Question 1041196
<pre><b>
{{{(x+6)^2=8(-3y+17)}}}

We have to get it in the standard form:

{{{x-h)^2=4p(y-k)}}}

where (h,k) is the vertex, and |p| is the distance
from the vertex to the focus and also the distance
from the vertex to the directrix.  If p is positive
the parabola opens upward and if p is negative the
parabola opens downward.

The left side of

{{{(x+6)^2=8(-3y+17)}}}

is already in that form.  We multiply out the right
side

{{{(x+6)^2=-24y+136)}}}

Now we factor out -24

{{{(x+6)^2=-24(y-136/24)}}}

Reduce the fraction:

{{{(x+6)^2=-24(y-17/3)}}}

Now we can compare it to

{{{x-h)^2=4p(y-k)}}}

and see that 
-h=+6 so h=-6
-k = -17/3 so k = 17/3
4p = -24, so p = -6

The vertex is (h,k) or (-6,17/3)
The parabola opens downward, since
p is a negative number.

We sketch the parabola by getting a few points
It goes through the vertex (-6,17/3) as well
as the points (-14,3), (-10,5), (-2,5), (2,3),
(10,-5)

{{{drawing(400,3400/13,-19,7,-3,14,
circle(-6,17/3,.15),
circle(-6,17/3,.2),
circle(-6,17/3,.25),
circle(-6,17/3,.3),

locate(-9,8,Vertex(matrix(1,3,-6,",",17/3))),
graph(400,3400/13,-19,7,-3,14,(-x^2-12x+100)/24)  )}}}
 
The focus (which is inside the parabola is |p| or 17/6
units below the vertex.  It has the same x-coordinate -6
So to find its y-coordinate we subtract 6 from the
y-coordinate of the vertex 17/3:

{{{17/3-6=17/3-18/3=-1/3}}}  

So the focus is (-6,-1/3)

{{{drawing(400,3400/13,-19,7,-3,14,
circle(-6,17/3,.15),
circle(-6,17/3,.2),
circle(-6,17/3,.25),
circle(-6,17/3,.3),

locate(-9,8,Vertex(matrix(1,3,-6,",",17/3))),

circle(-6,-1/3,.15),
circle(-6,-1/3,.2),
circle(-6,-1/3,.25),
circle(-6,-1/3,.3),

locate(-9,-.5,Focus(matrix(1,3,-6,",",-1/3))),




graph(400,3400/13,-19,7,-3,14,(-x^2-12x+100)/24)  )}}}

Finally we will draw the directrix, which is a horizontal
line (in green below) which is |p| = 6 units outside (above)
the vertex.  We determine how far above the x-axis that is
by adding 6 to the y-coordinate of the vertex 17/3:

{{{17/3+6=17/3+18/3=35/3}}}

So the directrix is a horizontal line y = 35/3

{{{drawing(400,3400/13,-19,7,-3,14,
circle(-6,17/3,.15),
circle(-6,17/3,.2),
circle(-6,17/3,.25),
circle(-6,17/3,.3),
green(line(-25,35/3,10,35/3)),
locate(-9,8,Vertex(matrix(1,3,-6,",",17/3))),
locate(-9,14,Directrix(y=35/3)),
circle(-6,-1/3,.15),
circle(-6,-1/3,.2),
circle(-6,-1/3,.25),
circle(-6,-1/3,.3),

locate(-9,-.5,Focus(matrix(1,3,-6,",",-1/3))),




graph(400,3400/13,-19,7,-3,14,(-x^2-12x+100)/24)  )}}}

  
The axis of symmetry is the vertical line (in blue below) 
through both the vertex and the focus which cuts the
parabola in two.  It is the line x=(the x-coordinate of
the vertex and focus).  In this case the equation of the
axis of symmetry is x=-6:

{{{drawing(400,3400/13,-19,7,-3,14,
circle(-6,17/3,.15),
circle(-6,17/3,.2),
circle(-6,17/3,.25),
circle(-6,17/3,.3),
green(line(-25,35/3,10,35/3)),
locate(-9,8,Vertex(matrix(1,3,-6,",",17/3))),
locate(-9,14,Directrix(y=35/3)),
circle(-6,-1/3,.15),
circle(-6,-1/3,.2),
circle(-6,-1/3,.25),
circle(-6,-1/3,.3),

locate(-9,-.5,Focus(matrix(1,3,-6,",",-1/3))),

blue(line(-6,17,-6,-5)),


graph(400,3400/13,-19,7,-3,14,(-x^2-12x+100)/24)  )}}}
  
Edwin</pre>