Question 24734
#1: Change to the form y-k = 4p(x-h)^2
which is a parabola opening up.
your equation becomes y+4=-3(x+5)^2
Vertex is h=-5; k=-4
Axis of symmetry is x=-5

#2: y=2(x^2+6x)-3
Create the form y-k=4p(x-h)^2
  y+3= 2(x^2+6x+9)-18
  y+21=2(x+3)^2
This is the "vertex form" you want.

#3: Use the determinant (b^2-4ac) to see how many real
number roots you have.  Find the roots by using the 
quadratic equation.

#4: x^2 becomes (x+6)^2 when you shift "6" to the left
    Multiplying by 1/2 will stretch the (x+6)^2
    Adding 12 will give the (1/2)(x+6)^2 a vertical shift up.

#5: Rewrite as y+1=(x+2)^2
    which has vertex at (-2,-1)
I'll leave the last part of #5 to you.

Cheers,
Stan H.