Question 22081
To begin with, you can tell by looking at this that it is a parabola, since it has one variable squared but not the other.  It opens either to the right or left because it has a {{{y^2}}}.  You can find the vertex by completing the square.
{{{y^2 - 12y + 4x + 4 =0}}}
{{{y^2 - 12y = -4x - 4 }}}


In order to make a perfect square trinomial on the left side of the equation, make sure you have a 1 for the coefficient of {{{y^2}}}, then take half of the -12 (which is -6) and square it (which is 36), and add + 36 to each side of the equation:
{{{y^2 - 12y ______= - 4x - 4 ______}}}
{{{y^2 - 12y +36 = - 4x - 4 + 36}}}
{{{ (y-6)^2 = - 4x + 32}}}


Factor out the -4 on the right side:
{{{ (y-6)^2 = - 4(x -8) }}}


Because the equation is in the form {{{y^2=-x}}}, it opens to the left.  To find the vertex, you must "zero out" the x and the y portions of the equation {{{ (y-6)^2 = - 4(x -8)}}} which means that y = 6 and x = 8, or the vertex is at (8,6).  


The graph should look like this parabola.  (I don't know what is making what looks like a vertical line in the graph--that should NOT be there!  Just the parabolic curve!!)


R^2 at SCC

{{{ graph(300,300, -10,10,-10,10, 6-sqrt(-4x+32), 6+sqrt(-4x+32)) }}}