Find all vertical asymptote(s) of the function \( y=\frac{1}{x^{2}-3} \). Justify your answers.

To find the vertical asymptotes of the function \( y = \frac{1}{x^2 - 3} \), we need to identify the values of \( x \) that make the denominator zero since vertical asymptotes occur where the function approaches infinity. Set the denominator equal to zero: \[ x^2 - 3 = 0 \] Solving for \( x \), we find: \[ x^2 = 3 \] \[ x = \pm\sqrt{3} \] Thus, \( x = \sqrt{3} \) and \( x = -\sqrt{3} \) are the points where the vertical asymptotes occur. To justify that these points are indeed vertical asymptotes, we can observe the behavior of the function as \( x \) approaches these values. As \( x \) approaches \( \sqrt{3} \) or \( -\sqrt{3} \), the denominator approaches zero, causing the function \( y \) to approach infinity (or negative infinity). Therefore, there are vertical asymptotes at \( x = \sqrt{3} \) and \( x = -\sqrt{3} \).