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**Unformatted text preview: **gain
[0, ∞). We recognize j (x) = f (9x) and by Theorem 1.6, we obtain the graph of j by dividing
the x-coordinates of the points on the graph of f by 9. From the graph, we see the range of
j is also [0, ∞). y y
(4, 2)
2 4
,2
9 1 2 1
,1
9 (1, 1)
1 (0, 0) (0, 0)
1 2 y = f (x) = 3. Solving x+3
2 3 √ 4 x horizontal scale by a factor of 1 1
9 2 3 4 −− − − − − − − − − −
− − − − − − − − − −→
x multiply each x-coordinate by 1
9 y = j (x) = f (9x) = √ x 9x ≥ 0 gives x ≥ −3, so the domain of m is [−3, ∞). To take advantage of what we know of transformations, we rewrite m(x) = − 1
2x + 3
2 + 1, or m(x) = −f 1
2x + 3
2
1
2x + 1. Focusing on the inputs ﬁrst, we note that the input to f in the formula for m(x) is
+ 3.
2
1
Multiplying the x by 2 corresponds to a horizontal stretch by a factor of 2, and adding the
3
3
2 corresponds to a shift to the left by 2 . As before, we resolve which to perform ﬁrst by
thinking about how we would ﬁnd the point on m corresponding to a point on f , in this case,
1
(4, 2). To use f (4) =...

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