My question concerns the steps and logic involved in graphing the square root of the negative of the variable x. Please see below.

The domain of your function includes all negative x-s and x=0  (-∞ < x ≤  0)(negative x semi-axis). This fucntion doesn't have an asymptote, it trends to infinity in the left (second) quadrant of the coordinate plane. For any negative x -x is positive and you can take square root of it.

Your confusion comes from misperception of the function: -x doesn't mean you are dealing only with negative numbers: "x" can be negative itself, but -x means number (-1) multiplied by x, which makes -x positive.

Excellent question.

What you are seeing being graphed is the real part of the function. Remember that all complex numbers are in the form a+bi which means a is the real part while bi is the imaginary. Now let's think of this in terms of x values. The reason the graph goes to the left is because x values to the left are negative. Thus if we plug in a negative x, the negatives in the function will cancel each other out. Ex) ƒ(-4)= √-(-4)= √4= 2. So your sources are showing you only the a part of the function.  

Now if we go ahead and plot this on the imaginary axis you will see the graph as you normal would a sqrt(x) function. Meaning, we start plugging in positive x values to the right of the imaginary axis. This will produce the bi part of the function. 

So remember, when dealing with imaginary and complex numbers, you must always make sure the graphs you are looking at are in the complex plane or else you'll lose a part of the function. Also, take a look at Wolfram Alpha. They have the correct graphs. 

Hope this helped! 

You are right that you can't get a real number when taking the square root of a negative number, but remember that not all values of x are positive. For all positive values of x, the function values will be imaginary; however, they will be real for all negative values of x. Let's say x = 5. f(5) = √(-5), which is imaginary. But now let's say x = -5, so f(-5) = √-(-5) = √(5), which is real. So that is why the real part of the graph goes to the left, because the real values will exist for all x less than or equal to 0.