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**Unformatted text preview: **f 2
Solution. We build up to a formula for g (x) using intermediate functions as we’ve seen in previous
examples. We let g1 take care of our ﬁrst step. Theorem 1.2 tells us g1 (x) = f (x)+2 = x2 +2. Next,
we reﬂect the graph of g1 about the x-axis using Theorem 1.4: g2 (x) = −g1 (x) = − x2 + 2 =
−x2 − 2. We shift the graph to the right 1 unit, according to Theorem 1.3, by setting g3 (x) =
g2 (x − 1) = −(x − 1)2 − 2 = −x2 + 2x − 3. Finally, we induce a horizontal stretch by a factor of 2
2
1
1
1
using Theorem 1.6 to get g (x) = g3 2 x = − 2 x + 2 2 x − 3 which yields g (x) = − 1 x2 + x − 3.
4
We use the calculator to graph the stages below to conﬁrm our result. shift up 2 units −− − − − −→
−−−−−−
add 2 to each y -coordinate y = f (x) = x2 y = g1 (x) = f (x) + 2 = x2 + 2 reﬂect across x-axis −− − − − −→
−−−−−− multiply each y -coordinate by −1 y = g1 (x) =
11 x2 +2 You really should do this once in your life. y = g2 (x) = −g1 (x) = −x2 − 2 1.8 Transformations 103 shift right 1...

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