Question 887887
<pre>
{{{"f(x)"}}}{{{""=""}}}{{{expr(1/2)(2^(3x))-4}}} to {{{"g(x)"}}}{{{""=""}}}{{{8^x}}}

The graph of {{{"f(x)"}}} is the red graph below

{{{graph(640/3,400,-5,3,-5,10, (1/2)(2^(3x))-4)}}}

Write that as:

{{{"f(x)"}}}{{{""=""}}}{{{expr(1/2)((2^3)^x)-4}}}

Then as

{{{"f(x)"}}}{{{""=""}}}{{{expr(1/2)(8^x)-4}}}

Then we form h(x) by adding 4 to the right side
which shifts the graph 4 units up

{{{"h(x)"}}}{{{""=""}}}{{{expr(1/2)(8^x)-4+red(4)}}}
{{{"h(x)"}}}{{{""=""}}}{{{expr(1/2)(8^x)}}}

The graph of h(x) is the green graph below

{{{graph(640/3,400,-5,3,-5,10, (1/2)(2^(3x))-4,(1/2)(2^(3x)))}}}

Finally we form g(x) by multiplying the right side of h(x) by 2.

{{{"g(x)"}}}{{{""=""}}}{{{red(2)*expr(1/2)(8^x)}}}
{{{"g(x)"}}}{{{""=""}}}{{{8^x}}}
  
which stretches the green graph by a factor of 2.  That's the same as 
if the graph of h(x) were on a rubber sheet and then we stretched the 
rubber sheet to double its size.

So the final graph of {{{"g(x)"}}}{{{""=""}}}{{{8^x}}} is the blue graph
below:

{{{graph(640/3,400,-5,3,-5,10, (1/2)(2^(3x))-4,(1/2)(2^(3x)),8^x)}}}

So the transformation is first to h(x), by adding 4 to the right side:

h(x) = f(x)+4

then the second transformation is from h(x) to g(x), by multiplying the
right side of h(x) by 2, or

g(x) = 2h(x)

then you can substitute f(x)+4 for h(x) and get

g(x) = 2[f(x)+4]

or 

g(x) = 2f(x)+8

Edwin</pre>