Answer
(a) $x = t+2$
$y = \frac{1}{t+2}$
(Note that $t \neq -2$)
We can see the graph below.
(b) $y = \frac{1}{x}$
$x\neq 0$
![](https://gradesaver.s3.amazonaws.com/uploads/solution/e574afab-6b98-4573-89ff-ba9af594e5b6/result_image/1531206589.jpg?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIAJVAXHCSURVZEX5QQ%2F20240908%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20240908T001819Z&X-Amz-Expires=900&X-Amz-SignedHeaders=host&X-Amz-Signature=0213daa04e8ac65eac92a68c1345396c701b3c28a6778b3387a6764733240e46)
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 $x\neq 0$
![](https://gradesaver.s3.amazonaws.com/uploads/solution/e574afab-6b98-4573-89ff-ba9af594e5b6/steps_image/small_1531206589.jpg?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIAJVAXHCSURVZEX5QQ%2F20240908%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20240908T001819Z&X-Amz-Expires=900&X-Amz-SignedHeaders=host&X-Amz-Signature=7e0d9069efb256ac83eaabfddd2c9390e7c255544ced8c3ebba4fca588c73e35)