Consider a rational function f that satisfies the following conditions.
lim 
x→±∞ 
f(x) = 0,	   
lim 
x→0 
f(x) = −∞,	   f(7) = 0,

 
lim 
x→8− 
f(x) = ∞,	   
lim 
x→8+ 
f(x) = −∞,	  
Find the factors of the denominator of f corresponding to the vertical asymptotes. (Enter your answers as a comma-separated list. Enter at least one expression for each vertical asymptote.)
  
Correct: Your answer is correct.
 Excellent!
Find the factors of the numerator of f corresponding to the x-intercepts. (Enter your answers as a comma-separated list. Enter at least one expression for each x-intercept.)
  
Correct: Your answer is correct.
 Good work!
Which of the following must be true?
The degrees of the numerator and the denominator are equal.
The degree of the denominator is greater than the degree of the numerator.    
The degree of the numerator is greater than the degree of the denominator.
Correct: Your answer is correct.
That's it!
Find an equation for f.
f(x) = 
  
Incorrect: Your answer is incorrect.

To find the equation for the rational function \( f(x) \) that satisfies the given conditions, let's carefully analyze each condition step by step:

### 1. **Horizontal Asymptote at \( y = 0 \)**:
\[
\lim_{x \to \pm \infty} f(x) = 0
\]
This tells us that the degree of the denominator must be greater than the degree of the numerator. This is consistent with the behavior of rational functions where the horizontal asymptote is at 0.

### 2. **Vertical Asymptotes at \( x = 0 \) and \( x = 8 \)**:
- We have:
  \[
  \lim_{x \to 0} f(x) = -\infty
  \]
  and
  \[
  \lim_{x \to 8^-} f(x) = \infty, \quad \lim_{x \to 8^+} f(x) = -\infty
  \]
  This implies that the function has vertical asymptotes at \( x = 0 \) and \( x = 8 \), which correspond to factors in the denominator of \( f(x) \). Therefore, the denominator will have the factors \( x \) and \( (x - 8) \).

  So, the denominator is:
  \[
  D(x) = x(x - 8)
  \]

### 3. **Zero at \( x = 7 \)**:
- We are told that \( f(7) = 0 \), which means that \( x = 7 \) is a root of the numerator. Therefore, the numerator must contain the factor \( (x - 7) \).

  So, the numerator will be:
  \[
  N(x) = k(x - 7)
  \]
  where \( k \) is some constant.

### 4. **Conclusion**:
- Since the horizontal asymptote is at \( y = 0 \), the degree of the numerator must be less than the degree of the denominator. The numerator is linear, and the denominator is quadratic, which satisfies this condition.
  
  Therefore, the rational function \( f(x) \) is:
  \[
  f(x) = \frac{k(x - 7)}{x(x - 8)}
  \]
  
### 5. **Determine \( k \)**:
To find \( k \), note that we don't have any more specific values given. Therefore, we can assume \( k = 1 \) as a general solution unless more information is provided.

Thus, the equation for \( f(x) \) is:
\[
f(x) = \frac{x - 7}{x(x - 8)}
\]

Would you like any further clarifications, or is there any part you'd like more details on?

---

Here are 5 related questions for further exploration:
1. What happens to the function \( f(x) \) near \( x = 7 \)?
2. How does the function behave at large positive and negative values of \( x \)?
3. How would the graph of \( f(x) \) look near the vertical asymptotes at \( x = 0 \) and \( x = 8 \)?
4. What would change if the horizontal asymptote was not at 0?
5. How do the limits at the vertical asymptotes affect the behavior of the graph?

**Tip:** The degree of the numerator and denominator in a rational function gives important clues about the horizontal asymptote behavior.

Consider a rational function f that satisfies the following conditions. lim x→±∞ f(x) = 0, lim x→0 f(x) = −∞, f(7) = 0, lim x→8− f(x) = ∞, lim x→8+ f(x) = −∞. Find the factors of the denominator of f corresponding to the vertical asymptotes. (Enter your answers as a comma-separated list. Enter at least one expression for each vertical asymptote).

Learn how to solve a rational function problem involving horizontal and vertical asymptotes, as well as limits. This problem covers concepts of rational functions, factoring, and asymptotic behavior. Suitable for students in Grades 11-12.

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