Question

all vertical asymptotes of 8) $f(x)=\frac{x + 9}{x^{2}-36}$

Explanation:

Step1: Factor the denominator

The denominator $x^{2}-36=(x + 6)(x - 6)$ using the difference - of - squares formula $a^{2}-b^{2}=(a + b)(a - b)$. So, $f(x)=\frac{x + 9}{(x + 6)(x - 6)}$.

Step2: Set the denominator equal to zero

To find the vertical asymptotes, we set $(x + 6)(x - 6)=0$.

Step3: Solve for x

Using the zero - product property, if $(x + 6)(x - 6)=0$, then $x+6 = 0$ or $x - 6=0$. Solving $x+6 = 0$ gives $x=-6$, and solving $x - 6=0$ gives $x = 6$.

Answer:

$x=-6$ and $x = 6$