now, let's take a peek at the graph, the vertex is at (-3,1) and the focus point is at (-1,1).
so, from x = -3, to x = -1, there are only 2 units, namely the focus point is 2 units away from the vertex, namely that "p" distance is 2 units. Because the parabola is opening to the right-hand-side, the "p" unit is positive, so p = +2.
[tex]\bf \textit{parabola vertex form with focus point distance} \\\\ \begin{array}{llll} 4p(x- h)=(y- k)^2 \end{array} \qquad \begin{array}{llll} vertex\ ( h, k)\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=-3\\ k=1\\ p=2 \end{cases}\implies 4(2)[x-(-3)]=(y-1)^2\implies 8(x+3)=(y-1)^2[/tex]
