1. Problem: Analyze the functions $f(x) = \frac{x^2 - 1}{x + 3}$ and $g(x) = \frac{2x - 5}{x - 2}$, their product $(f \cdot g)(x)$, and related rational expressions. 2. Find where $(f \cdot g)(x)$ is undefined. - $(f \cdot g)(x) = f(x) \times g(x) = \frac{x^2 - 1}{x + 3} \times \frac{2x - 5}{x - 2}$. - The product is undefined where either denominator is zero. - Denominators are $x + 3$ and $x - 2$. - So undefined at $x = -3$ and $x = 2$. - The problem states $(f \cdot g)(x)$ is always undefined at $x=2$, which aligns with the denominator of $g(x)$. 3. Simplify $\frac{x^{2} + 3x - 10}{x + 2} = \frac{2x - 5}{x + 2} + 4$. - Factor numerator: $x^2 + 3x - 10 = (x + 5)(x - 2)$. - Rewrite left side: $\frac{(x + 5)(x - 2)}{x + 2}$. - Combine right side over common denominator: $\frac{2x - 5}{x + 2} + \frac{4(x + 2)}{x + 2} = \frac{2x - 5 + 4x + 8}{x + 2} = \frac{6x + 3}{x + 2}$. - Equate numerators since denominators are same: $(x + 5)(x - 2) = 6x + 3$. - Expand left: $x^2 - 2x + 5x - 10 = x^2 + 3x - 10$. - So $x^2 + 3x - 10 = 6x + 3$. - Bring all to one side: $x^2 + 3x - 10 - 6x - 3 = 0 \implies x^2 - 3x - 13 = 0$. 4. Simplify $\frac{x^2 - 16}{x^2 - 4x}$. - Numerator: $x^2 - 16 = (x - 4)(x + 4)$ (difference of squares). - Denominator: $x^2 - 4x = x(x - 4)$. - So fraction is $\frac{(x - 4)(x + 4)}{x(x - 4)}$. - Cancel $(x - 4)$ (not zero): $\frac{x + 4}{x}$. 5. Understand vertical asymptote and hole. - Vertical asymptote at $x = 2$ means denominator zero and numerator nonzero there. For example, $g(x)$ denominator zero at $x=2$. - Hole at $x = -3$ means numerator and denominator both zero at $x=-3$ for $f(x)$. - Crosses $x$-axis at $x=1$ means function value zero there (numerator zero). Final answers and explanations given in the steps above clarify the undefined points, simplifications, and key features of the rational functions.