Answer
a)
Check graph. As t increases, y increases. As t increases, x initially increases and then decreases again back to its original position.
b)
$y=\sqrt(-x+1)-2$
or
$x=-(y+2)^{2}+1$
Work Step by Step
a)
$x=1-t^{2}$,
$y=t-2$,
$-2 \leq t \leq 4$
When,
t=-2 x=-3 y=-4
t=-1 x=0 y=-3
t=0 x=1 y=-2
t=1 x=0 y=-1
t=2 x=-3 y=0
t=3 x=-8 y=1
t=4 x=-15 y=1
b)
$x=1-t^{2}$
$\sqrt (-x+1)=t$
$y=t-2$
$y=\sqrt(-x+1)-2$
or
$x=-(y+2)^{2}+1$
From part A, we can infer that
$-4 \leq y \leq0$
