To find the vertical and horizontal asymptotes for the rational function:

$f(x) = \frac{2}{3x + 9}$

Step 1: Identify the Vertical Asymptote

The vertical asymptote occurs where the denominator is zero, as this would make $f(x)$ undefined.

1. Set the denominator equal to zero: $3x + 9 = 0$

2. Solve for $x$: $x = -3$

Therefore, there is a vertical asymptote at $x = -3$.

Step 2: Identify the Horizontal Asymptote

For rational functions, the horizontal asymptote depends on the degrees of the numerator and the denominator.

• Here, the degree of the numerator is 0 (constant 2), and the degree of the denominator is 1.
• Since the degree of the denominator is greater than the numerator, the horizontal asymptote is $y = 0$.

Summary

The asymptotes of $f(x) = \frac{2}{3x + 9}$ are:

• Vertical Asymptote: $x = -3$
• Horizontal Asymptote: $y = 0$

Graphing on the Provided Graph

1. Draw a vertical dashed line at $x = -3$ for the vertical asymptote.
2. Draw a horizontal dashed line at $y = 0$ for the horizontal asymptote.

Would you like further clarification on any part of this solution or additional steps for graphing?

Here are some related questions to expand understanding:

1. How would the vertical asymptote change if the denominator were $3x + 6$ instead of $3x + 9$?
2. What would the horizontal asymptote be if the function were $f(x) = \frac{3x + 2}{3x + 9}$?
3. Can a rational function have more than one vertical asymptote? Under what conditions?
4. How does the horizontal asymptote change if the degree of the numerator is higher than the degree of the denominator?
5. If the function were $f(x) = \frac{2x}{3x + 9}$, what would the horizontal asymptote be?

Tip: Vertical asymptotes occur where the function becomes undefined due to a zero in the denominator, while horizontal asymptotes are determined by comparing the degrees of the numerator and denominator.