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Graphing sqrt(x^2 + 1)

  1. Jun 20, 2010 #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 [tex]f(x) = \sqrt{x^{2} + 1}[/tex].

    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 [tex]x^{2} + 1[/tex], and i'm just not sure why.

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

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

    Thanks.
     
  2. jcsd
  3. Jun 20, 2010 #2

    Office_Shredder

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    Have you done calculus? You can use the second derivative to see what the concavity should be directly
     
  4. Jun 20, 2010 #3
    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 "[tex]y_{1}= f(x) + c[/tex]" shifts [tex]y = f(x)[/tex] up c units, and "[tex]y_{2}= f(x + c)[/tex]" shifts [tex]y = f(x)[/tex] left c units. Similarly, "[tex]y_{3} = c*f(x)[/tex]" stretches [tex]y = f(x)[/tex] 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...
     
  5. Jun 20, 2010 #4

    Mark44

    Staff: 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.
     
  6. Jun 20, 2010 #5

    HallsofIvy

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    Another way: square both sides of [itex]y= \sqrt{x^2+ 1}[/itex] to get [itex]y^2= x^2+ 1[/itex] or [itex]y^2- x^2= 1[/itex].

    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 [itex]y= \sqrt{x^2+ 1}[/itex] so that y must be positive, your graph is just the part above the x-axis.
     
  7. Jun 20, 2010 #6
    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)
     
  8. Jun 20, 2010 #7

    Mentallic

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    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.
     
  9. Jun 20, 2010 #8
    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?
     
  10. Jun 20, 2010 #9

    Mark44

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    True for x >= 0, but not true for y < 0.
     
  11. Jun 20, 2010 #10
    Thanks for clarifying. I probably should have explained why I put "basically" though.
     
  12. Jun 21, 2010 #11
    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.
     
  13. Jun 21, 2010 #12
    You can use the second derivative to see what the concavity should be directly
     
  14. Jun 21, 2010 #13

    Mark44

    Staff: Mentor

    The OP said he wanted to use non-calculus methods.
    .
     
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