Complete the table to vertically stretch the parent functionf(x)=x3

Question

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

Answer 1

To complete the tables for the transformations of the parent functions \( f(x) = x^3 \) and \( f(x) = x^4 \), we will apply the specified vertical stretch and compression.

Part A: Vertical Stretch of \( f(x) = x^3 \) by a factor of 3

The transformation is given by \( y = kf(x) \) where \( k = 3 \).

So, \( y = 3f(x) = 3x^3 \).

We can now calculate \( y \) for the given values of \( x \).

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

This gives us the completed table:

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

Part B: Vertical Compression of \( f(x) = x^4 \) by a factor of \( \frac{1}{2} \)

The transformation is given by \( y = kf(x) = \frac{1}{2}f(x) \) where \( k = \frac{1}{2} \).

So, \( y = \frac{1}{2}f(x) = \frac{1}{2}x^4 \).

Now, we calculate \( y \) for the given values of \( x \).

| x | y = \(\frac{1}{2} f(x) = \frac{1}{2} x^4\) | |---|--------------------| | 0 | \( \frac{1}{2}(0^4) = \frac{1}{2}(0) = 0 \) | | 1 | \( \frac{1}{2}(1^4) = \frac{1}{2}(1) = 0.5 \) | | 2 | \( \frac{1}{2}(2^4) = \frac{1}{2}(16) = 8 \) |

This gives us the completed table:

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

Final Results:

Part A Table (Vertical Stretch for \( f(x) = x^3 \)):

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

Part B Table (Vertical Compression for \( f(x) = x^4 \)):

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