1. Graphing relations

Graph the relation defined by the parametric equations when -2 ≤ t ≤ 3.
x(t)=t^2-5
y(t)=t-1

Well a point/tips in the right direction would be great, I am completely clueless on where to start with this, thanks!

2. If you want to graph, there are two ways:

(1) Assume arbitrary some values of t. Plot x and y corresponding to them.
e.g. if t=1, x=-4 and y=0

(2) The second method is to express y in terms of x..or x in terms of y.

Spoiler:

$x=t^2-5$
$y=t-1$........or $t=y+1$

$x=y^2+2y+1-5=y^2+2y-4$...either use this or

$y=\sqrt{x+5}-1$

3. Hello, Murphie!

You can simply plot-the-points . . .

Graph the relation defined by the parametric equations when: . $-2 \leq t \leq 3$

. . $\begin{array}{ccc}x(t)&=& t^2-5 \\ y(t)&=&t-1 \end{array}$

Let $t \:=\:\text{-}2,\text{-}1,0,1,2,3$
. . and we have some of the points of the graph.

. . $\begin{array}{|c||c|c||c|}\hline t & x & y & \text{Point}\\ \hline
\text{-}2 & \text{-}1 & \text{-}3 & (\text{-}1,\text{-}3)\\
\text{-}1 & \text{-}4 & \text{-}2 & (\text{-}4,\text{-}2)\\
0 & \text{-}5 & \text{-}1 & (\text{-}5,\text{-}1)\\
1 & \text{-}4 & 0 & (\text{-}4,0) \\
2 & \text{-}1 & 1 & (\text{-}1,1)\\
3 & 4 & 2 & (4,2) \\
\hline \end{array}$

Code:
                  |
|       *
* |
- - * - - - + - - - - - - -
*         |
*       |
* |
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We have a portion of the parabola: . $(y + 1)^2 \:=\:x+5$

Its vertex is (-5, -1) and it opens to the right.