The value of $x^2-6x+13$ can never be less than:
$ \textbf{(A)}\ 4 \qquad\textbf{(B)}\ 4.5 \qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 13 $
1. The given expression is \( x^2 - 6x + 13 \). This is a quadratic function of the form \( ax^2 + bx + c \), where \( a = 1 \), \( b = -6 \), and \( c = 13 \).

2. To find the minimum value of a quadratic function \( ax^2 + bx + c \), we use the vertex formula. The x-coordinate of the vertex is given by \( x = -\frac{b}{2a} \).

3. Substituting \( a = 1 \) and \( b = -6 \) into the vertex formula:
   \[
   x = -\frac{-6}{2 \cdot 1} = \frac{6}{2} = 3
   \]

4. Now, we substitute \( x = 3 \) back into the original quadratic expression to find the y-coordinate of the vertex, which represents the minimum value of the function:
   \[
   x^2 - 6x + 13 \bigg|_{x=3} = 3^2 - 6 \cdot 3 + 13
   \]
   \[
   = 9 - 18 + 13
   \]
   \[
   = 4
   \]

5. Therefore, the minimum value of the expression \( x^2 - 6x + 13 \) is 4.

The final answer is \( \boxed{4} \)