Graphing Question y=sqrt(x)

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Artimid

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Greetings, I have a question and I apologize in advance if I sound a bit annoyed. I have a highly paid college professor whose inability to understand English, and whose complete lack of ability to listen to what students are -actually- asking, make wandering to find competent people to answer the questions she is paid for.

Luckily, I know where to find some. ^_^

So, my question is this: if y=x[sup:326jp664]2[/sup:326jp664] is a parabola, where the range will always be positive because squared numbers are always positive as such, domain = -3,-2,1 0,1,2,3 and the range = 9,4,10,1,4,9
Then, why doesn't y=(sqrt)x lead to a parabola on the side, where x can never go into the negative. Leading the domain to be, for example: 4,9,16 and the range being -2,-3,-4, 2, 3, 4.

I ask this, because a perfect square of a positive number will always be +/- the number... if I am phrasing that right. So sqrt 4 = +/- 2. Why then doesn't this work the same on a graph as it does everywhere else?

Sorry if I didn't make this clear, it is easier to say when I am speaking it.

Btw, I understand that x>=0. I just don't understand why the y does, and this is where my teacher started saying how x >=0, and so forth, and then she just ignored my further attempts to seek out help, and went on with the next problem.
 
Basically there are two functions involved: \(\displaystyle y=\sqrt{x}, y = - \sqrt{x}\). If you plot them together you don't have a function but only a relation, since, for example, \(\displaystyle x=4\) would lead to two values \(\displaystyle y=2,-2\). In other words, this is just an artifact of the way we define functions. It's also true that square roots themselves are positive; in fact, we can define \(\displaystyle |x| = \sqrt{x^2}\).
 
Trying to make sure I understand, sorry if phrase this wrong...

So, it doesn't form the sideways parabola because it is a supposed to be a function? I think I am losing my way on why the square root property of +/- the root doesn't seem to apply here.
Or, in this case, we are applying the |x| = (sqrt) or x^2 rule? Just trying to make sure, because in the other sections we are doing, the square root of a number results in the +- square root.
For example: 5=x^2
x= +- sqrt of 5.

In the same way, y=5 which equals x^2, or to my thinking at least.. which appears to be wrong, so I apologize I will probably ask more questions as I try to grasp it.



btw, love how you add the actual math in here, compared to my text. :p
 
The square root of x is ALWAYS a positive number or zero (look up "principle square root").

Therefore, if we define y = sqrt(x), then y is ALWAYS a positive number or zero.

If you want to talk about the negative roots, instead, then you need to multiply sqrt(x) by -1.

y = -sqrt(x)

EG: x = 16

y = sqrt(16) = 4

The negative root is the opposite of sqrt(16).

y = -sqrt(16) = -4

We need to graph both of the following separately; we cannot graph both the positive and negative roots with a single equation.

y = sqrt(x)

y = -sqrt(x)
 

This is the way I have explained this difference to my students.


Case 1

If we are given a square root expression, \(\displaystyle \sqrt{9}\), we assume it is positive.
(Probably due to the invisible plus-sign in front.)

That is, we are asked "What is the square root of nine?" and we must answer "Three."


Case 2

In contrast, we could be asked, "What number, when squared, produces 9 ?"
. . We know that there are two possible answers: +3 and -3.
What is the difference?

In this problem, we have the equation: .\(\displaystyle x^2 \,=\,9\)
. . It is a quadratic equation . . . It has two answers.

In solving this quadratic, we could take the square root of both sides: .\(\displaystyle \sqrt{x^2} \:=\:\sqrt{9}\)
. . and get: .\(\displaystyle |x| \,=\,3 \quad\Rightarrow\quad x \:=\:\pm3\) . . . again, two answers.



In Case 1, we are given the square-root sign (radical).
. . So we assume a positive square root.


In Case 2, we introduced the square-root sign.
. . Hence, we are responsible for the double sign \(\displaystyle _{(\pm)}\).


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If we are given: .\(\displaystyle y = \sqrt{x}\), we assume the positive square root.

The graph is the upper half of a right-opening parabola.


If we are given: .\(\displaystyle y^2 \:=\:x\) and plot all the points satisfying the equation,
. . we get the entire right-opening parabola.

That is because we are graphing: .\(\displaystyle y \:=\:\pm\sqrt{x}\)
. . where we introduced the square root.


It is a very subtle distinction . . . I hope this helps.

 
I like Soroban's explanation, wherein the distinction is made between the radical sign being given to us versus being introduced by us. I read this difference as similar to the difference between evaluating an expression with x and solving an equation for x.

Yet, I would have chosen a different verb when making the statements about assumptions.

If we're given sqrt(9), then we do not need to assume that this number is positive. By definition of the radical sign (principle square root), we already know that this number is positive.
 
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