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 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 transformation for \( y = 2\sin(x) \)

The function \( y = 2\sin(x) \) is a vertical stretch of the parent sine function by a factor of \( 2 \). The period remains \( 360^\circ \) (since there is no horizontal scaling), the amplitude is \( 2 \) (the absolute value of the coefficient of \( \sin(x) \)).

Step3: Find key points

• x - intercepts: Set \( y = 0 \), so \( 2\sin(x)=0\Rightarrow\sin(x) = 0 \). 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 \( y = 2\times1=2 \). The \( x \) - value for maximum of \( \sin(x) \) is \( 90^\circ \). So the point is \( (90^\circ, 2) \).
• Minimum point: The minimum value of \( \sin(x) \) is \( - 1 \), so \( y=2\times(- 1)=-2 \). The \( x \) - value for minimum of \( \sin(x) \) is \( 270^\circ \). So the point is \( (270^\circ, - 2) \).

Step4: Plot the points and draw the curve

1. Draw the coordinate plane. Label the x - axis as "Angle (degrees)" with scale marks at \( 0^\circ, 90^\circ, 180^\circ, 270^\circ, 360^\circ \) and the y - axis as "y - value" with scale marks at \( - 2,-1,0,1,2 \).
2. Plot the points \( (0^\circ, 0) \), \( (90^\circ, 2) \), \( (180^\circ, 0) \), \( (270^\circ, - 2) \), \( (360^\circ, 0) \).
3. Draw a smooth curve passing through these points, 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 rising back to \( (360^\circ, 0) \).

Answer:

The graph of \( y = 2\sin(x) \) for \( 0^\circ\leq x\leq360^\circ \) is a sine - shaped curve with x - intercepts at \( (0^\circ, 0) \), \( (180^\circ, 0) \), \( (360^\circ, 0) \), a maximum at \( (90^\circ, 2) \) and a minimum at \( (270^\circ, - 2) \), plotted on a coordinate plane with x - axis labeled "Angle (degrees)" and y - axis labeled "y - value".