To find the vertical asymptotes (VA) of the given rational function:

$f(x) = \frac{x^3 + 27}{x + 3}$

Steps:

1. Factor the numerator if possible: The numerator $x^3 + 27$ is a sum of cubes. Using the formula: $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ with $a = x$ and $b = 3$, we factor $x^3 + 27$: $x^3 + 27 = (x + 3)(x^2 - 3x + 9)$

2. Simplify the rational function: Substituting the factorized form: $f(x) = \frac{(x + 3)(x^2 - 3x + 9)}{x + 3}$ For $x \neq -3$, the $(x + 3)$ terms cancel, leaving: $f(x) = x^2 - 3x + 9$

3. Identify vertical asymptotes: Vertical asymptotes occur when the denominator equals zero and the numerator does not simultaneously equal zero. Here: $\text{Denominator: } x + 3 = 0 \quad \Rightarrow \quad x = -3$ However, since $x + 3$ cancels with the numerator, $x = -3$ is not a vertical asymptote. Instead, it is a hole in the graph.

Final Answer:

This function has no vertical asymptotes, only a hole at $x = -3$.

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Related Questions:

1. What are the horizontal asymptotes of this function?
2. How do we find and represent the hole in the graph?
3. What is the end behavior of the simplified function $x^2 - 3x + 9$?
4. Can this method be used for other cubic polynomials?
5. What is the significance of removing common factors in rational functions?

Tip:

Always check if the numerator and denominator share common factors before concluding the presence of vertical asymptotes.