Certainly! Let's evaluate the function [tex]\( h(x) = \left(\frac{1}{10}\right)^x \)[/tex] for each given [tex]\( x \)[/tex] value step by step.
1. For [tex]\( x = -2 \)[/tex]:
[tex]\[
h(-2) = \left(\frac{1}{10}\right)^{-2}
\][/tex]
When a base with a negative exponent is found, we take the reciprocal of the base and change the exponent to positive:
[tex]\[
\left(\frac{1}{10}\right)^{-2} = 10^2 = 100
\][/tex]
Therefore,
[tex]\[
h(-2) = 100
\][/tex]
2. For [tex]\( x = -1 \)[/tex]:
[tex]\[
h(-1) = \left(\frac{1}{10}\right)^{-1}
\][/tex]
Similarly, taking the reciprocal of the base and changing the exponent to positive:
[tex]\[
\left(\frac{1}{10}\right)^{-1} = 10^1 = 10
\][/tex]
Therefore,
[tex]\[
h(-1) = 10
\][/tex]
3. For [tex]\( x = 0 \)[/tex]:
[tex]\[
h(0) = \left(\frac{1}{10}\right)^0
\][/tex]
Any non-zero number raised to the power of 0 is 1:
[tex]\[
\left(\frac{1}{10}\right)^0 = 1
\][/tex]
Therefore,
[tex]\[
h(0) = 1
\][/tex]
4. For [tex]\( x = 1 \)[/tex]:
[tex]\[
h(1) = \left(\frac{1}{10}\right)^1
\][/tex]
Any number raised to the power of 1 is the number itself:
[tex]\[
\left(\frac{1}{10}\right)^1 = \frac{1}{10} = 0.1
\][/tex]
Therefore,
[tex]\[
h(1) = 0.1
\][/tex]
5. For [tex]\( x = 2 \)[/tex]:
[tex]\[
h(2) = \left(\frac{1}{10}\right)^2
\][/tex]
Squaring the base 1/10:
[tex]\[
\left(\frac{1}{10}\right)^2 = \frac{1}{100} = 0.01
\][/tex]
Therefore,
[tex]\[
h(2) = 0.01
\][/tex]
Summarizing the results, we have:
[tex]\[
\begin{tabular}{|c|c|}
\hline
$x$ & $h(x)$ \\
\hline
-2 & 100 \\
\hline
-1 & 10 \\
\hline
0 & 1 \\
\hline
1 & 0.1 \\
\hline
2 & 0.01 \\
\hline
\end{tabular}
\][/tex]
