Question 810496
<pre>
The green line is the directrix x+y+1 = 0

{{{drawing(500,500,-1.5,3,-1.5,3,
graph(500,500,-1.5,3,-1.5,3,x-sqrt(4x+2)+1),
graph(500,500,-1.5,3,-1.5,3,x+sqrt(4x+2)+1),
green(line(-8,7,10,-11)), locate(.4,-1.3,"directrix"),
locate(.1,.2,FOCUS),locate(.5,.2,"(0,0)"),circle(-.25,-.25,.025),
circle(-.25,-.25,.02),circle(-.25,-.25,.015),circle(-.25,-.25,.028),
locate(-.42,-.18,V),
red(line(-.5,.3,-.5,.7))
  )}}}

The vertex V is half-way between the focus and the directrix on
a perpendicular from the focus to the directrix, the blue line
below:

{{{drawing(500,500,-1.5,3,-1.5,3,
graph(500,500,-1.5,3,-1.5,3,x-sqrt(4x+2)+1),
graph(500,500,-1.5,3,-1.5,3,x+sqrt(4x+2)+1),
green(line(-8,7,10,-11)), locate(.4,-1.3,"directrix"),
locate(.1,.2,FOCUS),locate(.5,.2,"(0,0)"),circle(-.25,-.25,.025),
circle(-.25,-.25,.02),circle(-.25,-.25,.015),circle(-.25,-.25,.028),
locate(-.42,-.18,V),
red(line(-.5,.3,-.5,.7)), blue(line(0,0,-1/2,-1/2))
  )}}}

Draw perpendiculars from the end of that blue line and from 
the vertex V to the x-axis:

{{{drawing(500,500,-1.5,3,-1.5,3,
graph(500,500,-1.5,3,-1.5,3,x-sqrt(4x+2)+1),
graph(500,500,-1.5,3,-1.5,3,x+sqrt(4x+2)+1),
green(line(-8,7,10,-11)), locate(.4,-1.3,"directrix"),
locate(.1,.2,FOCUS),locate(.5,.2,"(0,0)"),circle(-.25,-.25,.025),
circle(-.25,-.25,.02),circle(-.25,-.25,.015),circle(-.25,-.25,.028),
locate(-.4,-.18,V), 
red(line(-.5,.3,-.5,.7)), blue(line(0,0,-1/2,-1/2),line(-.5,-.5,-.5,0),line(-.25,-.25,-.25,0))
  )}}}

Now with a little bit of geometry knowledge, it's easy to 
see that the vertex is {{{(matrix(1,3,-1/4,",",-1/4))}}}

Edwin</pre>