Answer
$\begin{array}{|c|r|r|r|r|r|r|r|}
\hline
x & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\
\hline
h(x) &-54 & -18 & -6 & -2 & -\displaystyle \frac{2}{3} & -\displaystyle \frac{2}{9} & -\displaystyle \frac{2}{27} \\
\hline
\end{array}$
Technology formula:$\quad -2*3$^$(-x)\quad$or$ \quad-2*(1/3)$^$x$
Work Step by Step
$-2(3)=-2\displaystyle \times(\frac{1}{3})^{x}$
Technology formula:$\quad -2*3$^$(-x)\quad$or$ \quad-2*(1/3)$^$x$
$f(-3)=-2(\displaystyle \frac{1}{3})^{-3}=-2(3^{3})=-54\qquad $tech: $-2*(1/3)$^(-3)
$f(-2)=-2(\displaystyle \frac{1}{3})^{-2}=-2(3^{2})=-18\qquad $tech: $-2*(1/3)$^(-2)
$f(-1)=-2(\displaystyle \frac{1}{3})^{-1}=-2(3^{1})=-6\qquad $tech: $-2*(1/3)$^(-1)
$f(0)=-2(\displaystyle \frac{1}{3})^{0}=-2(1)=-2\qquad $tech: $-2*(1/3)$^0
$f(1)=-2(\displaystyle \frac{1}{3})^{1}=-2(\frac{1}{3})=-\frac{2}{3}\qquad $tech: $-2*(1/3)$^1
$f(2)=-2(\displaystyle \frac{1}{3})^{2}=-2(\frac{1}{3^{2}})=-\frac{2}{9}\qquad $tech: $-2*(1/3)$^2
$f(3)=-2(\displaystyle \frac{1}{3})^{3}=-2(\frac{1}{3^{3}})=-\frac{2}{27}\qquad $tech: $-2*(1/)$^3
