f(x)= 1 over 3x² + 3x - 18, solve for the vertical and Horizontal asymptotes

We have the function:

\[
f(x) = \frac{1}{3x^2 + 3x - 18}
\]

### Vertical Asymptotes
Vertical asymptotes occur where the denominator equals zero because the function is undefined at these points. To find the vertical asymptotes, set the denominator equal to zero and solve for \(x\):

\[
3x^2 + 3x - 18 = 0
\]

First, divide the entire equation by 3 to simplify:

\[
x^2 + x - 6 = 0
\]

Now, factor the quadratic equation:

\[
(x + 3)(x - 2) = 0
\]

So, the solutions are:

\[
x = -3 \quad \text{or} \quad x = 2
\]

Thus, the vertical asymptotes are at \(x = -3\) and \(x = 2\).

### Horizontal Asymptote
To find the horizontal asymptote, examine the degrees of the numerator and the denominator. The numerator is a constant (degree 0) and the denominator is a quadratic function (degree 2). 

Since the degree of the denominator is higher than the degree of the numerator, the horizontal asymptote is at:

\[
y = 0
\]

### Summary
- **Vertical asymptotes:** \(x = -3\) and \(x = 2\)
- **Horizontal asymptote:** \(y = 0\)

Would you like more details or further clarification?

Here are five related questions you might find interesting:
1. How do you determine horizontal asymptotes for rational functions in general?
2. What is the behavior of the function as \(x \to -3\) and \(x \to 2\)?
3. Can vertical asymptotes ever be crossed by the graph of the function?
4. How does the degree of the numerator and denominator influence the asymptotes?
5. What happens if the degree of the numerator is higher than that of the denominator?

**Tip:** For rational functions, always start by simplifying the expression to identify vertical and horizontal asymptotes easily.

f(x) = 1 / (3x² + 3x - 18), solve for the vertical and horizontal asymptotes

Learn how to solve for the vertical and horizontal asymptotes of the rational function f(x) = 1 / (3x² + 3x - 18). This problem involves factorizing a quadratic equation and understanding asymptote rules, suitable for students in Grades 9-12.

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