Question

question 7
find the vertical asymptote(s) of ( r(x) = \frac{(x - 3)(x + 5)}{(x^2 + 9)(x + 4)} )
choose all that apply.
( square x=-3 )
( square x=4 )
( square x=-4 )
( square x=-5 )
( square x=5 )
( square x=3 )

Explanation:

Step1: Identify denominator roots

To find vertical asymptotes, first solve for values that make the denominator of $R(x)$ equal to 0:
$$(x^2 + 9)(x + 4) = 0$$

Step2: Solve each factor

Set each factor equal to 0:

1. $x^2 + 9 = 0$ gives $x^2 = -9$, which has no real solutions.
2. $x + 4 = 0$ gives $x = -4$.

Step3: Check numerator at root

Verify that the numerator is not 0 at $x=-4$:

$$(-4 - 3)(-4 + 5) = (-7)(1) = -7 eq 0$$

A vertical asymptote exists here because the denominator is 0 and the numerator is non-zero.

Answer:

$\boldsymbol{x=-4}$