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Test for symmetry with respect to each axis and to the origin. y=xx2+1y=\frac{x}{x^{2}+1}

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Answered 9 months ago
Answered 9 months ago
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We have y=xx2+1y= \dfrac{x}{x^2+1}

First: we’ll test the equation for symmetry with respect to the x-axis\color{#4257b2}\text{First: we'll test the equation for symmetry with respect to the x-axis}:

(y)=xx2+1(-y) = \dfrac{x}{x^2+1}( Replace y by -y)

y=xx2+1\Rightarrow y=-\dfrac{x}{x^2+1}

The result is not an equivalent equation.

Therefore the graph is not symmetric with respect to the x-axis

Second: we’ll test the equation for symmetry with respect to the y-axis\color{#4257b2}\text{Second: we'll test the equation for symmetry with respect to the y-axis}:

y=x(x)2+1y =\dfrac{-x}{(-x)^2+1}( Replace x by -x)

y=xx2+1\Rightarrow y=-\dfrac{x}{x^2+1}

The result is not an equivalent equation.

Therefore the graph is not symmetric with respect to the y-axis

Third: we’ll test the equation for symmetry with respect to the origin\color{#4257b2}\text{Third: we'll test the equation for symmetry with respect to the origin}:

(y)=x(x)2+1(-y) = \dfrac{-x}{(-x)^2+1}( Replace y by -y and x by -x)

y=xx2+1\Rightarrow -y=-\dfrac{x}{x^2+1}

The result is an equivalent equation.

Therefore the graph is symmetric with respect to the origin

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