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This is a problem new to me. I need some guidance on what I should study/understand to be able to solve such problems.

Sketch the graph of the following absolute functions on $\mathbb{R}$ and state the range. $$g:x \rightarrow \sqrt {x^2 + 2}$$

I have done graphs of quadratic equations and square root functions. How do I do a graph of root function of a quadratic function?

Thanks for your help.

Edit: Clarified question.

I have tried plotting points. I got a curve that resembles the parabolic curve of a quadratic function. The quadratic function has a vertex formula, is open/closed parabola, etc.

I was wondering if there were any thing I need to study on the lines of the quadratic function? What if it is the nth root. Is there a way to figure this out for (fractional)powers of a quadratic?

The other approach I was looking at was via graph transformations. $$g(x) = \sqrt{x^2 + 2}$$ Let, $$h(x) = \sqrt{x}$$ Then, $$g(x) = h(x^2 + 2)$$

I not sure where to go further with this idea.

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How have you sketched the graphs of the functions whose graphs you have sketched? It is not unthinkable that the same strategy will work with these functions too! –  Mariano Suárez-Alvarez Jun 16 '11 at 13:48
    
Try some values first, just to see. –  Mitch Jun 16 '11 at 15:22
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2 Answers 2

up vote 7 down vote accepted

It sort of depends on how detailed a graph you want. Note first of all that the graph is symmetrical about the $y$-axis. This observation immediately cuts our work in half! Just do the graph for $x$ positive, and reflect the picture across the $y$-axis.

So we take $x \ge 0$. Near $x=0$, we have that $y \approx \sqrt{2}$. As $x$ increases, $\sqrt{x^2+2}$ increases.

For $x$ at all large positive, we have that $\sqrt{x^2+2}\approx x$. More precisely, it is a little bigger than $x$, but awfully close to $x$ when $x$ is large. (You may want to do a small number of calculations, using a calculator.)

Draw the half-line $y=x$, for $x \ge 0$. To say that $\sqrt{x^2+2}$ is awfully close to $x$, when $x$ is large, means that the curve $y=\sqrt{x^2+2}$ hugs the half-line $y=x$, for $x \ge 0$, when $x$ is large. So for large values, your curve will be roughly indistinguishable from $y=x$.

In technical language, the half-line is an (oblique) asymptote to the curve $y=\sqrt{x^2+2}$.

With this information, you should be able to get a good idea of the first-quadrant part of the curve. Then don't forget to reflect across the $y$-axis.

Wish I could supply a picture. There are (free) graphing calculator programs, or more generally graphing programs, available for download.

Look up also Wolfram Alpha. It will draw the curve for you. But first do it yourself.

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That helped a lot, Thanks. Good tip about Wolfram Alpha, excellent site. –  mathguy80 Jun 16 '11 at 16:23
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I realize you've accepted an answer. I just thought it might be helpful if you were to compare the equations:

$$y = \sqrt{x^2 + 2}$$ with $$y^2 = x^2 + 2 \quad\iff\quad y^2 - x^2 = 2$$

and their corresponding graphs. Do you see a relationship?

enter image description here

They are certainly related, but the range of the first equation (the graph of which is entirely above the $x$-axis, since $y\geq 2$ for all $y$) is half that of the first equation (entire graph of hyperbola includes both the negative and positive values in the range $y$).

I encourage you to explore with graphing the equations at WolframAlpha.

Have you studied conic sections or worked with graphs of conics such as parabolas, hyperbolas, ellipses and circles? If not, you might want to check out Wikipedia on Conic Sections for a sneak preview of what's to come!

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I think it is very often to post an answer even after one has been accepted. There is frequently, as in this case, useful material to add. –  André Nicolas Jun 16 '11 at 18:47
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@user6312: thanks. BTW: I've been wanting to tell you that I am heartened by your efforts to interact with OPs, taking them and their questions seriously, and gearing your answers to their varying levels of knowledge. It can get discouraging to come across, time and again, sarcastic comments/answers, and/or answers assuming far more background than the OP has...Anyway, keep up the good work! –  amWhy Jun 16 '11 at 19:11
    
Thank you, one tries. Presumably the idea is to take account of a student's level of experience, and try to guide them towards a more structured point of view. Answers on this site generally stay clear of sarcasm, but they can be unsuitable from a teaching point of view, since they present a highly polished product. And the student is left to wonder how (s)he could ever "come up with that." The intuition behind arguments is not given often enough. But the format is constraining, communication at the right level is much easier with live person and blackboard. –  André Nicolas Jun 16 '11 at 19:32
    
Thanks @amWhy. I have been peeking at wikipedia occasionally. It is overwhelming at times to see the amount of ground I have yet to cover, but then at least the roads I have to take are well documented. :) –  mathguy80 Jun 17 '11 at 5:10
    
@amWhy: Even a nice picture! =1 –  Amzoti May 11 '13 at 0:30
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