Answer :
Certainly! Let's go step-by-step to graph one cycle of the function [tex]\( y = 2\sin{x} - 3 \)[/tex].
### Step 1: Identify Key Properties of the Sine Function
The standard sine function, [tex]\( \sin{x} \)[/tex], has certain key properties:
- Amplitude: The maximum value is 1, and the minimum value is -1.
- Period: The function completes one cycle from [tex]\( 0 \)[/tex] to [tex]\( 2\pi \)[/tex].
- Midline: The middle of the wave is at [tex]\( y = 0 \)[/tex].
- Zero crossings: Occur at [tex]\( x = 0, \pi, 2\pi, \ldots \)[/tex].
### Step 2: Modify the Sine Function Based on Given Equation
The equation we need to graph is [tex]\( y = 2\sin{x} - 3 \)[/tex]. Let's break it down:
- Amplitude: The coefficient 2 in front of the sine function scales the amplitude. So, the new amplitude is 2.
- Vertical Shift: The -3 outside the sine function vertically shifts the entire graph down by 3 units.
- Period and Phase Shift: These properties are unchanged from the basic [tex]\( \sin{x} \)[/tex] function, as there are no coefficients modifying [tex]\( x \)[/tex].
### Step 3: Determine Key Points
#### Amplitude
The amplitude is the height from the midline to the maximum or minimum value of the function. For [tex]\( y = 2\sin{x} \)[/tex]:
- Maximum value = 2
- Minimum value = -2
With the vertical shift:
- New maximum value = [tex]\( 2 - 3 = -1 \)[/tex]
- New minimum value = [tex]\( -2 - 3 = -5 \)[/tex]
#### Midline
The midline of the function is [tex]\( y = -3 \)[/tex].
#### Zero Crossings
To find the zero crossings of [tex]\( y = 2\sin{x} - 3 \)[/tex]:
[tex]\[ 2\sin{x} - 3 = 0 \implies 2\sin{x} = 3 \implies \sin{x} = \frac{3}{2} \][/tex]
However, [tex]\( \sin{x} \)[/tex] cannot equal [tex]\(\frac{3}{2}\)[/tex] because the sine of an angle cannot be greater than 1. So, rather, we use the usual zero crossings of [tex]\( \sin{x} \)[/tex] and adjust accordingly:
[tex]\[ \sin{x} = 0 \text{ when } x = 0, \pi, 2\pi, \ldots \][/tex]
Thus, [tex]\( y = 2\sin{x} - 3 \)[/tex] will be -3 at these points.
### Step 4: Plotting the Graph
Let's plot the function from [tex]\( 0 \)[/tex] to [tex]\( 2\pi \)[/tex] and mark key points:
- At [tex]\( x = 0 \)[/tex]: [tex]\( y = 2\sin{0} - 3 = -3 \)[/tex]
- At [tex]\( x = \frac{\pi}{2} \)[/tex]: [tex]\( y = 2\sin{\frac{\pi}{2}} - 3 = 2 - 3 = -1 \)[/tex]
- At [tex]\( x = \pi \)[/tex]: [tex]\( y = 2\sin{\pi} - 3 = -3 \)[/tex]
- At [tex]\( x = \frac{3\pi}{2} \)[/tex]: [tex]\( y = 2\sin{\frac{3\pi}{2}} - 3 = -2 - 3 = -5 \)[/tex]
- At [tex]\( x = 2\pi \)[/tex]: [tex]\( y = 2\sin{2\pi} - 3 = -3 \)[/tex]
### Step 5: Sketch the Graph
1. Draw the x-axis from [tex]\( 0 \)[/tex] to [tex]\( 2\pi \)[/tex] and the y-axis.
2. Draw a horizontal line at [tex]\( y = -3 \)[/tex] to represent the midline.
3. Plot the key points and draw a smooth sinusoidal wave passing through these points. Remember that:
- The wave starts at (0, -3),
- Reaches a maximum of -1 at [tex]\( \frac{\pi}{2} \)[/tex],
- Returns to -3 at [tex]\(\pi\)[/tex],
- Reaches a minimum of -5 at [tex]\( \frac{3\pi}{2} \)[/tex],
- And returns to -3 at [tex]\( 2\pi \)[/tex].
### Final Graph
This results in a sinusoidal function with an amplitude of 2, vertically shifted down by 3 units, completing one full cycle from [tex]\( 0 \)[/tex] to [tex]\( 2\pi \)[/tex]. The maximum value is -1, the minimum value is -5, and it oscillates around the midline [tex]\( y = -3 \)[/tex].
