Stitz-Zeager_College_Algebra_e-book

This means we can cut our plug and plot time in half

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Unformatted text preview: (if any) of the graph of (x − 2)2 + y 2 = 1. Test for symmetry. Plot additional points as needed to complete the graph. 1.3 Graphs of Equations 25 Solution. To look for x-intercepts, we set y = 0 and solve: (x − 2)2 + y 2 (x − 2)2 + 02 (x − 2)2 (x − 2)2 x−2 x x = = = = = = = 1 1 1 √ 1 extract square roots ±1 2±1 3, 1 We get two answers for x which correspond to two x-intercepts: (1, 0) and (3, 0). Turning our attention to y -intercepts, we set x = 0 and solve: (x − 2)2 + y 2 (0 − 2)2 + y 2 4 + y2 y2 = = = = 1 1 1 −3 Since there is no real number which squares to a negative number (Do you remember why?), we are forced to conclude that the graph has no y -intercepts. Plotting the data we have so far, we get y 2 1 −1 (1, 0) 1 (3, 0) 2 3 4 x −2 Moving along to symmetry, we can immediately dismiss the possibility that the graph is symmetric about the y -axis or the origin. If the graph possessed either of these symmetries, then the fact that (1, 0) is on the graph would mean (−1, 0) would have to be on the graph. (Why?) Since (−1, 0) would be anoth...
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