Why Is There No Reflection of y=√x in the Fourth Quadrant?

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Homework Help Overview

The discussion revolves around the graph of the function y=√x and the absence of a reflection in the fourth quadrant. Participants explore the implications of the square root function and its definition, questioning why negative values are not included in the graph.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the nature of the square root function and its principal value, questioning the exclusion of negative results. Some express confusion about the reflection of the graph in the fourth quadrant and the implications of defining a function.

Discussion Status

The discussion includes various perspectives on the definition of the square root and its graphical representation. Some participants offer explanations regarding the conventions of the square root function, while others express dissatisfaction with the clarity of these explanations.

Contextual Notes

Participants reference the need for the graph to represent a function, which requires a single output for each input, leading to the exclusion of negative values from the graph of y=√x.

DryRun
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Homework Statement
Explain graph of y=√x

The attempt at a solution
I've done the graph on an online graphing calculator:
http://s2.ipicture.ru/uploads/20120117/7JtQ5VD2.png

But i don't understand why there is no equivalent reflection below the x-axis, in the 4th quadrant.

From y=√x,
if x=4, then y is either 2 or -2. Correct? Then, why is the value -2 discarded?
Furthermore, if i rewrite the equation y=√x as x=y^2, then the graph does indeed become like this:
http://s2.ipicture.ru/uploads/20120117/p9hbKFr6.gif
 
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I think it has something to do with the fact that including both answers would disqualify its graph from being a function, so by convention, you only use the graph of one of the answers, the positive ones. That is my guess, I could be wrong.
 
Unsatisfactory answer.:frown:
 
The graph of [itex]y = \sqrt{x}[/itex] is exactly as you show in post 1. Looking at the domain, x must be >= 0. For the range, the square root returns a value >= 0. In your example, if x = 4, y = 2, not -2, because the square root of 4 is 2, not -2.

It is true that 4 has two square roots, but the principal square root, which is indicated by the square root symbol, is nonnegative. For these reasons, the graph of this function appears only in the first quadrant.
 
sharks said:
Unsatisfactory answer.:frown:


Sorry about that. It's still a correct answer, in my opinion. Why else would the graph appear as you present it other than convention, as Mark much, much more precisely explained it. Hopefully I can get an equally precise answer to my most recent thread...I'm just trying to contribute what little I can to others' questions in the meantime :).
 
Just try out a few points.

y = sqrt(x). If you plug in x, you get y=1. Not y= +/- 1
 
sharks said:
But i don't understand why there is no equivalent reflection below the x-axis, in the 4th quadrant.
When you use a calculator to figure out a number's square root, are you bothered that it only gives you a single positive number for the result?
 
Mark44 said:
It is true that 4 has two square roots, but the principal square root, which is indicated by the square root symbol, is nonnegative. For these reasons, the graph of this function appears only in the first quadrant.
Redbelly98 said:
When you use a calculator to figure out a number's square root, are you bothered that it only gives you a single positive number for the result?

I'll use these explanations as they both justify the graph by the conventional value of a square root.
 
LearninDaMath said:
Sorry about that. It's still a correct answer, in my opinion. Why else would the graph appear as you present it other than convention, as Mark much, much more precisely explained it. Hopefully I can get an equally precise answer to my most recent thread...I'm just trying to contribute what little I can to others' questions in the meantime :).

I think you did well :smile:

sharks, do you know the quadratic formula?
 
  • #10
Hi Mentallic

$$-b \pm \sqrt{b^2 - 4ac} \over 2a$$
Yes, I'm familiar with solving unknowns using that formula. What point are you trying to make?:smile:
 
Last edited:
  • #11
sharks said:
What point are you trying to make?:smile:

Why does it have a [itex]\pm[/itex] symbol if the square root of 4 for example is both 2 and -2? We wouldn't need that symbol there if the square root already produced both values.
 
  • #12
Good point! I'll add it to the whoever-need-convincing list.:smile:
 
  • #13
sharks said:
Good point! I'll add it to the whoever-need-convincing list.:smile:

Spread the truth :biggrin:
 

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