Answer
(a) $y=x+2$, $(-1 \leq x \leq 4)$
(b) From $(-1,1)$ to $(4,6)$ (Graph 1(b))
(c) $$
\begin{array}{|c|r|l|l|l|l|l|}
\hline t & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline x & -1 & 0 & 1 & 2 & 3 & 4 \\
\hline y & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
\end{array}
$$
(d) Graph 1(d)
Work Step by Step
a) We can eliminate the parameter $t$ by solving for $t$ in terms of $x$ and $y$ from the given equations. Adding 1 to both sides of the equation for $x$, we get $t=x+1$. Similarly, subtracting 1 from both sides of the equation for $y$, we get $t=y-1$. Since both expressions are equal to $t$, we can set them equal to each other and solve for $y$ in terms of $x$ :
$$
x+1=y-1 \Rightarrow y=x+2
$$
So, the trajectory is given by the curve $y=x+2$.
b) When $t=0$, we find $(x,y)=(-1,1)$ and when $t=5$ we find $(x,y)=(4,6)$, so the direction of motion is from $(-1,1)$ to $(4,6)$..
c) To sketch the trajectory, we can plot a few points on the curve $y=x+2$ for various values of $x$. For example, when $x=0$, we have $y=2$, so the point $( 0,2 )$ lies on the curve. When $x=1$, we have $y=3$, so the point $(1,3)$ lies on the curve. Similarly, when $x=-1$, we have $y=1$, so the point $(-1,1)$ lies on the curve. Plotting these points (and any others you want to include) and connecting them with a smooth curve gives us the trajectory of the particle.
We organize the points in a table:
$$
\begin{array}{|c|r|l|l|l|l|l|}
\hline t & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline x & -1 & 0 & 1 & 2 & 3 & 4 \\
\hline y & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
\end{array}
$$
d) Here's a graph of the trajectory over the interval $0 \leq t \leq 5$ :
Note that the trajectory is a straight line with slope 1, since the equation for the trajectory is $y=x+2$.
