Stitz-Zeager_College_Algebra_e-book

11 how would you describe symmetry about the origin

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Unformatted text preview: s of the points on the graph of f . To find the x-coordinates of the corresponding points on g , we undo what has been done to x in the same way we would solve an equation. What’s happening to the output can be thought of as things happening ‘outside’ the function, f . Things happening outside affect the outputs or y -coordinates of the points on the graph of f . Here, we follow the usual order of operations agreement: we first multiply by A then add K to find the corresponding y -coordinates on the graph of g . Example 1.8.4. Below is the complete graph of y = f (x). Use it to graph g (x) = 4−3f (1−2x) . 2 y (0, 3) 3 2 1 (−2, 0) −4 −3 −2 (2, 0) −1 1 2 3 4 x −1 −2 −3 (−4, −3) (4, −3) Solution. We use Theorem 1.7 to track the five ‘key points’ (−4, −3), (−2, 0), (0, 3), (2, 0) and (4, −3) indicated on the graph of f to their new locations. We first rewrite g (x) in the form 3 presented in Theorem 1.7, g (x) = − 2 f (−2x + 1) + 2. We set −2x + 1 equal to the x-coordinates of the key points and solve. For example, solving −2x + 1 = −4, we first subtract 1 to get −2x = −5 5 then divide by −2 to get x = 2 . Subtracting the 1 is a horizontal shift to the left 1 unit. Dividing by...
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