Westonci.ca is the ultimate Q&A platform, offering detailed and reliable answers from a knowledgeable community. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

Which of the following is the graph of [tex]y=3 \sec \left[2\left(x-\frac{\pi}{2}\right)\right]+2[/tex]?

Sagot :

Certainly! Let's go through the process of understanding and graphing the function \( y = 3 \sec \left[2 \left( x - \frac{\pi}{2} \right) \right] + 2 \).

### Step-by-Step Solution:

1. Understanding the Secant Function:
The secant function, \(\sec(x)\), is the reciprocal of the cosine function, \(\cos(x)\). Thus, \(\sec(x) = \frac{1}{\cos(x)}\). The function \(\sec(x)\) has vertical asymptotes wherever \(\cos(x) = 0\), which occurs at odd multiples of \(\frac{\pi}{2}\) (i.e., \(\pm \frac{\pi}{2}, \pm \frac{3\pi}{2}, \ldots\)).

2. Transformations inside the Argument:
Consider the argument inside the secant function \(\left[2 \left( x - \frac{\pi}{2} \right)\right]\).
- The term \(2(x - \frac{\pi}{2})\) involves a horizontal shift and a horizontal scaling.
- The shift \(x - \frac{\pi}{2}\) shifts the function \(\sec(x)\) to the right by \(\frac{\pi}{2}\).
- The factor of 2 scales the function horizontally by a factor of \(\frac{1}{2}\), which means the period of the function is shortened by half.

3. Vertical Stretch and Shift:
The coefficient 3 outside the secant function, \(3 \sec(\cdots)\), stretches the graph vertically by a factor of 3.
The term \(+2\) outside the secant function shifts the graph upwards by 2 units.

### Key Features of the Graph:

- Period:
The basic period of \(\sec(x)\) is \(2\pi\). Since the argument has been scaled by 2 \(2 \left( x - \frac{\pi}{2} \right)\), the period of this secant function is \(\frac{2\pi}{2} = \pi\).

- Asymptotes:
The vertical asymptotes occur where \( \cos \left[2 \left( x - \frac{\pi}{2} \right)\right] = 0 \). Since \(\cos \theta = 0\) at \(\theta = \pm \frac{\pi}{2}, \pm \frac{3\pi}{2}, \ldots\), we solve:
[tex]\[ 2 \left( x - \frac{\pi}{2} \right) = \pm \frac{\pi}{2}, \pm \frac{3\pi}{2}, \ldots \][/tex]
Simplifying for \(x\), we get:
[tex]\[ x - \frac{\pi}{2} = \pm \frac{\pi}{4}, \pm \frac{3\pi}{4}, \ldots \][/tex]
[tex]\[ x = \frac{\pi}{2} \pm \frac{\pi}{4}, \frac{\pi}{2} \pm \frac{3\pi}{4}, \ldots \][/tex]
This simplifies to:
[tex]\[ x = \frac{3\pi}{4}, \frac{\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}, \ldots \][/tex]

- Behavior around Asymptotes:
Near each of these vertical asymptotes, the function will approach \(\pm \infty\).

- Midline and Amplitude:
The midline is \(y = 2\), and the maximum and minimum points (due to the vertical stretch) will oscillate around this line with an amplitude of 3.

Thus, the graph has vertical asymptotes at \(x = \frac{3\pi}{4}, \frac{\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}, \ldots\), a period of \(\pi\), and it oscillates with a vertical stretch by a factor of 3 and is shifted upward by 2 units.

Sketch the graph:

1. Draw the vertical asymptotes at \( x = \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}, \ldots\).
2. Plot the midline at \(y = 2\).
3. For \(y = 3 \sec \left[2 \left(x - \frac{\pi}{2}\right) \right] + 2\), around these asymptotes, the graph will have the shape similar to the sec function, stretched vertically by 3, oscillating around \( y = 2 \).

By following these steps, you can accurately draw or understand the graph of the function [tex]\(y = 3 \sec \left[2 \left(x - \frac{\pi}{2}\right) \right] + 2\)[/tex].