## College Algebra (10th Edition)

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