39_Cal_Solution of Calculus_6e

39_Cal_Solution of Calculus_6e - Thus the vertex of this...

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Unformatted text preview: Thus the vertex of this parabola is at the point (2, 4). C01S0M.041: The graph has a vertical asymptote at x = −5 and is shown next. 20 10 -8 -6 -4 -2 -10 -20 C01S0M.042: The graph has vertical asymptotes at x = ± 2 and is shown next. 6 4 2 -4 -2 2 4 -2 -4 -6 C01S0M.043: The graph of f is obtained by shifing the graph of g (x) = | x | three units to the right, so that the graph of f has its “vertex” at the point (3, 0). C01S0M.044: Given: f (x) = | x − 3 | + | x + 2 |. If x 3 then f (x) = x − 3 + x + 2 = 2x − 1, so the graph is the unbounded line segment with slope 2 and endpoint (3, 5) for x 3. If −2 x 3 then f (x) = 3 − x + x + 2 = 5, so another part of the graph is the horizontal line segment joining (−2, 5) with (3, 5). If x −2 then f (x) = 3 − x − x − 2 = −2x + 1, so the rest of the graph is the unbounded line segment with slope −2 and endpoint (−2, 5) for x −2. The graph is shown next. 8 6 4 2 -4 -2 2 6 4 ...
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This note was uploaded on 09/14/2011 for the course MATH 101 taught by Professor Cheng during the Spring '11 term at Ecole Hôtelière de Lausanne.

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