The described function has a discontinuity at \( x = 3 \). The behavior of the function indicates that as \( x \) approaches 3 from the left, the function approaches negative infinity, and from the right of \( x = 3 \), the function starts at the top and decreases. Since there is a gap there (the left and right hand limits do not converge to the same value at that point), this situation represents a vertical asymptote rather than a removable discontinuity.
Therefore, the correct conclusion is:
The function has a nonremovable discontinuity at \( x = 3 \).