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
x y=kf(x)
0 1
1 5
2 6

x y =kf(x)
0 -12
1 -15
2 -18

x y=kf(x)
0 12
1 15
2 18

x y=kf(x)
0 1.3
1 1.7
2 2

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 understand how the transformation works.

The transformation \( y = kf(x) \) means that every output of the function \( f(x) \) (in this case, \( |x + 4| \)) is multiplied by \( k \). For a vertical stretch by a factor of 3:

\[ y = 3|x + 4| \]

We will find the values of \( |x + 4| \) at the given x values and multiply each by 3 to see what \( kf(x) \) should look like.

Now, let’s compute \( |x + 4| \) for the values of \( x = 0, 1, 2 \):

- For \( x = 0 \):
\[
|0 + 4| = 4 \quad \Rightarrow \quad y = 3 \times 4 = 12
\]

- For \( x = 1 \):
\[
|1 + 4| = 5 \quad \Rightarrow \quad y = 3 \times 5 = 15
\]

- For \( x = 2 \):
\[
|2 + 4| = 6 \quad \Rightarrow \quad y = 3 \times 6 = 18
\]

Based on these calculations, we expect the outputs for \( y = kf(x) = 3|x + 4| \) to be:
- When \( x = 0 \), \( y = 12 \)
- When \( x = 1 \), \( y = 15 \)
- When \( x = 2 \), \( y = 18 \)

Now we compare this expected output with the choices:

1.
\[
\begin{array}{|c|c|}
\hline
x & y = kf(x) \\
\hline
0 & 1 \\
1 & 5 \\
2 & 6 \\
\hline
\end{array}
\]
(No matches)

2.
\[
\begin{array}{|c|c|}
\hline
x & y = kf(x) \\
\hline
0 & -12 \\
1 & -15 \\
2 & -18 \\
\hline
\end{array}
\]
(No matches)

3.
\[
\begin{array}{|c|c|}
\hline
x & y = kf(x) \\
\hline
0 & 12 \\
1 & 15 \\
2 & 18 \\
\hline
\end{array}
\]
(Matches perfectly!)

4.
\[
\begin{array}{|c|c|}
\hline
x & y = kf(x) \\
\hline
0 & 1.3 \\
1 & 1.7 \\
2 & 2 \\
\hline
\end{array}
\]
(No matches)

The correct table that shows the transformation of the parent function \( y = |x + 4| \) being vertically stretched by a factor of 3 is:

**Table 3:**
\[
\begin{array}{|c|c|}
\hline
x & y = kf(x) \\
\hline
0 & 12 \\
1 & 15 \\
2 & 18 \\
\hline
\end{array}
\]