Algebra 3 Quiz: Graphing Sine And Cosine

12 questions · exam conditions

0:00

Graphing Sine And CosineQuestion 1 of 12

A sinusoidal graph has amplitude $3$ , period $4\pi$ , midline $y=-1$ , and is shifted $\frac{\pi}{2}$ to the right relative to $y=\cos x$ . Which equation represents the graph?

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

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

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

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

← Back to quizzes

Algebra 3 Quiz

Algebra 3 Quiz: Graphing Sine And Cosine

Practice Graphing Sine And Cosine in Algebra 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Graphing Sine And Cosine, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 3.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A sinusoidal graph has amplitude $3$ , period $4\pi$ , midline $y=-1$ , and is shifted $\frac{\pi}{2}$ to the right relative to $y=\cos x$ . Which equation represents the graph?

$y=3\cos\left(\frac{1}{2}x-\frac{\pi}{2}\right)-1$
$y=3\cos\left(\frac{1}{2}x-\frac{\pi}{4}\right)-1$ (correct answer)
$y=3\cos\left(2x-\frac{\pi}{2}\right)+1$
$y=3\cos\left(\frac{1}{2}x+\frac{\pi}{4}\right)-1$

Explanation: Whenever you see a sinusoidal graph question, start by translating the key features into the general form

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

. Amplitude is

|A|=3

; midline

y=-1

gives

D=-1

; period

4\pi

means

B=\frac{2\pi}{4\pi}=\frac12

. The graph is shifted

\frac{\pi}{2}

right relative to cosine, so inside the parentheses you need

x-\frac{\pi}{2}

. Since

B=\frac12

, the argument becomes

\frac12\left(x-\frac{\pi}{2}\right)=\frac12 x-\frac{\pi}{4}

. Therefore the correct equation is

y=3\cos\left(\frac12 x-\frac{\pi}{4}\right)-1

. The choice with

y=3\cos\left(\frac12 x-\frac{\pi}{2}\right)-1

has the right period and midline, but it equals

3\cos\left(\frac12(x-\pi)\right)-1

, a right shift of

\pi

, not

\frac{\pi}{2}

. The choice

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

uses

B=2

, giving period

\pi

, and has midline

+1

instead of

-1

. The choice

y=3\cos\left(\frac12 x+\frac{\pi}{4}\right)-1

equals

3\cos\left(\frac12(x+\frac{\pi}{2})\right)-1

, which shifts left, not right. A solid strategy: always factor out

B

before reading the phase shift. Writing

\cos(Bx-C)

\cos\left(B\left(x-\frac CB\right)\right)

reveals the true horizontal shift and helps you avoid sign and scaling traps.

Question 2

Which statement about $y=-2\cos(4x-\pi)+3$ is true?

Amplitude $2$ , period $2\pi$ , midline $y=3$ , maximum value $5$ .
Amplitude $-2$ , period $\frac{\pi}{2}$ , midline $y=-3$ , minimum value $-5$ .
Amplitude $4$ , period $\frac{\pi}{2}$ , midline $y=3$ , maximum value $7$ .
Amplitude $2$ , period $\frac{\pi}{2}$ , midline $y=3$ , maximum value $5$ . (correct answer)

Explanation: Whenever you see a transformed cosine like

y=-2\cos(4x-\pi)+3

, think of the standard form

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

. The amplitude is

|A|

, the period is

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

, the midline is

y=D

, and the maximum and minimum values are

D+|A|

and

D-|A|

. Here,

A=-2

, so the amplitude is

2

. Since

B=4

, the period is

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

. The midline is

y=3

, because

D=3

. Therefore the maximum is

3+2=5

and the minimum is

3-2=1

. So the true statement is: amplitude

2

, period

\frac{\pi}{2}

, midline

y=3

, maximum value

5

. Now look at the incorrect statements. The one with amplitude

2

, period

2\pi

, midline

y=3

, maximum value

5

uses the wrong period: it ignores the horizontal compression caused by the coefficient

4

. The one with amplitude

-2

, period

\frac{\pi}{2}

, midline

y=-3

, minimum value

-5

has two issues: amplitude is never negative, and the midline should be

y=3

, not

y=-3

. The statement with amplitude

4

, period

\frac{\pi}{2}

, midline

y=3

, maximum value

7

mistakes the x-coefficient

4

for the amplitude; the amplitude comes from the coefficient in front of cosine, which is

2

after taking absolute value. Your takeaway: rewrite the function in standard form, then compute amplitude, period, and midline systematically. Always use

|A|

for amplitude, and remember the maximum and minimum are just the midline plus or minus the amplitude.

