Use the image to answer the question.

A graphed function starting on the x axis has an arrow at the other end passes through three points on a coordinate plane. The x-axis ranges from negative 9 to 9 in unit increments and the y-axis ranges from negative 11 to 11 in unit increments.

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 negative 1 negative 6 negative 2

x −f(x)
2 0
3 −1
6 −2
x f left parenthesis x right parenthesis 2 0 3 negative 1 6 negative 2

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

x −f(x)
2 0
3 1
6 2

1 answer

To find the correct table that shows \( y = -f(x) \), we need to consider how the values of \( f(x) \) are transformed by multiplying them by -1. This transformation reflects the graph over the x-axis.

The values of \( y = -f(x) \) should be the negative of the corresponding values of \( f(x) \).

Let's analyze the given tables:

  1. First Table:

    • \( x \) | \( -f(x) \)
    • -2 | ?
    • 0 | ?
    • -3 | ?
    • -1 | ?
    • -6 | ?
    • -2 | ?

    Here the transformation is not clear since \( f(x) \) values are not provided.

  2. Second Table:

    • \( x \) | \( -f(x) \)
    • 2 | 0
    • 3 | -1
    • 6 | -2

    Here, potential \( f(x) \) values could be:

    • \( f(2) = 0 \)
    • \( f(3) = 1 \) (since -f(3) = -1)
    • \( f(6) = 2 \) (since -f(6) = -2)
  3. Third Table:

    • \( x \) | \( -f(x) \)
    • -2 | ?
    • 0 | ?
    • -3 | 1
    • -6 | 2

    Here also \( f(x) \) values are uncertain.

  4. Fourth Table:

    • \( x \) | \( -f(x) \)
    • 2 | 0
    • 3 | 1
    • 6 | 2

    Here, potential \( f(x) \) values could be:

    • \( f(2) = 0 \) (since -f(2) = 0)
    • \( f(3) = -1 \) (since -f(3) = 1)
    • \( f(6) = -2 \) (since -f(6) = 2)

After analyzing the tables, it seems that the second table correctly shows the relationship of \( y = -f(x) \) because the \( -f(x) \) values are indeed the negatives of reasonable \( f(x) \) values.

Thus, the correct answer is the second table.