Question 1048415
y = -x2-4x+5
y = -(x^2 + 4x) + 5
y = -[(x+2)^2 -4] + 5
y = -(x+2)^2 +9 Parabola opening Downward -1 < 0  
V(-2, 9) when y = 0 (x+2) = ±3, x = 1 or x = -5
x = -2 the Line of symmetry Blue
{{{drawing(300,300,  -10, 10, -10, 10,  grid(1), blue(line(-2,10,-2,-10)) ,

circle(-2, 9,0.4),
circle(1, 0,0.4),
circle(-5, 0,0.4),
graph( 300, 300,  -10, 10, -10, 10,0,-(x+2)^2 +9))}}}
|
Need to Know
the vertex form of a Parabola opening up(a>0) or down(a<0), 
{{{y=a(x-h)^2 +k}}}. V(h, k)