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 transformed function \( y = 2\sin(x) \)

For the function \( y = A\sin(x) \), where \( A = 2 \) in our case, the amplitude is \( |A| \). So the amplitude of \( y = 2\sin(x) \) is \( 2 \). The period remains the same as the parent sine function, i.e., \( 360^\circ \) since there is no horizontal transformation (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 \( 0^\circ\leq x\leq360^\circ \), \( x = 0^\circ, 180^\circ, 360^\circ \). So the points 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., \( x = 90^\circ \). So the maximum point 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., \( x = 270^\circ \). 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".
• Mark 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 \( 360^\circ \). The curve starts at \( (0^\circ, 0) \), rises to \( (90^\circ, 2) \), falls back to \( (180^\circ, 0) \), continues to fall to \( (270^\circ, - 2) \) and then rises back to \( (360^\circ, 0) \).

Answer:

The key points for the graph of \( y = 2\sin(x) \) in the interval \( 0^\circ\leq x\leq360^\circ \) 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 and connecting them with a smooth curve.