AP Precalculus Practice Test: Unit 3 Question #38 Find the Vertical Asymptotes of f(x) = 2sec(x) - 5

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**AP Precalculus Practice Test: Unit 3, Question #38** asks you to find the vertical asymptotes of the function \( f(x) = 2\sec(x) - 5 \).

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### **Key Concepts**
1. **Secant Function**:
- The secant function, \( \sec(x) \), is the reciprocal of the cosine function:
\[
\sec(x) = \frac{1}{\cos(x)}
\]
- The secant function has vertical asymptotes wherever \( \cos(x) = 0 \) because the reciprocal of zero is undefined.

2. **Finding Vertical Asymptotes**:
- To find the vertical asymptotes of \( f(x) = 2\sec(x) - 5 \), we need to determine where \( \sec(x) \) (or equivalently \( \cos(x) \)) is undefined, i.e., where \( \cos(x) = 0 \).

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### **Solving the Problem**

1. **Set the cosine function equal to 0**:
- \( \sec(x) \) is undefined wherever \( \cos(x) = 0 \).
- The values of \( x \) where \( \cos(x) = 0 \) occur at:
\[
x = \frac{\pi}{2} + n\pi, \quad \text{where} \quad n \in \mathbb{Z}
\]
This is because cosine is 0 at odd multiples of \( \frac{\pi}{2} \).

2. **Find the vertical asymptotes**:
- The vertical asymptotes of \( f(x) = 2\sec(x) - 5 \) occur at the same values of \( x \) where \( \cos(x) = 0 \), i.e., the values where \( x = \frac{\pi}{2} + n\pi \).

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### **Summary**:
The vertical asymptotes of \( f(x) = 2\sec(x) - 5 \) occur at \( x = \frac{\pi}{2} + n\pi \), where \( n \) is any integer. These are the points where \( \cos(x) = 0 \), causing the secant function to be undefined.

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Nick Perich
Norristown Area High School
Norristown Area School District
Norristown, Pa

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