+206 Find the largest value of $c$ such that $1$ is in the range of $f(x)=x^2-5x+c$.
+1223 Complete the square: \(f(x) = x^2 - 5x + c = (x - \frac{5}{2})^2 + (c - \frac{25}{4})\)
So, \(c - \frac{25}{4} = 1\) which makes the minimum value of c equal to 29/4.
