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The task involves finding the **zeros** of each original function \( f(x) \) and identifying the **vertical asymptotes** of its corresponding reciprocal function \( g(x) \), given by \( g(x) = \frac{1}{f(x)} \) where possible.

### Step-by-Step Analysis for Each Pair

For each part (a) to (f):

1. **Find the Zeros of \( f(x) \):** Set \( f(x) = 0 \) and solve for \( x \).
   - These values represent the zeros of \( f(x) \).

2. **Identify Vertical Asymptotes for \( g(x) = \frac{1}{f(x)} \):**
   - Vertical asymptotes for \( g(x) \) occur where \( f(x) = 0 \) since \( g(x) \) becomes undefined at these points.
   - Hence, the vertical asymptotes are at the \( x \)-values where \( f(x) = 0 \).

Let’s go through each pair:

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### Solution Outline

#### a) \( f(x) = x - 6 \), \( g(x) = \frac{1}{x - 6} \)
- **Zeros of \( f(x) \):** \( x - 6 = 0 \Rightarrow x = 6 \)
- **Vertical Asymptote for \( g(x) \):** \( x = 6 \)

#### b) \( f(x) = 3x + 4 \), \( g(x) = \frac{1}{3x + 4} \)
- **Zeros of \( f(x) \):** \( 3x + 4 = 0 \Rightarrow x = -\frac{4}{3} \)
- **Vertical Asymptote for \( g(x) \):** \( x = -\frac{4}{3} \)

#### c) \( f(x) = x^2 - 2x - 15 \), \( g(x) = \frac{1}{x^2 - 2x - 15} \)
- **Zeros of \( f(x) \):** Factor \( x^2 - 2x - 15 = (x - 5)(x + 3) = 0 \)
  - \( x = 5 \) and \( x = -3 \)
- **Vertical Asymptotes for \( g(x) \):** \( x = 5 \) and \( x = -3 \)

#### d) \( f(x) = 4x^2 - 25 \), \( g(x) = \frac{1}{4x^2 - 25} \)
- **Zeros of \( f(x) \):** Factor \( 4x^2 - 25 = (2x - 5)(2x + 5) = 0 \)
  - \( x = \frac{5}{2} \) and \( x = -\frac{5}{2} \)
- **Vertical Asymptotes for \( g(x) \):** \( x = \frac{5}{2} \) and \( x = -\frac{5}{2} \)

#### e) \( f(x) = x^2 + 4 \), \( g(x) = \frac{1}{x^2 + 4} \)
- **Zeros of \( f(x) \):** \( x^2 + 4 = 0 \Rightarrow x^2 = -4 \), which has no real roots.
  - Therefore, \( f(x) \) has **no real zeros**.
- **Vertical Asymptotes for \( g(x) \):** Since \( f(x) \) has no real zeros, \( g(x) \) has **no vertical asymptotes**.

#### f) \( f(x) = 2x^2 + 5x + 3 \), \( g(x) = \frac{1}{2x^2 + 5x + 3} \)
- **Zeros of \( f(x) \):** Factor \( 2x^2 + 5x + 3 = (2x + 3)(x + 1) = 0 \)
  - \( x = -\frac{3}{2} \) and \( x = -1 \)
- **Vertical Asymptotes for \( g(x) \):** \( x = -\frac{3}{2} \) and \( x = -1 \)

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Please let me know if you would like further clarification or a deeper explanation for any of these solutions.

### Follow-up Questions

1. How do you determine the zeros of a polynomial function?
2. Why do vertical asymptotes occur at the zeros of \( f(x) \) in a reciprocal function \( g(x) = \frac{1}{f(x)} \)?
3. What does it mean for a function to have no real zeros, and how does that affect the reciprocal function?
4. Can a reciprocal function have both vertical and horizontal asymptotes? If so, when?
5. How would you sketch the graph of a function and its reciprocal?

### Tip

For reciprocal functions, always start by analyzing the zeros of the denominator, as these determine the locations of the vertical asymptotes.

For each pair of functions, determine where the zeros of the original function occur and state the equations of the vertical asymptotes of the reciprocal function, if possible. Then, sketch the graph of each function, and use it to help sketch the graph of the reciprocal function.

Learn how to find the zeros and vertical asymptotes of reciprocal functions with a step-by-step solution for each pair of functions. This math problem covers concepts like algebra, functions, and asymptotes, making it ideal for students in Grades 10-12.

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