Answer
(a) $x = t$
$y = \sqrt{t^2+2}$
We can see the graph below.
(b) $y = \sqrt{x^2+2}$
Work Step by Step
(a) $x = t$
$y = \sqrt{t^2+2}$
When $t = -3$:
$x = -3$
$y = \sqrt{(-3)^2+2} = \sqrt{11}$
When $t = -2$:
$x = -2$
$y = \sqrt{(-2)^2+2} = \sqrt{6}$
When $t = -1$:
$x = -1$
$y = \sqrt{(-1)^2+2} = \sqrt{3}$
When $t = 0$:
$x = 0$
$y = \sqrt{(0)^2+2} = \sqrt{2}$
When $t = 1$:
$x = 1$
$y = \sqrt{(1)^2+2} = \sqrt{3}$
When $t = 2$:
$x = 2$
$y = \sqrt{(2)^2+2} = \sqrt{6}$
When $t = 3$:
$x = 3$
$y = \sqrt{(3)^2+2} = \sqrt{11}$
We can see the graph below.
(b) $x = t$
$y = \sqrt{t^2+2}$
Therefore: $~~y = \sqrt{x^2+2}$
