Question 983231
Use the definition of a parabola and the distance formula.


Distance of points (x,y) are equally distant from (0,-2) as from (x,2).
{{{sqrt((x-0)^2+(y-(-2))^2)=sqrt((x-x)^2+(y-2)^2)}}}
{{{sqrt(x^2+(y+2)^2)=sqrt(0+(y-2)^2)}}}
{{{x^2+(y+2)^2=(y-2)^2}}}
{{{x^2+y^2+4y+4=y^2-4y+4}}}
{{{x^2+4y=-4y}}}
{{{x^2=-4y-4y}}}
{{{-8y=x^2}}}, not yet standard form but useful for understanding the derivation.
{{{highlight(y=-(1/8)x^2)}}}-----standard form.


That can also be shown as {{{y=-(1/8)(x-0)^2+0}}} to help show how you can read the standard form equation.  Vertex is (h,k) same as  (0,0).  The parabola has y-axis as its axis of symmetry and the parabola opens downward; the vertex is a maximum point.