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Symmetry about x-axis, y-axis or origin

  1. Aug 6, 2015 #1
    1. The problem statement, all variables and given/known data

    determine whether the graph of the function has symmetry about the x-axis, the y-axis or the origin. Check work by graphing:
    x^(2/3) + y^(2/3) = 1
    2. Relevant equations


    3. The attempt at a solution
    [ x^(2/3) + y^(2/3) = 1 ] ^(3/2)

    y = -x+1

    Its a straight line with y-intercept at 1. I don't understand how it has any symmetry about either x or y axis?
    thanks
     
  2. jcsd
  3. Aug 6, 2015 #2

    SammyS

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    Bad algebra.

    ##\displaystyle \ (a+b)^{p}\ne a^p+b^p \ ## unless p=1 .
     
  4. Aug 6, 2015 #3
    Well, you could solve for y:

    [tex]f(-x)=f(x)? [/tex] or [tex]f(-x)=-f(x)?[/tex]
     
  5. Aug 6, 2015 #4

    Mark44

    Staff: Mentor

    This is not helpful. The equation given does not have y as a function of x.
     
  6. Aug 7, 2015 #5
    And why can't you just solve for y? I know you will have a plus or minus...so just analyze both parts. Same thing-This will show its symmetric about the origin.
     
    Last edited: Aug 7, 2015
  7. Aug 7, 2015 #6
    Keep in mind that a function will have symmetry about the x-axis if you can replace all the y's with y-'s and obtain the same function, and it will have symmetry about the y-axis if you can do the same thing to the x's in the function and obtain the same function. If you change both y's with y-'s and x's with x-'s and the function remains the same, then the function will have symmetry about the origin.
     
  8. Aug 7, 2015 #7

    Mark44

    Staff: Mentor

    If you solve for y, you don't get a function. Some of the points on the graph of the given equation are (-1, 0), (1, 0), (0, 1), and (0, -1). The latter two points show that this graph is not that of a function, so writing f(x) and f(x) isn't applicable here.
     
  9. Aug 7, 2015 #8

    Mark44

    Staff: Mentor

    If you can do this, then the graph you're looking at is not a function.

    Corrected, the above would say "a function graph will have symmetry about the x-axis if you can replace all the y's with y-'s their opposites and obtain the same function graph"

     
  10. Aug 7, 2015 #9
    So, strictly speaking, one would have to label it as an equation? Or is my proposed method incorrect?
     
  11. Aug 7, 2015 #10

    Mark44

    Staff: Mentor

    See my previous post.
     
  12. Aug 7, 2015 #11
    Ah, okay. Thanks for clarifying.
     
  13. Aug 7, 2015 #12
    So you are saying that [tex] y=(1-x^{2/3})^{3/2} [/tex] and [tex]y=-(1-x^{2/3})^{3/2}[/tex] are not functions?

    And how do (0,1) and (0,-1) make it not a function? The top and bottom half of the unit circle contain those points - this is analogous to saying that the graph of the top half of a circle is not a function because it contains these points.
     
    Last edited: Aug 7, 2015
  14. Aug 7, 2015 #13

    Mark44

    Staff: Mentor

    No, I'm not. Those are functions. However, for almost every value of x in the domain of the original equation, there are two values of y. For a function, each value in the domain is paired with a single value in the codomain.

    For a simpler example, consider the equation ##x^2 + y^2 = 1##. If you solve for y you get ##y = \pm \sqrt{1 - x^2}##. Again, there are two y values for each x value in the domain (with exceptions for x = 1 and x = -1. The graph of this equation is a circle of radius 1, with center at the origin. The graph does not pass the vertical line test, another clue that the equation does not represent a function. This is the same situation that is present in the equation of this thread.
     
    Last edited: Aug 7, 2015
  15. Aug 7, 2015 #14
    No, for every x value in the domain there is one y if you analyze just the top or bottom half. The graph passes the verticals line test as well. I do not understand why you are saying there are two y values for one x value if you analyze just the top or bottom half. I know that a functions inverse will also be a function if it's one to one. Maybe I'm not understand what you are saying. It sounds to me right know that you are saying something like y=x^2-1 is not a function because it contains (0,1) and (0,-1)

    All I was originally saying was that you can break the function into two separate functions of x and test for symmetry.
     
    Last edited by a moderator: Aug 7, 2015
  16. Aug 7, 2015 #15

    Mark44

    Staff: Mentor

    That's true, but the original equation includes both halves.

    If we go back to the simpler example I gave, ##x^2 + y^2 = 1##, if you solve for y you get ##y = \pm \sqrt{1 - x^2}##. The top half (with the pos. square root) is a function, and the bottom half (neg. sq. root) is also a function, but both together do not represent a function.
    No it doesn't. As I said before, the graph of the equation at the start of this thread includes (0, 1) and (0, -1), as well as many other pairs of points that are on a vertical line.

    The graph of ##x^2 + y^2 = 1## also includes (0, 1) and (0, -1) and a lot more pairs that are on vertical lines. The equations ##x^{2/3} + y^{2/3} = 1## and ##x^2 + y^2 = 1## do not represent functions.
    First, I am not saying this, and second, it's not true. This equation, which is different from the one in my example, is a function. If x = 0, then y = -1, so the point (0, -1) is on the graph. The point (0, 1) is not on the graph of this parabola.
     
  17. Aug 7, 2015 #16
    Ok, I feel we have just mis understood each others posts. Obviously (0,1) is not on the graph of a porabola 1-x^2. We disagreed because I was talking about the graph of ONE of the two functions while you are talking about the entire graph (+/-). I was saying it sounded like you were implying that (0,1) AND (0,-1) were on the graph of THE TOP HALF of the graph ( which is what I was talking about all along). No more needs to be said here. Simple mis communication.
     
  18. Aug 7, 2015 #17

    Mark44

    Staff: Mentor

    Yes, throughout I was talking about the graph of the given equation and that of my example, not part of the graph of either equation.
     
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