To show that the function \( y = kf(x) \) results in the parent function being vertically compressed when \( k = \frac{1}{12} \), we start with the function \( y = x^4 \) and then apply the compression factor \( k = \frac{1}{12} \).
Given the parent function \( y = x^4 \):
\[ \begin{array}{|c|c|} \hline x & y = x^4 \ \hline 0 & 0 \ 1 & 1 \ 2 & 16 \ \hline \end{array} \]
Now, let’s calculate \( y = \frac{1}{12} x^4 \):
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For \( x = 0 \): \[ y = \frac{1}{12} \cdot 0^4 = 0 \]
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For \( x = 1 \): \[ y = \frac{1}{12} \cdot 1^4 = \frac{1}{12} \]
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For \( x = 2 \): \[ y = \frac{1}{12} \cdot 2^4 = \frac{1}{12} \cdot 16 = \frac{16}{12} = \frac{4}{3} \approx 1.33 \]
Now, we can complete the table for \( y = \frac{1}{12} x^4 \):
\[ \begin{array}{|c|c|} \hline x & y = \frac{1}{12} x^4 \ \hline 0 & 0 \ 1 & \frac{1}{12} \ 2 & \frac{4}{3} \ \hline \end{array} \]
So, summarizing the completed tables:
Parent Function \( y = x^4 \)
\[ \begin{array}{|c|c|} \hline x & y = x^4 \ \hline 0 & 0 \ 1 & 1 \ 2 & 16 \ \hline \end{array} \]
Compressed Function \( y = \frac{1}{12} x^4 \)
\[ \begin{array}{|c|c|} \hline x & y = \frac{1}{12} x^4 \ \hline 0 & 0 \ 1 & \frac{1}{12} \ 2 & \frac{4}{3} \ \hline \end{array} \]
This shows that the output values for \( y = \frac{1}{12} x^4 \) are less than those for \( y = x^4 \), confirming that the function \( y = \frac{1}{12} x^4 \) is a vertical compression of the parent function \( y = x^4 \).