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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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