Consider the function [tex]$f(x)=\frac{x-9}{x^3-81x}$[/tex].

Find the vertical asymptote(s) of [tex]$f(x)$[/tex].

A. [tex][tex]$x=0, -9$[/tex][/tex]
B. [tex]$x=-9$[/tex]
C. [tex]$x=0, 9$[/tex]
D. [tex][tex]$x=9$[/tex][/tex]



Answer :

To find the vertical asymptotes of the function [tex]\( f(x) = \frac{x-9}{x^3 - 81x} \)[/tex], we need to determine where the function becomes undefined. This occurs when the denominator equals zero, but the numerator does not simultaneously equal zero.

First, let's analyze the denominator:
[tex]\[ x^3 - 81x = 0 \][/tex]

We can factor this expression:
[tex]\[ x(x^2 - 81) = 0 \][/tex]
[tex]\[ x(x - 9)(x + 9) = 0 \][/tex]

Setting each factor equal to zero gives us:
[tex]\[ x = 0 \][/tex]
[tex]\[ x - 9 = 0 \implies x = 9 \][/tex]
[tex]\[ x + 9 = 0 \implies x = -9 \][/tex]

Therefore, the potential points where the function could have vertical asymptotes are [tex]\( x = 0 \)[/tex], [tex]\( x = 9 \)[/tex], and [tex]\( x = -9 \)[/tex].

Next, we need to check if the numerator, [tex]\( x - 9 \)[/tex], equals zero at any of these values:
[tex]\[ x - 9 = 0 \implies x = 9 \][/tex]

At [tex]\( x = 9 \)[/tex], both the numerator and the denominator equal zero. This indicates a hole in the graph at [tex]\( x = 9 \)[/tex] rather than a vertical asymptote.

Now, let's revisit our other candidates for vertical asymptotes where the denominator is zero, but the numerator is not:
- For [tex]\( x = 0 \)[/tex], the denominator is zero ([tex]\( 0 \cdot (-81) = 0 \)[/tex]), but the numerator is [tex]\( 0 - 9 = -9 \)[/tex].
- For [tex]\( x = -9 \)[/tex], the denominator is zero ([tex]\( -9 \cdot 0 \cdot (-18) = 0 \)[/tex]), but the numerator is [tex]\( -9 - 9 = -18 \)[/tex].

Since the numerator does not equal zero at [tex]\( x = 0 \)[/tex] and [tex]\( x = -9 \)[/tex], these values create vertical asymptotes.

Therefore, the vertical asymptotes of the function [tex]\( f(x) \)[/tex] are at:
[tex]\[ \boxed{x = 0 \text{ and } x = -9} \][/tex]