What is the period of the function f(x)=-6\sin(3\pi x+4)-2f(x)=−6sin(3πx+4)−2f, left parenthesis, x, right parenthesis, equals, minus, 6, sine, left parenthesis, 3, pi, x, plus, 4, right parenthesis, minus, 2?

geez, repeat yourself much?

Recall that the period of sin(kx) is 2π/k
so, the period of sin(3π x) is 2π/(3π) = 2/3
The phase shift does not affect the period.

To find the period of the function f(x) = -6sin(3πx+4) - 2, we need to determine the value of period.

The general formula to find the period of a sinusoidal function is given by:

Period = 2π / |B|

In this case, the coefficient of x is 3π. Therefore, |B| = 3π.

Now, substitute this value into the formula:

Period = 2π / (3π)

Simplifying further:

Period = 2/3

Hence, the period of the function f(x) = -6sin(3πx+4) - 2 is 2/3.

To find the period of a function, you need to determine the length of one complete cycle of the function. In the case of a sine function, the period is the distance between two consecutive peaks or troughs.

For the given function f(x) = -6sin(3πx + 4) - 2, the period can be found by examining the coefficient of x inside the sine function. In this case, the coefficient is 3π.

The formula to calculate the period of a sine function with respect to x is:

Period (P) = (2π) / |coefficient of x|

Therefore, for the given function, the period is:

P = (2π) / |3π| = 2/3

Hence, the period of the function f(x) = -6sin(3πx + 4) - 2 is 2/3.