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Let's break down the problem and construct the rational function \( r(x) \) step by step.

### Given Information:
- **x-intercept** at \( (8, 0) \): This means \( r(8) = 0 \), so \( x = 8 \) is a root of the numerator.
- **Vertical asymptote** at \( x = 4 \): This means the denominator will have a factor of \( x - 4 \).
- **Horizontal asymptote** at \( y = 3 \): This implies the degrees of the numerator and the denominator are the same, and the ratio of the leading coefficients is 3.
- The degree of both the numerator and the denominator is 1, so the rational function is in the form \( r(x) = \frac{ax + b}{cx + d} \).

### Steps:
1. **Form of the Function**: 
   Since the degree of both numerator and denominator is 1, write \( r(x) \) in the form \( r(x) = \frac{ax + b}{cx + d} \).

2. **Use the x-intercept (8,0)**: 
   At \( x = 8 \), \( r(8) = 0 \), so the numerator must be zero when \( x = 8 \). This gives:
   \[
   a(8) + b = 0 \quad \Rightarrow \quad 8a + b = 0 \quad \Rightarrow \quad b = -8a
   \]

3. **Vertical Asymptote (x = 4)**:
   The vertical asymptote occurs when the denominator equals zero. Therefore, set \( cx + d = 0 \) at \( x = 4 \):
   \[
   c(4) + d = 0 \quad \Rightarrow \quad 4c + d = 0 \quad \Rightarrow \quad d = -4c
   \]

4. **Horizontal Asymptote (y = 3)**: 
   The horizontal asymptote \( y = 3 \) means that the ratio of the leading coefficients \( \frac{a}{c} = 3 \), so \( a = 3c \).

### Now, let's substitute these values back:

- From the x-intercept condition: \( b = -8a = -8(3c) = -24c \),
- From the vertical asymptote condition: \( d = -4c \),
- And from the horizontal asymptote condition: \( a = 3c \).

Thus, the function becomes:
\[
r(x) = \frac{3cx - 24c}{cx - 4c}
\]
Factoring out \( c \) from both the numerator and the denominator:
\[
r(x) = \frac{c(3x - 24)}{c(x - 4)}
\]
Canceling \( c \) from the numerator and denominator (assuming \( c \neq 0 \)):
\[
r(x) = \frac{3x - 24}{x - 4}
\]

Thus, the rational function is:
\[
r(x) = \frac{3x - 24}{x - 4}
\]

Would you like any further details on this or have any questions? Here are 5 related questions:

1. What happens to the graph of \( r(x) \) near the vertical asymptote at \( x = 4 \)?
2. How would the function change if the horizontal asymptote was \( y = 2 \) instead of \( y = 3 \)?
3. Why is the degree of the numerator and denominator important for determining the horizontal asymptote?
4. How can you confirm that the x-intercept is indeed at \( (8,0) \)?
5. What if the vertical asymptote occurred at \( x = -2 \) instead of \( x = 4 \)?

**Tip**: Always check both the behavior near the asymptotes and the intercepts to verify your rational function.

Find a rational function r(x) that has an x-intercept at (8, 0), a vertical asymptote at x = 4, and a horizontal asymptote y = 3. The degree of both the numerator and the denominator of r(x) is 1. Write the function in this form: r(x) = (ax + b) / (cx + d).

Learn how to find a rational function with an x-intercept at (8, 0), a vertical asymptote at x = 4, and a horizontal asymptote at y = 3. This solution involves algebraic techniques for rational functions, asymptotes, and intercepts. Suitable for Grades 10-12.

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