Graphing help

[Steph Annie]

Greetings!

I have a question about graphing. It's a problem from a lesson on radical expressions and functions.
I get the feeling that I am overthinking these graphing problems. Could you look at my work below and give me any suggestions or corrections? Do I have way too many points? You will have to project my points onto your own graph as I can't copy and paste my graph here.

Graph g(x)= √x + 2

Give its domain and range.

I plotted these points (2,0), (2,2), (6,8), (-1,1), (4,6), (1,3.)

x g(x)
2 0
2 2
6 8
-1 1
4 6
1 3

g(x) = √x+2
g(-2) = √-2+2
= √0
= 0

g(x) = √x+2
g(2) = √2+2
=√4
= 2

g(x) = √x+2
g(6) = √x+6
= √6+2
= √8
= 2.83

g(x) = √x+2
g(-1) = √x+2
= √-1+2
= √1
= 1

g(x) = √x+2
g(4) = √x+2
= √4+2
= √6
= 2.44

g(x) = √x+2
g(1) = √x+2
= √1+2
= √3
= 1.73

Domain:[0, ∞).
Range: [2, ∞).

Is this correct?

Thank you very much!!

 

[Romsek]

Greetings!

I have a question about graphing. It's a problem from a lesson on radical expressions and functions.
I get the feeling that I am overthinking these graphing problems. Could you look at my work below and give me any suggestions or corrections? Do I have way too many points? You will have to project my points onto your own graph as I can't copy and paste my graph here.

Graph g(x)= √x + 2

Give its domain and range.

I plotted these points (2,0), (2,2), (6,8), (-1,1), (4,6), (1,3.)

x g(x)
2 0
2 2
6 8
-1 1
4 6
1 3

g(x) = √x+2
g(-2) = √-2+2
= √0
= 0

this isn't right. First off since you are dealing with real numbers you can't be taking the square root of a negative number.

g(x) = √x+2
g(2) = √2+2
=√4
= 2

this isn't right either. sqrt(2) + 2 != sqrt(2+2)

Domain:[0, ∞).
Range: [2, ∞).

Your domain and range are correct.
Your life would be a lot easier if you chose x's that happened to be perfect squares. Such as 1 4 9 16 25 36

then your table is
x g(x)
---------
1 3
4 4
9 5
16 6
25 7
36 8

etc. Then just trace your curve using those points. That's a lot easier than trying to deal with all those square roots.

 

[Steph Annie]

Okay. So, no negative square roots?

But there is this other graph that is f(x) = x to the third power (...3 is the index)
and they use a bunch of negative x values.

And as far as using perfect squares, yes, my life would probably be a lot easier if I used these.
But drawing a graph with 36 points is also not my idea of easier when I am just trying to get through ONE SINGLE PROBLEM
in a realistic amount of time.

And I don't get what your saying about the square root of 2+2 not being right. I thought it would = 4 and the square root of 4 is 2.


I have about 20 of these to get through and I'm feeling grumpy.

Thanks anyways.

this isn't right. First off since you are dealing with real numbers you can't be taking the square root of a negative number.



this isn't right either. sqrt(2) + 2 != sqrt(2+2)



Your domain and range are correct.
Your life would be a lot easier if you chose x's that happened to be perfect squares. Such as 1 4 9 16 25 36

then your table is
x g(x)
---------
1 3
4 4
9 5
16 6
25 7
36 8

etc. Then just trace your curve using those points. That's a lot easier than trying to deal with all those square roots.

 

[Romsek]

Okay. So, no negative square roots?

But there is this other graph that is f(x) = x to the third power (...3 is the index)
and they use a bunch of negative x values.

And as far as using perfect squares, yes, my life would probably be a lot easier if I used these.
But drawing a graph with 36 points is also not my idea of easier when I am just trying to get through ONE SINGLE PROBLEM
in a realistic amount of time.

And I don't get what your saying about the square root of 2+2 not being right. I thought it would = 4 and the square root of 4 is 2.


I have about 20 of these to get through and I'm feeling grumpy.

Thanks anyways.

x³ can take negative values, x² can't. This should be pretty clear. You need to understand this.

What 36 points? I listed 6 points. You could probably make a decent sketch using 3 of them.

I think this was maybe a problem with formatting. You seemed to be saying that sqrt(2) + 2 = sqrt(2+2). that's not right.

but if you meant sqrt(2+2) = sqrt(4) = 2, then yes, that is correct.

 

[Steph Annie]

Denis,

I have no idea what you're trying to say here. Most likely it has to do with an error in the way I wrote the problem.
I spent 20 minutes writing the problem down, making columns for my x values, and copying and pasting a square root symbol for each one, all so that it would be as clear as possible. But I'm obviously just confusing everyone, and getting vague replies that are only frustrating me and confusing me further.

Thanks anyways.

Back to math.

√(x) + 2 or √(x+2) ?

x => 0 or x => -2

 

[Romsek]

Denis,

I have no idea what you're trying to say here. Most likely it has to do with an error in the way I wrote the problem.
I spent 20 minutes writing the problem down, making columns for my x values, and copying and pasting a square root symbol for each one, all so that it would be as clear as possible. But I'm obviously just confusing everyone, and getting vague replies that are only frustrating me and confusing me further.

Thanks anyways.

Back to math.

yeah the problem was that the top line of the square root symbol you used didn't cover all the pieces it should so it was confusing. No worries.

Sorry you're having such a rough time.

 

[Steph Annie]

Thanks. I see.

But why can't x² take negative x values? I know this seems obvious but I'm missing it.
And yep, I meant the square root of 2 +2 is 4 and the square root of 4 is 2.
That square root symbol I copied and pasted from Google must have not come through. *figures!*

x³ can take negative values, x² can't. This should be pretty clear. You need to understand this.

What 36 points? I listed 6 points. You could probably make a decent sketch using 3 of them.

I think this was maybe a problem with formatting. You seemed to be saying that sqrt(2) + 2 = sqrt(2+2). that's not right.

but if you meant sqrt(2+2) = sqrt(4) = 2, then yes, that is correct.

 

[Steph Annie]

Oh and by the 36 points, I meant that one of the values of x is 36, so I was thinking that I'd have to make a graph that had 36 numbers.
I usually have like 1-10 on the x and y axis of my graph.

Thanks. I see.

But why can't x² take negative x values? I know this seems obvious but I'm missing it.
And yep, I meant the square root of 2 +2 is 4 and the square root of 4 is 2.
That square root symbol I copied and pasted from Google must have not come through. *figures!*

 

[JeffM]

Thanks. I see.

But why can't x² take negative x values? I know this seems obvious but I'm missing it.
And yep, I meant the square root of 2 +2 is 4 and the square root of 4 is 2.
That square root symbol I copied and pasted from Google must have not come through. *figures!*

The equation you originally gave was

\(\displaystyle y = \sqrt{x} + 2.\) THAT function cannot accept negative values of x if y is considered a real number.

If, as I guess is the case, the equation is \(\displaystyle y = \sqrt{x + 2}\), then \(\displaystyle x \ge -2\) for y to be real.

 

[Romsek]

Thanks. I see.

But why can't x² take negative x values? I know this seems obvious but I'm missing it.
And yep, I meant the square root of 2 +2 is 4 and the square root of 4 is 2.
That square root symbol I copied and pasted from Google must have not come through. *figures!*

i meant that x² will never be negative (if x is a real number)

x³ on the other hand can be, -1³ = -1 for example