**Unformatted text preview: **−−−−−−− 3
2 multiply each x-coordinate by 2 y = m2 (x) = m1 1
x
2 = 1
x
2 + 3
2 1
We now examine what’s happening to the outputs. From m(x) = −f 2 x + 3 + 1, we see the
2
output from f is being multiplied by −1 (a reﬂection about the x-axis) and then a 1 is added
(a vertical shift up 1). As before, we can determine the correct order by looking at how the
point (4, 2) is moved. We have already determined that to make use of the equation f (4) = 2,
1
we need to substitute x = 5. We get m(5) = −f 2 (5) + 3 + 1 = −f (4) + 1 = −2 + 1 = −1.
2
We see that f (4) (the output from f ) is ﬁrst multiplied by −1 then the 1 is added meaning we
ﬁrst reﬂect the graph about the x-axis then shift up 1. Theorem 1.4 tells us m3 (x) = −m2 (x)
will handle the reﬂection. y y
(5, 2) 2 2 (−1, 1) (−3, 0) −2 −1 (−3, 0) 1 2 3 4 5 1 −2 −1 x −1 1 2 3 4 x 5 (−1, −1) −2 −2 (5, −2)
reﬂect across x-axis y = m2 (x) = 1
x
2 + 3
2 −− − − − −→
−−−−−− multiply each y -coordinate by −1 y = m3 (x) = −m2 (x) = − 1
x
2 + 3
2 1.8 Transformations 99 Finally, to handle the vertical shift, Theorem 1.2 gives m(x) = m3 (x) + 1, and we see that
the range of m is (−∞, 1].
y y
(−3, 1) 2 (−3, 0) 2 1 (−1, 0) −2 −1 1 2 3...

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