Summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of \( f(x)=\frac{8 x}{x^{2}-1} \) Find the domain of \( f(x) \). Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The domain is all real \( x \), except \( x=\square \). (Type an integer or a decimal. Use a comma to separate answers as needed.) B. The domain is all real \( x \). Find the \( x \)-intercepts of \( f(x) \). Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The \( x \)-intercept(s) is/are at \( x=\square \). (Type an integer or a decimal. Use a comma to separate answers as needed.) B. There are no \( x \)-intercepts.

To find the domain of \( f(x) = \frac{8x}{x^2 - 1} \), we need to identify where the denominator is not equal to zero. The denominator \( x^2 - 1 = 0 \) when \( x^2 = 1 \), or \( x = 1 \) and \( x = -1 \). Therefore, the domain is all real \( x \), except \( x = 1, -1 \). For the \( x \)-intercepts, we set \( f(x) = 0 \). This occurs when the numerator is zero, which is when \( 8x = 0 \) or \( x = 0 \). Thus, the \( x \)-intercept is at \( x = 0 \). The answers are: Domain: A. The domain is all real \( x \), except \( x = 1, -1 \). \( x \)-intercept: A. The \( x \)-intercept is at \( x = 0 \). --- Ever wondered how graphs can tell a story? Let's sketch the function \( f(x) = \frac{8x}{x^2 - 1} \)! First, plot the \( x \)-intercept at \( (0, 0) \) and vertical asymptotes at \( x = 1 \) and \( x = -1 \). The graph approaches these asymptotes but never touches them, showcasing its dramatic dance between quadrants. Feeling a bit overwhelmed? A common mistake is forgetting to check vertical asymptotes when determining the domain! Just remember, always look for values that make the denominator zero and don't forget to include those pesky exceptions. Happy graphing!