Question

the function h is defined by the following rule.\\( h(x) = 9^x \\)\
find \\( h(x) \\) for each x - value in the table.\

x	h(x)	\
-2	\\( \square \\)	\
-1	\\( \square \\)	\
0	\\( \square \\)	\
1	\\( \square \\)	\
2	\\( \square \\)

Explanation:

Step1: For \( x = -2 \)

Recall the negative exponent rule \( a^{-n}=\frac{1}{a^{n}} \). So \( h(-2) = 9^{-2}=\frac{1}{9^{2}}=\frac{1}{81} \).

Step2: For \( x = -1 \)

Using the negative exponent rule, \( h(-1)=9^{-1}=\frac{1}{9^{1}}=\frac{1}{9} \).

Step3: For \( x = 0 \)

Any non - zero number to the power of 0 is 1. So \( h(0)=9^{0} = 1 \).

Step4: For \( x = 1 \)

\( h(1)=9^{1}=9 \).

Step5: For \( x = 2 \)

\( h(2)=9^{2}=81 \).

Answer:

When \( x=-2 \), \( h(x)=\frac{1}{81} \); when \( x = - 1 \), \( h(x)=\frac{1}{9} \); when \( x = 0 \), \( h(x)=1 \); when \( x = 1 \), \( h(x)=9 \); when \( x = 2 \), \( h(x)=81 \)

The completed table is:

\( x \)	\( h(x) \)
-1	\(\frac{1}{9}\)
0	1
1	9
2	81