Precalculus (6th Edition)

$y=-(2^{x+2}-3 )$
The graph of the original function $f(x)=b^{x}$ passes through (0,1), and has the x-axis as the asymptote. The shape of this graph suggests there was a reflection about the x-axis. We reflect this graph, and get new points: (0,1), (-1,-1) and (-2,-2), and the asymptote is y=-3. To move the asymptote back to y=0, we raise this graph by three units. New points: (0,4), (-1,2) and (-2,1). Since f(x) passes through (0,1), we need to move (-2,1) to the right by 2 units.New points: (2,4), (1,2) and (0,1) The original function was $f(x)=2^{x},$ which was shifted left 2 units $(2^{x+2}),$ lowered by 3, ($2^{x+2}-3$ ), and reflected about the x-axis, resulting in $y=-(2^{x+2}-3 )$