Question

Part A: Complete the table to vertically stretch the parent functionf(x)=x3
by a factor of 3.

x y=kf(x)=

x3
0
1
2
Part B: Complete the table to show that y=kf(x)
results in the parent function being vertically compressed when k=12
x y=kf(x)=

x4
0
1
2

Answer 1

To complete the tables, we can follow the instructions for each part.

### Part A: Vertically Stretching the Parent Function \( f(x) = x^3 \) by a Factor of 3

For a vertical stretch, the transformation is given by \( y = kf(x) \), where \( k = 3 \). Thus, \( y = 3f(x) = 3x^3 \).

We can calculate the values for the given \( x \):

| x | \( y = 3f(x) = 3x^3 \) |
|---|---------------------------|
| 0 | \( 3(0^3) = 0 \) |
| 1 | \( 3(1^3) = 3 \) |
| 2 | \( 3(2^3) = 3(8) = 24 \) |

So the table for Part A is:

| x | \( y = kf(x) = 3x^3 \) |
|---|--------------------------|
| 0 | 0 |
| 1 | 3 |
| 2 | 24 |

### Part B: Vertically Compressing the Parent Function \( f(x) = x^4 \) when \( k = \frac{1}{2} \)

For a vertical compression, the transformation is given by \( y = kf(x) \), where \( k = \frac{1}{2} \). Thus, \( y = \frac{1}{2}f(x) = \frac{1}{2}x^4 \).

We can calculate the values for the given \( x \):

| x | \( y = \frac{1}{2}f(x) = \frac{1}{2}x^4 \) |
|---|-----------------------------------------------|
| 0 | \( \frac{1}{2}(0^4) = 0 \) |
| 1 | \( \frac{1}{2}(1^4) = \frac{1}{2} \) |
| 2 | \( \frac{1}{2}(2^4) = \frac{1}{2}(16) = 8 \) |

So the table for Part B is:

| x | \( y = kf(x) = \frac{1}{2}x^4 \) |
|---|----------------------------------|
| 0 | 0 |
| 1 | 0.5 |
| 2 | 8 |

### Summary

**Part A Table (Vertical Stretch)**:

| x | \( y = 3x^3 \) |
|---|----------------|
| 0 | 0 |
| 1 | 3 |
| 2 | 24 |

**Part B Table (Vertical Compression)**:

| x | \( y = \frac{1}{2}x^4 \) |
|---|--------------------------|
| 0 | 0 |
| 1 | 0.5 |
| 2 | 8 |