Question

An open endpoint is plotted at left parenthesis 0 comma negative 5 right parenthesis and has a line extending upward and to the right. The line passes through left parenthesis 5 comma 0 right parenthesis and ends at left parenthesis 10 comma 5 right parenthesis. A closed endpoint is plotted at left parenthesis 0 comma 2 right parenthesis and has a line extending upward and to the left. The line passes through left parenthesis 3 comma 5 right parenthesis and ends at left parenthesis 5 comma 7.5 right parenthesis.


Which of the following piecewise functions matches the graph?

(1 point)
Responses

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)={x+5 if x≤−22x+3 if x>−2
f left parenthesis x right parenthesis equals left brace Start Rows 1st Row x plus 5 if x less-than-or-equal-to negative 2 2nd Row 2 x plus 3 if x greater than negative 2 End Rows

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)={−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

Answer 1

The third option.

Reason: For x>0 the line is y = x − 5 (gives the open point (0, −5) and passes through (5,0), (10,5)). For x≤0 the value is y = 2 − x (gives the closed point (0,2) and the leftward/upward line). So f(x) = {2 − x if x ≤ 0; x − 5 if x > 0}.