Find the vertical asymptotes, if any, and the values of x corresponding to holes , if any, of the graph of f(x)=(x)/(x-4) Select the correct choice below and, if necessary fill in the answer box to complete your choice. (Type A. The vertical asymptote(s) is(are) square There are no holes. B. There are no vertical asymptotes but there is(are) hole(s corresponding to square C. The vertical asymptote(s) is(are) square and hole(s)corresponding to square D. There are no discontinuities.

To determine the vertical asymptotes and holes of the function \( f(x) = \frac{x}{x-4} \), we need to analyze the denominator and numerator.<br /><br />1. **Vertical Asymptotes:**<br /> Vertical asymptotes occur where the denominator is zero, provided that the numerator is not also zero at those points (which would indicate a hole instead).<br /><br /> The denominator of \( f(x) \) is \( x - 4 \). Setting the denominator equal to zero gives:<br /> \[<br /> x - 4 = 0 \implies x = 4<br /> \]<br /> Since the numerator \( x \) is not zero when \( x = 4 \), there is a vertical asymptote at \( x = 4 \).<br /><br />2. **Holes:**<br /> Holes occur where both the numerator and the denominator are zero simultaneously, indicating a common factor that can be canceled out. <br /><br /> In this case, the numerator \( x \) is zero when \( x = 0 \), but the denominator \( x - 4 \) is not zero at \( x = 0 \). Therefore, there are no common factors to cancel out, and thus no holes in the graph.<br /><br />Based on this analysis, the correct choice is:<br /><br />**A. The vertical asymptote(s) is(are) \( x = 4 \). There are no holes.**