Question

1. A sine function has the following key features:

Frequency = 16π
Amplitude = 2
Midline: y = 3
y-intercept: (0, 3)
The function is not a reflection of its parent function over the x-axis.
Use the sine tool to graph the function. The first point must be on the midline and the second point must be a maximum or minimum value on the graph closest to the first point.

2. A sine function has the following key features:
Period = 12
Amplitude = 4
Midline: y = 1
y-intercept: (0, 1)
The function is not a reflection of its parent function over the x-axis.
Use the sine tool to graph the function. The first point must be on the midline and the second point must be a maximum or minimum value on the graph closest to the first point.

3. A sine function has the following key features:
Period = 4π
Amplitude = 2
Midline: y = 3
y-intercept: (0, 3)
The function is a reflection of its parent function over the x-axis.
Use the sine tool to graph the function. The first point must be on the midline and the second point must be a maximum or minimum value on the graph closest to the first point.

4. picture

5. picture

Answer 1

1. Picture Below

2. Ordered Pair : (0,1) (3,5) (6,1) - Function: f(x)=4sin(pi/6x)+1

3. Ordered Pair : (0,3) (3.14,5) (6.27,3) - Function: f(x)= 2sin(1/2x)+3

4. I have no idea.

5. Plot the first point at (0, 20) and the second point at (2, 22)

Explanation:

Answer 2

Problem 1

See the attached image (figure 1)

16pi seems like a typo. I'm going to assume that it's a fraction and it is 1/(6pi)
f = 1/(6pi) = frequency
T = 1/f = 1/(1/(6pi)) = 6pi
Amplitude = 2
a = 2
b = 2pi/T = 2pi/(6pi) = 1/3
Midline: y = 3
d = 3

The function is
y = a*sin(bx-c)+d
y = 2*sin(1/3*x-0)+3
y = 2*sin(x/3)+3

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Problem 2

See the attached image (figure 2)

T = 12 is the period
a = 4 is the amplitude
b = 2pi/T = 2pi/12 = pi/6
y = 1 is the midline so d = 1
The y intercept is (0,1) which is the midline, which indicates no phase shifts have occurred so c = 0

The function is
y = a*sin(bx-c)+d
y = 4*sin((pi/6)x-0)+1
y = 4*sin((pi/6)x)+1

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Problem 3

See the attached image (figure 3)

Period = 4pi
T = 4pi
b = 2pi/T = 2pi/(4pi) = 1/2 = 0.5
Amplitude = 2
a = 2
Midline: y = 3
d = 3
y-intercept: (0,3)
The function is a reflection of its parent function over the x-axis, so 'a' is negative meaning a = -2 instead of a = 2

The function is
y = a*sin(bx-c)+d
y = -2*sin(0.5x-0)+3
y = -2*sin(0.5x)+3

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Problem 4

See the attached image (figure 4)

a = 10 which is half of the distance between the highest and lowest points
T = 8 is the period
b = 2pi/T = 2pi/8 = pi/4
c = -pi/2 is the phase shift since its really a cosine graph
d = 0 is the midline

The function is
y = a*sin(bx-c)+d
y = 10*sin((pi/4)*x+(-pi/2))+0
y = 10*sin((pi/4)*x+pi/2)

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Problem 5

See the attached image (figure 5)

a = 2 is the amplitude since it bobs up and down this distance from the midline
T = 8 seconds is the period (double that of the time it takes for it to go from the highest to the lowest point)
b = 2pi/T = 2pi/8 = pi/4
c = 0 is the phase shift as the buoy starts at normal depth of 20 meters
d = 20 is the midline

The function is
y = a*sin(bx-c)+d
y = 2*sin((pi/4)x-0)+20
y = 2*sin((pi/4)x)+20

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