0

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

According to the publication I've been using and an on-line function-graphing website, the graph of the function f(x)=the square root of (minus x) starts at the origin and extends up and asymtotically left. The problem is that the axes of the graph represent real numbers, and you can't get a real number answer when taking the square root of a negative number. Please show step by step the logic involved in obtaining such a graph. I believe my sources to be reputable. Thank you.

### 4 Answers by Expert Tutors

Grigori S. | Certified Physics and Math Teacher G.S.Certified Physics and Math Teacher G.S.
2

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.

2

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!

Christopher G. | Math Tutor - Algebra, Trig, Calculus, SAT/ACT MathMath Tutor - Algebra, Trig, Calculus, SA...
5.0 5.0 (3 lesson ratings) (3)
2

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.

Stanton D. | Tutor to Pique Your Sciences InterestTutor to Pique Your Sciences Interest
4.6 4.6 (42 lesson ratings) (42)
0
Another way to say this is: in order to graph (-x)^0.5 for negative-valued x, there's no problem since the result is a positive real number : i.e. you can do it on regular 2-D graph paper. But for positive-valued x, now you need to show the result on the imaginary portion of the complex plane. So, in order to unify these two portions of the graph, you could show x as a continous axis in 3-D (say, as the x-axis!), then have the other 2 axes show the complex plane for the function result. The two portions of the function graph change from lying (let's say) in the x-y plane, for x negative, to lying in the x-z plane, for x positive.
Now, here's an interesting thought: what's the angle formed between the two pieces of the graph in 3-D, as x-> 0+ and 0- respectively? For what (real) values of z is that angle independent of the power z for f(x)=x^z ?