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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We are given the following conditions:

1. $$ y $$-intercept at $$-1$$, which means $$ f(0) = -1 $$.
2. Amplitude of $$2$$, meaning the absolute value of the coefficient in front of $$\sin(x)$$ or $$\cos(x)$$ should be $$2$$.

Let's analyze each option.

**Option 1:** $$ f(x) = -\sin(x) - 1 $$  
- $$ f(0) = -\sin(0) - 1 = 0 - 1 = -1 $$.  
- The amplitude is $$ | -1 | = 1 $$, not $$2$$.

**Option 2:** $$ f(x) = -2 \sin(x) - 1 $$  
- $$ f(0) = -2\sin(0) - 1 = 0 - 1 = -1 $$.  
- The amplitude is $$ | -2 | = 2 $$, which meets the requirement.

**Option 3:** $$ f(x) = -\cos(x) $$  
- $$ f(0) = -\cos(0) = -1 $$.  
- The amplitude is $$ | -1 | = 1 $$, not $$2$$.  
- Also, there is no $$-1$$ constant term for the $$ y $$-intercept adjustment.

**Option 4:** $$ f(x) = -2 \cos(x) - 1 $$  
- $$ f(0) = -2\cos(0) - 1 = -2 - 1 = -3 $$, which does not meet the $$ y $$-intercept requirement.

Since only **Option 2** meets both conditions, the correct function is:

$$
f(x) = -2 \sin(x) - 1
$$
The correct function is $$ f(x) = -2 \sin(x) - 1 $$.

Which function will have a -intercept at -1 and an amplitude of 2 ?

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