Answer :
To solve the problem of graphing the function [tex]\( y = 3 \sin(x) - 1 \)[/tex], let’s break it down step-by-step:
1. Understand the Basic Function: The sine function, [tex]\( y = \sin(x) \)[/tex], oscillates between -1 and 1 with a period of [tex]\( 2\pi \)[/tex].
2. Apply the Amplitude Change: The coefficient 3 in [tex]\( 3\sin(x) \)[/tex] changes the amplitude of the sine wave. Instead of oscillating between -1 and 1, it now oscillates between -3 and 3. So the new function is [tex]\( y = 3\sin(x) \)[/tex].
3. Apply the Vertical Shift: The "-1" in the expression [tex]\( 3\sin(x) - 1 \)[/tex] shifts the entire graph down by 1 unit. This means the new range of the function is from [tex]\(-4\)[/tex] to [tex]\(2\)[/tex] instead of [tex]\(-3\)[/tex] to [tex]\(3\)[/tex].
4. Determine Critical Points: Let’s identify some key points in one period (from [tex]\( -2\pi \)[/tex] to [tex]\( 2\pi \)[/tex]):
- At [tex]\( x = 0 \)[/tex], [tex]\( y = 3\sin(0) - 1 = -1 \)[/tex]
- At [tex]\( x = \pi/2 \)[/tex], [tex]\( y = 3\sin(\pi/2) - 1 = 3 - 1 = 2 \)[/tex]
- At [tex]\( x = \pi \)[/tex], [tex]\( y = 3\sin(\pi) - 1 = -1 \)[/tex]
- At [tex]\( x = 3\pi/2 \)[/tex], [tex]\( y = 3\sin(3\pi/2) - 1 = -3 - 1 = -4 \)[/tex]
- At [tex]\( x = 2\pi \)[/tex], [tex]\( y = 3\sin(2\pi) - 1 = -1 \)[/tex]
5. Sketch the Graph: Now, knowing the key points, we can plot these values and the other points between them to form the smooth, sinusoidal curve.
In the interval from [tex]\( -2\pi \)[/tex] to [tex]\( 2\pi \)[/tex], you can repeat this pattern:
- From [tex]\( x = -2\pi \)[/tex] to [tex]\( x = 0 \)[/tex]: the function will behave symmetrically as it does from [tex]\( 0 \)[/tex] to [tex]\( 2\pi \)[/tex] due to the periodic nature of sine.
- Calculate and plot the points:
- At [tex]\( x = \frac{-3\pi}{2} \)[/tex]: [tex]\( y = -4 \)[/tex]
- At [tex]\( x = \pi \)[/tex]: [tex]\( y = -1 \)[/tex]
The graph of [tex]\( y = 3\sin(x) - 1 \)[/tex] is a vertically shifted sine wave with an amplitude of 3, oscillating between -4 and 2, and a vertical shift down by 1 unit.
The numerical output confirms the values:
```
(array([-6.28318531, -6.25169064, -6.22019598, ... , 6.22019598, 6.25169064, 6.28318531]), array([-1.00000000e+00, -9.05531630e-01, -8.11156956e-01, ... , -8.11156956e-01, -9.05531630e-01, -1.00000000e+00]))
```
To visualize this graph:
1. Start at -6.28318531 (approximately [tex]\( -2\pi \)[/tex]) and note the [tex]\( y \)[/tex]-value is approximately -1.
2. As you move rightwards (increase [tex]\( x \)[/tex]), the value of [tex]\( y \)[/tex] increases, reaching a peak at [tex]\( \pi/2 \)[/tex] (approximately 1.5708) with [tex]\( y \)[/tex] near 2.
3. The function decreases again, reaching around -4 at [tex]\( 3\pi/2 \)[/tex] (approximately 4.7124), then increases back to around -1 at [tex]\( 2\pi \)[/tex] (approximately 6.2832).
By plotting multiple such points and their corresponding y-values, you can draw a smooth sinusoidal curve. The overall graph will show waves oscillating vertically between -4 and 2, centered around -1, with a period of [tex]\( 2\pi \)[/tex].
