Answer
$\begin{array}{|c|r|r|r|r|r|r|r|}
\hline
x & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\
\hline
h(x) &-24 & -12 & -6 & -3 & -\displaystyle \frac{3}{2} & -\displaystyle \frac{3}{4} & -\displaystyle \frac{3}{8} \\
\hline
\end{array}$
Technology formula:$\quad -3*2$^$(-x)\quad$or$ \quad-3*(1/2)$^$x$
Work Step by Step
$-3(2^{-x})=-3\displaystyle \times(\frac{1}{2})^{x}$
Technology formula:$\quad -3*2$^$(-x)\quad$or$ \quad-3*(1/2)$^$x$
$f(-3)=-3(\displaystyle \frac{1}{2})^{-3}=-3(2^{3})=-24\qquad $tech: $-3*(1/2)$^(-3)
$f(-2)=-3(\displaystyle \frac{1}{2})^{-2}=-3(2^{2})=-12\qquad $tech: $-3*(1/2)$^(-2)
$f(-1)=-3(\displaystyle \frac{1}{2})^{-1}=-3(2)=-6\qquad $tech: $-3*(1/2)$^(-1)
$f(0)=-3(\displaystyle \frac{1}{2})^{0}=-3(1)=-3\qquad $tech: $-3*(1/2)$^0
$f(1)=-3(\displaystyle \frac{1}{2})^{1}=-3(\frac{1}{2})=-\frac{3}{2}\qquad $tech: $-3*(1/2)$^1
$f(2)=-3(\displaystyle \frac{1}{2})^{2}=-3(\frac{1}{4})=-\frac{3}{4}\qquad $tech: $-3*(1/2)$^2
$f(3)=-3(\displaystyle \frac{1}{2})^{3}=-3(\frac{1}{2^{3}})=-\frac{3}{8}\qquad $tech: $-3*(1/2)$^3
