Question

find all vertical asymptotes of the following function.
$f(x) = \frac{4x^2 + 18x}{6x + 27}$
answer attempt 1 out of 2
no vertical asymptotes
no vertical asymptotes
one vertical asymptote
two vertical asymptotes

Explanation:

Step1: Simplify the function

First, factor the numerator and the denominator. The numerator \(4x^2 + 18x = 2x(2x + 9)\), and the denominator \(6x + 27 = 3(2x + 9)\). So we can rewrite the function as:

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

We note that \(2x + 9
eq0\) (to avoid division by zero in the original function before simplification), so \(x
eq-\frac{9}{2}\). But when we simplify the function (canceling out the non - zero factor \(2x + 9\)), we get \(f(x)=\frac{2x}{3}\) for \(x
eq-\frac{9}{2}\). This means the function has a hole at \(x =-\frac{9}{2}\) (a removable discontinuity) rather than a vertical asymptote. A vertical asymptote occurs when the denominator is zero and the numerator is non - zero at that point after simplifying the rational function. Since after canceling the common factor, the denominator of the simplified function is a non - zero constant (3), there are no vertical asymptotes.

Step2: Conclusion about vertical asymptotes

Since the only point where the original denominator is zero (\(x =-\frac{9}{2}\)) is a removable discontinuity (hole) and not a point where the numerator is also zero in a non - cancelable way, the function \(f(x)\) has no vertical asymptotes.

Answer:

No Vertical Asymptotes