To determine which graph accurately represents the function \( f(x) = 12x + 3 \) with the specified inputs of \( -4, -2, 0, 2, \) and \( 4 \), we will first calculate the outputs for these inputs:
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For \( x = -4 \): \[ f(-4) = 12(-4) + 3 = -48 + 3 = -45 \]
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For \( x = -2 \): \[ f(-2) = 12(-2) + 3 = -24 + 3 = -21 \]
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For \( x = 0 \): \[ f(0) = 12(0) + 3 = 0 + 3 = 3 \]
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For \( x = 2 \): \[ f(2) = 12(2) + 3 = 24 + 3 = 27 \]
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For \( x = 4 \): \[ f(4) = 12(4) + 3 = 48 + 3 = 51 \]
Now we summarize the input-output table:
| \( x \) | \( f(x) \) | |---------|------------| | -4 | -45 | | -2 | -21 | | 0 | 3 | | 2 | 27 | | 4 | 51 |
Next, we can plot these points on a coordinate plane:
- Point (-4, -45)
- Point (-2, -21)
- Point (0, 3)
- Point (2, 27)
- Point (4, 51)
Given that the outputs of this linear function result in relatively large negative values and positive values, we can observe that the graph would have to range from a low value (around -45) up to a high value (around 51).
Based on the descriptions of the images you provided, the graph that accurately represents this function will likely be the one that:
- Covers the y-axis values from -50 to 60 (to include the range of outputs)
- Markets points in the appropriate range for both x and \( f(x) \).
Since I do not have direct access to the images, I recommend choosing the graph that covers a wider range of y-values (extending down to -45 and up to 51) and matches the corresponding points from the table. Since all the other responses include graphs with limited ranges, the correct graph would almost certainly be one reflecting the required range and points accurately.