Find the vertical asymptote. \( f(x)=\frac{x^{2}-2 x-8}{3 x-2} \) \( x=2 / 3 \) \( x=3 / 2 \) \( y=2 / 3 \)

To find the vertical asymptote of the function \( f(x) = \frac{x^{2}-2x-8}{3x-2} \), we need to look for values of \( x \) that make the denominator zero, as these points indicate where the function may tend toward infinity. Setting the denominator \( 3x - 2 = 0 \) gives us \( x = \frac{2}{3} \). So, the vertical asymptote of this function is at \( x = \frac{2}{3} \). Additionally, remember that vertical asymptotes only occur where a function is undefined due to division by zero and where there is no corresponding zero in the numerator. In this case, the numerator \( x^{2}-2x-8 \) does not equal zero when \( x = \frac{2}{3} \), confirming that this indeed is a vertical asymptote!