Find the best solutions to your questions at Westonci.ca, the premier Q&A platform with a community of knowledgeable experts. Join our Q&A platform and get accurate answers to all your questions from professionals across multiple disciplines. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
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.
Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.
