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

Explanation:

Step1: Recall the general form of a sinusoidal function

The general form of a sinusoidal 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 $y = \sin(x)$ is $[-1,1]$. For $y=-2\sin(x)$, the range is $[-2,2]$ (since we multiply the output of $\sin(x)$ by $-2$). Then for $y=-2\sin(x)-3$, we shift the range of $y = - 2\sin(x)$ down by 3 units. So the range 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 up.

Step4: Analyze the amplitude

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

Step5: Analyze the period

The period of a sinusoidal 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.