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

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 of \( y = 2\sin(x) \) is 2. The period remains \( 360^\circ \) since there is no horizontal scaling (the coefficient of \( x \) is 1).

Step3: Find key points

• x - intercepts: Set \( y = 0 \), then \( 2\sin(x)=0\Rightarrow\sin(x) = 0 \). In the interval \( 0^\circ\leq x\leq360^\circ \), \( x = 0^\circ, 180^\circ, 360^\circ \). The coordinates are \( (0^\circ, 0) \), \( (180^\circ, 0) \), \( (360^\circ, 0) \).
• Maximum point: The maximum value of \( \sin(x) \) is 1, so the maximum value of \( y = 2\sin(x) \) is \( 2\times1=2 \). This occurs when \( \sin(x) = 1 \), i.e., \( x = 90^\circ \). The coordinate is \( (90^\circ, 2) \).
• Minimum point: The minimum value of \( \sin(x) \) is - 1, so the minimum value of \( y = 2\sin(x) \) is \( 2\times(- 1)=-2 \). This occurs when \( \sin(x)=-1 \), i.e., \( x = 270^\circ \). The coordinate is \( (270^\circ, - 2) \).

Step4: Plot the graph

• Draw the x - axis labeled "Angle (degrees)" with values from \( 0^\circ \) to \( 360^\circ \) and the y - axis labeled "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:

The graph of \( y = 2\sin(x) \) for \( 0^\circ\leq x\leq360^\circ \) has x - intercepts at \( (0^\circ, 0) \), \( (180^\circ, 0) \), \( (360^\circ, 0) \), a maximum at \( (90^\circ, 2) \) and a minimum at \( (270^\circ, - 2) \). The graph is a sine - wave with amplitude 2, period \( 360^\circ \) passing through these key points.