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]
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]
