Answer
See graph and explanations.
Work Step by Step
Step 1. Given the equation $y=\frac{1}{2}x^2+1, x\geq0$, we have $x^2=2(y-1)$. Thus $4p=2, p=\frac{1}{2}$, with the parabola opening upward and a vertex at $(0,1)$
Step 2. We can identify its focus at $(0,\frac{3}{2})$ and directrix as $y=\frac{1}{2}$
Step 3. We can graph the equation $y=\frac{1}{2}x^2+1, x\geq0$ as shown in the figure.
