Let's solve this step by step.

1. Domain of $f(x) = \frac{5 + x}{x^3 - 125}$:

The domain of a function is defined as the set of all values of $x$ for which the function is defined. Since $f(x)$ is a rational function, the function will be undefined where the denominator is zero. So, we need to find the values of $x$ that make the denominator $x^3 - 125 = 0$.

Step 1: Factor the denominator

$x^3 - 125$ is a difference of cubes: $x^3 - 125 = (x - 5)(x^2 + 5x + 25)$ Now, set the denominator equal to zero: $x^3 - 125 = (x - 5)(x^2 + 5x + 25) = 0$

Step 2: Solve for $x$

1. $x - 5 = 0$ gives $x = 5$.
2. The quadratic $x^2 + 5x + 25 = 0$ has no real solutions because the discriminant $\Delta = 5^2 - 4(1)(25) = 25 - 100 = -75$, which is negative.

So, the only value of $x$ that makes the denominator zero is $x = 5$.

Domain Conclusion:

The domain of $f(x)$ is all real numbers except $x = 5$.

2. Vertical Asymptote(s):

Vertical asymptotes occur where the function becomes undefined, i.e., where the denominator is zero but the numerator is not zero.

From the domain analysis, we know that $x = 5$ makes the denominator zero. Since the numerator $5 + x$ does not equal zero at $x = 5$, there is a vertical asymptote at $x = 5$.

Vertical Asymptote:

$x = 5$

3. Horizontal Asymptote(s):

Horizontal asymptotes are found by analyzing the behavior of the function as $x \to \infty$ or $x \to -\infty$.

For a rational function like $f(x) = \frac{5 + x}{x^3 - 125}$, the degree of the numerator (which is 1) is less than the degree of the denominator (which is 3). When the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is $y = 0$.

Horizontal Asymptote:

$y = 0$

Final Answer:

• Domain: All real numbers except $x = 5$.
• Vertical Asymptote(s): $x = 5$.
• Horizontal Asymptote(s): $y = 0$.

Do you have any questions or need further details?

Related Questions:

1. How do you find the domain of a rational function in general?
2. What is the method to factor a difference of cubes like $x^3 - a^3$?
3. What happens to the function values as $x \to \infty$ and $x \to -\infty$?
4. Can a function have more than one vertical asymptote? Why or why not?
5. Why does a negative discriminant indicate no real solutions for a quadratic?

Tip:

To find vertical asymptotes, always factor the denominator and solve for values of $x$ where the denominator equals zero, while making sure the numerator does not cancel out at those points.