Stitz-Zeager_College_Algebra_e-book

# 33 a y x2 1 b y x2 2x 8 c y x3 x d y e

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Unformatted text preview: s happens, we move on and try another point. This is another drawback of the ‘plug-and-plot’ approach to graphing equations. Luckily, we will devote much of the remainder of this book developing techniques which allow us to graph entire families of equations quickly.1 Second, it is instructive to show what would have happened had we tested the equation in the last example for symmetry about the y -axis. Substituting (−x, y ) into the equation yields (x − 2)2 + y 2 = 1 ? (−x − 2)2 + y 2 = 1 ? ((−1)(x + 2))2 + y 2 = 1 ? (x + 2)2 + y 2 = 1. This last equation does not appear to be equivalent to our original equation. However, to prove it is not symmetric about the y -axis, we need to ﬁnd a point (x, y ) on the graph whose reﬂection (−x, y ) is not. Our x-intercept (1, 0) ﬁts this bill nicely, since if we substitute (−1, 0) into the equation we get ? (x − 2)2 + y 2 = 1 (−1 − 2)2 + 02 = 1 9 = 1. This proves that (−1, 0) is not on the graph. 1 Without the use of a calculat...
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