Answer :
Certainly! Let's construct and graph the sine function step by step, given the parameters:
1. Amplitude (A): The amplitude is the distance from the midline to the maximum or minimum value of the function. Here, the amplitude is 5.
2. Period (B): The period of the sine function is the distance (in the x-axis) over which the function completes one full cycle. It's given as \(6\pi\). The formula for period \(T\) of the function \(y = A \sin(Bx)\) is \(T = \frac{2\pi}{|B|}\):
[tex]\[ 6\pi = \frac{2\pi}{|B|} \][/tex]
Solving for \(B\):
[tex]\[ |B| = \frac{2\pi}{6\pi} = \frac{1}{3} \][/tex]
3. Midline (D): The midline is the horizontal axis around which the sine function oscillates. Here, it’s given as -2. So, \(D = -2\).
4. Reflection: The graph is a reflection over the x-axis, which means we multiply the sine function by -1. If the original sine function is \(y = \sin(x)\), the reflected function is \(y = -\sin(x)\).
Considering all of these together, our function takes the form:
[tex]\[ y = -5 \sin\left(\frac{1}{3}x\right) - 2 \][/tex]
To graph this function, follow these steps:
### Finding Key Points:
1. Y-Intercept: At \(x = 0\),
[tex]\[ y = -5 \sin\left(\frac{1}{3} \cdot 0\right) - 2 = -5 \sin(0) - 2 = -2 \][/tex]
So the point (0, -2) is on the midline.
2. Next Key Point (Minimum or Maximum):
The function is:
[tex]\[ y = -5 \sin\left(\frac{1}{3}x\right) - 2 \][/tex]
The minimum value of \(\sin(x)\) is -1 and the maximum value is 1. For this reflected and shifted function:
- Maximum: When \(\sin\left(\frac{1}{3}x\right) = -1\),
[tex]\[ y = -5(-1) - 2 = 5 - 2 = 3 \][/tex]
- Minimum: When \(\sin\left(\frac{1}{3}x\right) = 1\),
[tex]\[ y = -5(1) - 2 = -5 - 2 = -7 \][/tex]
3. Period and Quarter Points:
- Quarter points are intervals at \(\frac{period}{4}\). Period is \(6\pi\), therefore \(\frac{6\pi}{4} = \frac{3\pi}{2}\).
At each quarter interval from \(x = 0\):
- \(x = 0\): \(y = -2\)
- \(x = \frac{3\pi}{2}\): \(y\)-minimum \((=-7)\)
- \(x = 3\pi\): \(y = -2\)
- \(x = \frac{9\pi}{2}\): \(y\)-maximum \((=3)\)
- \(x = 6\pi\): \(y = -2\) (one full period)
### Plotting Points
1. Plot \((0, -2)\)
2. Plot \(\left(\frac{3\pi}{2}, -7\right)\)
3. Plot \((3\pi, -2)\)
4. Plot \(\left(\frac{9\pi}{2}, 3\right)\)
5. Plot \((6\pi, -2)\)
### Sketch the Curve
Connect these points smoothly to form the sinusoidal curve.
Using the sine tool, you can plot the function by placing your first point at (0, -2), and your close point for maximum or minimum nearby, following the instructions provided.
1. Amplitude (A): The amplitude is the distance from the midline to the maximum or minimum value of the function. Here, the amplitude is 5.
2. Period (B): The period of the sine function is the distance (in the x-axis) over which the function completes one full cycle. It's given as \(6\pi\). The formula for period \(T\) of the function \(y = A \sin(Bx)\) is \(T = \frac{2\pi}{|B|}\):
[tex]\[ 6\pi = \frac{2\pi}{|B|} \][/tex]
Solving for \(B\):
[tex]\[ |B| = \frac{2\pi}{6\pi} = \frac{1}{3} \][/tex]
3. Midline (D): The midline is the horizontal axis around which the sine function oscillates. Here, it’s given as -2. So, \(D = -2\).
4. Reflection: The graph is a reflection over the x-axis, which means we multiply the sine function by -1. If the original sine function is \(y = \sin(x)\), the reflected function is \(y = -\sin(x)\).
Considering all of these together, our function takes the form:
[tex]\[ y = -5 \sin\left(\frac{1}{3}x\right) - 2 \][/tex]
To graph this function, follow these steps:
### Finding Key Points:
1. Y-Intercept: At \(x = 0\),
[tex]\[ y = -5 \sin\left(\frac{1}{3} \cdot 0\right) - 2 = -5 \sin(0) - 2 = -2 \][/tex]
So the point (0, -2) is on the midline.
2. Next Key Point (Minimum or Maximum):
The function is:
[tex]\[ y = -5 \sin\left(\frac{1}{3}x\right) - 2 \][/tex]
The minimum value of \(\sin(x)\) is -1 and the maximum value is 1. For this reflected and shifted function:
- Maximum: When \(\sin\left(\frac{1}{3}x\right) = -1\),
[tex]\[ y = -5(-1) - 2 = 5 - 2 = 3 \][/tex]
- Minimum: When \(\sin\left(\frac{1}{3}x\right) = 1\),
[tex]\[ y = -5(1) - 2 = -5 - 2 = -7 \][/tex]
3. Period and Quarter Points:
- Quarter points are intervals at \(\frac{period}{4}\). Period is \(6\pi\), therefore \(\frac{6\pi}{4} = \frac{3\pi}{2}\).
At each quarter interval from \(x = 0\):
- \(x = 0\): \(y = -2\)
- \(x = \frac{3\pi}{2}\): \(y\)-minimum \((=-7)\)
- \(x = 3\pi\): \(y = -2\)
- \(x = \frac{9\pi}{2}\): \(y\)-maximum \((=3)\)
- \(x = 6\pi\): \(y = -2\) (one full period)
### Plotting Points
1. Plot \((0, -2)\)
2. Plot \(\left(\frac{3\pi}{2}, -7\right)\)
3. Plot \((3\pi, -2)\)
4. Plot \(\left(\frac{9\pi}{2}, 3\right)\)
5. Plot \((6\pi, -2)\)
### Sketch the Curve
Connect these points smoothly to form the sinusoidal curve.
Using the sine tool, you can plot the function by placing your first point at (0, -2), and your close point for maximum or minimum nearby, following the instructions provided.
