Answer
The potential at $x = 3.0~m$ is $-550~V$
Work Step by Step
We can write an expression for the potential difference:
$\Delta V = -\Delta x~E_x$
The potential difference is equal to the the negative of the area under the E versus x graph. We can find the area under the E versus x graph between x = 0 m and x = 3.0 m:
$area = (2.0~m)(200~V/m)+\frac{1}{2}(1.0~m)(200~V/m)$
$area = 500~V$
The potential difference between $x_i = 0~m$ and $x_f = 3.0~m$ is $-500~V$. Since the potential at the origin is $-50~V$, then the potential at $x = 3.0~m$ is $-550~V$
