Answer :
Let's analyze the given function [tex]\( y = 3 \sec \left[2\left(x - \frac{\pi}{2}\right)\right] + 2 \)[/tex].
### Step-by-Step Analysis:
1. Understand the Basic Function:
- The secant function [tex]\(\sec(u)\)[/tex] is the reciprocal of the cosine function, i.e., [tex]\(\sec(u) = \frac{1}{\cos(u)}\)[/tex].
- The function [tex]\( y = \sec(x) \)[/tex] has vertical asymptotes where [tex]\(\cos(x) = 0\)[/tex], which occurs at [tex]\( x = \frac{\pi}{2} + k\pi \)[/tex] for [tex]\( k \in \mathbb{Z} \)[/tex].
2. Phase Shift and Scaling:
- Inside the secant function, we have a transformation involving [tex]\( x - \frac{\pi}{2} \)[/tex].
- This indicates a phase shift of [tex]\( \frac{\pi}{2} \)[/tex] to the right.
- The argument [tex]\( 2\left(x - \frac{\pi}{2}\right) \)[/tex] introduces a horizontal scaling by a factor of 2, which means the period of the secant function is halved. The period of [tex]\( \sec(2x) \)[/tex] is [tex]\( \pi \)[/tex] instead of [tex]\( 2\pi \)[/tex].
3. Vertical Stretch and Translation:
- The factor 3 before the secant function indicates a vertical stretch by a factor of 3.
- The "+2" shifts the entire graph upward by 2 units.
### Key Characteristics of the Graph:
- Period: The period of [tex]\( \sec \left[2\left(x - \frac{\pi}{2}\right)\right] \)[/tex] is [tex]\( \pi \)[/tex] because [tex]\(\sec(kx)\)[/tex] has a period of [tex]\( \frac{2\pi}{k} \)[/tex], so [tex]\(\sec(2x)\)[/tex] has [tex]\( \frac{2\pi}{2} = \pi \)[/tex].
- Vertical Asymptotes: These occur where [tex]\( \cos \left[2\left(x - \frac{\pi}{2}\right)\right] = 0 \)[/tex]. Therefore, the asymptotes are at points where:
[tex]\[ 2\left(x - \frac{\pi}{2}\right) = \frac{\pi}{2} + k\pi \quad \Rightarrow \quad x - \frac{\pi}{2} = \frac{\pi}{4} + \frac{k\pi}{2} \quad \Rightarrow \quad x = \frac{3\pi}{4} + \frac{k\pi}{2} \][/tex]
This gives vertical asymptotes at [tex]\( x = \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}, \ldots \)[/tex].
- Intersections and Maximum/Minimum Points:
- The secant function has its invariants at [tex]\( \sec(\theta) = 1 \)[/tex] or [tex]\(\sec(\theta) = -1\)[/tex].
- Applying the vertical stretch and shift, the midline of the graph is [tex]\( y = 2 \)[/tex].
- Peaks are at [tex]\( y = 2 + 3 = 5 \)[/tex] and valleys at [tex]\( y = 2 - 3 = -1 \)[/tex].
### Summary:
The correct graph will have:
- A periodic pattern with a period of [tex]\( \pi \)[/tex].
- Vertical asymptotes located at [tex]\( x = \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}, \ldots \)[/tex].
- Peaks at [tex]\( y = 5 \)[/tex] and valleys at [tex]\( y = -1 \)[/tex].
- The midline of the graph is at [tex]\( y = 2 \)[/tex], around which the function oscillates.
Having understood these characteristics, the description of the function [tex]\( y = 3 \sec \left[2\left(x - \frac{\pi}{2}\right)\right] + 2 \)[/tex] matches these properties correctly.
### Step-by-Step Analysis:
1. Understand the Basic Function:
- The secant function [tex]\(\sec(u)\)[/tex] is the reciprocal of the cosine function, i.e., [tex]\(\sec(u) = \frac{1}{\cos(u)}\)[/tex].
- The function [tex]\( y = \sec(x) \)[/tex] has vertical asymptotes where [tex]\(\cos(x) = 0\)[/tex], which occurs at [tex]\( x = \frac{\pi}{2} + k\pi \)[/tex] for [tex]\( k \in \mathbb{Z} \)[/tex].
2. Phase Shift and Scaling:
- Inside the secant function, we have a transformation involving [tex]\( x - \frac{\pi}{2} \)[/tex].
- This indicates a phase shift of [tex]\( \frac{\pi}{2} \)[/tex] to the right.
- The argument [tex]\( 2\left(x - \frac{\pi}{2}\right) \)[/tex] introduces a horizontal scaling by a factor of 2, which means the period of the secant function is halved. The period of [tex]\( \sec(2x) \)[/tex] is [tex]\( \pi \)[/tex] instead of [tex]\( 2\pi \)[/tex].
3. Vertical Stretch and Translation:
- The factor 3 before the secant function indicates a vertical stretch by a factor of 3.
- The "+2" shifts the entire graph upward by 2 units.
### Key Characteristics of the Graph:
- Period: The period of [tex]\( \sec \left[2\left(x - \frac{\pi}{2}\right)\right] \)[/tex] is [tex]\( \pi \)[/tex] because [tex]\(\sec(kx)\)[/tex] has a period of [tex]\( \frac{2\pi}{k} \)[/tex], so [tex]\(\sec(2x)\)[/tex] has [tex]\( \frac{2\pi}{2} = \pi \)[/tex].
- Vertical Asymptotes: These occur where [tex]\( \cos \left[2\left(x - \frac{\pi}{2}\right)\right] = 0 \)[/tex]. Therefore, the asymptotes are at points where:
[tex]\[ 2\left(x - \frac{\pi}{2}\right) = \frac{\pi}{2} + k\pi \quad \Rightarrow \quad x - \frac{\pi}{2} = \frac{\pi}{4} + \frac{k\pi}{2} \quad \Rightarrow \quad x = \frac{3\pi}{4} + \frac{k\pi}{2} \][/tex]
This gives vertical asymptotes at [tex]\( x = \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}, \ldots \)[/tex].
- Intersections and Maximum/Minimum Points:
- The secant function has its invariants at [tex]\( \sec(\theta) = 1 \)[/tex] or [tex]\(\sec(\theta) = -1\)[/tex].
- Applying the vertical stretch and shift, the midline of the graph is [tex]\( y = 2 \)[/tex].
- Peaks are at [tex]\( y = 2 + 3 = 5 \)[/tex] and valleys at [tex]\( y = 2 - 3 = -1 \)[/tex].
### Summary:
The correct graph will have:
- A periodic pattern with a period of [tex]\( \pi \)[/tex].
- Vertical asymptotes located at [tex]\( x = \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}, \ldots \)[/tex].
- Peaks at [tex]\( y = 5 \)[/tex] and valleys at [tex]\( y = -1 \)[/tex].
- The midline of the graph is at [tex]\( y = 2 \)[/tex], around which the function oscillates.
Having understood these characteristics, the description of the function [tex]\( y = 3 \sec \left[2\left(x - \frac{\pi}{2}\right)\right] + 2 \)[/tex] matches these properties correctly.
