Question 750691
Identify the coordinates fo the vertex and focus, the equations of the axis of symmetry and directrix and the direction of opening of the parabola with equation
y= -2x^2-16x-27
complete the square:
y=-2(x^2+8x+16)+32-27
y=-2(x+4)^2+5
parabola opens downward:
vertex: (-4,5)
axis of symmetry: x=-4
standard (vertex) form of equation: {{{y=A(x-h)^2+k}}}, (h,k)=(x,y) coordinates of the vertex
basic form of equation: {{{(x-h)^2=-4p(y-k)}}}
y=-2(x+4)^2+5
-2(x+4)^2=(y-5)
(x+4)^2=-(1/2)(y-5)
4p=1/2
p=1/8
focus:(-4,39/8) (p-distance below vertex on the axis of symmetry)
directrix: y=41/8 (p-distance above vertex on the axis of symmetry)

see graph below as a visual check:

{{{ graph( 300, 300, -10, 10, -10, 10, -2x^2-16x-27) }}}