## Precalculus (10th Edition)

$(y-1)^2=\dfrac{1}{3}(x+2),-2\leq x\leq 25$ See graph
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: $\sqrt t=1-y$ $t=(1-y)^2$ $x=3(y-1)^2-2$ $3(y-1)^2=x+2$ $(y-1)^2=\dfrac{1}{3}(x+2)$