Hello, Murphie!
You can simply plotthepoints . . .
Graph the relation defined by the parametric equations when: .$\displaystyle 2 \leq t \leq 3$
. . $\displaystyle \begin{array}{ccc}x(t)&=& t^25 \\ y(t)&=&t1 \end{array}$
Let $\displaystyle t \:=\:\text{}2,\text{}1,0,1,2,3$
. . and we have some of the points of the graph.
. . $\displaystyle \begin{array}{cccc}\hline t & x & y & \text{Point}\\ \hline
\text{}2 & \text{}1 & \text{}3 & (\text{}1,\text{}3)\\
\text{}1 & \text{}4 & \text{}2 & (\text{}4,\text{}2)\\
0 & \text{}5 & \text{}1 & (\text{}5,\text{}1)\\
1 & \text{}4 & 0 & (\text{}4,0) \\
2 & \text{}1 & 1 & (\text{}1,1)\\
3 & 4 & 2 & (4,2) \\
\hline \end{array}$
Code:

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We have a portion of the parabola: .$\displaystyle (y + 1)^2 \:=\:x+5$
Its vertex is (5, 1) and it opens to the right.