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:

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

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

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

3. Hello, Murphie!

You can simply plot-the-points . . .

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

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

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

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

Code:
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- - * - - - + - - - - - - -
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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.