We can do this by completing the square:
#x^2+10x+8y=-17#
#(x^2+10x+25)+8y=-17+25#
#(x+5)^2+8y=8#
#8y=-(x+5)^2+8#
#y=-1/8(x+5)^2+1#
The vertex is #(-5,1)# because the vertex, in this form of the equation, is #(-h, k)#. #-5# is #-h# and #1# is #k#.
To find the focus and the directrix, we need to know #p#, which is the distance from the vertex to both the focus and the directrix. We can find #p# by setting the scale factor (#-1/8#) equal to #-1/(4p)#:
#-1/8=-1/(4p)#
#8=4p#
#2=p#
Before we find the focus and the directrix, we must remember that the graph faces down and that the vertex is #(-5,1)#. The focus will be 2 units down from the vertex since the focus is always within the area covered by the parabola. The focus, therefore, is #(-5,-1)#. The directrix is the line #2# units away from the vertex in the opposite direction. Since the graph faces down, the line will be horizontal and at #y=3#. Here's the graph of our equation in case you're confused:
graph{-1/8(x+5)^2+1 [-15, 5, -6, 4]}
