We are asked to write a rational function that satisfies the following conditions:

1. The only zero is at $x = 4$.
2. Vertical asymptotes at $x = -2$ and $x = 3$.
3. A horizontal asymptote at $y = \frac{5}{3}$.

Step-by-Step Construction:

1. Zero at $x = 4$: The function must have a factor of $(x - 4)$ in the numerator, since the zero occurs when $x = 4$.

2. Vertical asymptotes at $x = -2$ and $x = 3$: The denominator must have factors of $(x + 2)$ and $(x - 3)$ because vertical asymptotes occur where the denominator is zero but the numerator is not.

3. Horizontal asymptote at $y = \frac{5}{3}$: To achieve a horizontal asymptote of $\frac{5}{3}$, the degrees of the numerator and denominator should be the same, and the ratio of the leading coefficients of the numerator to the denominator should be $\frac{5}{3}$.

Rational Function:

A function satisfying these conditions is:

$f(x) = \frac{\frac{5}{3}(x - 4)}{(x + 2)(x - 3)}$

This function meets all the given conditions:

• It has a zero at $x = 4$ because the numerator becomes zero when $x = 4$.
• It has vertical asymptotes at $x = -2$ and $x = 3$ because the denominator becomes zero at these points.
• The horizontal asymptote is $y = \frac{5}{3}$, because for large values of $x$, the function behaves like $\frac{\frac{5}{3}x}{x^2}$, which simplifies to $\frac{5}{3}$ as $x \to \infty$.

Would you like more details on the graph or the function's behavior?

Here are some related questions you might explore:

1. How does the degree of the numerator affect the horizontal asymptote?
2. What happens if we change the location of the vertical asymptotes?
3. How would the function change if the horizontal asymptote were $y = 2$ instead?
4. How does the multiplicity of zeros and poles (asymptotes) affect the shape of the graph?
5. What if there were no horizontal asymptote, but instead a slant asymptote?

Tip: When graphing rational functions, always start by identifying the asymptotes and zeros for a clearer understanding of the overall behavior.