Question

Which of the following tables shows that y=kf(x)
results in the parent function y=|x+4|
being vertically stretched by a k factor of 3?(1 point)
Responses

x y=kf(x)
0 4
1 5
2 6
x y equals k f left parenthesis x right parenthesis 0 4 1 5 2 6

x y=kf(x)
0 1.3
1 1.7
2 2
x y equals k f left parenthesis x right parenthesis 0 1.3 1 1.7 2 2

x y=kf(x)
0 12
1 15
2 18
x y equals k f left parenthesis x right parenthesis 0 12 1 15 2 18

x y=kf(x)
0 −12
1 −15
2 −18
x y equals k f left parenthesis x right parenthesis 0 negative 12 1 negative 15 2 negative 18
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Answer 1

To determine which table shows that \( y = kf(x) \) results in the parent function \( y = |x + 4| \) being vertically stretched by a factor of \( k = 3 \), we first need to find the values of \( f(x) \).

The parent function is \( f(x) = |x + 4| \). Let's calculate \( f(x) \) for the values of \( x \) in the tables provided:

1. For \( x = 0 \):
\[
f(0) = |0 + 4| = 4
\]

2. For \( x = 1 \):
\[
f(1) = |1 + 4| = 5
\]

3. For \( x = 2 \):
\[
f(2) = |2 + 4| = 6
\]

When \( y = kf(x) \) and \( k = 3 \), we have:
- For \( x = 0 \): \( y = 3 \cdot 4 = 12 \)
- For \( x = 1 \): \( y = 3 \cdot 5 = 15 \)
- For \( x = 2 \): \( y = 3 \cdot 6 = 18 \)

Thus, the matching table would yield:
\[
\begin{aligned}
x & \quad y = kf(x) \\
0 & \quad 12 \\
1 & \quad 15 \\
2 & \quad 18 \\
\end{aligned}
\]

Looking at the options provided, the correct table is:
\[
x \quad y = kf(x) \\
0 \quad 12 \\
1 \quad 15 \\
2 \quad 18
\]

So the correct response is the third table:
- \( x = 0 \), \( y = 12 \)
- \( x = 1 \), \( y = 15 \)
- \( x = 2 \), \( y = 18 \)