Testing Graphs for Symmetry

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Jason
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You can test for x or y symmetry, respectively, by replacing the variable name opposite of what symmetry you're looking for with something negative. If the equation is the same after plugging in, then it's symmetric.

\(ex.\,\,a.\)

\(Does\,\,this\,\,parabola\,\,have\,\,x\,\,symmetry?\)

\(y = 5x^{2} - 2 \)

\(-y = 5x^{2} - 2\)

\(\dfrac{-y}{-1} = \dfrac{5x^{2}}{-1} - \dfrac{2}{-1}\)

\(y = -5x^{2} + 2\)

\(No. \,\,Above\,\,equation\,\,is\,\, a \,\,different \,\,equation.\)


\(ex.\,\,b\)

\(Does\,\,this\,\,parabola\,\,have\,\,y\,\,symmetry?\)

\(y = -x^{2} + 14\)

\(y = -(-x)^{2} + 14\)

\(y = -x^{2} + 14 \)

\(Yes.\,\,The\,\,equation\,\,is\,\,the\,\,same\,\,equation.\)

Replacing both x and y with negative equivalents let's us test for origin symmetry because if the equation doesn't change, it has it.

\(ex.\,\,c.\)

\(Does\,\,this\,\,parabola\,\,have\,\,origin\,\,symmetry?\)

\(y = 13x^{2} - 6x + 1\)

\(-y = 13(-x)^{2} - 6(-x) + 1\)

\(-y = 13x^{2} + 6x + 1\)

\(\dfrac{-y}{-1} = \dfrac{13x^{2}}{-1} + \dfrac{6x}{-1} + \dfrac{1}{-1}\)

\(y = -13x^{2} - 6x - 1\)

\(No.\,\,The\,\, above\,\,equation \,\,is\,\, different.\)
 

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