Graph y=5sin(2x)

Problem

y=5*sin(2*x)

Solution

1. Identify the amplitude by looking at the coefficient in front of the sine function. The amplitude A is |5| which means the graph oscillates between y=5 and y=−5

2. Determine the period using the formula P=(2*π)/b where b is the coefficient of x Here b=2 so the period is (2*π)/2=π

3. Calculate the phase shift and vertical shift. Since there are no constants added inside or outside the sine function, the phase shift is 0 and the vertical shift is 0

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

5. Evaluate the function at these key points to find the ycoordinates.

y(0)=5*sin(0)=0

y(π/4)=5*sin(π/2)=5

y(π/2)=5*sin(π)=0

y((3*π)/4)=5*sin((3*π)/2)=−5

y(π)=5*sin(2*π)=0

6. Plot the points (0,0) (π/4,5) (π/2,0) ((3*π)/4,−5) and (π,0) and connect them with a smooth sine wave.

Final Answer

y=5*sin(2*x)* has amplitude *5* and period *π

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