Answer
$(y-1)^2=\dfrac{1}{3}(x+2),-2\leq x\leq 25$
See graph
Work Step by Step
We are given the parametric equations:
$\begin{cases}
x=3t-2\\
y=1-\sqrt t
\end{cases}$
with $0\leq t\leq 9$
Build a table of values:
$t=0; x=-2; y=1; (x,y)=(-2,1)$
$t=1; x=1; y=0; (x,y)=(1,0)$
$t=4; x=10; y=-1; (x,y)=(10,-1)$
$t=9; x=25; y=-2; (x,y)=(25,-2)$
Graph the curve:
Find the rectangular equation of the curve:
$\sqrt t=1-y$
$t=(1-y)^2$
$x=3(y-1)^2-2$
$3(y-1)^2=x+2$
$(y-1)^2=\dfrac{1}{3}(x+2)$
