Question

consider the function $y = 2\sin(x)$ for $0^\circ \leq x \leq 360^\circ$.

1. graph the function:

plot the function $y = 2\sin(x)$ on a coordinate plane. label the x-axis as \angle (degrees)\ and the y-axis as \y-value\. mark the coordinates of all key points where the graph intersects the x-axis, reaches its maximum, or minimum within the given interval.

Explanation:

Step1: Recall the parent sine function

The parent function is \( y = \sin(x) \), which has a period of \( 360^\circ \), amplitude 1, x - intercepts at \( 0^\circ, 180^\circ, 360^\circ \), maximum at \( (90^\circ, 1) \) and minimum at \( (270^\circ, - 1) \) in the interval \( 0^\circ\leq x\leq360^\circ \).

Step2: Analyze the transformed function \( y = 2\sin(x) \)

For the function \( y = A\sin(x) \), the amplitude is \( |A| \). Here \( A = 2 \), so the amplitude is 2. The period remains \( 360^\circ \) (since there is no horizontal stretch or compression, \( B = 1 \) in \( y = A\sin(Bx) \)).

Step3: Find key points

• X - intercepts: The graph of \( y=\sin(x) \) crosses the x - axis at \( x = 0^\circ,180^\circ,360^\circ \). For \( y = 2\sin(x) \), when \( y = 0 \), \( 2\sin(x)=0\Rightarrow\sin(x) = 0\Rightarrow x = 0^\circ,180^\circ,360^\circ \). So the points are \( (0^\circ,0) \), \( (180^\circ,0) \), \( (360^\circ,0) \).
• Maximum point: The maximum value of \( \sin(x) \) is 1, which occurs at \( x = 90^\circ \). For \( y = 2\sin(x) \), when \( x = 90^\circ \), \( y=2\times1 = 2 \). So the maximum point is \( (90^\circ,2) \).
• Minimum point: The minimum value of \( \sin(x) \) is - 1, which occurs at \( x = 270^\circ \). For \( y = 2\sin(x) \), when \( x = 270^\circ \), \( y = 2\times(- 1)=-2 \). So the minimum point is \( (270^\circ,-2) \).

Step4: Plot the graph

• Draw the x - axis and label it "Angle (degrees)" with values from \( 0^\circ \) to \( 360^\circ \) and the y - axis and label it "y - value" with values from - 2 to 2.
• Plot the points \( (0^\circ,0) \), \( (90^\circ,2) \), \( (180^\circ,0) \), \( (270^\circ,-2) \), \( (360^\circ,0) \).
• Connect the points with a smooth curve, following the shape of a sine wave, starting at \( (0^\circ,0) \), rising to \( (90^\circ,2) \), falling to \( (180^\circ,0) \), falling further to \( (270^\circ,-2) \), and then rising back to \( (360^\circ,0) \).

Answer:

To graph \( y = 2\sin(x) \) for \( 0^\circ\leq x\leq360^\circ \):

• X - axis label: "Angle (degrees)" (scale from \( 0^\circ \) to \( 360^\circ \))
• Y - axis label: "y - value" (scale from - 2 to 2)
• Key points:
• Intersections with x - axis: \( (0^\circ,0) \), \( (180^\circ,0) \), \( (360^\circ,0) \)
• Maximum point: \( (90^\circ,2) \)
• Minimum point: \( (270^\circ,-2) \)

Connect these points with a smooth sinusoidal curve.