Answer
$\displaystyle \quad y^{2}=\frac{1}{2}(x+2)$
Work Step by Step
Opens $right$. By Table 2,
$\begin{array}{|c|c|c|c|} \hline
{focus}&{directrix}&{equation}& ... opens\\ \hline
{(h+a, k)}&{x=h-a}&{(y-k)^{2}=4a(x-h)}& right\\ \hline \end{array}$
Read from the graph: $\quad (h,k)=(-2,0),$
so the equation has form $\quad y^{2}=4a(x+2)$
Find $4a$ by using the point on the graph (insert its coordinates into the equation)
$1^{2}=4a(0+2)$
$1=4a\cdot 2$
$4a=\displaystyle \frac{1}{2}$
The equation is $\displaystyle \quad y^{2}=\frac{1}{2}(x+2)$
