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 standard sine function

The standard sine function \( y = \sin(x) \) has a period of \( 360^\circ \), amplitude \( 1 \), key points at \( (0^\circ, 0) \), \( (90^\circ, 1) \), \( (180^\circ, 0) \), \( (270^\circ, -1) \), \( (360^\circ, 0) \).

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

The function \( y = 2\sin(x) \) is a vertical stretch of the standard sine function by a factor of \( 2 \). This means the amplitude becomes \( 2 \) (since the amplitude of \( A\sin(x) \) is \( |A| \), here \( A = 2 \)). The period remains \( 360^\circ \) because there is no horizontal scaling (the coefficient of \( x \) is \( 1 \)).

Step3: Find key points for \( y = 2\sin(x) \)

• At \( x = 0^\circ \): \( y = 2\sin(0^\circ)= 0 \), so the point is \( (0^\circ, 0) \).
• At \( x = 90^\circ \): \( y = 2\sin(90^\circ)= 2\times1 = 2 \), so the point is \( (90^\circ, 2) \) (this is the maximum point).
• At \( x = 180^\circ \): \( y = 2\sin(180^\circ)= 2\times0 = 0 \), so the point is \( (180^\circ, 0) \).
• At \( x = 270^\circ \): \( y = 2\sin(270^\circ)= 2\times(-1)= - 2 \), so the point is \( (270^\circ, - 2) \) (this is the minimum point).
• At \( x = 360^\circ \): \( y = 2\sin(360^\circ)= 2\times0 = 0 \), so the point is \( (360^\circ, 0) \).

Step4: Plot the graph

1. Draw the coordinate plane with the x - axis labeled "Angle (degrees)" and the y - axis labeled "y - value".
2. Mark the key points \( (0^\circ, 0) \), \( (90^\circ, 2) \), \( (180^\circ, 0) \), \( (270^\circ, - 2) \), \( (360^\circ, 0) \).
3. Connect these points with a smooth curve, following the shape of a sine wave, keeping in mind the period and amplitude. The curve should start at \( (0^\circ, 0) \), rise to \( (90^\circ, 2) \), fall back to \( (180^\circ, 0) \), fall to \( (270^\circ, - 2) \), and then rise back to \( (360^\circ, 0) \).

Answer:

The key points for the function \( y = 2\sin(x) \) on \( 0^\circ\leq x\leq360^\circ \) are \( (0^\circ, 0) \), \( (90^\circ, 2) \), \( (180^\circ, 0) \), \( (270^\circ, - 2) \), \( (360^\circ, 0) \). To graph the function, plot these points on a coordinate plane with the x - axis as "Angle (degrees)" and y - axis as "y - value" and connect them with a smooth sine - shaped curve.