*This preview shows
page 1. Sign up
to
view the full content.*

**Unformatted text preview: **er x-intercept (and we’ve found all of these), the graph can’t have y -axis or
origin symmetry. The only symmetry left to test is symmetry about the x-axis. To that end, we
substitute (x, −y ) into the equation and simplify
(x − 2)2 + y 2 = 1
? (x − 2)2 + (−y )2 = 1
(x − 2)2 + y 2 = 1
Since we have obtained our original equation, we know the graph is symmetric about the x-axis.
This means we can cut our ‘plug and plot’ time in half: whatever happens below the x-axis is
reﬂected above the x-axis, and vice-versa. Proceeding as we did in the previous example, we obtain 26 Relations and Functions
y
2
1 −1 1 2 3 x 4 −2 A couple of remarks are in order. First, it is entirely possible to choose a value for x which does
not correspond to a point on the graph. For example, in the previous example, if we solve for y as
is our custom, we get:
y = ± 1 − (x − 2)2 .
Upon substituting x = 0 into the equation, we would obtain
√
√
y = ± 1 − (0 − 2)2 = ± 1 − 4 = ± −3,
which is not a real number. This means there are no points on the graph with an x-coordinate
of 0. When thi...

View
Full
Document