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Stitz-Zeager_College_Algebra_e-book

# We summarize the results in the table below 18

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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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