Question 973817
<pre>
The standard form for a parabola with a y<sup>2</sup> term is

{{{a(y-k)^2}}}{{{""=""}}}{{{x-h}}} where (h,k) is the vertex 
   
We can get 

{{{y^2}}}{{{""=""}}}{{{2x+4}}} 

in standard form: 

Write y as (y-0)

{{{(y-0)^2}}}{{{""=""}}}{{{2x+4}}}

Factor out 2 on the right:

{{{(y-0)^2}}}{{{""=""}}}{{{2(x+2)}}}
 
Multiply both sides by {{{1/2}}}

{{{expr(1/2)(y-0)^2}}}{{{""=""}}}{{{expr(1/2)2(x+2)}}}

{{{expr(1/2)(y-0)^2}}}{{{""=""}}}{{{expr(1/cross(2))cross(2)(x+2)}}}

{{{expr(1/2)(y-0)^2}}}{{{""=""}}}{{{x+2}}}

Compare to

{{{a(y-k)^2}}}{{{""=""}}}{{{x-h}}}

-k=-0,  a=1/2,  -h=+2
 k=0             h=-2

So the vertex is (h,k) = (-2,0), the point where the big dot is
in the graph below:

{{{drawing(400,400,-5,5,-5,5,
graph(400,400,-5,5,-5,5,sqrt(2x+4)),
graph(400,400,-5,5,-5,5,-sqrt(2x+4)),


circle(-2,0,0.15),circle(-2,0,0.13),circle(-2,0,0.11),circle(-2,0,0.09),circle(-2,0,0.07),circle(-2,0,0.05),circle(-2,0,0.03),circle(-2,0,0.01) )}}}

Edwin</pre>