Question 558739
<pre>
The standard form of a parabola whose axis is symmetry is vertical is

(x - h)² = 4p(y - k)      [Some books use "a" or "c" instead of "p"]

Where (h,k) is the vertex.  |p| is the diatance from the vertex to
the focus (which is a point inside the parabola on its axis of symmetry),
and also to the directrix, which is a line outside the parabola 
perpendicular to its line of symmetry.  If p is positive the parabola
opens upward, and if p is negative it opens downward.

We compare your equation to that one:

(x - 2)² = y + 3

To get it looking like 

(x - h)² = 4p(y - k)

we put parentheses around the right side and a 1 infront

(x - 2)² = 1(y + 3)

So we see that h = 2, k = -3, and 4p = 1 which makes p = {{{1/4}}}

So the vertex is (h,k) = (2,-3).

We plot the vertex (2,-3), and draw a green axis of symmetry through
it.

{{{drawing(400,400,-2,7,-5,4, graph(400,400,-2,7,-5,4),
circle(2,-3,.08), green(line(2,20,2,-20))  )}}}

That green axis of symmetry goes through x = 2, so that's its equation.

The vertex is a point p or {{{1/4}}} of a unit above the vertex. It is 
on the axis of symmetry so it's x-coordinate is the same as the x-coordinate
of the vertex, which is 2, but its y-coordinate is {{{1/4}}} of a unit
more, so we add {{{1/4}}} to the y-coordinate of the vertex:

-3+{{{1/4}}} = {{{-12/4}}}+{{{1/4}}} = {{{-11/4}}}, 
So the focus has the coordinates (2,{{{-11/4}}})

We draw the focus:

{{{drawing(400,400,-2,7,-5,4, graph(400,400,-2,7,-5,4),
circle(2,-3,.08), green(line(2,20,2,-20)), circle(2,-11/4,.08)  )}}}

The directrix is a horizontal line p or {{{1/4}}} of a unit below the vertex    

We draw it in blue:


{{{drawing(400,400,-2,7,-5,4, graph(400,400,-2,7,-5,4),
circle(2,-3,.08), green(line(2,20,2,-20)), circle(2,-11/4,.08)
blue(line(-20,-3-1/4,20,-3-1/4)) 

  )}}}


Since the line is {{{1/4}}} unit below the vertex, we subtract {{{1/4}}} from
its y-coordinate -3-{{{1/4}}} = {{{-12/4}}}-{{{1/4}}} = {{{-13/4}}},

so the equation of the directrix is y = {{{-13/4}}}

We draw two adjacent squares, with a common side from the directrix
to the focus, like this:

{{{drawing(400,400,-2,7,-5,4, graph(400,400,-2,7,-5,4),
circle(2,-3,.08), green(line(2,20,2,-20)), circle(2,-11/4,.08)
blue(line(-20,-3-1/4,20,-3-1/4)), rectangle(2-1/2,-3-1/4,2+1/2,-3+1/4) 

  )}}}

and sketch in the parabola through the upper corners of those squares and
through the vertex:

{{{drawing(400,400,-2,7,-5,4, graph(400,400,-2,7,-5,4, (x-2)^2-3),
circle(2,-3,.08), green(line(2,20,2,-20)), circle(2,-11/4,.08)
blue(line(-20,-3-1/4,20,-3-1/4)), rectangle(2-1/2,-3-1/4,2+1/2,-3+1/4) 

  )}}}

Edwin</pre>