We can break the term [tex]2^{1-x}[/tex] apart using the following law of exponents:
[tex]a^{x+y}=a^xa^y[/tex]
Applying that, we find that
[tex]2^{1-x}=2^{1+(-x)}=2^1\cdot2^{-x}=2\cdot \frac{1}{2^x} [/tex]
Substituting that into our original function, we have
[tex]f(x)=-4\cdot2^{1-x}\\
f(x)=-4\cdot2\cdot \frac{1}{2^x} [/tex]
Which we can rewrite in the form [tex]f(x)=ab^x[/tex] as
[tex]f(x)=-8\cdot\big( \frac{1}{2}\big)^x[/tex]
