Question

question
consider the graph of the function $f(x)=\frac{x^{3}-2x^{2}-15x}{x^{2}-9}$.
what are the vertical asymptotes? just list the $x$-values, separated by commas.
provide your answer below:
$x = \square$

Explanation:

Step1: Find the denominator of the function

The function is \( f(x)=\frac{x^{3}-2x^{2}-15x}{x^{2}-9} \). The denominator is \( x^{2}-9 \).

Step2: Factor the denominator

Using the difference - of - squares formula \( a^{2}-b^{2}=(a + b)(a - b) \), where \( a=x \) and \( b = 3 \), we have \( x^{2}-9=(x + 3)(x - 3) \).

Step3: Set the denominator equal to zero

Set \( (x + 3)(x - 3)=0 \). By the zero - product property, if \( AB = 0 \), then \( A=0 \) or \( B = 0 \). So \( x+3=0 \) gives \( x=-3 \), and \( x - 3=0 \) gives \( x = 3 \).

Step4: Check for common factors in numerator and denominator

Factor the numerator: \( x^{3}-2x^{2}-15x=x(x^{2}-2x - 15)=x(x - 5)(x+3) \).
Since there is a common factor of \( (x + 3) \) in the numerator and denominator, \( x=-3 \) is a hole (removable discontinuity), not a vertical asymptote.

Answer:

\(x = 3\)