Question 3

Which equation is equivalent to $y=4\sin\left(2x+\frac{\pi}{2}\right)-3$ ?

$y=-4\cos(2x)-3$
$y=4\cos(2x)+3$
$y=4\cos(2x)-3$ (correct answer)
$y=4\cos\left(2x-\frac{\pi}{2}\right)-3$

Explanation: When you see a sine expression with a phase shift of

\frac{\pi}{2}

, you're really being tested on cofunction identities. The key fact is

\sin(\theta+\frac{\pi}{2})=\cos\theta

. Applying it with

\theta=2x

gives

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

so the equivalent equation is the one reading

y=4\cos(2x)-3

. The choice

y=-4\cos(2x)-3

gets the sign wrong:

\sin(\theta+\frac{\pi}{2})

is positive

\cos\theta

, not its negative. The choice

y=4\cos(2x)+3

keeps the correct cosine term but shifts the graph in the wrong vertical direction; the original has

-3

, so the vertical shift must stay

-3

. The choice

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

is a different expression:

\cos(2x-\frac{\pi}{2})=\sin 2x

, so that option equals

4\sin 2x-3

, not the original

4\sin(2x+\frac{\pi}{2})-3

. It confuses which sign inside the argument produces the cofunction shift. Study tip: memorize the two core cofunction shifts:

\sin(x+\frac{\pi}{2})=\cos x

and

\cos(x-\frac{\pi}{2})=\sin x

. On this exam, a

\frac{\pi}{2}

phase shift is almost always a signal to swap sine and cosine — just check the sign and preserve the vertical shift exactly.

Question 4

A cosine graph has range $[-5,3]$ and period $\pi$ . Which equation could represent it?

$y=4\cos(2x)+1$
$y=8\cos(2x)-1$
$y=4\cos(x)-1$
$y=4\cos(2x)-1$ (correct answer)

Explanation: Whenever you see a cosine graph described by its range and period, think of the general form

y=A\cos(Bx)+D

. Your job is to extract the amplitude

|A|

, the midline

D

, and the frequency factor

B

from those two facts. The midpoint of the range

[-5,3]

D=\frac{-5+3}{2}=-1

, so the graph is shifted down 1. The distance from midline to either endpoint is

4

, so

A=4

. And for cosine, the period is

\frac{2\pi}{B}

; a period of

\pi

means

B=2

. Therefore the equation is

y=4\cos(2x)-1

. Each wrong answer misses one of these pieces:

y=4\cos(2x)+1

has the right amplitude and period but shifts up, giving range

[-3,5]

y=8\cos(2x)-1

has the right shift and period but amplitude 8, giving range

[-9,7]

y=4\cos(x)-1

has the right amplitude and shift but period

2\pi

, since

B=1

, not

\pi

. These distractors test whether you can confidently connect range to midline/amplitude and period to

B

. When you see a trig graph question, first find midline: it is the average of max and min. Then find amplitude: half the distance between max and min. Finally get

B

from period using $$B=\frac{2\pi}{\text{period}}

Question 5

For $y=3\cos(2x-\pi)+1$ , what is the $x$ -coordinate of the first maximum to the right of the $y$ -axis?

$x=\pi$
$x=\frac{\pi}{4}$
$x=\frac{\pi}{2}$ (correct answer)
$x=0$

Explanation: Whenever you see a cosine function like

y=3\cos(2x-\pi)+1

, remember that a maximum occurs when the cosine argument equals a multiple of

2\pi

, because

\cos(0)=1

is the largest cosine value. Set the inside equal to

2\pi k

2x-\pi = 2\pi k \quad \Rightarrow \quad x=\frac{\pi}{2}+\pi k

For the first maximum to the right of the

y

-axis, choose the smallest nonnegative solution. With

k=0

, you get

x=\frac{\pi}{2}

, so that is the correct coordinate. The choice

x=\pi

is a trap: substituting gives

2\pi-\pi=\pi

, and

\cos\pi=-1

, so that point is actually a minimum, not a maximum. Similarly,

x=0

gives

2(0)-\pi=-\pi

, so

\cos(-\pi)=-1

, also a minimum. The choice

x=\frac{\pi}{4}

makes the argument

2(\frac{\pi}{4})-\pi=-\frac{\pi}{2}

, where

\cos=0

, so that point is on the midline, not at a peak. The key takeaway: for maximums of cosine, solve

\text{argument}=2\pi k

, not zero generically. Here, solving

2x-\pi=0

directly already gives

x=\frac{\pi}{2}

, but if there were several peaks, you would add multiples of the period,

\pi

, and then choose the first one to the right of the

y

-axis.

