\(\bf \textit{vertex form of a parabola}\\\\
\boxed{y=a(x-{{ h}})^2+{{ k}}}\\\\
x=a(y-{{ k}})^2+{{ h}}\qquad\qquad vertex\ ({{ h}},{{ k}})\\\\
-----------------------------\\\\
x^2=8y\implies (x-0)^2=8y\implies \cfrac{(x-0)^2}{8}=(y-0)
\\\\\\
\cfrac{1}{8}(x-0)^2=(y-0)\implies \boxed{\cfrac{1}{8}(x-0)^2+0=y}\)
