To determine which table shows that \( y = kf(x) \) results in the parent function \( y = |x + 4| \) being vertically stretched by a factor of \( k = 3 \), we first need to understand how the transformation works.
The transformation \( y = kf(x) \) means that every output of the function \( f(x) \) (in this case, \( |x + 4| \)) is multiplied by \( k \). For a vertical stretch by a factor of 3:
\[ y = 3|x + 4| \]
We will find the values of \( |x + 4| \) at the given x values and multiply each by 3 to see what \( kf(x) \) should look like.
Now, let’s compute \( |x + 4| \) for the values of \( x = 0, 1, 2 \):
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For \( x = 0 \): \[ |0 + 4| = 4 \quad \Rightarrow \quad y = 3 \times 4 = 12 \]
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For \( x = 1 \): \[ |1 + 4| = 5 \quad \Rightarrow \quad y = 3 \times 5 = 15 \]
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For \( x = 2 \): \[ |2 + 4| = 6 \quad \Rightarrow \quad y = 3 \times 6 = 18 \]
Based on these calculations, we expect the outputs for \( y = kf(x) = 3|x + 4| \) to be:
- When \( x = 0 \), \( y = 12 \)
- When \( x = 1 \), \( y = 15 \)
- When \( x = 2 \), \( y = 18 \)
Now we compare this expected output with the choices:
\[ \begin{array}{|c|c|} \hline x & y = kf(x) \ \hline 0 & 1 \ 1 & 5 \ 2 & 6 \ \hline \end{array} \] (No matches)
\[ \begin{array}{|c|c|} \hline x & y = kf(x) \ \hline 0 & -12 \ 1 & -15 \ 2 & -18 \ \hline \end{array} \] (No matches)
\[ \begin{array}{|c|c|} \hline x & y = kf(x) \ \hline 0 & 12 \ 1 & 15 \ 2 & 18 \ \hline \end{array} \] (Matches perfectly!)
\[ \begin{array}{|c|c|} \hline x & y = kf(x) \ \hline 0 & 1.3 \ 1 & 1.7 \ 2 & 2 \ \hline \end{array} \] (No matches)
The correct table that shows the transformation of the parent function \( y = |x + 4| \) being vertically stretched by a factor of 3 is:
Table 3: \[ \begin{array}{|c|c|} \hline x & y = kf(x) \ \hline 0 & 12 \ 1 & 15 \ 2 & 18 \ \hline \end{array} \]