Consider the following functions:
i) [tex]y=4 \sin \left[\frac{1}{3}\left(x+30^{\circ}\right)\right]-1[/tex]
ii) [tex]y=-\frac{1}{2} \cos \left[4\left(x+135^{\circ}\right)\right]+2[/tex]

a) What is the amplitude of each function?

b) What is the period of each function?

c) Describe the phase shift of each function.

d) Describe the vertical shift of each function.

Sure, I'd be happy to walk you through the detailed solution for each part of the given question.

### Part (a): Amplitude

The amplitude of a trigonometric function [tex]$ y = a \sin(bx + c) + d $[/tex] or [tex]$ y = a \cos(bx + c) + d $[/tex] is given by the absolute value of the coefficient [tex]$ a $[/tex].

1. For the function [tex]$ y = 4 \sin \left[ \frac{1}{3} (x + 30°) \right] - 1 $[/tex]:
- The coefficient of [tex]$\sin$[/tex] is [tex]$4$[/tex].
- Therefore, the amplitude is [tex]$4$[/tex].

2. For the function [tex]$ y = -\frac{1}{2} \cos \left[ 4 (x + 135°) \right] + 2 $[/tex]:
- The coefficient of [tex]$\cos$[/tex] is [tex]$-\frac{1}{2}$[/tex].
- Therefore, the amplitude is the absolute value of [tex]$-\frac{1}{2}$[/tex], which is [tex]$\frac{1}{2}$[/tex].

### Part (b): Period

The period of a sine or cosine function [tex]$ y = a \sin(bx + c) + d $[/tex] or [tex]$ y = a \cos(bx + c) + d $[/tex] is given by the formula [tex]$ \frac{2π}{|b|} $[/tex].

1. For the function [tex]$ y = 4 \sin \left[ \frac{1}{3} (x + 30°) \right] - 1 $[/tex]:
- The coefficient inside the sine function (b) is [tex]$\frac{1}{3}$[/tex].
- The period is [tex]$ \frac{2π}{|\frac{1}{3}|} $[/tex].
- Simplifying this, we get [tex]$ 2π \times 3 = 6π $[/tex].

2. For the function [tex]$ y = -\frac{1}{2} \cos \left[ 4 (x + 135°) \right] + 2 $[/tex]:
- The coefficient inside the cosine function (b) is [tex]$4$[/tex].
- The period is [tex]$ \frac{2π}{|4|} $[/tex].
- Simplifying this, we get [tex]$ \frac{2π}{4} = \frac{π}{2} $[/tex].

### Part (c): Phase Shift

The phase shift of a trigonometric function [tex]$ y = a \sin(bx + c) + d $[/tex] or [tex]$ y = a \cos(bx + c) + d $[/tex] is given by solving [tex]$ bx + c = 0 $[/tex] for [tex]$ x $[/tex].

1. For the function [tex]$ y = 4 \sin \left[ \frac{1}{3} (x + 30°) \right] - 1 $[/tex]:
- Set the inside of the sine function to zero: [tex]$ \frac{1}{3} (x + 30°) = 0 $[/tex].
- Solve for [tex]$ x $[/tex]: [tex]$ x + 30° = 0 $[/tex] [tex]$\Rightarrow x = -30°$[/tex].
- Therefore, the phase shift is [tex]$-30°$[/tex].

2. For the function [tex]$ y = -\frac{1}{2} \cos \left[ 4 (x + 135°) \right] + 2 $[/tex]:
- Set the inside of the cosine function to zero: [tex]$ 4 (x + 135°) = 0 $[/tex].
- Solve for [tex]$ x $[/tex]: [tex]$ x + 135° = 0 $[/tex] [tex]$\Rightarrow x = -135°$[/tex].
- Therefore, the phase shift is [tex]$-135°$[/tex].

### Part (d): Vertical Shift

The vertical shift of a trigonometric function [tex]$ y = a \sin(bx + c) + d $[/tex] or [tex]$ y = a \cos(bx + c) + d $[/tex] is given by the constant [tex]$ d $[/tex] at the end of the expression.

1. For the function [tex]$ y = 4 \sin \left[ \frac{1}{3} (x + 30°) \right] - 1 $[/tex]:
- The vertical shift is [tex]$-1$[/tex].

2. For the function [tex]$ y = -\frac{1}{2} \cos \left[ 4 (x + 135°) \right] + 2 $[/tex]:
- The vertical shift is [tex]$2$[/tex].

### Summary of Results

a) Amplitude:
- Function (i): [tex]$4$[/tex]
- Function (ii): [tex]$\frac{1}{2}$[/tex]

b) Period:
- Function (i): [tex]$6π$[/tex] (approximately [tex]$18.85$[/tex])
- Function (ii): [tex]$\frac{π}{2}$[/tex] (approximately [tex]$1.57$[/tex])

c) Phase Shift:
- Function (i): [tex]$-30°$[/tex]
- Function (ii): [tex]$-135°$[/tex]

d) Vertical Shift:
- Function (i): [tex]$-1$[/tex]
- Function (ii): [tex]$2$[/tex]