How do you graph #sqrtx", if "x>0 and -1/2x+3", if "x≤ 0#?

Consider #sqrt(x)" where "x>0#

Set #y=sqrt(x)# where #x>0=> x!=0#

Any number that is negative times its self will give a positive answer.

Any number that is positive times itself will give a positive answer.

So in this case #y# can be either positive or negative so we write:
#y=+-sqrt(x)#

This is a roundabout way of writing #(+-y)^2=x" where "y!=0#

So really this is a quadratic in #y# which is of the shape #uu# but 'rotated' #90^o# to the right.

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Consider: #y=-1/2x+3# where #x<0 =>x!=0#

Firstly this is an equation of a straight line: form #y=mx+c#
The gradient (#m->-1/2#) is negative so the line slopes downward reading left to right on the x-axis.

The constant #c# is of value #+3# so the line crosses the y-axis (y-intercept) at #y=3#. However, it is not permitted for #x# to be greater than or equal to 0 so the actual y-intercept is an 'excluded value'.

The plot well terminate at the y-axis so there is no line on and to the right of the y-axis.

Pick any value for #x#: I* choose #10#

then #y=1/2x+3color(white)("ddd") ->color(white)("ddd") y=-1/2(-10)+3 = +8#

Mark the point #(x,y)=(-10,8)#
Using a ruler line this up with the value of 3 on the x-axis and draw your line making sure you do not go beyond the y-axis into the zone #x>=0#