[tex]\bf \textit{vertex of a horizonal parabola, using f(y) for "x"} \\\\ x=\stackrel{\stackrel{a}{\downarrow }}{a}y^2\stackrel{\stackrel{b}{\downarrow }}{+b}y\stackrel{\stackrel{c}{\downarrow }}{+c} \qquad \left(f\left(-\cfrac{ b}{2 a}\right)~~~~ ,~~~~ -\cfrac{ b}{2 a} \right) \\\\[-0.35em] \rule{34em}{0.25pt}[/tex]
[tex]\bf 2x+y^2=0\implies 2x=-y^2\implies x=\cfrac{-y^2}{2}\implies x=\stackrel{\stackrel{a}{\downarrow }}{-\cfrac{1}{2}}y^2\stackrel{\stackrel{b}{\downarrow }}{+0}y\stackrel{\stackrel{c}{\downarrow }}{+0} \\\\\\ -\cfrac{b}{2a}\implies -\cfrac{0}{2\left(-\frac{1}{2} \right)}\implies 0\qquad therefore\qquad (f(0)~~,~~0)\implies \stackrel{vertex}{(0,0)}[/tex]
you can see it this way, x = -(1/2)y² is just a horizontal parabola opening to the left-hand-side, the -1/2 is just a stretch transformation of the parent function x = y², but as much as it stretches, their vertex is the same, at the origin.
