Question 670606
f(x)={-2-x if x is less than or equal to 1
     {-2+2x if x is > 1


f(x) = {{{system(matrix(2,3,

-2-x,if,x<=1,
-2+2x,if,x>1))}}}

We plot the graphs of y = -2-x and the graph of y = -2+x 

{{{drawing(400,400,-4,5,-5,4, graph(400,400,-4,5,-5,4,(-2-x)



),
graph(400,400,-4,5,-5,4,-2+2x), locate(-4,2.3,y=-2-x),
locate(3,4,y=-2+2x)
 )}}}

We chop off the line y = -2-x at x=2 and erase the part beyond that on the
right because of x <u><</u> 1 tells us we can use only the portion of that line which is left of x=1

We chop off the line y = -2+2x at x=2 and erase the part beyond that on the
left because of x > 1 tells us we can use only the portion of that line which is right of x = 1  
{{{drawing(400,400,-4,5,-5,4, graph(400,400,-4,5,-5,4,(-2-x)*sqrt(1-x)/sqrt(1-x)



),
graph(400,400,-4,5,-5,4,-2+2x*sqrt(x-1)/sqrt(x-1)), locate(-4,2.3,y=(-2-x)


),
locate(3,4,y=-2+2x)
 )}}}

But that's not all.  We must place a CLOSED circle at the end point of
the graph of y=-2-x because of the inequality x<u><</u>1 and we must 
place an OPEN circle at the wnd point of of the graph of y = -2+2x because
of the inequality x>1 

{{{drawing(400,400,-4,5,-5,4, graph(400,400,-4,5,-5,4,(-2-x)*sqrt(.96-x)/sqrt(.96-x)), circle(1,-3,.1),circle(1,-3,.08),circle(1,-3,.07),circle(1,-3,.06),circle(1,-3,.02),
graph(400,400,-4,5,-5,4,-2+2x*sqrt(x-1.05)/sqrt(x-1.05)), locate(-4,2.3,y=(-2-x)),
circle(1,0,.13),    locate(3,4,y=-2+2x)
 )}}}