The problem asks to find a rational function $f(x)$ that satisfies the following conditions:

1. $\lim_{x \to +\infty} f(x) = 0$
2. $\lim_{x \to 0} f(x) = -\infty$
3. $f(5) = 0$
4. $\lim_{x \to 8^-} f(x) = \infty$
5. $\lim_{x \to 8^+} f(x) = -\infty$

Step-by-step analysis of the conditions:

1. $\lim_{x \to +\infty} f(x) = 0$: This implies the function is a rational function with the degree of the denominator greater than the degree of the numerator.

2. $\lim_{x \to 0} f(x) = -\infty$: This indicates there is a vertical asymptote at $x = 0$, suggesting the denominator has $x$ as a factor.

3. $f(5) = 0$: This suggests that $f(x)$ has a root at $x = 5$.

4. $\lim_{x \to 8^-} f(x) = \infty$ and $\lim_{x \to 8^+} f(x) = -\infty$: This suggests there is a vertical asymptote at $x = 8$, with different behaviors on the left and right of $x = 8$.

Proposed function structure:

To satisfy these conditions, we need a rational function of the form: $f(x) = \frac{N(x)}{D(x)}$ Where $N(x)$ is the numerator and $D(x)$ is the denominator.

• For the vertical asymptote at $x = 0$, the denominator should include $x$ as a factor.
• For the root at $x = 5$, the numerator should include $(x - 5)$.
• For the vertical asymptote at $x = 8$ with different behavior on either side, the denominator should include $(x - 8)^2$, because the sign change between $x \to 8^-$ and $x \to 8^+$ suggests a factor of $(x - 8)^1$ rather than $(x - 8)^2$.

Putting this together:

A possible rational function that satisfies these conditions is: $f(x) = \frac{(x - 5)}{x(x - 8)}$

Verifying the conditions:

1. $\lim_{x \to +\infty} f(x) = 0$:

   • As $x \to +\infty$, both the numerator and denominator grow linearly, but the denominator grows faster due to the higher degree. Hence, $\lim_{x \to +\infty} f(x) = 0$.

2. $\lim_{x \to 0} f(x) = -\infty$:

   • As $x \to 0$, the function behaves like $\frac{1}{x}$, which tends to $-\infty$.

3. $f(5) = 0$:

   • Substituting $x = 5$, we get $f(5) = \frac{(5 - 5)}{5(5 - 8)} = 0$.

4. $\lim_{x \to 8^-} f(x) = \infty$ and $\lim_{x \to 8^+} f(x) = -\infty$:

   • As $x \to 8^-$, $f(x)$ approaches $+\infty$ because the numerator is negative and the denominator is negative, making the overall value positive.
   • As $x \to 8^+$, $f(x)$ approaches $-\infty$ because the numerator is negative and the denominator is positive, making the overall value negative.

Thus, the rational function $f(x) = \frac{(x - 5)}{x(x - 8)}$ satisfies all the given conditions.

Would you like more details or have any questions? Here are some follow-up questions for deeper understanding:

1. How do vertical asymptotes affect the behavior of rational functions?
2. What is the role of the numerator in determining the zeros of a rational function?
3. How can we determine the behavior of a rational function near an asymptote?
4. What happens to a rational function as $x \to \infty$ based on the degrees of the numerator and denominator?
5. How do limits help in analyzing the end behavior of functions?

Tip: Always check the degree of the numerator and denominator in rational functions to predict the end behavior (horizontal asymptotes).