Graph y=2sin(1/4x)

Problem

y=2*sin(1/4*x)

Solution

1. Identify the amplitude by looking at the coefficient a in the general form y=a*sin(b*x) Here, a=2 so the amplitude is |2|=2 This means the graph oscillates between y=2 and y=−2

2. Determine the period using the formula P=(2*π)/b In this equation, b=1/4

3. Calculate the period value:

P=(2*π)/(1/4)

P=8*π

4. Find the key points by dividing the period into four equal intervals. The increment is (8*π)/4=2*π The xcoordinates for one cycle starting at the origin are 0,2*π,4*π,6*π,8*π

5. Evaluate the function at these key xvalues to find the ycoordinates:

• At x=0 y=2*sin(0)=0

• At x=2*π y=2*sin(π/2)=2

• At x=4*π y=2*sin(π)=0

• At x=6*π y=2*sin(3*π/2)=−2

• At x=8*π y=2*sin(2*π)=0

6. Plot the points (0,0),(2*π,2),(4*π,0),(6*π,−2),(8*π,0) and connect them with a smooth sine wave.

Final Answer

y=2*sin(1/4*x)* has amplitude *2* and period *8*π

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