**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
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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
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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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