## Trigonometry (11th Edition) Clone

(a) $x = t+2$ $y = \frac{1}{t+2}$ (Note that $t \neq -2$) We can see the graph below. (b) $y = \frac{1}{x}$ $x\neq 0$
(a) $x = t+2$ $y = \frac{1}{t+2}$ $t \neq -2$ t = -4: $x = (-4)+2 = -2$ $y = \frac{1}{(-4)+2} = -\frac{1}{2}$ t = -3: $x = (-3)+2 = -1$ $y = \frac{1}{(-3)+2} = -1$ t = -\frac{5}{2}: $x = (-\frac{5}{2})+2 = -\frac{1}{2}$ $y = \frac{1}{(-\frac{5}{2})+2} = -2$ t = -\frac{3}{2}: $x = (-\frac{3}{2})+2 = \frac{1}{2}$ $y = \frac{1}{(-\frac{3}{2})+2} = 2$ t = -1: $x = (-1)+2 = 1$ $y = \frac{1}{(-1)+2} = 1$ t = 0: $x = (0)+2 = 2$ $y = \frac{1}{(0)+2} = \frac{1}{2}$ t = 1: $x = (1)+2 = 3$ $y = \frac{1}{(1)+2} = \frac{1}{3}$ t = 2: $x = (2)+2 = 4$ $y = \frac{1}{(2)+2} = \frac{1}{4}$ We can see the graph below. (b) $x = t+2$ $y = \frac{1}{t+2}$ Therefore: $~~y = \frac{1}{x}$ Since $t \neq -2$, then $x\neq 0$