Question

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

Explanation:

Identify the denominator of the function

Using the Rational Functions knowledge point

$$ f(x) = \frac{x^2 + 1}{3(x - 8)} $$

The denominator is \(3(x - 8)\).

Find the values that make the denominator zero

Using the Vertical Asymptotes knowledge point

$$
LATEXBLOCK0
$$

Since \(x = 8\) does not make the numerator \(x^2 + 1\) zero, \(x = 8\) is a vertical asymptote.

Answer:

Identify the vertical asymptote of the function.
\(f(x) = \frac{x^2+1}{3(x-8)}\)
The vertical asymptote is at \(x =\) <blank>8</blank>