[Notice that I had to change your asymptotes' equations, because
the equations you gave (y=±.5x) for asymptotes were not possible 
with a hyperbola with center C(-2,0), for the two asymptotes
must intersect at the center of the hyperbola, so I changed
them to have the equations y=±.5(x+2) instead, for either you
copied something wrong, or there was a typo in the book you
got the problem from.  But rather than do nothing, I went
ahead and changed the asymptotes.]  

First we plot what is given, the vertex, the center, and the two
asymptotes, whose equations are  and  
 

 
We see that the hyperbola opens upward and downward.
 
So we know its standard equation is this:
 
, where the center is C(h,k),
and so we know that the center is (-2,0), then h=-2, and k=0.
 
The center is halway between the vertices, so you are right
that the other vertex is (-2,-3). We plot it:




We connect the vertices to find the transverse axis



 
We can see that the transverse axis is 6 units long, and since the
transverse axis is  units long, then  and 
 
Next we draw the top and bottom of the defining rectangle, which
are the horizontal line segments between the two intercepts which
passes through the vertex:
 


It appears that the corners of the defining rectangle are at 
(-8,3), (4,3), (-8,-3), and (4,-3).  But we can't just look and see.

We know the y-coordinate of the corners of the defining rectangle
are 3 and -3.  So we substitute 3 and -3 for y into each of the asymptotes' 
equations:



We finish drawing in the defining rectangle, by
drawing its two vertical sides:



Now we can sketch in the hyperbola:




The conjugate axis is along the x-axis, and extends from (-8,0) to (4,0)
Tha conjugate axis has length 12, and since b = one-half of the
conjugate axis's length, b = 6.

Therefore the equation of the hyperbola is

, where h=-2, and k=0, a=3, b=6

, or

.

Edwin