[tex]\bf ~~~~~~\textit{parabola vertex form}
\\\\
\begin{array}{llll}
\boxed{y=a(x- h)^2+ k}\\\\
x=a(y- k)^2+ h
\end{array}
\qquad\qquad
vertex~~(\stackrel{}{ h},\stackrel{}{ k})\\\\
-------------------------------\\\\
y=\cfrac{1}{2}(x-\stackrel{h}{3})^2+\stackrel{k}{5}\qquad \qquad vertex~(3,5)[/tex]
since the squared variable in this case is the "x", that means is a vertical parabola, and so the axis of symmetry is the vertical line running over the x-coordinate of the vertex.
since the x-coordinate is 3, then x = 3.
