Stitz-Zeager_College_Algebra_e-book

In other words to graph g we divide the x coordinates

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Unformatted text preview: √ have j (x) = j1 (−x). Putting it all together, we have j (x) = j1 (−x) = f (−x + 3) = −x + 3, 5 Or divide - it amounts to the same thing. 92 Relations and Functions which is what we want.6 From the graph, we confirm the domain of j is (−∞, 3] and we get the range is [0, ∞). y y (1, 2) 2 2 (−2, 1) (2, 1) (−1, 2) 1 (3, 0) (−3, 0) −4 −3 −2 −1 1 y = j1 (x) = √ 2 3 4 x reflect across y -axis −− − − − −→ −−−−−− multiply each x-coordinate by −1 x+3 −4 −3 −2 −1 1 2 y = j (x) = j1 (−x) = 3 √ 4 x −x + 3 √ 3. The domain of m works out to be the domain of f , [0, ∞). Rewriting m(x) = − x + 3, we see m(x) = −f (x) + 3. Since we are multiplying the output of f by −1 and then adding 3, we once again have two transformations to deal with: a reflection across the x-axis and a vertical shift. To determine the correct order in which to apply the transformations, we imagine trying to determine the point on the graph of m which corresponds to (4, 2) on the graph of f . Since in the formula for m(x), the input to f is just x, we substitute to find m(4) = ...
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