Substituting x y into the equation yields x 22 y 2

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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...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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