1. Understand the Basic Function: The sine function, [tex]\( y = \sin(x) \)[/tex], oscillates between -1 and 1 with a period of [tex]\( 2\pi \)[/tex].
2. Apply the Amplitude Change: The coefficient 3 in [tex]\( 3\sin(x) \)[/tex] changes the amplitude of the sine wave. Instead of oscillating between -1 and 1, it now oscillates between -3 and 3. So the new function is [tex]\( y = 3\sin(x) \)[/tex].
3. Apply the Vertical Shift: The "-1" in the expression [tex]\( 3\sin(x) - 1 \)[/tex] shifts the entire graph down by 1 unit. This means the new range of the function is from [tex]\(-4\)[/tex] to [tex]\(2\)[/tex] instead of [tex]\(-3\)[/tex] to [tex]\(3\)[/tex].
4. Determine Critical Points: Let’s identify some key points in one period (from [tex]\( -2\pi \)[/tex] to [tex]\( 2\pi \)[/tex]):
- At [tex]\( x = 0 \)[/tex], [tex]\( y = 3\sin(0) - 1 = -1 \)[/tex]
- At [tex]\( x = \pi/2 \)[/tex], [tex]\( y = 3\sin(\pi/2) - 1 = 3 - 1 = 2 \)[/tex]
- At [tex]\( x = \pi \)[/tex], [tex]\( y = 3\sin(\pi) - 1 = -1 \)[/tex]
- At [tex]\( x = 3\pi/2 \)[/tex], [tex]\( y = 3\sin(3\pi/2) - 1 = -3 - 1 = -4 \)[/tex]
- At [tex]\( x = 2\pi \)[/tex], [tex]\( y = 3\sin(2\pi) - 1 = -1 \)[/tex]
5. Sketch the Graph: Now, knowing the key points, we can plot these values and the other points between them to form the smooth, sinusoidal curve.
In the interval from [tex]\( -2\pi \)[/tex] to [tex]\( 2\pi \)[/tex], you can repeat this pattern:
- From [tex]\( x = -2\pi \)[/tex] to [tex]\( x = 0 \)[/tex]: the function will behave symmetrically as it does from [tex]\( 0 \)[/tex] to [tex]\( 2\pi \)[/tex] due to the periodic nature of sine.
- Calculate and plot the points:
- At [tex]\( x = \frac{-3\pi}{2} \)[/tex]: [tex]\( y = -4 \)[/tex]
- At [tex]\( x = \pi \)[/tex]: [tex]\( y = -1 \)[/tex]
The graph of [tex]\( y = 3\sin(x) - 1 \)[/tex] is a vertically shifted sine wave with an amplitude of 3, oscillating between -4 and 2, and a vertical shift down by 1 unit.
The numerical output confirms the values:
```
(array([-6.28318531, -6.25169064, -6.22019598, ... , 6.22019598, 6.25169064, 6.28318531]), array([-1.00000000e+00, -9.05531630e-01, -8.11156956e-01, ... , -8.11156956e-01, -9.05531630e-01, -1.00000000e+00]))
```
To visualize this graph:
1. Start at -6.28318531 (approximately [tex]\( -2\pi \)[/tex]) and note the [tex]\( y \)[/tex]-value is approximately -1.
2. As you move rightwards (increase [tex]\( x \)[/tex]), the value of [tex]\( y \)[/tex] increases, reaching a peak at [tex]\( \pi/2 \)[/tex] (approximately 1.5708) with [tex]\( y \)[/tex] near 2.
3. The function decreases again, reaching around -4 at [tex]\( 3\pi/2 \)[/tex] (approximately 4.7124), then increases back to around -1 at [tex]\( 2\pi \)[/tex] (approximately 6.2832).
By plotting multiple such points and their corresponding y-values, you can draw a smooth sinusoidal curve. The overall graph will show waves oscillating vertically between -4 and 2, centered around -1, with a period of [tex]\( 2\pi \)[/tex].
