Question

ifying multiple vertical asymptotes
what are the vertical asymptotes of $f(x) = \frac{10}{x^2 - 1}$? choose two correct answers.
$x = -1$
$x = 1$
$x = 10$

Explanation:

Step1: Recall Vertical Asymptote Rule

Vertical asymptotes of a rational function occur where the denominator is zero (and numerator is non - zero). For \(f(x)=\frac{10}{x^{2}-1}\), factor the denominator: \(x^{2}-1=(x - 1)(x + 1)\) (using the difference of squares formula \(a^{2}-b^{2}=(a - b)(a + b)\) where \(a=x\) and \(b = 1\)).

Step2: Find Denominator Zeros

Set the denominator equal to zero: \((x - 1)(x + 1)=0\). Using the zero - product property (if \(ab = 0\), then \(a = 0\) or \(b = 0\)), we get \(x-1=0\) or \(x + 1=0\). Solving \(x-1=0\) gives \(x = 1\), and solving \(x + 1=0\) gives \(x=-1\). The numerator \(10
eq0\) for all real \(x\), so these are the vertical asymptotes. \(x = 10\) does not make the denominator zero, so it is not a vertical asymptote.

Answer:

A. \(x=-1\), B. \(x = 1\)