Graph y=2sin(3x)

Problem

y=2*sin(3*x)

Solution

1. Identify the amplitude by looking at the coefficient of the sine function. In the form y=A*sin(B*x) the amplitude is |A|=|2|=2 This means the graph oscillates between y=2 and y=−2

2. Determine the period using the formula P=(2*π)/B Here, B=3 so the period is P=(2*π)/3 This is the horizontal distance required for the function to complete one full cycle.

3. Calculate the key points by dividing the period into four equal intervals. The interval width is P/4=(2*π/3)/4=π/6 The xcoordinates for one cycle starting at the origin are 0 π/6 π/3 π/2 and (2*π)/3

4. Evaluate the function at these key xvalues to find the ycoordinates.

y(0)=2*sin(0)=0

y(π/6)=2*sin(π/2)=2

y(π/3)=2*sin(π)=0

y(π/2)=2*sin((3*π)/2)=−2

y((2*π)/3)=2*sin(2*π)=0

5. Plot the points (0,0) (π/6,2) (π/3,0) (π/2,−2) and ((2*π)/3,0) on a coordinate plane and connect them with a smooth sine wave.

Final Answer

y=2*sin(3*x)* has amplitude *2* and period (2*π)/3

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