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(a) Sketch the curve by using the parametric equations to plot points. Indicat with an arrow the direction in which the curve is traced as $ t $ increases.

(b) Eliminate the parameter to find a Cartesian equation of the curve.

$ x = t^2 - 3 $, $ \quad y = t + 2 $, $ \quad -3 \leqslant t \leqslant 3 $

(a) $$\begin{array}{|c|rrrr|}

\hline t & -3 & -1 & 1 & 3 \\

\hline x & 6 & -2 & -2 & 6 \\

\hline y & -1 & 1 & 3 & 5 \\

\hline

\end{array}$$

(b) $$\begin{array}{l}

y=t+2 \Rightarrow t=y-2, \text { so } \\

x=t^{2}-3=(y-2)^{2}-3=y^{2}-4 y+4-3 \Rightarrow \\

x=y^{2}-4 y+1,-1 \leq y \leq 5

\end{array}$$

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Campbell University

Oregon State University

University of Michigan - Ann Arbor

the problem is scarce occurred by using the parametric equations to plot points indicated with an error. The direction in which security is traced as T increases at a what T is equal to negative three X is equal to six wise. You could meet one when T is equal to negative two. X is equal to one, and why you go to zero equals 91 actually go to negative two. Why equal to one equal to zero X automated? Three. Why would, too the equal to one X equal to negative two. One equal to three t e Go to to Mexico to one. Why do you go to or my T equal to three? Actually go to six and y CO. Two off, then that scares the curve. It's in six negative one. Yeah, and 10 negative. 21 negative three two and negative two three. Negative one or so. The curve is like this. The arrow is the direction in which the curve is traced as T increases are to be eliminated. The perimeter to find a partition equation of the curve. So from here we have key is equal to why minus two and plugging the first equation we have X is equal to y minus two square minus three. So this is a Cartesian equation of the truth.