Question

identifying the vertical asymptote
identify the vertical asymptote of the function.
$f(x) = \frac{x^2 + 1}{3(x - 8)}$
the vertical asymptote is at $x = \square$.

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).

Step2: Find Denominator Zero Point

Given \( f(x)=\frac{x^2 + 1}{3(x - 8)} \), set denominator \( 3(x - 8)=0 \).
Solve for \( x \): \( x - 8 = 0\implies x = 8 \).
Check numerator at \( x = 8 \): \( 8^2+1=64 + 1 = 65
eq0 \), so vertical asymptote at \( x = 8 \).

Answer:

\( 8 \)