Question

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 basic sine function

The basic sine function \( y = \sin(x) \) has a period of \( 360^\circ \), amplitude \( 1 \), x - intercepts at \( x = 0^\circ, 180^\circ, 360^\circ \), maximum value \( 1 \) at \( x = 90^\circ \), and minimum value \( - 1 \) at \( x = 270^\circ \).

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

For the function \( y = 2\sin(x) \), the amplitude is \( 2 \) (since the amplitude of \( A\sin(x) \) is \( |A| \), here \( A = 2 \)) and the period remains \( 360^\circ \) (because there is no horizontal scaling, the coefficient of \( x \) inside the sine function is \( 1 \)).

Step3: Find key points

• X - intercepts: Set \( y=0 \), then \( 2\sin(x)=0\Rightarrow\sin(x) = 0 \). In the interval \( 0^\circ\leq x\leq360^\circ \), \( x = 0^\circ, 180^\circ, 360^\circ \). So the points are \( (0^\circ, 0) \), \( (180^\circ, 0) \), \( (360^\circ, 0) \).
• Maximum point: The sine function reaches its maximum value of \( 1 \) at \( x = 90^\circ \) for \( y=\sin(x) \). For \( y = 2\sin(x) \), when \( x = 90^\circ \), \( y=2\times1 = 2 \). So the maximum point is \( (90^\circ, 2) \).
• Minimum point: The sine function reaches its minimum value of \( - 1 \) at \( x = 270^\circ \) for \( y=\sin(x) \). 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 coordinate plane. Label the x - axis as "Angle (degrees)" and the y - axis as "y - value".
• Plot the key 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 the sine wave, with amplitude \( 2 \) and period \( 360^\circ \).

Answer:

The key points of the function \( y = 2\sin(x) \) for \( 0^\circ\leq x\leq360^\circ \) are \( (0^\circ, 0) \), \( (90^\circ, 2) \), \( (180^\circ, 0) \), \( (270^\circ, - 2) \), \( (360^\circ, 0) \). The graph is a sine - wave with amplitude \( 2 \), period \( 360^\circ \), passing through these key points, with the x - axis labeled "Angle (degrees)" and the y - axis labeled "y - value".