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:

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 $x = -1$

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.