Answer
$(x,y)=(2,\frac{1}{3})$ for $t=-2$
$(0,1)$ for $t=-1$
$(0,3)$ for $t=0$
$(2,9)$ for $t=1$
$(6,27)$ for $t=2$
Work Step by Step
$x=t^{2}+t$
$y=3^{t+1}$
When $t=-2$,
$x=(-2)^{2}+(-2)=4-2=2$
$y=3^{(-2)+1}=3^{-1}=\frac{1}{3}$
$(x,y)=(2,\frac{1}{3})$ for $t=-2$
When $t=-1$,
$x=(-1)^{2}+(-1)=1-1=0$
$y=3^{(-1)+1}=3^{0}=1$
$(x,y)=(0,1)$ for $t=-1$
When $t=0$,
$x=(0)^{2}+(0)=0+0=0$
$y=3^{(0)+1}=3^{1}=3$
$(x,y)=(0,3)$ for $t=0$
When $t=1$,
$x=(1)^{2}+(1)=1+1=2$
$y=3^{(1)+1}=3^{2}=9$
$(x,y)=(2,9)$ for $t=1$
When $t=2$,
$x=(2)^{2}+(2)=4+2=6$
$y=3^{(2)+1}=3^{3}=27$
$(x,y)=(6,27)$ for $t=2$