Answer
$\quad (y-1)^{2}=x$
Work Step by Step
1. $\quad $ Axis of symmetry is parallel to the $x$ -axis, opens right
2. $\quad $By table 2,
$\begin{array}{|c|c|c|c|}
\hline{focus}&{directrix}&{equation}& ... opens\\\hline
{(h+a, k)}&{x=h-a}&{(y-k)^{2}=4a(x-h)}& right \\\hline \end{array}$
3. $\quad $Read from the graph: $\quad (h,k)=(0,1),$
so the equation has form $(y-1)^{2}=4ax$
4. $\quad $find $a$ by using the point on the graph (insert its coordinates into the equation of step 3:
$(2-1)^{2}=4a(1)$
$1=4a$
$a=\displaystyle \frac{1}{4}$
$4a=1\quad $so the equation (3) is $\quad (y-1)^{2}=x$
