To find the vertical asymptotes of the function [tex]\( f(x) = \frac{x^2 - 81}{x^3 + 7x^2 - 18x} \)[/tex], we need to determine the values of [tex]\( x \)[/tex] that make the denominator zero, since vertical asymptotes occur where the function goes to infinity, which happens when the denominator approaches zero while the numerator does not also go to zero at the same rate.
Given the function [tex]\( f(x) = \frac{x^2 - 81}{x^3 + 7x^2 - 18x} \)[/tex]:
1. Find the denominator:
[tex]\[
\text{Denominator} = x^3 + 7x^2 - 18x
\][/tex]
2. Set the denominator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[
x^3 + 7x^2 - 18x = 0
\][/tex]
3. Factor out the common term [tex]\( x \)[/tex]:
[tex]\[
x(x^2 + 7x - 18) = 0
\][/tex]
4. Set each factor equal to zero:
[tex]\[
x = 0 \quad \text{or} \quad x^2 + 7x - 18 = 0
\][/tex]
5. Solve the quadratic equation [tex]\( x^2 + 7x - 18 = 0 \)[/tex] using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 7 \)[/tex], and [tex]\( c = -18 \)[/tex]:
[tex]\[
x = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 1 \cdot (-18)}}{2 \cdot 1}
\][/tex]
[tex]\[
x = \frac{-7 \pm \sqrt{49 + 72}}{2}
\][/tex]
[tex]\[
x = \frac{-7 \pm \sqrt{121}}{2}
\][/tex]
[tex]\[
x = \frac{-7 \pm 11}{2}
\][/tex]
6. Calculate the two solutions:
[tex]\[
x = \frac{4}{2} = 2 \quad \text{or} \quad x = \frac{-18}{2} = -9
\][/tex]
Therefore, the values of [tex]\( x \)[/tex] that make the denominator zero, and thus the vertical asymptotes of the function [tex]\( f(x) \)[/tex], are [tex]\( x = -9 \)[/tex], [tex]\( x = 0 \)[/tex], and [tex]\( x = 2 \)[/tex].
The vertical asymptotes are located at the [tex]\( x \)[/tex]-values: [tex]\(\boxed{-9, 0, 2}\)[/tex].
