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Graph f(x) = sqrt(a^2 + x^2)?

  1. Jul 29, 2010 #1
    1. The problem statement, all variables and given/known data

    I'm unsure how to do this with the two variables please help
    sorry it's actually f(x) = sqrt(a^2 - x^2)
    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jul 29, 2010 #2
    Hi GreenPrint.

    What are your initial thoughts? Also, I assume a is an arbituary constant and not specified as anything else in your original question.

    The Bob
  4. Jul 29, 2010 #3


    Staff: Mentor

    There's really only one variable: x. You should take a to be a constant, albeit one that is not known.

    If you let y = f(x), then your equation is y = sqrt(a2 - x2).
    What is the domain of allowed values for x?
    If you square both sides of the equation just above, you might recognize the equation as that of a familiar geometric object. Keep in mind, though, that you need to graph y = sqrt(a2 - x2), not the one you get by squaring both sides. They are different.
  5. Jul 29, 2010 #4
    oh it's a circle with a center at the origin but how do I deal with the fact that I'm not graphing y^2 but just y
  6. Jul 29, 2010 #5


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    Homework Helper

    If you have an equation of a circle with radius a:
    [tex]x^2 + y^2 = a^2[/math]
    ... and you solve for y, how many equations will you actually get?

  7. Jul 29, 2010 #6
    so it's f(x) = -sqrt(a^2 + x^2)
    and f(x) = sqrt(a^2 + x^2)
  8. Jul 29, 2010 #7
    so should I just draw a circle with center at the origin and draw in a radius and put "a" above it or something
  9. Jul 29, 2010 #8


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    Staff Emeritus
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    No, because f(x) isn't the equation of a complete circle. Don't forget that the radical sign gives you only the positive square root of what's inside.

    On your drawing, you should label where the graph intersects the two axes.
  10. Jul 29, 2010 #9
    Ok so it would be a semi circle on the positive acess with center at the orgin and would cross the y intercept at (o,a) the x axis at (a,0) (-a,0)?
  11. Jul 29, 2010 #10
    Yep, you know this either from realizing that x^2<=a^2 or by saying that there is no way for y to be negative because sqrts never return negative values, and preferably you thought a little bit of both.
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