Question 6

Which function has a frequency of $\frac{1}{4}$ cycle per unit?

$y=3\cos(2x)$
$y=3\cos\left(\frac{\pi}{2}x\right)$ (correct answer)
$y=3\cos(4x)$
$y=3\cos\left(\frac{\pi}{4}x\right)$

Explanation: Whenever you see a question about trigonometric functions, the key is to connect the coefficient of

x

to the period and frequency. For a function like

y = A\cos(Bx)

, the period is

\frac{2\pi}{B}

, and the frequency is the reciprocal:

\frac{B}{2\pi}

cycles per unit. Here, you need that frequency to be

\frac{1}{4}

, so set

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

, which gives

B = \frac{\pi}{2}

. Therefore, the function

y=3\cos\left(\frac{\pi}{2}x\right)

has the correct

B

. The amplitude

3

does not affect frequency. Now check the other choices. The function

y=3\cos(2x)

has

B=2

, giving frequency

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

, not

\frac{1}{4}

. Similarly,

y=3\cos(4x)

has

B=4

, so its frequency is

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

, far too large. The function

y=3\cos\left(\frac{\pi}{4}x\right)

has

B=\frac{\pi}{4}

, yielding frequency

\frac{\pi/4}{2\pi} = \frac{1}{8}

, which is half of what you want. Each of these traps comes from confusing the coefficient

B

with the frequency itself—remember that

B

is related to the angular frequency, but the number of cycles per unit requires dividing by

2\pi

. A quick study tip: to avoid this mistake, always write the formula

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

and solve for

B

given the desired frequency. Also, note that multiplying by a constant like

3

only changes the amplitude, not the frequency or period. Practice recognizing that the argument inside cosine is

\text{frequency} \times 2\pi \times x

when you need a specific cycle count.

Question 7

A sine graph has midline $y=-1$ and amplitude $3$ . It increases through its midline at $x=1$ and reaches its first maximum at $x=3$ . Which equation represents the graph?

$y=3\sin\left(\frac{\pi}{2}(x-1)\right)-1$
$y=3\cos\left(\frac{\pi}{4}(x-1)\right)-1$
$y=3\sin\left(\frac{\pi}{4}(x+1)\right)-1$
$y=3\sin\left(\frac{\pi}{4}(x-1)\right)-1$ (correct answer)

Explanation: Whenever you see a sine graph described in words, translate the clues into the standard form

y=A\sin(B(x-h)))+k

. The midline

y=-1

tells you

k=-1

; the amplitude

3

tells you the vertical coefficient is

3

. Now locate horizontal behavior: a sine curve that increases through its midline acts like

\sin x

x=0

. Since this graph increases through midline at

x=1

, the phase shift is

x-1

. From an upward midline crossing to the first maximum is one-quarter cycle. Here that distance is

3-1=2

, so one-quarter period

=2

and the full period

=8

. For

y=3\sin(B(x-1))-1

\frac{2\pi}{B}=8

, so

B=\frac{\pi}{4}

. Thus the matching equation is

y=3\sin\left(\frac{\pi}{4}(x-1)\right)-1

. The choice using

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

has the right midline, amplitude, and shift, but its period is only

4

, so by

x=3

it has already cycle back to midline rather than reaching a maximum. The cosine choice

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

has the correct period, but a cosine curve starts at a maximum at

x=1

, not at the midline going up. The choice with

x+1

shifts the curve in the wrong direction: it crosses midline at

x=-1

and reaches its maximum at

x=1

, opposite the given condition. A useful check: test the two points

x=1

and

x=3

in each equation. The correct graph must give

-1

x=1

and

2

x=3

; this quickly eliminates the incorrect periods, signs, and cosine setup.

Question 8

Which sequence of transformations maps $y=\cos x$ to $y=3\cos(2x-\pi)-1$ ?

