Start with the graph of #y=sqrtx#. Remember that this means our domain is #x>0#, and we can graph the points #(0,0),(1,1),(4,2),(9,3),(16,4)#, and so on.
graph{sqrtx [-4.73, 35.82, -8.76, 11.51]}
The next step is to graph #y=sqrt(x-4)#. Notice that at #x=4#, this gives us #sqrt0=0#. At #x=13#, we have #sqrt(13-4)=sqrt9=3#. What adding the #-4# within the function does is actually shift the function #4# units to the right. (Think about comparing the point on #y=sqrtx# of #(9,3)# with the point on #y=sqrt(x-4)# of #(13,3)#--that's the shift.)
graph{sqrt(x-4) [-4.73, 35.82, -8.76, 11.51]}
Next, we can graph #y=1/2sqrt(x-4)#. This means you can take whatever #y# value previously existed, and halve it. So, the point at #(20,4)# will become #(20,2)# and the point at #(5,1)# will become #(5,1/2)#.
graph{1/2sqrt(x-4) [-4.73, 35.82, -8.76, 11.51]}
For #y=-1/2sqrt(x-4)#, just reflect all the points over the #x#-axis, which is the same as taking the negative versions of all the existing #y#-values.
graph{-1/2sqrt(x-4) [-4.73, 35.82, -8.76, 11.51]}
Finally, for #y=-1/2sqrt(x-4)+1#, take the graph and shift it upwards #1# point:
graph{-1/2sqrt(x-4)+1 [-4.73, 35.82, -8.76, 11.51]}
