Normally, for graphing linear equations, I'd state to equate both parameters to #0#, but this here isn't a linear equation, so let's do it a bit differently.
The given equation is a parabola which has its vertex at #(3,0)#. Wanna know why? Because, if we take the equation and square it on both sides, we get what is an equation that is similar to a parabola, i.e #y=\sqrt{x-3}\impliesy^2=x-3#. Since the #x# parameter is positive, this means the parabola grows towards the positive x-axis. Also, since the original function is that of square root, the graph will exist only in the first quadrant (square roots by definition only give the positive value of the square root).
From the original equation, we now know that the points of the function cannot be previous to that of #x=3# in the x-axis. With that done, we also know how the graph will look like. But we now need to know how the original equation will be like.
So, from #y=\sqrt{x-3}#, if we substitute #x=4#, we get #y=1# (we could take #+-1# but square roots give us only positive numbers as defined). So taking the vertex and this point into consideration, we can graph the equation, keeping in mind how the line would change its curve as one progresses onward, and get a graph somewhat similar to this.
graph{y=(x-3)^{1/2} [-10, 10, -5, 5]}
One can see that the graph originates from #(3,0)# and also makes a point at #(4,1)#.