Vertically stretch by $3$ , horizontally compress by a factor of $\frac{1}{2}$ , shift right $\pi$ , shift down $1$ .
Vertically stretch by $3$ , horizontally compress by a factor of $\frac{1}{2}$ , shift right $\frac{\pi}{2}$ , shift down $1$ . (correct answer)
Vertically stretch by $3$ , horizontally stretch by a factor of $2$ , shift right $\frac{\pi}{2}$ , shift down $1$ .
Vertically stretch by $3$ , horizontally compress by a factor of $\frac{1}{2}$ , shift left $\frac{\pi}{2}$ , shift down $1$ .

Explanation: Whenever you are asked to describe transformations from a base trig function, rewrite the equation as

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

. Then each parameter maps directly: amplitude

|a|

, period

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

, horizontal shift

c

, vertical shift

d

. For

y=3\cos(2x-\pi)-1

, factor the inside:

2x-\pi=2(x-\frac{\pi}{2})

, so the function is

3\cos[2(x-\frac{\pi}{2})]-1

. Thus

a=3

means vertical stretch by

3

;

b=2

means horizontal compression by a factor of

\frac12

;

c=\frac{\pi}{2}

means shift right

\frac{\pi}{2}

;

d=-1

means shift down

1

. That matches the choice with vertical stretch by

3

, horizontal compression by

\frac12

, shift right

\frac{\pi}{2}

, and shift down

1

. The distractor with "shift right

\pi

" comes from forgetting to factor out the

2

: the phase shift is

\frac{\pi}{2}

, not

\pi

. The one with "horizontally stretch by

2

" reverses the effect of

b>1

, which compresses the graph horizontally. The one with "shift left

\frac{\pi}{2}

" confuses the sign inside:

x-\frac{\pi}{2}

moves the graph right, not left. Study tip: always rewrite as

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

first. If

b

is not factored out, you cannot correctly read the phase shift.

Question 9

In $y=a\cos(bx-c)+d$ , assume $a>0$ and $b>0$ . If $a$ is halved and $b$ is doubled while $c$ and $d$ stay the same, which statement about the new graph is true?

Amplitude is halved, period is doubled, and phase shift is halved.
Amplitude is halved, period is halved, and phase shift is halved. (correct answer)
Amplitude is halved, period is halved, and phase shift is doubled.
Amplitude is unchanged, period is unchanged, and phase shift is unchanged.

Explanation: Whenever you see transformations of

y=a\cos(bx-c)+d

, identify each parameter's job: amplitude is

a

, period is

2\pi/b

, and phase shift is

c/b

. Here

a>0

and

b>0

, with

c

and

d

fixed. If

a

is halved, the amplitude becomes half of its original value. If

b

is doubled, the period becomes

2\pi/(2b)=\pi/b

, which is half the original period

2\pi/b

. The phase shift becomes

c/(2b)

, which is also half the original

c/b

. The vertical shift

d

stays the same. So the true statement is: amplitude is halved, period is halved, and phase shift is halved. The choice saying "amplitude is halved, period is doubled, and phase shift is halved" gets the period backwards: doubling

b

shrinks the period, not lengthens it. The choice saying "amplitude is halved, period is halved, and phase shift is doubled" treats phase shift as if it moved with

b

, but since phase shift is

c/b

, doubling

b

actually halves it. The choice saying "amplitude is unchanged, period is unchanged, and phase shift is unchanged" ignores that changing

a

and

b

directly changes amplitude and period. A reliable strategy is to rewrite the function as

y=a\cos\left(b\left(x-\frac{c}{b}\right)\right)+d

before comparing. That instantly shows how amplitude, period, and phase shift each respond to changes in

a

and

b

Question 10

What is the phase shift of $y=2\cos\left(\pi x+\frac{\pi}{2}\right)$ relative to $y=2\cos(\pi x)$ ?

$\frac{1}{2}$ unit to the left (correct answer)
$\frac{1}{2}$ unit to the right
$\frac{\pi}{2}$ units to the left
$\frac{\pi}{2}$ units to the right

Explanation: Whenever you see a cosine function with a constant added inside the argument, your first instinct should be to factor out the coefficient of

x

. The phase shift is the horizontal translation of one graph relative to another, and the number

2

out front only affects amplitude, not left/right movement. For

y=2\cos(\pi x+\frac{\pi}{2}

, factor the

\pi

\cos\left(\pi x+\frac{\pi}{2}\right)=\cos\left(\pi\left(x+\frac{1}{2}\right)\right).

