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# Draw the graph y = sin(3pi x)

The graph of y = A sin Bx has the property
(1). amplitude = |A|
(2). period = 2pi/B
For the function of y = sin(3pi x),
since A = 1, so its amplitude is |1|
since B = 3pi, so its period = 2pi/B = 2pi ÷ 3pi = 2/3
Thus, its amplitude is 1 and period is 2/3
Find the five points in one period
one period is 2/3, half period is 1/3, quarter period is 1/6
divide the five points equally in a period [0, 2/3]
so the five points in x-axis are:
x1 = 0
x2 = 1/6
x3 = 1/3
x4 = 1/3 + 1/6 = 2/6 + 1/6 = 3/6 = 1/2
x5 = 1/2 + 1/6 = 3/6 + 1/6 = 4/6 = 2/3
so the five points in xy-plane are (0, ?), (1/6, ?), (1/3, ?), (1/2, ?), (2/3, ?)
Now find the values of the function y = sin(3pi x) in the five points
when x = 0, y = sin(3pi x) = sin(3pi × 0) = sin(0) = 0, so the point is (0, 0)
when x = 1/6, y = sin(3pi x) = sin(3pi × 1/6) = sin(pi/2) = 1, so the point is (1/6, 1)
when x = 1/3, y = sin(3pi × 1/3) = sin(pi) = 0, so the point is (1/3, 0)
when x = 1/2, y = sin(3pi × 1/2) = sin(3pi/2) = -1, so the point is (1/2, -1)
when x = 2/3, y = sin(3pi × 2/3) = sin(2pi) = 0, so the point is (2/3, 0)
The five points are (0, 0), (1/6, 1), (1/3, 0), (1/2, -1), (2/3, 0)
Draw the graph of y = sin(3pi x) based on the five points The general form of Sine function is:
y = A sin ( Bx - C ), C is the initial phase shift.
When C = 0, y = A sin Bx.
For the graph of y = sin(3pi x), its one period is from x = 0 to x = 2/3.
Analysis of the graph:
x = 0, y = 0.
x = 1/6 is its quarter period, at this point, its value of y is 1 which is maximum.
x = 1/3 is its half period, at this point, its value of y is 0.
x = 1/2 is its three-fourth period, at this point, its value of y is -1 which is minimum.
x = 2/3 is its end point of the first period, at this point, its value of y is 0.
The curve of y = sin 3 pi x is continuous, its second period is from x = 2/3 to x = 4/3. 