Graph y=3cos(2/3x)

Problem

y=3*cos(2/3*x)

Solution

1. Identify the amplitude by looking at the coefficient A in the form y=A*cos(B*x) Here, A=3 so the amplitude is 3 This means the graph oscillates between y=3 and y=−3

2. Determine the period using the formula P=(2*π)/B In this function, B=2/3

P=(2*π)/2/3

P=3*π

3. Find the key points by dividing the period into four equal intervals. The interval width is (3*π)/4 The xcoordinates for one cycle starting at x=0 are 0 (3*π)/4 (3*π)/2 (9*π)/4 and 3*π

4. Evaluate the function at these key xvalues to find the ycoordinates.
   For x=0 y=3*cos(0)=3
   For x=(3*π)/4 y=3*cos(π/2)=0
   For x=(3*π)/2 y=3*cos(π)=−3
   For x=(9*π)/4 y=3*cos((3*π)/2)=0
   For x=3*π y=3*cos(2*π)=3

5. Sketch the curve by plotting the points (0,3) ((3*π)/4,0) ((3*π)/2,−3) ((9*π)/4,0) and (3*π,3) and connecting them with a smooth wave.

Final Answer

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

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