To analyze the values given in the table, we need to find the missing values of \( y \) corresponding to the specified \( x \) values and see which of the options (A, B, C, or D) fits best.
Let's summarize the data in the provided table:
\[ \begin{array}{|c|c|} \hline x & y \ \hline 0 & ? \ 2 & 18 \ ? & 19 \ 6 & ? \ 8 & 21 \ \hline \end{array} \]
Now, we can determine the missing values:
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For \( x = 0 \): Using the change from \( x = 2 \) (where \( y = 18 \)) to \( x = 8 \) (where \( y = 21 \)): \[ \text{Rate of change} = \frac{21 - 18}{8 - 2} = \frac{3}{6} = 0.5 \] This indicates the slope \( m = 0.5 \).
So, to find \( y \) when \( x = 0 \): \[ y = mx + b \] At \( x = 2 \), \( y = 18 \): \[ 18 = 0.5(2) + b \implies 18 = 1 + b \implies b = 17 \]
Now we can find \( y \) at \( x = 0 \): \[ y = 0.5(0) + 17 = 17 \]
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For \( x \) where \( y = 19 \): If we interpolate between \( 18 \) (at \( x = 2 \)) and \( 21 \) (at \( x = 8 \)): \[ 19 \text{ would occur when } x = 4. \] Thus: \[ y = 19 \text{ at } x = 4. \]
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For \( x = 6 \): From the data: \[ y = 0.5(6) + 17 = 3 + 17 = 20. \]
Now the table after filling in the blanks should look like this:
\[ \begin{array}{|c|c|} \hline x & y \ \hline 0 & 17 \ 2 & 18 \ 4 & 19 \ 6 & 20 \ 8 & 21 \ \hline \end{array} \]
Now let's match these findings to the options given:
- Option A: Initial value \( b = \frac{1}{2} \), rate of change \( m = 17 \) -> not a match.
- Option B: Initial value \( b = 17 \), rate of change \( m = \frac{1}{2} \) -> matches with our findings!
- Option C: Initial value \( b = 18 \), rate of change \( m = 2 \) -> not a match.
- Option D: Initial value \( b = 2 \), rate of change \( m = 18 \) -> not a match.
Therefore, the correct answer is B: initial value \( b = 17 \), rate of change \( m = \frac{1}{2} \).