To determine which function has a [tex]\( y \)[/tex]-intercept of -1 and an amplitude of 2, we need to analyze each function individually:
1. Finding the [tex]\( y \)[/tex]-intercept:
- The [tex]\( y \)[/tex]-intercept of a function occurs when [tex]\( x = 0 \)[/tex].
For each function:
[tex]\[ f(x) = -\sin(x) - 1 \][/tex]
When [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -\sin(0) - 1 = 0 - 1 = -1 \][/tex]
[tex]\[ f(x) = -2 \sin(x) - 1 \][/tex]
When [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -2 \sin(0) - 1 = 0 - 1 = -1 \][/tex]
[tex]\[ f(x) = -\cos(x) \][/tex]
When [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -\cos(0) = -1 \][/tex]
[tex]\[ f(x) = -2 \cos(x) - 1 \][/tex]
When [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -2 \cos(0) - 1 = -2 \times 1 - 1 = -2 - 1 = -3 \][/tex]
Comparing these intercepts:
- [tex]\(-\sin(x) - 1\)[/tex]: [tex]\( y \)[/tex]-intercept = -1
- [tex]\(-2 \sin(x) - 1\)[/tex]: [tex]\( y \)[/tex]-intercept = -1
- [tex]\(-\cos(x)\)[/tex]: [tex]\( y \)[/tex]-intercept = -1
- [tex]\(-2 \cos(x) - 1\)[/tex]: [tex]\( y \)[/tex]-intercept = -3
The functions with a [tex]\( y \)[/tex]-intercept of -1 are [tex]\( -\sin(x) - 1 \)[/tex], [tex]\( -2 \sin(x) - 1 \)[/tex], and [tex]\( -\cos(x) \)[/tex].
2. Finding the amplitude:
- The amplitude of a function [tex]\( A \cos(x) + C \)[/tex] or [tex]\( A \sin(x) + C \)[/tex] is the absolute value of [tex]\( A \)[/tex].
For each function:
[tex]\[ f(x) = -\sin(x) - 1 \][/tex]
- Amplitude: [tex]\( \left| -1 \right| = 1 \)[/tex]
[tex]\[ f(x) = -2 \sin(x) - 1 \][/tex]
- Amplitude: [tex]\( \left| -2 \right| = 2 \)[/tex]
[tex]\[ f(x) = -\cos(x) \][/tex]
- Amplitude: [tex]\( \left| -1 \right| = 1 \)[/tex]
[tex]\[ f(x) = -2 \cos(x) - 1 \][/tex]
- Amplitude: [tex]\( \left| -2 \right| = 2 \]
The functions with amplitude 2 are \( -2 \sin(x) - 1 \)[/tex] and [tex]\( -2 \cos(x) - 1 \)[/tex].
3. Combining both conditions:
The function that has both a [tex]\( y \)[/tex]-intercept of -1 and an amplitude of 2 is:
[tex]\[ \boxed{-2 \sin(x) - 1} \][/tex]
