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

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 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 scaling.

  • 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 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 point is \( (270^\circ, - 2) \).

Step3: Plot the graph

  • Draw the x - axis (labeled "Angle (degrees)") and y - axis (labeled "y - value").
  • Mark the x - intercepts \( (0^\circ, 0) \), \( (180^\circ, 0) \), \( (360^\circ, 0) \), the maximum point \( (90^\circ, 2) \) and the minimum point \( (270^\circ, - 2) \).
  • Connect these points with a smooth curve, following the shape of the sine wave, with amplitude 2 and period \( 360^\circ \).

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, with x - axis labeled "Angle (degrees)" and y - axis labeled "y - value".