We are given the rational function:

$f(x) = \frac{x^2 - x - 6}{3x^2 + 3x - 6}$

Horizontal Asymptotes:

To find the horizontal asymptotes, we compare the degrees of the polynomials in the numerator and denominator:

• Degree of the numerator $(x^2 - x - 6)$ is 2.
• Degree of the denominator $(3x^2 + 3x - 6)$ is also 2.

Since the degrees are equal, the horizontal asymptote is given by the ratio of the leading coefficients of the numerator and denominator. The leading coefficient of the numerator is 1, and the leading coefficient of the denominator is 3.

Thus, the equation of the horizontal asymptote is:

$y = \frac{1}{3}$

Vertical Asymptotes:

To find the vertical asymptotes, we set the denominator equal to zero and solve for $x$:

$3x^2 + 3x - 6 = 0$

Dividing the equation by 3 simplifies to:

$x^2 + x - 2 = 0$

Now, factor the quadratic:

$(x - 1)(x + 2) = 0$

So, the vertical asymptotes occur where $x - 1 = 0$ or $x + 2 = 0$, giving:

$x = 1 \quad \text{and} \quad x = -2$

Final Answer:

• Horizontal asymptote: $y = \frac{1}{3}$
• Vertical asymptotes: $x = 1, -2$

Would you like further details on this solution, or do you have any other questions?

Here are 5 related questions:

1. How do you determine the horizontal asymptotes for other types of rational functions?
2. What happens if the degree of the numerator is less than the degree of the denominator?
3. How do you handle rational functions with higher degrees in the numerator than the denominator?
4. Can you explain what happens at the vertical asymptotes graphically?
5. How do slant asymptotes arise in rational functions?

Tip: Always factor the denominator carefully when finding vertical asymptotes, as it helps locate points where the function is undefined.