Answer
See graph, bounded, $\left(2,0\right),\left(4,0\right),\left(0,1\right),\left(0,8\right)$.
Work Step by Step
See graph for $\begin{cases}x\ge0\\y\ge0 \\ 2x+y\le8\\ x+2y\ge2 \end{cases}$
It is bounded, and we can identify the corner(s) as $\left(2,0\right),\left(4,0\right),\left(0,1\right),\left(0,8\right)$.
