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.