Question

How do you find the vertex of 2x+y^2=0

Answer 1

`\bf \textit{vertex of a horizonal parabola, using f(y) for`

`\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(-(1)/(2) \right)}\implies 0\qquad therefore\qquad (f(0)~~,~~0)\implies \stackrel{vertex}{(0,0)}`

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.