"y=\\left\\{\n\\begin{matrix}\n \\sqrt{-x} , x<0 \\\\\n 3-x, 0\\leq x<3\\\\\n(x-3)^2, x>3\n\\end{matrix}\\right."
"x=0\\\\\n\\lim\\limits_{x\\to0-0}\\sqrt{-x}=0\\\\\n\\lim\\limits_{x\\to0+0}(3-x)=3\\\\"
 The function at the point "x=0" has a jump discontinuity
"x=3\\\\\n\\lim\\limits_{x\\to3-0}(3-x)=0\\\\\n\\lim\\limits_{x\\to3+0}(x-3)^2=0\\\\"
The function  is a continuous at the point "x=3" .
The function  is discontinuous at "x=0".
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