Question

Part A: Complete the table to vertically stretch the parent functionf(x)=x3
by a factor of 3. Write your answer as a whole number of decimal

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
. Write your answer as a whole number of decimal

x y=kf(x)=

x4
0
1
2
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Answer 1

Let's complete both parts of the problem step by step.

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

When we vertically stretch a function by a factor of \( k \), we multiply the function by \( k \). In this case, \( k = 3 \).

So the modified function will be:
\[
y = kf(x) = 3f(x) = 3x^3
\]

Now, we will calculate the values of \( y \) for \( x = 0, 1, 2 \):

1. **For \( x = 0 \)**:
\[
y = 3(0^3) = 3(0) = 0
\]

2. **For \( x = 1 \)**:
\[
y = 3(1^3) = 3(1) = 3
\]

3. **For \( x = 2 \)**:
\[
y = 3(2^3) = 3(8) = 24
\]

### Completed Table for Part A:

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

---

### Part B: Vertically Compress the Function \( f(x) = x^4 \) by a Factor of \( \frac{1}{2} \)

When we vertically compress a function by a factor of \( k \), we also multiply the function by \( k \). In this case, \( k = \frac{1}{2} \).

So the modified function will be:
\[
y = kf(x) = \frac{1}{2}f(x) = \frac{1}{2}x^4
\]

Now, we will calculate the values of \( y \) for \( x = 0, 1, 2 \):

1. **For \( x = 0 \)**:
\[
y = \frac{1}{2}(0^4) = \frac{1}{2}(0) = 0
\]

2. **For \( x = 1 \)**:
\[
y = \frac{1}{2}(1^4) = \frac{1}{2}(1) = 0.5
\]

3. **For \( x = 2 \)**:
\[
y = \frac{1}{2}(2^4) = \frac{1}{2}(16) = 8
\]

### Completed Table for Part B:

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

---

These tables successfully represent the vertically stretched and compressed functions. If you have any further questions or need additional explanations, feel free to ask!