Chapter 1.5, Problem 6E

Solution

a.

To find : The points determined by $-3,-2,-1,0,1,2,3$ ,where

  $x=t^2+1,y=t^2-2t$

$t$	$x=t^2+1$	$y=t^2-2t$	$\left(x,y\right)$
$-3$	$10$	$15$	$\left(10,15\right)$
$-2$	$5$	$8$	$\left(5,8\right)$
$-1$	$2$	$3$	$\left(2,3\right)$
$0$	$1$	$0$	$\left(1,0\right)$
$1$	$2$	$-1$	$\left(2,-1\right)$
$2$	$5$	$0$	$\left(5,0\right)$
$3$	$10$	$4$	$\left(10,4\right)$

Given information:

The given points are $-3,-2,-1,0,1,2,3$ and $x=t^2+1,y=t^2-2t$

Concept used:

By substitution method can find the values of $t$ for $x\text{and }y$ .

Calculations:

Substitute each value of $t$ for $x\text{and }y$:

$t$	$x=t^2+1$	$y=t^2-2t$	$\left(x,y\right)$
$-3$	$10$	$15$	$\left(10,15\right)$
$-2$	$5$	$8$	$\left(5,8\right)$
$-1$	$2$	$3$	$\left(2,3\right)$
$0$	$1$	$0$	$\left(1,0\right)$
$1$	$2$	$-1$	$\left(2,-1\right)$
$2$	$5$	$0$	$\left(5,0\right)$
$3$	$10$	$4$	$\left(10,4\right)$

This is the table of value $t$ for the for $x\text{and }y$.

b.

To find : The direct algebraic relation between $x\text{and}y$ , where

  $x=t^2+1,y=t^2-2t$ .

The relation between $x\text{and}y\text{is}\left(\sqrt{x-1}\right)\left(\sqrt{x-1}-2\right)$

Given information:

The given $x=t^2+1,y=t^2-2t$ .

Concept used :

By using the substitution, method can find the relation.

Calculations:

Here

  $\begin{array}{l}x=t^2+1\\ t=\sqrt{x-1}\end{array}$

Substitute $t$ into the equation of $y$ .

  $\begin{array}{c}y={\left(\sqrt{x-1}\right)}^2-2\left(\sqrt{x-1}\right)\\ =\left(\sqrt{x-1}\right)\left(\sqrt{x-1}-2\right)\end{array}$

So the relation between $x\text{and}y\text{is}\left(\sqrt{x-1}\right)\left(\sqrt{x-1}-2\right)$

c.

To plot: the relation between $x\text{and}y\text{is}\left(\sqrt{x-1}\right)\left(\sqrt{x-1}-2\right)$

Given information:

The given equation is $y=\sqrt{x-1}\left(\sqrt{x-1}-2\right)$ .

Plot the graph:

Graph the relation $y=\sqrt{x-1}\left(\sqrt{x-1}-2\right)$ is given below.