Answer :
To graph the function \( y = 3 \sec \left[2\left(x-\frac{\pi}{2}\right)\right] + 2 \), follow these steps:
1. Understand the Secant Function: Recall that the secant function, \( \sec x \), is the reciprocal of the cosine function, \( \sec x = \frac{1}{\cos x} \). The basic shape of the secant function involves vertical asymptotes where the cosine function is zero (because division by zero is undefined), and it has a pattern of up-and-down curves (the secant function has values going to \(\pm\infty\) as it approaches these asymptotes).
2. Identify Transformations: The function \( y = 3 \sec \left[2(x - \frac{\pi}{2})\right] + 2 \) includes several transformations of the basic secant function.
- Horizontal Shift: The expression \( x - \frac{\pi}{2} \) inside the secant function represents a horizontal shift to the right by \(\frac{\pi}{2}\).
- Horizontal Stretch/Compression: The factor of 2 inside the secant function represents a horizontal compression by a factor of \(\frac{1}{2}\). This means that the period of the secant function is halved. For the basic secant function, the period is \( 2\pi \), so after compression, the period becomes \(\pi\).
- Vertical Stretch: The coefficient 3 in front of the secant function indicates a vertical stretch by a factor of 3.
- Vertical Shift: The addition of 2 outside the secant function indicates a vertical shift upward by 2 units.
3. Locate the Asymptotes: The asymptotes of \( \sec(x) \) occur where \( \cos(x) = 0 \). For \( \cos (2(x - \frac{\pi}{2})) = 0 \), solve:
[tex]\[ 2 \left( x - \frac{\pi}{2} \right) = \frac{\pi}{2} + k\pi \quad \text{ for } k \in \mathbb{Z} \][/tex]
Simplify to find:
[tex]\[ 2x - \pi = \frac{\pi}{2} + k\pi \][/tex]
[tex]\[ 2x = \frac{3\pi}{2} + k\pi \][/tex]
[tex]\[ x = \frac{3\pi}{4} + \frac{k\pi}{2} \][/tex]
Hence, the vertical asymptotes are at \( x = \frac{3\pi}{4} + \frac{k\pi}{2} \).
4. Plot Key Points: Between each pair of asymptotes, the secant function will have a pattern similar to a vertically stretched cosine function shifted upward. Identify the midpoints of intervals between asymptotes (\( \frac{3\pi}{4} + \frac{k\pi}{2} \)) to calculate specific points.
5. Graph the Function:
- Draw the vertical asymptotes at \( x = \frac{3\pi}{4} + \frac{k\pi}{2} \).
- Note that the function reaches its minimum or maximum values halfway between the asymptotes.
- The basic shape is stretched vertically by a factor of 3 and shifted upward by 2. So, if \( \sec(t) \) would normally be 1 or -1, it will instead be 3 + 2 or -3 + 2.
In conclusion, your graph will exhibit repeated vertical asymptotes at [tex]\( x = \frac{3\pi}{4} + \frac{k\pi}{2} \)[/tex], with sections of the graph rising rapidly toward [tex]\(\infty\)[/tex] or [tex]\(-\infty\)[/tex] near each vertical asymptote, vertically shifted up by 2 units and stretched by a factor of 3.
1. Understand the Secant Function: Recall that the secant function, \( \sec x \), is the reciprocal of the cosine function, \( \sec x = \frac{1}{\cos x} \). The basic shape of the secant function involves vertical asymptotes where the cosine function is zero (because division by zero is undefined), and it has a pattern of up-and-down curves (the secant function has values going to \(\pm\infty\) as it approaches these asymptotes).
2. Identify Transformations: The function \( y = 3 \sec \left[2(x - \frac{\pi}{2})\right] + 2 \) includes several transformations of the basic secant function.
- Horizontal Shift: The expression \( x - \frac{\pi}{2} \) inside the secant function represents a horizontal shift to the right by \(\frac{\pi}{2}\).
- Horizontal Stretch/Compression: The factor of 2 inside the secant function represents a horizontal compression by a factor of \(\frac{1}{2}\). This means that the period of the secant function is halved. For the basic secant function, the period is \( 2\pi \), so after compression, the period becomes \(\pi\).
- Vertical Stretch: The coefficient 3 in front of the secant function indicates a vertical stretch by a factor of 3.
- Vertical Shift: The addition of 2 outside the secant function indicates a vertical shift upward by 2 units.
3. Locate the Asymptotes: The asymptotes of \( \sec(x) \) occur where \( \cos(x) = 0 \). For \( \cos (2(x - \frac{\pi}{2})) = 0 \), solve:
[tex]\[ 2 \left( x - \frac{\pi}{2} \right) = \frac{\pi}{2} + k\pi \quad \text{ for } k \in \mathbb{Z} \][/tex]
Simplify to find:
[tex]\[ 2x - \pi = \frac{\pi}{2} + k\pi \][/tex]
[tex]\[ 2x = \frac{3\pi}{2} + k\pi \][/tex]
[tex]\[ x = \frac{3\pi}{4} + \frac{k\pi}{2} \][/tex]
Hence, the vertical asymptotes are at \( x = \frac{3\pi}{4} + \frac{k\pi}{2} \).
4. Plot Key Points: Between each pair of asymptotes, the secant function will have a pattern similar to a vertically stretched cosine function shifted upward. Identify the midpoints of intervals between asymptotes (\( \frac{3\pi}{4} + \frac{k\pi}{2} \)) to calculate specific points.
5. Graph the Function:
- Draw the vertical asymptotes at \( x = \frac{3\pi}{4} + \frac{k\pi}{2} \).
- Note that the function reaches its minimum or maximum values halfway between the asymptotes.
- The basic shape is stretched vertically by a factor of 3 and shifted upward by 2. So, if \( \sec(t) \) would normally be 1 or -1, it will instead be 3 + 2 or -3 + 2.
In conclusion, your graph will exhibit repeated vertical asymptotes at [tex]\( x = \frac{3\pi}{4} + \frac{k\pi}{2} \)[/tex], with sections of the graph rising rapidly toward [tex]\(\infty\)[/tex] or [tex]\(-\infty\)[/tex] near each vertical asymptote, vertically shifted up by 2 units and stretched by a factor of 3.
