$(y-1)^2=-12(x-3)$, vertex $(3,1)$, focus $(0,1)$, directrix $x=6$, see graph.
Work Step by Step
Step 1. Rewriting the equation as $y^2-2x+1=-12x+35+1$ or $(y-1)^2=-12(x-3)$, we have $4p=-12$ and $p=-3$ with the parabola opening to the left and vertex $(3,1)$. Step 2. We can find the focus at $(0,1)$ and directrix as $x=6$ Step 3. We can graph the parabola as shown in the figure.