Which function will have a $$ y $$-intercept at -1 and an amplitude of 2 ? $$ f(x)=-\sin (x)-1 $$ $$ f(x)=-2 \sin (x)-1 $$ $$ f(x)=-\cos (x) $$ $$ f(x)=-2 \cos (x)-1 $$

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To determine which function has a $$ y $$-intercept at -1 and an amplitude of 2, we need to analyze each function based on these criteria.

1. **Amplitude**: The amplitude of a sine or cosine function $$ A \sin(x) $$ or $$ A \cos(x) $$ is given by the absolute value of the coefficient $$ A $$. Therefore, the amplitude is $$ |A| $$.

2. **Y-intercept**: The $$ y $$-intercept of a function $$ f(x) $$ is found by evaluating $$ f(0) $$.

Now, let's analyze each function:

1. **Function**: $$ f(x) = -\sin(x) - 1 $$
   - Amplitude: $$ |A| = |-1| = 1 $$
   - Y-intercept: $$ f(0) = -\sin(0) - 1 = -0 - 1 = -1 $$

2. **Function**: $$ f(x) = -2 \sin(x) - 1 $$
   - Amplitude: $$ |A| = |-2| = 2 $$
   - Y-intercept: $$ f(0) = -2 \sin(0) - 1 = -2(0) - 1 = -1 $$

3. **Function**: $$ f(x) = -\cos(x) $$
   - Amplitude: $$ |A| = |-1| = 1 $$
   - Y-intercept: $$ f(0) = -\cos(0) = -1 $$

4. **Function**: $$ f(x) = -2 \cos(x) - 1 $$
   - Amplitude: $$ |A| = |-2| = 2 $$
   - Y-intercept: $$ f(0) = -2 \cos(0) - 1 = -2(1) - 1 = -2 - 1 = -3 $$

Now, summarizing the results:

- $$ f(x) = -\sin(x) - 1 $$: Amplitude = 1, Y-intercept = -1
- $$ f(x) = -2 \sin(x) - 1 $$: Amplitude = 2, Y-intercept = -1
- $$ f(x) = -\cos(x) $$: Amplitude = 1, Y-intercept = -1
- $$ f(x) = -2 \cos(x) - 1 $$: Amplitude = 2, Y-intercept = -3

The function that has a $$ y $$-intercept at -1 and an amplitude of 2 is:

$$
\boxed{f(x) = -2 \sin(x) - 1}
$$
$$ f(x) = -2 \sin(x) - 1 $$ has a $$ y $$-intercept at -1 and an amplitude of 2.

Which function will have a \( y \)-intercept at -1 and an amplitude of 2 ? \( f(x)=-\sin (x)-1 \) \( f(x)=-2 \sin (x)-1 \) \( f(x)=-\cos (x) \) \( f(x)=-2 \cos (x)-1 \)

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