ind the vertical asymptotes, if any, and the values of x corresponding to holes , if any, of the graph h(x)=(x)/(x(x-9)) 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. The vertical asymptote(s) is(are) square and hole(s) corresponding to square C. There are no vertical asymptotes but there is(are) hole(s)corresponding to square D. There are no discontinuities

To find the vertical asymptotes and holes of the function \( h(x) = \frac{x}{x(x-9)} \), we need to analyze the denominator, as vertical asymptotes occur where the denominator is zero and the numerator is not zero, while holes occur where both the numerator and denominator are zero.<br /><br />1. **Identify where the denominator is zero:**<br /><br /> The denominator is \( x(x-9) \). Set it equal to zero:<br /><br /> \[<br /> x(x-9) = 0<br /> \]<br /><br /> This gives us the solutions:<br /><br /> \[<br /> x = 0 \quad \text{or} \quad x = 9<br /> \]<br /><br />2. **Check the numerator at these points:**<br /><br /> The numerator is \( x \).<br /><br /> - At \( x = 0 \), the numerator is also zero (\( x = 0 \)).<br /> - At \( x = 9 \), the numerator is not zero (\( x = 9 \)).<br /><br />3. **Determine the type of discontinuity:**<br /><br /> - At \( x = 0 \), since both the numerator and denominator are zero, there is a hole.<br /> - At \( x = 9 \), since only the denominator is zero, there is a vertical asymptote.<br /><br />Based on this analysis, the correct choice is:<br /><br />**B. The vertical asymptote(s) is(are) \( x = 9 \) and hole(s) corresponding to \( x = 0 \).**