To determine which table corresponds to the function \( y = kf(x) \), where the parent function is \( f(x) = |x + 4| \) and \( k = 3 \), we need to evaluate how that affects the output values.
The parent function \( f(x) = |x + 4| \) produces certain output values that will then be scaled by a factor of 3. Let's first compute the outputs of the parent function for the provided \( x \) values:
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For \( x = 0 \): \[ f(0) = |0 + 4| = |4| = 4 \quad \Rightarrow \quad y = 3 \cdot 4 = 12 \]
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For \( x = 1 \): \[ f(1) = |1 + 4| = |5| = 5 \quad \Rightarrow \quad y = 3 \cdot 5 = 15 \]
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For \( x = 2 \): \[ f(2) = |2 + 4| = |6| = 6 \quad \Rightarrow \quad y = 3 \cdot 6 = 18 \]
The resulting values we computed from the function \( y = 3f(x) \) for our chosen \( x \) values are:
- At \( x = 0 \), \( y = 12 \)
- At \( x = 1 \), \( y = 15 \)
- At \( x = 2 \), \( y = 18 \)
Now, let's analyze the given options to find a table that provides these output values:
- \( y = kf(x) \)
- \( x=0, y=4 \)
- \( x=1, y=5 \)
- \( x=2, y=6 \)
- \( y = kf(x) \)
- \( x=0, y=1.3 \)
- \( x=1, y=1.7 \)
- \( x=2, y=2 \)
- \( y = kf(x) \)
- \( x=0, y=-12 \)
- \( x=1, y=-15 \)
- \( x=2, y=-18 \)
- \( y = kf(x) \)
- \( x=0, y=12 \)
- \( x=1, y=15 \)
- \( x=2, y=18 \)
The fourth table (the last option) is the only one that shows \( y \) values of \( 12, 15, \) and \( 18 \) corresponding to \( x \) values of \( 0, 1, 2 \), respectively, which matches the expected outputs.
Thus, the correct response is:
The fourth table: \( x \) \( y = kf(x) \)
- \( 0 \quad 12 \)
- \( 1 \quad 15 \)
- \( 2 \quad 18 \)