8 8 88. 2 1 0     x2 x 83. 2 2 2 1 0  i x2  x   1  i i i 8 b  b 4ac 2  2 i x 2a i (2) 8i (2)2  4(1)(1) x  i 2(1) 2 8  8i 2 i   2 2  i 2  i 16i 8i 2 2 2 2 4  i2 2  1  2 16i 8 4  1 The solution set is {1 2}. 8 16i [0, range: 2] 5 8 16 89. The graph of is the graph of shifted 1 unit up and g f   i 5 5 3 units to the left. [0, 84. domain: 2) 85. ( 1 at f x) and 3 . = 2 1 86. 2 3)2 8  2(   x 87. 2( 3)2 8 x   0  3)2  4 x  3   4 x  3  2 x  1, 5 x
8 8 88.  x2  2 x  1  0 83. 2  i 2 1  x2  2 x  1  0 i i i 8  2 i b  b 2 4ac x 2a i 8i  i (2) x 2  8 (2)2  4(1)(1) 2(1) 8i 2 i   2 2  i 2  i 16i 8i 2 4  i2 2 2 2 2  1  2 16i 8 4  1 The solution set is {1 2}. 8 16i 5 8 16   i 5 5 84. domain: [0, 2) range: [0, 2] 89. The graph of g is the graph of f shifted 1 unit up and 3 units to the left. 85. 86. f ( x) = 1 at 1 2 and 3 . 2 87. 0  2( x  3)2  8 2( x  3)2  8 x  3)2  4 x  3   4 x  3  2 x  1, 5
Precalculus-Essentials-5th-Edition-Blitzer-Solutions-Manual.docx