Question

identify the vertical asymptotes of the function. choose two correct answers.
$f(x) = \frac{x - 8}{x^2 - 3x + 2}$
$x = -2$ $x = 1$ $x = -8$
$x = 8$ $x = 2$ $x = -1$

Explanation:

Step1: Factor the denominator

The denominator is \(x^2 - 3x + 2\). We factor it: \(x^2 - 3x + 2=(x - 1)(x - 2)\) (since \(1\times2 = 2\) and \(1+2 = 3\), but with a negative sign for the middle term).

Step2: Find vertical asymptotes

Vertical asymptotes occur where the denominator is zero (and the numerator is not zero at those points). Set the denominator equal to zero: \((x - 1)(x - 2)=0\). Solving for \(x\), we get \(x - 1 = 0\) or \(x - 2 = 0\), so \(x = 1\) or \(x = 2\). The numerator \(x - 8\) is not zero at \(x = 1\) or \(x = 2\) (when \(x = 1\), numerator is \(1 - 8=-7
eq0\); when \(x = 2\), numerator is \(2 - 8=-6
eq0\)).

Answer:

B. \(x = 1\), E. \(x = 2\) (assuming the options are labeled as: A. \(x=-2\), B. \(x = 1\), C. \(x=-8\), D. \(x = 8\), E. \(x = 2\), F. \(x=-1\))