Chapter 6.3, Problem 5E

Solution

a)

To calculate: The values in the table for a parametric equation.

The complete table is as follows.

$t$	$-2$	$-1$	$0$	$1$	$2$
$x$	$0$	$1$	$2$	$3$	$4$
$y$	$-\frac{1}{2}$	$-2$	undefined	$4$	$\frac{5}{2}$

Given information:

The parametric equations are $x=t+2$ and $y=1+\frac{3}{t}$ .

Calculation:

Find the values of $x$ and $y$ for the different values of $t$ .

Substitute $-2$ for $t$ in equation $x=t+2$ .

  $\begin{array}{c}x=-2+2\\ =0\end{array}$

Substitute $-2$ for $t$ in equation $y=1+\frac{3}{t}$ .

  $\begin{array}{c}y=1+\frac{3}{-2}\\ =-\frac{1}{2}\end{array}$

Substitute $-1$ for $t$ in equation $x=t+2$ .

  $\begin{array}{c}x=-1+2\\ =1\end{array}$

Substitute $-1$ for $t$ in equation $y=1+\frac{3}{t}$ .

  $\begin{array}{c}y=1+\frac{3}{-1}\\ =-2\end{array}$

Substitute $0$ for $t$ in equation $x=t+2$ .

  $\begin{array}{c}x=0+2\\ =2\end{array}$

Substitute $0$ for $t$ in equation $y=1+\frac{3}{t}$ .

  $\begin{array}{c}y=1+\frac{3}{0}\\ =\text{undefined}\end{array}$

Substitute $1$ for $t$ in equation $x=t+2$ .

  $\begin{array}{c}x=1+2\\ =3\end{array}$

Substitute $1$ for $t$ in equation $y=1+\frac{3}{t}$ .

  $\begin{array}{c}y=1+\frac{3}{1}\\ =4\end{array}$

Substitute $2$ for $t$ in equation $x=t+2$ .

  $\begin{array}{c}x=2+2\\ =4\end{array}$

Substitute $2$ for $t$ in equation $y=1+\frac{3}{t}$ .

  $\begin{array}{c}y=1+\frac{3}{2}\\ =\frac{5}{5}\end{array}$

Complete the table as follows.

$t$	$-2$	$-1$	$0$	$1$	$2$
$x$	$0$	$1$	$2$	$3$	$4$
$y$	$-\frac{1}{2}$	$-2$	undefined	$4$	$\frac{5}{2}$

Therefore, the complete table shown above.

b)

To graph: The calculated points in part (a).

The points are shown in figure (1).

Given Information:

From part (a), the values given in the table are as follows.

$t$	$-2$	$-1$	$0$	$1$	$2$
$x$	$0$	$1$	$2$	$3$	$4$
$y$	$-\frac{1}{2}$	$-2$	undefined	$4$	$\frac{5}{2}$

Calculation:

Plot the points as follows.

  

Figure (1)

The value of the table is shown in the figure (1).