Question

Complete the tables to show the parent function y=f(x)
𝑦
=
𝑓
(
𝑥
)
and its translation y=f(x+5)
𝑦
=
𝑓
(
𝑥
+
5
)
. (2 points)
Original Function
x
𝑥
y=f(x)
𝑦
=
𝑓
(
𝑥
)
−2
−
2
5
−1
−
1
2
0 1
1 2
2 5

Translated Function
x
𝑥

Answer 1

To complete the translations for the parent function \( y = f(x) \) to its translated version \( y = f(x + 5) \), we need to understand how the translation affects the values of \( x \) and \( y \).

The function \( y = f(x + 5) \) represents a horizontal shift of the original function \( y = f(x) \) to the left by 5 units. This means that for every \( x \) value we have for the original function, we will subtract 5 from that \( x \) value to find the corresponding \( x \) value for the translated function.

Let's create the tables.

### Original Function
\[
\begin{array}{|c|c|}
\hline
x & y = f(x) \\
\hline
-2 & -2 \\
-1 & -1 \\
0 & 1 \\
1 & 2 \\
2 & 5 \\
\hline
\end{array}
\]

### Translated Function
For \( y = f(x + 5) \), we take \( x \) from the original function and subtract 5.

\[
\begin{array}{|c|c|}
\hline
x & y = f(x + 5) \\
\hline
-7 & -2 \quad (y = f(-7 + 5) = f(-2)) \\
-6 & -1 \quad (y = f(-6 + 5) = f(-1)) \\
-5 & 1 \quad (y = f(-5 + 5) = f(0)) \\
-4 & 2 \quad (y = f(-4 + 5) = f(1)) \\
-3 & 5 \quad (y = f(-3 + 5) = f(2)) \\
\hline
\end{array}
\]

### Final Tables

#### Original Function
\[
\begin{array}{|c|c|}
\hline
x & y = f(x) \\
\hline
-2 & -2 \\
-1 & -1 \\
0 & 1 \\
1 & 2 \\
2 & 5 \\
\hline
\end{array}
\]

#### Translated Function
\[
\begin{array}{|c|c|}
\hline
x & y = f(x + 5) \\
\hline
-7 & -2 \\
-6 & -1 \\
-5 & 1 \\
-4 & 2 \\
-3 & 5 \\
\hline
\end{array}
\]

That's the completed translation of the original function.