Given that $ \minus{} 4\le x\le \minus{} 2$ and $ 2\le y\le 4$, what is the largest possible value of $ (x \plus{} y)/x$?

$ \textbf{(A)}\ \minus{}\!1\qquad
\textbf{(B)}\ \minus{}\!\frac {1}{2}\qquad
\textbf{(C)}\ 0\qquad
\textbf{(D)}\ \frac {1}{2}\qquad
\textbf{(E)}\ 1$
1. We are given the ranges for \( x \) and \( y \):
   \[
   -4 \leq x \leq -2 \quad \text{and} \quad 2 \leq y \leq 4
   \]
2. We need to find the largest possible value of the expression \(\frac{x + y}{x}\).

3. Rewrite the expression:
   \[
   \frac{x + y}{x} = \frac{x}{x} + \frac{y}{x} = 1 + \frac{y}{x}
   \]
   Therefore, we need to maximize \(1 + \frac{y}{x}\).

4. Since \( x \) is negative, \(\frac{y}{x}\) will be negative if \( y \) is positive. To maximize \(1 + \frac{y}{x}\), we need to minimize the magnitude of the negative term \(\frac{y}{x}\).

5. To minimize \(\frac{y}{x}\), we need to choose the smallest possible value of \( y \) and the largest (least negative) value of \( x \):
   \[
   y = 2 \quad \text{and} \quad x = -2
   \]
6. Substitute these values into the expression:
   \[
   \frac{x + y}{x} = \frac{-2 + 2}{-2} = \frac{0}{-2} = 0
   \]

7. Check other combinations to ensure this is the maximum:
   - For \( y = 4 \) and \( x = -2 \):
     \[
     \frac{x + y}{x} = \frac{-2 + 4}{-2} = \frac{2}{-2} = -1
     \]
   - For \( y = 2 \) and \( x = -4 \):
     \[
     \frac{x + y}{x} = \frac{-4 + 2}{-4} = \frac{-2}{-4} = \frac{1}{2}
     \]
   - For \( y = 4 \) and \( x = -4 \):
     \[
     \frac{x + y}{x} = \frac{-4 + 4}{-4} = \frac{0}{-4} = 0
     \]

8. The largest value among these is \(\frac{1}{2}\).

The final answer is \(\boxed{\frac{1}{2}}\)