To determine the vertical asymptotes of the function [tex]\( f(x) = \frac{x-9}{x^3 - 81x} \)[/tex], we need to find the values of [tex]\( x \)[/tex] that make the denominator zero. These values of [tex]\( x \)[/tex] will potentially be vertical asymptotes as long as they do not cancel out with zeros in the numerator.
Let's go through the process step-by-step:
1. Identify the function and its components: - The function is [tex]\( f(x) = \frac{x-9}{x^3 - 81x} \)[/tex]. - The numerator is [tex]\( x-9 \)[/tex]. - The denominator is [tex]\( x^3 - 81x \)[/tex].
2. Set the denominator equal to zero and solve for [tex]\( x \)[/tex]: - [tex]\( x^3 - 81x = 0 \)[/tex]. - Factor out the common term in the denominator: [tex]\[
x(x^2 - 81) = 0
\][/tex] - Notice that [tex]\( x^2 - 81 \)[/tex] is a difference of squares. Factor it further: [tex]\[
x(x - 9)(x + 9) = 0
\][/tex]
3. Solve for [tex]\( x \)[/tex]: - Set each factor equal to zero: [tex]\[
x = 0
\][/tex] [tex]\[
x - 9 = 0 \implies x = 9
\][/tex] [tex]\[
x + 9 = 0 \implies x = -9
\][/tex]
4. Determine which of these values are vertical asymptotes: - Vertical asymptotes occur where the denominator is zero but the numerator is not zero. - For [tex]\( x = 9 \)[/tex]: - Substitute [tex]\( x = 9 \)[/tex] into the numerator: [tex]\( 9 - 9 = 0 \)[/tex]. The numerator is zero, so [tex]\( x = 9 \)[/tex] is not a vertical asymptote. - For [tex]\( x = -9 \)[/tex]: - Substitute [tex]\( x = -9 \)[/tex] into the numerator: [tex]\( -9 - 9 = -18 \)[/tex]. The numerator is not zero, so [tex]\( x = -9 \)[/tex] is a vertical asymptote. - For [tex]\( x = 0 \)[/tex]: - Substitute [tex]\( x = 0 \)[/tex] into the numerator: [tex]\( 0 - 9 = -9 \)[/tex]. The numerator is not zero, so [tex]\( x = 0 \)[/tex] is a vertical asymptote.
Therefore, the vertical asymptotes of [tex]\( f(x) = \frac{x-9}{x^3 - 81x} \)[/tex] are [tex]\( x = 0 \)[/tex] and [tex]\( x = -9 \)[/tex].
So the answer is:
- [tex]\( x = -9 \)[/tex] - [tex]\( x = 0 \)[/tex]
These correspond to the vertical asymptotes correctly.
