Yes, \( F(x) = \frac{x + 9}{x - 7} \) is a rational function. Here's a breakdown of its key features:

1. **Domain**: The function is undefined where the denominator equals zero. So, set \( x - 7 = 0 \), which gives \( x = 7 \). Thus, the domain is all real numbers except \( x = 7 \):
   \[
   \text{Domain}: \{ x \in \mathbb{R} \mid x \neq 7 \}
   \]

2. **Vertical Asymptote**: Since the denominator becomes zero at \( x = 7 \), there is a vertical asymptote at \( x = 7 \).

3. **Horizontal Asymptote**: Since the degrees of the numerator and denominator are the same (both linear, degree 1), the horizontal asymptote is determined by the ratio of the leading coefficients. In this case, both are 1, so the horizontal asymptote is:
   \[
   y = 1
   \]

4. **X-intercept**: To find the x-intercept, set the numerator equal to zero: \( x + 9 = 0 \), so \( x = -9 \). Therefore, the x-intercept is \( (-9, 0) \).

5. **Y-intercept**: To find the y-intercept, set \( x = 0 \):
   \[
   F(0) = \frac{0 + 9}{0 - 7} = \frac{9}{-7} = -\frac{9}{7}
   \]
   Thus, the y-intercept is \( \left(0, -\frac{9}{7}\right) \).

Would you like more details or have any questions?

Here are five related questions:
1. How do you determine the vertical asymptotes for other rational functions?
2. Can a rational function have more than one vertical asymptote?
3. What happens to the horizontal asymptote if the degree of the numerator exceeds the degree of the denominator?
4. How do you compute slant (oblique) asymptotes for rational functions?
5. How would the graph of this function change if the numerator had a higher degree than the denominator?

**Tip**: Always check for holes in the graph by factoring both the numerator and denominator!

Explore the rational function F(x) = (x+9)/(x-7) with a breakdown of its key features, including vertical and horizontal asymptotes, x-intercepts, and y-intercepts. Suitable for Grades 10-12, this page provides detailed analysis and examples.

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