To complete the table for the exponential function \( h(x) \), we can start by observing the values provided for \( h(x) \) at \( x = -4 \) and \( x = -3 \).
Given:
- \( h(-4) = 486 \)
- \( h(-3) = 162 \)
An exponential function typically has the form \( h(x) = a \cdot b^x \). We can use the values we have to find constants \( a \) and \( b \).
First, we calculate the ratio of \( h(-3) \) to \( h(-4) \):
\[ \frac{h(-3)}{h(-4)} = \frac{162}{486} = \frac{1}{3} \]
Since the function is exponential, this ratio can be expressed as follows:
\[ \frac{h(-3)}{h(-4)} = \frac{b^{-3}}{b^{-4}} = b^{-3 + 4} = b^1 = b \]
Thus, we find that \( b = \frac{1}{3} \).
Next, we can use one of the known values to solve for \( a \). We use \( h(-4) \):
\[ h(-4) = a \cdot b^{-4} = a \cdot \left(\frac{1}{3}\right)^{-4} = a \cdot 81 \]
Setting this equal to 486:
\[ a \cdot 81 = 486 \]
Solving for \( a \):
\[ a = \frac{486}{81} = 6 \]
Now we have the function:
\[ h(x) = 6 \cdot \left(\frac{1}{3}\right)^x \]
Next, we can calculate \( h(x) \) for the missing values of \( x \):
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For \( x = -2 \): \[ h(-2) = 6 \cdot \left(\frac{1}{3}\right)^{-2} = 6 \cdot 9 = 54 \]
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For \( x = -1 \): \[ h(-1) = 6 \cdot \left(\frac{1}{3}\right)^{-1} = 6 \cdot 3 = 18 \]
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For \( x = 0 \): \[ h(0) = 6 \cdot \left(\frac{1}{3}\right)^{0} = 6 \cdot 1 = 6 \]
Now we can fill in the completed table:
| x | h(x) | |-----|------| | -4 | 486 | | -3 | 162 | | -2 | 54 | | -1 | 18 | | 0 | 6 |
So the completed values for \( h(x) \) are:
- \( h(-2) = 54 \)
- \( h(-1) = 18 \)
- \( h(0) = 6 \)
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