To find the period of the function [tex]\( y = -2 \sin(3\pi x) \)[/tex], we need to consider the general form of a sine function, which is [tex]\( y = A \sin(Bx + C) + D \)[/tex]. In this function:
- [tex]\( A \)[/tex] represents the amplitude.
- [tex]\( B \)[/tex] affects the period of the function.
- [tex]\( C \)[/tex] represents the phase shift.
- [tex]\( D \)[/tex] represents the vertical shift.
The period of a sine function is determined by the coefficient [tex]\( B \)[/tex]. For a function in the form [tex]\( y = \sin(Bx) \)[/tex], the period [tex]\( T \)[/tex] is given by the formula:
[tex]\[ T = \frac{2\pi}{|B|} \][/tex]
In our specific function [tex]\( y = -2 \sin(3\pi x) \)[/tex], we have [tex]\( B = 3\pi \)[/tex].
Plug [tex]\( B = 3\pi \)[/tex] into the period formula:
[tex]\[ T = \frac{2\pi}{|3\pi|} \][/tex]
Calculate the absolute value of [tex]\( B \)[/tex]:
[tex]\[ |3\pi| = 3\pi \][/tex]
Now, substitute this value back into the formula:
[tex]\[ T = \frac{2\pi}{3\pi} \][/tex]
Simplify the fraction by canceling [tex]\(\pi\)[/tex] in the numerator and the denominator:
[tex]\[ T = \frac{2}{3} \][/tex]
Therefore, the period of the function [tex]\( y = -2 \sin(3\pi x) \)[/tex] is:
[tex]\[ \boxed{\frac{2}{3}} \][/tex]
