Let's begin with the line that goes from minus infinity to zero
the equation of this line is

because of the filled circle we know that this interval touch the zero

in inequality, this will be

then for the line that is between 0 and 2, we have a filled circle in the number two therefore we can touch this number
the interval will be

in inequality notation
](https://img.qammunity.org/2023/formulas/mathematics/high-school/dyrs04p4cufml91jvufir7vmf4s7sf0mwm.png)
in inequality notation

Therefore the piecewise function i

the correct option is the second one