Drawing this graph will provide a visual representation of the function [tex]\( y = 2\sin{x} - 3 \)[/tex].
### Step 1: Identify Key Properties of the Sine Function
The standard sine function, [tex]\( \sin{x} \)[/tex], has certain key properties:
- Amplitude: The maximum value is 1, and the minimum value is -1.
- Period: The function completes one cycle from [tex]\( 0 \)[/tex] to [tex]\( 2\pi \)[/tex].
- Midline: The middle of the wave is at [tex]\( y = 0 \)[/tex].
- Zero crossings: Occur at [tex]\( x = 0, \pi, 2\pi, \ldots \)[/tex].
### Step 2: Modify the Sine Function Based on Given Equation
The equation we need to graph is [tex]\( y = 2\sin{x} - 3 \)[/tex]. Let's break it down:
- Amplitude: The coefficient 2 in front of the sine function scales the amplitude. So, the new amplitude is 2.
- Vertical Shift: The -3 outside the sine function vertically shifts the entire graph down by 3 units.
- Period and Phase Shift: These properties are unchanged from the basic [tex]\( \sin{x} \)[/tex] function, as there are no coefficients modifying [tex]\( x \)[/tex].
### Step 3: Determine Key Points
#### Amplitude
The amplitude is the height from the midline to the maximum or minimum value of the function. For [tex]\( y = 2\sin{x} \)[/tex]:
- Maximum value = 2
- Minimum value = -2
With the vertical shift:
- New maximum value = [tex]\( 2 - 3 = -1 \)[/tex]
- New minimum value = [tex]\( -2 - 3 = -5 \)[/tex]
#### Midline
The midline of the function is [tex]\( y = -3 \)[/tex].
#### Zero Crossings
To find the zero crossings of [tex]\( y = 2\sin{x} - 3 \)[/tex]:
[tex]\[ 2\sin{x} - 3 = 0 \implies 2\sin{x} = 3 \implies \sin{x} = \frac{3}{2} \][/tex]
However, [tex]\( \sin{x} \)[/tex] cannot equal [tex]\(\frac{3}{2}\)[/tex] because the sine of an angle cannot be greater than 1. So, rather, we use the usual zero crossings of [tex]\( \sin{x} \)[/tex] and adjust accordingly:
[tex]\[ \sin{x} = 0 \text{ when } x = 0, \pi, 2\pi, \ldots \][/tex]
Thus, [tex]\( y = 2\sin{x} - 3 \)[/tex] will be -3 at these points.
### Step 4: Plotting the Graph
Let's plot the function from [tex]\( 0 \)[/tex] to [tex]\( 2\pi \)[/tex] and mark key points:
- At [tex]\( x = 0 \)[/tex]: [tex]\( y = 2\sin{0} - 3 = -3 \)[/tex]
- At [tex]\( x = \frac{\pi}{2} \)[/tex]: [tex]\( y = 2\sin{\frac{\pi}{2}} - 3 = 2 - 3 = -1 \)[/tex]
- At [tex]\( x = \pi \)[/tex]: [tex]\( y = 2\sin{\pi} - 3 = -3 \)[/tex]
- At [tex]\( x = \frac{3\pi}{2} \)[/tex]: [tex]\( y = 2\sin{\frac{3\pi}{2}} - 3 = -2 - 3 = -5 \)[/tex]
- At [tex]\( x = 2\pi \)[/tex]: [tex]\( y = 2\sin{2\pi} - 3 = -3 \)[/tex]
### Step 5: Sketch the Graph
1. Draw the x-axis from [tex]\( 0 \)[/tex] to [tex]\( 2\pi \)[/tex] and the y-axis.
2. Draw a horizontal line at [tex]\( y = -3 \)[/tex] to represent the midline.
3. Plot the key points and draw a smooth sinusoidal wave passing through these points. Remember that:
- The wave starts at (0, -3),
- Reaches a maximum of -1 at [tex]\( \frac{\pi}{2} \)[/tex],
- Returns to -3 at [tex]\(\pi\)[/tex],
- Reaches a minimum of -5 at [tex]\( \frac{3\pi}{2} \)[/tex],
- And returns to -3 at [tex]\( 2\pi \)[/tex].
### Final Graph
This results in a sinusoidal function with an amplitude of 2, vertically shifted down by 3 units, completing one full cycle from [tex]\( 0 \)[/tex] to [tex]\( 2\pi \)[/tex]. The maximum value is -1, the minimum value is -5, and it oscillates around the midline [tex]\( y = -3 \)[/tex].
Drawing this graph will provide a visual representation of the function [tex]\( y = 2\sin{x} - 3 \)[/tex].
