which of the following is true for (f(x)=-2sin(x)-3)?\nthe range of the function is the set of real numbers…

Explanation:

Step1: Recall sine - function properties

The general form of a sine function is (y = A\sin(Bx - C)+D), where (A) is the amplitude, (B) affects the period ((T=\frac{2\pi}{|B|})), (C) is the phase - shift, and (D) is the vertical shift. For the function (f(x)=- 2\sin(x)-3), we have (A=-2), (B = 1), (C = 0), and (D=-3).

Step2: Analyze the range

The range of the basic sine function (y = \sin(x)) is ([-1,1]). When we multiply by (A=-2), the range of (y=-2\sin(x)) becomes ([-2,2]). Then, when we subtract 3 ((y=-2\sin(x)-3)), we shift the range down by 3 units. So the range of (y=-2\sin(x)-3) is ([-2 - 3,2 - 3]=[-5,-1]).

Step3: Analyze the vertical shift

The graph of (y=-2\sin(x)-3) is the graph of (y=-2\sin(x)) shifted 3 units down (because (D=-3)), not 3 units up.

Step4: Analyze the amplitude

The amplitude of a sine function (y = A\sin(Bx - C)+D) is (|A|). For (y=-2\sin(x)-3), (|A| = |-2|=2).

Step5: Analyze the period

The period of a sine function (y = A\sin(Bx - C)+D) is (T=\frac{2\pi}{|B|}). Since (B = 1), the period (T=\frac{2\pi}{|1|}=2\pi), not (4\pi).

Answer:

The amplitude of the function is 2.