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Thread: Symmetry of y = sqrt(2 - a^2)

  1. #1
    Junior Member jonnygill's Avatar
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    Symmetry of y = sqrt(2 - a^2)

    Hi!

    i am trying to determine if this equation is symmetrical to the x-axis and/or the y-axis. or both. the equation is...

    $\displaystyle y=\sqrt{2-a^2}$

    which i changed to...

    $\displaystyle y^2=2-a^2$

    the book states that this equation is ONLY symmetrical to the y-axis. i agree it is symmetrical to the y-axis, but it also seems to me to be symmetrical to the x-axis.

    since...

    $\displaystyle (-y)^2=2-a^2 \Rightarrow y^2=2-a^2$

    Because a square root has both a positive and negative value, doesn't this make sense?

    what am i missing/misunderstanding?

    p.s. is this algebra or pre-calc? i wasn't entirely sure
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  2. #2
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    e^(i*pi)'s Avatar
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    Note that your expression is only real for $\displaystyle -\sqrt{2} \leq a \leq \sqrt{2}$

    Underneath the radical it's the difference of two squares, which iirc, are always symmetrical in the y axis. Plus your graph is a semi-circle.

    No doubt someone can put it more elegantly than I though
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  3. #3
    Junior Member jonnygill's Avatar
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    Quote Originally Posted by e^(i*pi) View Post
    Note that your expression is only real for $\displaystyle -\sqrt{2} \leq a \leq \sqrt{2}$

    Underneath the radical it's the difference of two squares, which iirc, are always symmetrical in the y axis. Plus your graph is a semi-circle.

    No doubt someone can put it more elegantly than I though
    thanks. but i still don't get it.

    is $\displaystyle y=\sqrt{2-a^2}$ another expression of $\displaystyle y^2=2-a^2$

    edit: ok obviously it's not.

    something about the a^2 under the radical i gather. suppose i wanted to square $\displaystyle y=\sqrt{2-a^2}$ ...how would i do this?
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  4. #4
    Junior Member jonnygill's Avatar
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    ok lets say a=.5

    then...

    $\displaystyle y=\sqrt{2-.25}\approx1.3229 and 1.3229^2=1.75=y^2=2-.25$

    is this totally wrong?

    my graphing calculator does show that it is a semi-circle.

    edit: ok, so you can never have a negative number as the answer for a square root. so is this just something that one has to remember? i guess i'm struggling with the fact that the method i learned from the book i'm reading explained how to determine symmetry using specific rules.

    it states that you can substitute in -y or -b in place of y or b respectively when trying to see if a graph or function displays symmetry with the x-axis. (y=b). If the resulting equation is the same as the original, then it is symmetrical with respect to the x-axis.

    so when i plug in -y i get $\displaystyle -y=\sqrt{2-a^2}$

    so is it as simple as saying that $\displaystyle -y\neq y$ ?

    this inequality isn't always true though, since y could be zero.
    Last edited by jonnygill; Feb 18th 2011 at 03:14 PM.
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  5. #5
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by jonnygill View Post
    thanks. but i still don't get it.

    is $\displaystyle y=\sqrt{2-a^2}$ another expression of $\displaystyle y^2=2-a^2$

    edit: ok obviously it's not.

    something about the a^2 under the radical i gather. suppose i wanted to square $\displaystyle y=\sqrt{2-a^2}$ ...how would i do this?
    You correctly squared the equation, the issue with squaring is that it introduces possible extraneous solutions (solutions not in the original domain). For example if you had $\displaystyle y = 2$ and squared it you'd get $\displaystyle y^2 =4 $. Solutions here are $\displaystyle \pm 2$


    In your case if you wanted to solve $\displaystyle y^2 = 2-a^2$ for y then you'd have $\displaystyle y = \pm \sqrt{2-a^2}$ and the negative solution is extraneous.

    Your equation will hold for $\displaystyle |a| \leq \sqrt{2}$

    Using your example of a = 0.5

    $\displaystyle y = \sqrt{2-0.5^2} = \dfrac{\sqrt{7}}{2}$ and $\displaystyle y^2 = 2-0.5 = \dfrac{7}{4} = \left(\dfrac{\sqrt{7}}{2}\right)^2$

    Say if you were only given the second equation $\displaystyle y^2 = \dfrac{7}{4}$ then you'd deduce that $\displaystyle y = \pm \dfrac{\sqrt{7}}{2}$ - but where has that negative solution come from?

    Wiki: http://en.wikipedia.org/wiki/Extrane...sing_solutions
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  6. #6
    Junior Member jonnygill's Avatar
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    i guess it would come from squaring the radical and then taking the square root of the answer you get.
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by jonnygill View Post
    Hi!

    i am trying to determine if this equation is symmetrical to the x-axis and/or the y-axis. or both. the equation is...

    $\displaystyle y=\sqrt{2-a^2}$

    which i changed to...

    $\displaystyle y^2=2-a^2$

    the book states that this equation is ONLY symmetrical to the y-axis. i agree it is symmetrical to the y-axis, but it also seems to me to be symmetrical to the x-axis.

    since...

    $\displaystyle (-y)^2=2-a^2 \Rightarrow y^2=2-a^2$

    Because a square root has both a positive and negative value, doesn't this make sense?

    what am i missing/misunderstanding?

    p.s. is this algebra or pre-calc? i wasn't entirely sure
    Your $\displaystyle $$y$ is a constant depending on the parameter $\displaystyle $$a$, so assuming $\displaystyle $$a$ is such that $\displaystyle $$y$ is real the curve is a horizontal line in the $\displaystyle $$x-y$ plane, and so symmetric about the $\displaystyle $$y$ axis

    CB
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  8. #8
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    Squaring both sides of an equation in general is a "non-reversible" step. When you "undo" the squaring operation you don't get back what you started with. For example, if $\displaystyle x=3$, and you square both sides you get $\displaystyle x^2=9$. Reversing this gives $\displaystyle x=\pm 3$. An extra solution has been generated that wasn't there before.
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