Let's solve each part of the question for the given function $f(x) = \frac{5x^2}{x^2 - 9}$:

Part (a): Domain of $f(x)$

The function is undefined where the denominator equals zero. Set the denominator equal to zero: $x^2 - 9 = 0 \implies x^2 = 9 \implies x = \pm 3$ Thus, the domain of $f(x)$ is all real numbers except $x = 3$ and $x = -3$.

Domain in interval notation: $(-\infty, -3) \cup (-3, 3) \cup (3, \infty)$

Part (b): Vertical Asymptotes

Vertical asymptotes occur where the denominator equals zero and the numerator is not zero. As determined earlier, the denominator is zero at $x = \pm 3$.

For $x = \pm 3$: $\text{Denominator: } x^2 - 9 = 0, \quad \text{Numerator: } 5x^2 \neq 0$ Since the numerator is nonzero at these points, there are vertical asymptotes at: $x = 3 \quad \text{and} \quad x = -3$

Part (c): Horizontal Asymptotes

Horizontal asymptotes are determined by the degrees of the numerator and denominator.

• Degree of numerator: 2 (from $5x^2$)
• Degree of denominator: 2 (from $x^2 - 9$)

Since the degrees are equal, the horizontal asymptote is found by dividing the leading coefficients: $y = \frac{\text{Leading coefficient of numerator}}{\text{Leading coefficient of denominator}} = \frac{5}{1} = 5$ Thus, the horizontal asymptote is: $y = 5$

Final Summary:

1. Domain: $(-\infty, -3) \cup (-3, 3) \cup (3, \infty)$
2. Vertical asymptotes: $x = 3$ and $x = -3$
3. Horizontal asymptote: $y = 5$

Would you like further clarification or detailed steps for any part?
Here are 5 related questions for practice:

1. What happens to the graph of $f(x)$ as $x \to 3^+$ and $x \to 3^-$?
2. Determine the slant asymptote if the degrees of numerator and denominator differed by 1.
3. For what values of $x$ does $f(x) = 0$?
4. How does $f(x)$ behave as $x \to \pm \infty$?
5. How would the domain change if the denominator were $x^2 - 9x$?

Tip: Always check the numerator when identifying vertical asymptotes to ensure it's nonzero!