Period}}

.

b = \frac{2\pi}{8} = \frac{\pi}{4}

.

For a trigonometric function of the form

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

, the relationship between the period and

b

is

\text{Period} = \frac{2\pi}{b}

.

Key Concept

Determining Period and Frequency Coefficient of a Cosine Function

Question 13Question →

In the standard $(x,y)$ coordinate plane, a sinusoidal function $f(x) = A \sin(Bx - C) + D$ reaches a maximum value of $9$ at $x = 1$ and its immediate next minimum value of $-1$ at $x = 4$ . What is the period of $f(x)$ ?

Show answer & explanation

Answer: 6

Answer

The period of the sinusoidal function is 6.

For any sinusoidal graph, the horizontal distance between a maximum point and the immediate next minimum point corresponds to one-half of the period. Given that the maximum occurs at

x = 1

and the next minimum occurs at

x = 4

, the half-period is

4 - 1 = 3

. Multiplying this half-period by

2

gives the full period of

6

.

Step-by-Step Solution

1

Identify the horizontal distance between the consecutive maximum and minimum points.

The horizontal distance is

4 - 1 = 3

.

The maximum occurs at

x = 1

and the consecutive minimum occurs at

x = 4

.

2

Relate the horizontal distance between consecutive extrema to the period of the function.

Half of the period is equal to

3

.

In any sinusoidal function, the horizontal distance between a peak (maximum) and the adjacent trough (minimum) represents exactly one-half of a complete cycle.

3

Calculate the full period of the function.

Period =

2 \times 3 = 6

.

Multiplying the half-period by 2 yields the full period of the function.

Key Concept

The horizontal distance between consecutive maximum and minimum points of a sinusoidal graph is half of the function's period.

Estimated Time:1m 15s

Question 14Question →

Which of the following values represents the period, in radians, of the trigonometric function $f(x) = 3 \tan\left(\frac{2}{3}x - \frac{\pi}{6}\right) + 5$ ?

\frac{3\pi}{2}

3\pi

\frac{2\pi}{3}

\frac{\pi}{6}

\frac{3\pi}{4}

Show answer & explanation

Answer: $\frac{3\pi}{2}$

Answer

The period of the given tangent function is

\frac{3\pi}{2}

radians.

The parent function

y = \tan(x)

repeats every

\pi

radians. For a function in the form

f(x) = A \tan(Bx - C) + D

, the horizontal scale factor

B

alters the period according to

\text{Period} = \frac{\pi}{|B|}

. With

B = \frac{2}{3}

, dividing

\pi

by

\frac{2}{3}

gives

\frac{3\pi}{2}

.

Step-by-Step Solution

1

Identify the standard form of the transformed tangent function and its parameters.

For

f(x) = A \tan(Bx - C) + D

, the coefficient of

x

is

B = \frac{2}{3}

.

The horizontal stretch/compression factor

B

determines the period of the function.

2

Apply the period formula for the tangent function.

\text{Period} = \frac{\pi}{|B|} = \frac{\pi}{\frac{2}{3}} = \frac{3\pi}{2}

Unlike sine and cosine functions which have a fundamental period of

2\pi

, the parent tangent function

y = \tan(x)

has a period of

\pi

radians.

Key Concept

Period of Transformed Tangent Functions

Estimated Time:1m 0s

Question 15Question →

Match each trigonometric function on the left with its correct combination of amplitude and period on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

f(x) = 4 \sin(3x)

matches Amplitude = 4, Period =

\frac{2\pi}{3}

;

g(x) = 2 \cos\left(\frac{1}{2}x\right)

matches Amplitude = 2, Period =

4\pi

;

h(x) = -3 \sin(2x)

matches Amplitude = 3, Period =

\pi

.

Each trigonometric function is correctly evaluated using the general properties: Amplitude equals

|A|

and Period equals

\frac{2\pi}{|B|}

.

Step-by-Step Solution

1

Identify the standard trigonometric form parameters.

For equations of the form

y = A \sin(Bx)

or

y = A \cos(Bx)

, Amplitude

= |A|

and Period

= \frac{2\pi}{|B|}

.

Applying the definitions of amplitude and period for sine and cosine functions.

2

Calculate amplitude and period for

f(x) = 4 \sin(3x)

.

Amplitude

= |4| = 4

, Period

= \frac{2\pi}{3}

.

Here

A = 4

and

B = 3

.

3

Calculate amplitude and period for

g(x) = 2 \cos\left(\frac{1}{2}x\right)

.

Amplitude

= |2| = 2

, Period

= \frac{2\pi}{1/2} = 4\pi

.

Here

A = 2

and

B = \frac{1}{2}

.

4

Calculate amplitude and period for

h(x) = -3 \sin(2x)

.

Amplitude

= |-3| = 3

, Period

= \frac{2\pi}{2} = \pi

.

Here

A = -3

(so

|A| = 3

) and

B = 2

.

Key Concept

Amplitude and Period of Sine and Cosine Graphs

Question 16Question →

In the standard $(x,y)$ coordinate plane, the graph of a trigonometric function is given by $f(x) = 4 \sin\left(3x - \frac{\pi}{2}\right) + 2$ . What is the horizontal distance between any two consecutive points where the graph intersects its midline $y = 2$ ?

\frac{\pi}{6}

\frac{\pi}{3}

\frac{\pi}{2}

\frac{2\pi}{3}

\frac{4\pi}{3}

Show answer & explanation

Answer: $\frac{\pi}{3}$

Answer

The horizontal distance between consecutive midline intersections is

\frac{\pi}{3}

.

The midline of

f(x) = 4 \sin\left(3x - \frac{\pi}{2}\right) + 2

is the horizontal line

y = 2

. Intersections with this line occur when

\sin\left(3x - \frac{\pi}{2}\right) = 0

. Since the sine function equals zero at integer multiples of

\pi

, the difference in the argument between consecutive zero points is

\pi

. Setting

3\Delta x = \pi

gives

\Delta x = \frac{\pi}{3}

, which is half the period of the function.

Step-by-Step Solution

1

Identify the frequency parameter

B

from the function

f(x) = 4 \sin\left(3x - \frac{\pi}{2}\right) + 2

.

The frequency coefficient inside the sine expression is

B = 3

.

The standard form of a transformed sine function is

f(x) = A \sin(Bx - C) + D

.

2

Calculate the full period

T

of the function.

T = \frac{2\pi}{|B|} = \frac{2\pi}{3}

.

The standard period

2\pi

of a sine function is scaled horizontally by a factor of

\frac{1}{B}

.

3

Determine the horizontal distance between consecutive intersections with the midline

y = 2

.

\text{Distance} = \frac{T}{2} = \frac{\frac{2\pi}{3}}{2} = \frac{\pi}{3}.

A sinusoidal wave completes one full period over length

T

and intersects its midline twice during each full cycle, making consecutive midline crossings separated by half of the period.

Key Concept

Distance between consecutive midline intersections of a sinusoidal function

Estimated Time:1m 0s

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