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Thread: find intercepts, test for symmetry

  1. #1
    Super Member bigwave's Avatar
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    find intercepts, test for symmetry

    a cardiod microphone pattern is given by the following equation:

    $\displaystyle \left{({x}^{2}+{y}^{2}-x\right)}^{2}={x}^{2}+{y}^{2} $

    the expanded form of this equation is
    $\displaystyle x^4-2 x^3+2 x^2 y^2+x^2-2 x y^2+y^4 = x^2+y^2$

    (a) Find the intercepts of the graph of the equation

    to find $\displaystyle {x}_{intercepts}$ I set $\displaystyle y = 0$

    thus resulting in

    $\displaystyle x^4-2x^3=0\Rightarrow x^3\left( x-2 \right) $

    so $\displaystyle {x}_{intercepts} = 0, 2$

    to find $\displaystyle {y}_{intercepts}$ I set $\displaystyle x = 0$

    thus resulting in

    $\displaystyle Y^2=1$ so $\displaystyle {y}_{intercepts}$ are $\displaystyle \pm 1$

    (b) Test for symmetry with respect to the x-axis

    my questions is that i do not how the symmetry is found except by observation from the graph. so symmetry is only on x-axis not on y-axis.



    the books answers were (a)$\displaystyle \left( 0,0 \right),\left( 2,0 \right) \left( 0,1\right) \left(0,-1 \right)$(b)x=axis symmetry

    but wolfframalpha gave something else

    btw like the upgrades to latex
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by bigwave View Post
    a cardiod microphone pattern is given by the following equation:

    $\displaystyle \left{({x}^{2}+{y}^{2}-x\right)}^{2}={x}^{2}+{y}^{2} $

    the expanded form of this equation is
    $\displaystyle x^4-2 x^3+2 x^2 y^2+x^2-2 x y^2+y^4 = x^2+y^2$

    (a) Find the intercepts of the graph of the equation

    to find $\displaystyle {x}_{intercepts}$ I set $\displaystyle y = 0$

    thus resulting in

    $\displaystyle x^4-2x^3=0\Rightarrow x^3\left( x-2 \right) $

    so $\displaystyle {x}_{intercepts} = 0, 2$

    to find $\displaystyle {y}_{intercepts}$ I set $\displaystyle x = 0$

    thus resulting in

    $\displaystyle Y^2=1$ so $\displaystyle {y}_{intercepts}$ are $\displaystyle \pm 1$

    (b) Test for symmetry with respect to the x-axis

    my questions is that i do not how the symmetry is found except by observation from the graph. so symmetry is only on x-axis not on y-axis.



    the books answers were (a)$\displaystyle \left( 0,0 \right),\left( 2,0 \right) \left( 0,1\right) \left(0,-1 \right)$(b)x=axis symmetry

    but wolfframalpha gave something else

    btw like the upgrades to latex
    There are three typical symmetries looked at...Reflection over the x axis, reflection over the y axis and reflection over the line y = x.

    If the substitution y --> -y gives the same equation, then the equation is symmetric over the x axis.

    If the substitution x --> -x gives the same equation, then the equation is symmetric over the y axis.

    If the substituions x --> y and y --> x (ie switch x and y in the equation) gives the same equation, then the equation is symmetric over the x axis.

    This is a long-winded way to say that I agree with your answer for b).

    -Dan
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  3. #3
    Member eXist's Avatar
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    In order to test for symmetry, check these conditions:

    If f(x) = f(-x), then the graph is symmetric on the Y-axis.

    If f(y) = f(-y), then the graph is symmetric on the X-axis.

    Does this make sense why these two conditions give you symmetry?
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  4. #4
    Super Member bigwave's Avatar
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    this was a little confusing because it appears to be an implicit equation. so doesn't mean we really don't have a function since it fails the vertical line test?
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by bigwave View Post
    this was a little confusing because it appears to be an implicit equation. so doesn't mean we really don't have a function since it fails the vertical line test?
    You are correct...your equation is not a function. However the symmetry tests still work.

    -Dan
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