is niether because = has the domian and for that domain is defined · bx, f(x) x f(–x) ≥ 0, only for = 0. Thus neither equals nor x f(–x) f(x) –f(x). 2-125. See below: Neither Neither Even 2-126. See below: See graph below. See graph below. Neither function is odd nor even. 2-127. y = 2)2 + 3 –(x – 2-128. See below: (–1, 0), (0, 2), (–1, 0), = 2(x + 1)2. x: y: V: y (0, 0), (2, 0), (0, 0), (1, 1) = 1)2 + 1 x: y: V: y –(x – 2-129. See below: = y x (, ) (, ) The solution to the system is the point at which the lines intersect. 2-130. See below: = 40()n or 10()n–1 t(n) = + 4 t(n) –6n 2-131. See below: (2, 0), (6, 0) (0, 2), vertex: (4, D: all real numbers, R: x: y: y –2), ≥ –2 (–4, 0), (2, 0), (0, 2), vertex (–1, 3), D: all real numbers, R: x: y: y ≤ 3
bx, · is niether because f(x) = has the domian x ≥ 0, and for that domain f(–x) is defined only for x = 0. Thus f(–x) neither equals f(x) nor –f(x). 2-125. See below: Neither Neither Even 2-126. See below: See graph below. See graph below. Neither function is odd nor even. 2-127. y = –(x – 2)2 + 3 2-128. See below: x: (–1, 0), y: (0, 2), V: (–1, 0), y = 2(x + 1)2. x: (0, 0), (2, 0), y: (0, 0), V: (1, 1) y = –(x – 1)2 + 1 2-129. See below: y = x (, ) (, ) The solution to the system is the point at which the lines intersect. 2-130. See below: t(n) = 40()n or 10()n–1 t(n) = –6n + 4 2-131. See below: x: (2, 0), (6, 0) y: (0, 2), vertex: (4, –2), D: all real numbers, R: y ≥ –2 x: (–4, 0), (2, 0), y: (0, 2), vertex (–1, 3), D: all real numbers, R: y ≤ 3