Comparing

\cos(\pi(x+\frac12))

\cos(\pi x)

, the input

x

has been replaced by

x+\frac12

. That replacement moves every point

\frac12

unit to the left, because a point that used to occur at

x=0

now occurs at

x=-\frac12

. So the phase shift is

\frac12

unit left. The distractor "

\frac12

unit to the right" recognizes the correct magnitude but forgets that

x+\frac12

means left, not right; a right shift would require

x-\frac12

inside. The choices with

\frac{\pi}{2}

units come from grabbing the added constant

\frac{\pi}{2}

directly as the shift without dividing by the frequency

\pi

. Since

\frac{\pi}{2}\div\pi=\frac12

, the magnitude cannot remain

\frac{\pi}{2}}

. And "

\frac{\pi}{2}

units to the right" combines both the wrong magnitude and the wrong direction. Study takeaway: always rewrite

Bx+C

B\left(x+\frac{C}{B}\right)

. The phase shift is

-\frac{C}{B}

, with the sign automatically telling you left or right.

Question 11

Consider $y=4\sin(3x+\pi)-2$ . Which statement correctly describes its graph?

The period is $2\pi$ , the phase shift is $\frac{\pi}{3}$ to the right, and the midline is $y=-2$ .
The period is $\frac{2\pi}{3}$ , the phase shift is $\pi$ to the left, and the midline is $y=2$ .
The period is $\frac{2\pi}{3}$ , the phase shift is $\frac{\pi}{3}$ to the left, and the midline is $y=-2$ . (correct answer)
The period is $\frac{2\pi}{3}$ , the phase shift is $\frac{\pi}{3}$ to the right, and the midline is $y=-2$ .

Explanation: When you see

y=4\sin(3x+\pi)-2

, compare it to the standard form

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

. The value of

B

controls the period,

C

controls the phase shift, and

D

controls the midline. The key move is to factor the coefficient of

x

out of the argument first:

3x+\pi = 3\left(x+\frac{\pi}{3}\right)

B=3

, which means the period is

\frac{2\pi}{3}

. The expression

x+\frac{\pi}{3}

x-\left(-\frac{\pi}{3}\right)

, so the phase shift is

\frac{\pi}{3}

to the left. The vertical shift is

-2

, so the midline is

y=-2

. Therefore the correct description is period

\frac{2\pi}{3}

, phase shift

\frac{\pi}{3}

to the left, midline

y=-2

. The other choices each contain a common mistake. Saying the period is

2\pi

forgets to divide by

B=3

. Saying the phase shift is

\frac{\pi}{3}

to the right confuses the sign: a plus sign inside the argument shifts the graph left, not right. Saying the phase shift is

\pi

to the left uses the unsimplified

+\pi

instead of dividing by 3 after factoring. Saying the midline is

y=2

drops the negative sign on the vertical shift. Strategy: for any sinusoidal function, rewrite the inside as

B(x-C)

before reading period, phase shift, and midline. Factor first, then interpret.

Question 12

A cosine curve has a maximum at $x=1$ , a minimum at $x=4$ , midline $y=2$ , and amplitude $3$ . Which equation could represent it?

$y=3\cos\left(\frac{\pi}{6}(x-1)\right)+2$
$y=3\cos\left(\frac{\pi}{3}(x-4)\right)+2$
$y=3\cos\left(\frac{\pi}{3}(x-1)\right)+2$ (correct answer)
$y=3\sin\left(\frac{\pi}{3}(x-1)\right)+2$

Explanation: When you see a cosine curve described by its maximum, minimum, midline, and amplitude, connect each feature to the general form

y=A\cos(B(x-h))+k

. The midline

y=2

gives

k=2

. The amplitude

3

gives

A=3

. A positive cosine curve reaches its maximum when

\cos=1

, so the location of the maximum tells you the horizontal shift: maximum at

x=1

, so

h=1

. That means the correct form must be

y=3\cos(B(x-1))+2

. Now find

B

: the distance from a maximum to the next minimum is half a period. Here that distance is

4-1=3

, so the full period is

6

. Since

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

B=\frac{2\pi}{6}=\frac{\pi}{3}

. Therefore the equation is

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

. The choice with

\frac{\pi}{6}(x-1)

has the right shift and midline, but its period is

12

, not

6

, so the minimum would occur at

x=7

, not

x=4

. The choice with

\frac{\pi}{3}(x-4)

has the correct period but shifts the maximum to

x=4

. The sine version

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

gives the midline at

x=1

, not a maximum. For any trig graph question, use the maximum and minimum x-values to find both the horizontal shift and half the period. That single step usually determines the correct equation.