Question
Which table shows that y=−f(x)
results in a reflection over the x-axis?
(1 point)
Responses
x −f(x)
−2
0
−3
1
−6
2
x f left parenthesis x right parenthesis negative 2 0 negative 3 1 negative 6 2
tabel 1
x −f(x)
−2
0
−3
−1
−6
−2
x f left parenthesis x right parenthesis negative 2 0 negative 3 negative 1 negative 6 negative 2
table2
x −f(x)
2 0
3 1
6 2
x f left parenthesis x right parenthesis 2 0 3 1 6 2
table3
x −f(x)
2 0
3 −1
6 −2
table 4
goes (2,0) to (3,1) stopping at (6,2)
results in a reflection over the x-axis?
(1 point)
Responses
x −f(x)
−2
0
−3
1
−6
2
x f left parenthesis x right parenthesis negative 2 0 negative 3 1 negative 6 2
tabel 1
x −f(x)
−2
0
−3
−1
−6
−2
x f left parenthesis x right parenthesis negative 2 0 negative 3 negative 1 negative 6 negative 2
table2
x −f(x)
2 0
3 1
6 2
x f left parenthesis x right parenthesis 2 0 3 1 6 2
table3
x −f(x)
2 0
3 −1
6 −2
table 4
goes (2,0) to (3,1) stopping at (6,2)
Answers
GPT-4o mini
To determine which table shows that \( y = -f(x) \) results in a reflection over the x-axis, we need to ensure that the output values for \( -f(x) \) are the negative counterparts of the corresponding output values of \( f(x) \).
In other words, for a given \( x \) value, if \( f(x) \) yields some value \( y \), then \( -f(x) \) should yield \( -y \).
Let's analyze the tables provided to find the one that reflects this property:
1. **Table 1**
\[
\begin{array}{c|c}
x & -f(x) \\
\hline
-2 & 0 \\
-3 & 1 \\
-6 & 2 \\
\end{array}
\]
Corresponding \( f(x) \) values are \( f(-2) = 0, f(-3) = -1, f(-6) = -2 \).
2. **Table 2**
\[
\begin{array}{c|c}
x & -f(x) \\
\hline
-2 & 0 \\
-3 & -1 \\
-6 & -2 \\
\end{array}
\]
Corresponding \( f(x) \) values are \( f(-2) = 0, f(-3) = 1, f(-6) = 2 \).
3. **Table 3**
\[
\begin{array}{c|c}
x & -f(x) \\
\hline
2 & 0 \\
3 & 1 \\
6 & 2 \\
\end{array}
\]
Corresponding \( f(x) \) values are \( f(2) = 0, f(3) = -1, f(6) = -2 \).
4. **Table 4**
\[
\begin{array}{c|c}
x & -f(x) \\
\hline
2 & 0 \\
3 & -1 \\
6 & -2 \\
\end{array}
\]
Corresponding \( f(x) \) values are \( f(2) = 0, f(3) = 1, f(6) = 2 \).
**Analysis**:
- In **Table 1**, the values do not correspond to the negative of \( f(x) \). It does not show reflection over the x-axis.
- In **Table 2**, the values \( 0, 1, 2 \) are not consistently negated in \( -f(x) \); hence, it is incorrect.
- In **Table 3**, only the \( f(x) = 0 \) at \( x=2 \) reflects correctly, but fails for \( x=3 \) and \( x=6 \).
- **Table 4** has the same \( f(x) = 0 \), but \( f(x) = 1 \) at \( x=3 \) becomes \( -1 \), and \( f(x) = 2 \) at \( x=6 \) becomes \( -2 \) correctly showing reflection.
Thus, **Table 4** correctly represents that \( y = -f(x) \) results in a reflection over the x-axis.
**Final Answer**: **Table 4**
In other words, for a given \( x \) value, if \( f(x) \) yields some value \( y \), then \( -f(x) \) should yield \( -y \).
Let's analyze the tables provided to find the one that reflects this property:
1. **Table 1**
\[
\begin{array}{c|c}
x & -f(x) \\
\hline
-2 & 0 \\
-3 & 1 \\
-6 & 2 \\
\end{array}
\]
Corresponding \( f(x) \) values are \( f(-2) = 0, f(-3) = -1, f(-6) = -2 \).
2. **Table 2**
\[
\begin{array}{c|c}
x & -f(x) \\
\hline
-2 & 0 \\
-3 & -1 \\
-6 & -2 \\
\end{array}
\]
Corresponding \( f(x) \) values are \( f(-2) = 0, f(-3) = 1, f(-6) = 2 \).
3. **Table 3**
\[
\begin{array}{c|c}
x & -f(x) \\
\hline
2 & 0 \\
3 & 1 \\
6 & 2 \\
\end{array}
\]
Corresponding \( f(x) \) values are \( f(2) = 0, f(3) = -1, f(6) = -2 \).
4. **Table 4**
\[
\begin{array}{c|c}
x & -f(x) \\
\hline
2 & 0 \\
3 & -1 \\
6 & -2 \\
\end{array}
\]
Corresponding \( f(x) \) values are \( f(2) = 0, f(3) = 1, f(6) = 2 \).
**Analysis**:
- In **Table 1**, the values do not correspond to the negative of \( f(x) \). It does not show reflection over the x-axis.
- In **Table 2**, the values \( 0, 1, 2 \) are not consistently negated in \( -f(x) \); hence, it is incorrect.
- In **Table 3**, only the \( f(x) = 0 \) at \( x=2 \) reflects correctly, but fails for \( x=3 \) and \( x=6 \).
- **Table 4** has the same \( f(x) = 0 \), but \( f(x) = 1 \) at \( x=3 \) becomes \( -1 \), and \( f(x) = 2 \) at \( x=6 \) becomes \( -2 \) correctly showing reflection.
Thus, **Table 4** correctly represents that \( y = -f(x) \) results in a reflection over the x-axis.
**Final Answer**: **Table 4**