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

Explanation:

Step1: Recall range of sine function

The range of $y = \sin(x)$ is $- 1\leqslant\sin(x)\leqslant1$. For $y=-2\sin(x)-3$, when $\sin(x)=-1$, $y=-2\times(-1)-3=-1$; when $\sin(x) = 1$, $y=-2\times1 - 3=-5$. So the range of $y=-2\sin(x)-3$ is $-5\leqslant y\leqslant - 1$, not $-2\leqslant y\leqslant2$.

Step2: Analyze vertical - shift

The graph of $y = f(x)+k$ is the graph of $y = f(x)$ shifted $k$ units vertically. For $y=-2\sin(x)-3$, it is the graph of $y=-2\sin(x)$ shifted 3 units down, not up.

Step3: Determine the amplitude

For a function of the form $y = A\sin(x)+B$, the amplitude is $|A|$. For $y=-2\sin(x)-3$, $A=-2$, so the amplitude is $| - 2|=2$.

Step4: Recall the period of sine function

The period of $y = \sin(x)$ is $2\pi$, and for $y = A\sin(Bx)+C$, the period is $T=\frac{2\pi}{|B|}$. For $y=-2\sin(x)-3$, $B = 1$, so the period is $T = 2\pi$, not $4\pi$.

Answer:

The amplitude of the function is 2.