To complete the table for the linear function \( h(x) \), we first need to identify the values for each entry based on the linearity of the function. We can generate a linear equation in the form of \( h(x) = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept.
The first step is to identify some points from the table based on the given data. From the available information, we see the following filled entries:
- \( h(3) = 25 \)
- \( h(-6) = 7 \)
We can use these two points to determine the slope (\( m \)):
- The coordinates from the known points are (3, 25) and (-6, 7).
- The slope \( m \) is calculated as:
\[ m = \frac{h(3) - h(-6)}{3 - (-6)} = \frac{25 - 7}{3 + 6} = \frac{18}{9} = 2 \]
Using the slope \( m = 2 \), we can use one of the points to find the y-intercept \( b \). We'll use the point (3, 25):
\[ h(x) = 2x + b \] \[ 25 = 2(3) + b \rightarrow b = 25 - 6 = 19 \]
Now we have the linear function:
\[ h(x) = 2x + 19 \]
Now we can fill in the table using this function for the other values of \( x \):
- For \( x = -7 \): \[ h(-7) = 2(-7) + 19 = -14 + 19 = 5 \]
- For \( x = -5 \): \[ h(-5) = 2(-5) + 19 = -10 + 19 = 9 \]
- For \( x = -4 \): \[ h(-4) = 2(-4) + 19 = -8 + 19 = 11 \]
- For \( x = -3 \): \[ h(-3) = 2(-3) + 19 = -6 + 19 = 13 \]
Now we can complete the table:
\[ \begin{array}{c|c} x & h(x) \ \hline -7 & 5 \ -6 & 7 \ -5 & 9 \ -4 & 11 \ -3 & 13 \ 3 & 25 \ \end{array} \]
So the completed table is:
\[ \begin{array}{c|c} x & h(x) \ \hline -7 & 5 \ -6 & 7 \ -5 & 9 \ -4 & 11 \ -3 & 13 \ 3 & 25 \ \end{array} \]
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