Ask a New Question
Menu

Use the image to answer the question.

A coordinate plane with two lines, one with one open endpoint and one with a closed endpoint. The x-axis ranges from negative 5 to 10 and the y- axis ranges from negative 8 to 10. Both axes are in unit increments and are labeled in increments of 5.

Which of the following piecewise functions matches the graph?

(1 point)
Responses

f(x)={2−x if x≤0x−5 if x>0
f left parenthesis x right parenthesis equals left brace Start Rows 1st Row 2 minus x if x less-than-or-equal-to 0 2nd Row x minus 5 if x greater than 0 End Rows

f(x)={−x−1 if x<1−x+2 if x≥1
f left parenthesis x right parenthesis equals left brace Start Rows 1st Row negative x minus 1 if x less than 1 2nd Row negative x plus 2 if x greater-than-or-equal-to 1 End Rows

f(x)={−5 if x<0−x+1 if x≥0
f left parenthesis x right parenthesis equals left brace Start Rows 1st Row negative 5 if x less than 0 2nd Row negative x plus 1 if x greater-than-or-equal-to 0 End Rows

f(x)={x+5 if x≤−22x+3 if x>−2

1 answer

To determine the piecewise function that matches the graph described, we need to analyze the behavior of the function based on the conditions provided.

  1. For the first piecewise function:

    • \( f(x) = 2 - x \) if \( x \leq 0 \)
    • \( f(x) = x - 5 \) if \( x > 0 \)

    This function decreases from (0, 2) when \( x = 0 \) and drops as \( x \) increases. The first piece is valid at \( x \leq 0 \) with a closed endpoint at (0, 2) since \( f(0) = 2 \).

  2. For the second piecewise function:

    • \( f(x) = -x - 1 \) if \( x < 1 \)
    • \( f(x) = -x + 2 \) if \( x \geq 1 \)

    This function has a different intersection point and does not match the initial condition for \( f(0) \).

  3. For the third piecewise function:

    • \( f(x) = -5 \) if \( x < 0 \)
    • \( f(x) = -x + 1 \) if \( x \geq 0 \)

    Here, \( f(0) = 1 \) which does not match any \( y \)-intercept provided.

  4. For the fourth piecewise function:

    • \( f(x) = x + 5 \) if \( x \leq -2 \)
    • \( f(x) = 2x + 3 \) if \( x > -2 \)

    This function does not seem to match the behavior at or around the defined boundaries.

Based on your description of the graph, the only option with a closed endpoint at \( (0, 2) \) and an open endpoint for positive \( x \) which fits the given criteria is the first piecewise function:

Answer: f(x) = { 2 − x if x ≤ 0, x − 5 if x > 0 }