Question 189649
A) Here's the graph. The red line is the graph of {{{y=2-x}}}, the green line is {{{y=2}}} and the blue line is {{{y=8-2x}}}. 


{{{ drawing(500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,(2-x)((sqrt(-x))/(sqrt(-x))),2(sqrt(sqrt(9/2)+.2-(x-3/2)^2)/sqrt(sqrt(9/2)+.2-(x-3/2)^2)),(8-2x)(sqrt(x-2.8)/sqrt(x-2.8)))

)}}}


Note: there is an open circle at (0,2) and (3,2). Also, there's a closed circle at (0,2) and (3,2).


B)

Since there isn't symmetry about the y-axis or the origin, this means that the function is neither even nor odd.



C)


This function is known as a <u>piece</u>wise function