Answer
(a) $x = \sqrt{t}$
$y = t^2-1$
We can see the graph below.
(b) $y = x^4-1$
for $x$ in $[0,\infty)$
Work Step by Step
(a) $x = \sqrt{t}$
$y = t^2-1$
When $t = 0$:
$x = \sqrt{0} = 0$
$y = (0)^2 - 1 = -1$
When $t = 1$:
$x = \sqrt{1} = 1$
$y = (1)^2 - 1 = 0$
When $t = 2$:
$x = \sqrt{2}$
$y = (2)^2 - 1 = 3$
When $t = 3$:
$x = \sqrt{3} = $
$y = (3)^2 - 1 = 8$
When $t = 4$:
$x = \sqrt{4} = 2$
$y = (4)^2 - 1 = 15$
When $t = 5$:
$x = \sqrt{5}$
$y = (5)^2 - 1 = 24$
We can see the graph below.
(b) $x = \sqrt{t}$
$t = x^2$
We can replace this expression for $t$ in the equation for $y$:
$y = t^2-1$
$y = (x^2)^2-1$
$y = x^4-1$
Since $t$ in $[0,\infty)$, then $x$ in $[0,\infty)$
