# Graphing sqrt(x^2 + 1)

Hi all,

I have recently begun to try and graph most of the functions I see by hand without resorting to the use of my graphing calculator. However, I am having a problem trying to begin to graph $$f(x) = \sqrt{x^{2} + 1}$$.

The problem is all I really know is that f(0) = 1. Trying to solve for roots shows that there are none. My graphing calculator essentially graphs this as a horizontally stretched out $$x^{2} + 1$$, and i'm just not sure why.

At first I thought that it would just be a stretched graph similar to that of just $$\sqrt{x}$$, but now I see that $$\sqrt{x^{2} + 1}$$ has values on both sides of the y-axis. But what's confusing is that the graph seems to be concave up (like $$x^{2}$$) instead of concave up (like $$\sqrt{x}$$).

Anyways, does anyone have any recommendations as to how to approach graphing this by hand?

Thanks.

## Answers and Replies

Office_Shredder
Staff Emeritus
Gold Member
Have you done calculus? You can use the second derivative to see what the concavity should be directly

Have you done calculus? You can use the second derivative to see what the concavity should be directly
Yes. I'm sorry I should have been more explicit in my first post. While I have taken calculus, I was just curious to know if there were any methods of graphing this without falling back on taking the derivatives (because that would just be too easy ;p).

What I mean by "methods of graphing without falling back on taking the derivatives" would be something akin to realizing that "$$y_{1}= f(x) + c$$" shifts $$y = f(x)$$ up c units, and "$$y_{2}= f(x + c)$$" shifts $$y = f(x)$$ left c units. Similarly, "$$y_{3} = c*f(x)$$" stretches $$y = f(x)$$ vertically by a factor of c units. And so on, and so forth.

So without resorting to any real "calculus" you can graph alot of algebraic equations as long as you know how to graph just a small handful of other equations (like simple power functions, square root functions, etc.).

But maybe I'm asking for too much here, and there is no way of actually going about doing this without studying the function's derivatives...

Mark44
Mentor
To sketch the graph of y = sqrt(x2 + 1), first sketch a graph of y = x2, then translate upward by 1 unit to get y = x2 + 1.

Finally, sketch the graph of y = sqrt(x2 + 1) by plotting the square root of the y values of the previous graph (y = x2 + 1). I.e., instead of plotting (1, 2), plot (1, sqrt(2)) and so on. Your final graph will be roughly similar to the graph of y = x2 + 1 except that as x gets large (or very negative) the graph has a slant asymptote, which the graph of y = x2 + 1 doesn't have. This shouldn't be too surprising, since the graph you want is actually the upper half of a hyperbola. y = sqrt(x2 + 1) ==> y2 = x2 + 1 ==> y2 - x2 = 1, with y >= 0.

HallsofIvy
Homework Helper
Another way: square both sides of $y= \sqrt{x^2+ 1}$ to get $y^2= x^2+ 1$ or $y^2- x^2= 1$.

You might be able to recognize that as a "hyperbola" with axis of symmetry the x-axis. A hyperbola has two parts- here one is above the x-axis and one below. Since your problem was originally $y= \sqrt{x^2+ 1}$ so that y must be positive, your graph is just the part above the x-axis.

Another method:
First, observe that the function is even (meaning y(x)=y(-x)), so the function will be symmetric about the y-axis.

Next, notice, as you said,
y(0)=1.

Now notice that if x is relatively large (larger than 1), the sqrt[x^2+1] barely depends on the "1".
Sqrt[3^2+1]=3.16
sqrt[4^2+1]=4.1
Sqrt[5^2+1]=5.099

And so on. As x gets larger, y(x) is approximately equal to x. In fact, approximating Sqrt[x^2+1] as Abs[x] works really well for x>2 (and extremely well for x>>1)

Mentallic
Homework Helper
What djeikyb said.

Now that you have these 3 tools to work with, f(0)=1, y=|x| asymptote, f(x)=f(-x), it should be quite obvious what is going to happen.

However, if you don't see this as obvious yet, make use of your calculus which shows it's concave up (but don't neglect the asymptote) and maybe even try to see if it cuts the asymptote anywhere, i.e. check to see if y=x is a solution - but obviously from looking at the function, this won't be the case.

If you haven't taken calculus, just think about y = \sqrt{x^2}, which is basically y=x. If that's just linear throughout, and \sqrt{x^2 +1} is very similar, wouldn't \sqrt{x^2 +1} approach a linear slope at very large numbers of x?

Mark44
Mentor
If you haven't taken calculus, just think about y = \sqrt{x^2}, which is basically y=x.
True for x >= 0, but not true for y < 0.
If that's just linear throughout, and \sqrt{x^2 +1} is very similar, wouldn't \sqrt{x^2 +1} approach a linear slope at very large numbers of x?

Thanks for clarifying. I probably should have explained why I put "basically" though.

Mark44, HallsofIvy, and djeikyb, thanks for your posts. It really helped clear this up for me, and I followed everything you guys said.

Thanks again.

You can use the second derivative to see what the concavity should be directly

Mark44
Mentor
You can use the second derivative to see what the concavity should be directly
The OP said he wanted to use non-calculus methods.