Answer: First Option is correct.
Step-by-step explanation:
Since we have given that
[tex]f(x)=-x+1,x<0\\\\f(x)=x+1,x\geq 0[/tex]
If we let f(x)=y
then, y=x+1 and y=-x+1
Put x=0,
[tex]y=0+1\\\\y=1[/tex]
so, there must be coordinate on y -axis at 1.
and it must take Ist and II nd quadrant .
Hence, First Option is correct.
