Answer
(a) $x = t+2$
$y = \frac{1}{t+2}$
(Note that $t \neq -2$)
We can see the graph below.
(b) $y = \frac{1}{x}$
$x\neq 0$
Work Step by Step
(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 $x\neq 0$
