Question 1090782
<pre>
I'll assume you meant {{{ y=-3(2^(x+6))-1}}} and not {{{ cross(y=-3(2^x+6)-1)}}},
which is what you wrote meant because you didn't put the 
exponent in parentheses and so the +6 didn't get included 
in the exponent when you don't put parentheses to include it.

Start with {{{y=2^x}}}.  I'll mark the point (1,2)
and show what happens to that point:
Domain (-oo,oo), Range (0,oo)

{{{drawing(400,400,-15,5,-10,10,
circle(1,2,.15),locate(1,2,"(1,2)"),
graph(400,400,-15,5,-10,10,2^x))}}}

Next replace x by x+6 which shifts the graph 6 units left.
This is the graph is of {{{y=2^(x+6)}}}.
Notice that the point (1,2) shifts horizontally 6 units left 
to (-5,2): 
Domain (-oo,oo), Range (0,oo)
{{{drawing(400,400,-15,5,-10,10,
circle(-5,2,.15),locate(-5,2,"(-5,2)"),
graph(400,400,-15,5,-10,10,2^(x+6)))}}}

Next multiply by -1, which reflects it into (across) the x-axis
This is the graph of {{{y=-(2^(x+6))}}}
Notice that the point (-5,2) reflects across the x-axis to (-5,-2): 
Domain (-oo,oo), Range (-oo,0)
{{{drawing(400,400,-15,5,-10,10,
circle(-5,-2,.15),locate(-5,-2,"(-5,-2)"),
graph(400,400,-15,5,-10,10,-2^(x+6)))}}}

Next multiply by 3 which stretches it vertically by a factor of 3.  
This is as if the graph were on a rubber sheet and we stretched the 
rubber sheet three times vertically.
This is the graph of {{{y=-3(2^(x+6))}}}
Notice that the point (-5,-2) stretches vertically by a factor of
3 to (-5,-6): 
Domain (-oo,oo), Range (-oo,0)
{{{drawing(400,400,-15,5,-10,10,
circle(-5,-6,.15),locate(-5,-6,"(-5,-6)"),
graph(400,400,-15,5,-10,10,(0-3)(2^(x+6))))}}}

Finally we add -1 which shifts it down 1 unit.
Notice that the point (-5,-6) shifts vertically down to (-5,-7) 
This is the final graph:
Domain (-oo,oo), Range (-oo,-1).  Notice that the horizontal
asymptote did not move until now. It moved from the x-axis (y=0)
to the green dotted line y=-1:
{{{drawing(400,400,-15,5,-10,10,
circle(-5,-7,.15),locate(-5,-7,"(-5,-7)"),
graph(400,400,-15,5,-10,10,-3(2^(x+6))-1,-sqrt(sin(9x))/sqrt(sin(9x))))}}}

Edwin</pre>