Question 910751
 <pre><font face = "Tohoma" size = 3 color = "indigo"><b> 
Hi
the vertex form of a Parabola opening up(a>0) or down(a<0), {{{y=a(x-h)^2 +k}}} 
where(h,k) is the vertex  and  x = h  is the Line of Symmetry
Y= -4x^2-12x-3   |completing Square of ax^2 + bx + c
y = -4(x - (-3/2))^2 + 9 - 3  |Note: -3/2 = -b/2a
 V(-3/2, 6), a=-4<0. Opens Downward, axis of symmetry x = -3/2
y = -4(x+3/2)^2 +6 
0 = -4(x+3/2)^2 +6
4(x+3/2)^2 = 6
(x+3/2)^2 = 6/4
x = -3/2 ± &#8730;6 / 2  roots
domain All real Numbers
range [6,&#8722;&#8734;) 
{{{drawing(300,300,   -10,10,-10,10,  blue(line(-1.5,10,-1.5,-10))  ,  
 grid(1),
circle(-1.5,6,0.4),
graph( 300, 300, -10,10,-10,10,0,-4x^2-12x-3))}}}