Question

the function $f(x) = \frac{1}{x^2 - 9}$ is a rational function. answer parts (a) - (i).

a. the x-intercept(s) is(are) $x = \square$.
(simplify your answer. type an integer or a simplified fraction. use a comma to separate answers as needed.)
b. the function has no x-intercept.

f. find the equation(s) of all vertical asymptotes. select the correct choice and, if necessary, fill in the answer box(es) to complete your choice.
a. the function has one vertical asymptote, $\square$.
(simplify your answer. type an equation. use integers or fractions for any numbers in the equation.)
b. the function has two vertical asymptotes. the leftmost asymptote is $\square$ and the rightmost asymptote is $\square$.
(simplify your answers. type equations. use integers or fractions for any numbers in the equation.)
c. the function has no vertical asymptotes.

Explanation:

Step1: Recall vertical asymptote rule

For a rational function \( f(x)=\frac{N(x)}{D(x)} \), vertical asymptotes occur where \( D(x) = 0 \) and \( N(x)
eq0 \) at those points.
Here, \( N(x) = 1 \), \( D(x)=x^{2}-9 \).

Step2: Solve \( D(x) = 0 \)

Solve \( x^{2}-9 = 0 \). Factor the difference of squares: \( x^{2}-9=(x - 3)(x + 3)=0 \).
Set each factor equal to zero: \( x - 3=0\Rightarrow x = 3 \); \( x + 3=0\Rightarrow x=-3 \).
Check \( N(x) \) at \( x = 3 \) and \( x=-3 \): \( N(3)=1
eq0 \), \( N(-3)=1
eq0 \). So vertical asymptotes at \( x = - 3 \) and \( x = 3 \).

Answer:

B. The function has two vertical asymptotes. The leftmost asymptote is \( x=-3 \) and the rightmost asymptote is \( x = 3 \).