Question

Complete the tables to show that y=kf(x)
results in the parent function being vertically compressed when k=12
.(1 point)
x y=x4
0 0
1 1
2 16
x y=12x4
0 0
1
2
Remaining Attempts : 3
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Answer 1

To show that \( y = kf(x) \) results in a vertical compression when \( k = \frac{1}{12} \), we need to compute the values of \( y = 12x^4 \) for the given \( x \) values (0, 1, and 2).

First, let's complete the table by calculating \( y \) when using \( y = 12x^4 \):

### Original Table (for \( y = x^4 \))
| x | y = x^4 |
|---|---------|
| 0 | 0 |
| 1 | 1 |
| 2 | 16 |

### New Table (for \( y = 12x^4 \))
To calculate the values for \( y = 12x^4 \):
- For \( x = 0 \):
\( y = 12(0^4) = 12(0) = 0 \)

- For \( x = 1 \):
\( y = 12(1^4) = 12(1) = 12 \)

- For \( x = 2 \):
\( y = 12(2^4) = 12(16) = 192 \)

### Completed Table
| x | y = 12x^4 |
|---|-----------|
| 0 | 0 |
| 1 | 12 |
| 2 | 192 |

### Summary
This shows:
- For \( x = 1 \), \( y = 1 \) (from the original function) vs. \( y = 12 \) (from the compressed function).
- For \( x = 2 \), \( y = 16 \) (from the original function) vs. \( y = 192 \) (from the compressed function).

You can see that as \( k = 12 \) increases the output values significantly, showing that the function \( y = 12x^4 \) is vertically stretched rather than compressed.

If we reduce \( k \) (for example, \( k = \frac{1}{12} \)), we would see a vertical compression. But with \( k = 12 \), the output values are larger, indicating a vertical stretch.