Question

identify the vertical asymptotes of the function. choose two correct answers.
$f(x) = \frac{3x}{x^2 - 16}$
$x = 16$
$x = 4$
$x = -16$
$x = -4$
$x = 0$

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, \( D(x)=x^{2}-16 \).

Step2: Solve \( x^{2}-16 = 0 \)

Factor the denominator: \( x^{2}-16=(x - 4)(x + 4) \) (difference of squares: \( a^{2}-b^{2}=(a - b)(a + b) \), here \( a = x \), \( b = 4 \)).
Set each factor to zero:

• \( x - 4 = 0\Rightarrow x = 4 \)
• \( x + 4 = 0\Rightarrow x=-4 \)

Answer:

B. \( x = 4 \), D. \( x=-4 \) (assuming the options are labeled as: A. \( x = 16i \), B. \( x = 4 \), C. \( x=-16 \), D. \( x=-4 \), E. \( x = 0 \))