Let’s solve the problems step by step for each given function:

Problem 1: $g(x) = \frac{3x - 6}{x^2 + x - 6}$

Step 1: Horizontal intercepts (set numerator = 0)

• The numerator is $3x - 6$.
• Solve $3x - 6 = 0$: $x = 2$
• Horizontal intercept: $(2, 0)$

Step 2: Vertical intercept (evaluate $g(0)$)

• Substitute $x = 0$ into $g(x)$: $g(0) = \frac{3(0) - 6}{0^2 + 0 - 6} = \frac{-6}{-6} = 1$
• Vertical intercept: $(0, 1)$

Step 3: Vertical asymptotes (set denominator = 0)

• The denominator is $x^2 + x - 6$. Factorize: $x^2 + x - 6 = (x + 3)(x - 2)$
• Solve $x + 3 = 0$ and $x - 2 = 0$: $x = -3, \quad x = 2$
• Since $x = 2$ is canceled as it is a root of both numerator and denominator, the vertical asymptote is: $x = -3$

Step 4: Horizontal asymptote (compare degrees of numerator and denominator)

• Degree of numerator = 1, degree of denominator = 2.
• Since the numerator's degree is less than the denominator's degree, the horizontal asymptote is: $y = 0$

Step 5: Domain

• The denominator is undefined at $x = -3$ and $x = 2$.
• Domain: $x \in (-\infty, -3) \cup (-3, 2) \cup (2, \infty)$

Problem 2: $f(x) = \frac{2x^2 - 2x - 12}{x^2 + 3x - 18}$

Step 1: Horizontal intercepts (set numerator = 0)

• The numerator is $2x^2 - 2x - 12$. Factorize: $2x^2 - 2x - 12 = 2(x^2 - x - 6) = 2(x - 3)(x + 2)$
• Solve $(x - 3)(x + 2) = 0$: $x = 3, \quad x = -2$
• Horizontal intercepts: $(3, 0)$ and $(-2, 0)$

Step 2: Vertical intercept (evaluate $f(0)$)

• Substitute $x = 0$ into $f(x)$: $f(0) = \frac{2(0)^2 - 2(0) - 12}{(0)^2 + 3(0) - 18} = \frac{-12}{-18} = \frac{2}{3}$
• Vertical intercept: $(0, \frac{2}{3})$

Step 3: Vertical asymptotes (set denominator = 0)

• The denominator is $x^2 + 3x - 18$. Factorize: $x^2 + 3x - 18 = (x + 6)(x - 3)$
• Solve $x + 6 = 0$ and $x - 3 = 0$: $x = -6, \quad x = 3$
• Vertical asymptotes: $x = -6$, $x = 3$

Step 4: Horizontal asymptote (compare degrees of numerator and denominator)

• Degree of numerator = 2, degree of denominator = 2.
• Since the degrees are the same, divide the leading coefficients: $y = \frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}} = \frac{2}{1} = 2$
• Horizontal asymptote: $y = 2$

Step 5: Domain

• The denominator is undefined at $x = -6$ and $x = 3$.
• Domain: $x \in (-\infty, -6) \cup (-6, 3) \cup (3, \infty)$

Would you like a detailed breakdown of any specific part? Here are 5 related questions:

1. How are horizontal asymptotes determined for rational functions?
2. What is the process of identifying vertical asymptotes in more complex functions?
3. Can you explain why the horizontal intercepts occur at numerator roots?
4. How does the degree of the polynomial affect the behavior of a rational function?
5. Would you like additional practice problems with rational functions?

Tip: Always check for removable discontinuities (holes) when a factor is shared between numerator and denominator!