Stitz-Zeager_College_Algebra_e-book

3 we could equally have chosen the convention outputs

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Unformatted text preview: ause the domain of f is only [0, 5]. The same thing happens when we attempt to find g (5). What we need here is a new strategy. We know, for instance, f (0) = 1. To determine the corresponding point on the graph of g , we need to figure out what value of x we must substitute into g (x) = f (x + 2) so that the quantity x + 2, works out to be 0. Solving x + 2 = 0 gives x = −2, and g (−2) = f ((−2) + 2) = f (0) = 1 so (−2, 1) on the graph of g . To use the fact f (2) = 3, we set x + 2 = 2 to get x = 0. Substituting gives g (0) = f (0 + 2) = f (2) = 3. Continuing in this fashion, we get x −2 0 2 3 x + 2 g (x) = f (x + 2) (x, g (x)) 0 g (−2) = f (0) = 1 (−2, 1) 2 g (0) = f (2) = 3 (0, 3) 4 g (2) = f (4) = 3 (2, 3) 5 g (3) = f (5) = 5 (3, 5) In summary, the points (0, 1), (2, 3), (4, 3) and (5, 5) on the graph of y = f (x) give rise to the points (−2, 1), (0, 3), (2, 3) and (3, 5) on the graph of y = g (x), respectively. In general, if (a, b) is on the graph of y = f (x), then f (a) = b. Solvi...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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