Chapter 1.5, Problem 8E

Solution

a.

To find : The points determined by $-3,-2,-1,0,1,2,3$ ,where $x=\sqrt{t},y=2t-5$ .

$t$	$x=\sqrt{t}$	$y=2t-5$	$\left(x,y\right)$
$-3$	$\sqrt{-3}$	$-11$	$\left(\sqrt{-3},-11\right)$
$-2$	$\sqrt{-2}$	$-9$	$\left(\sqrt{-2},-9\right)$
$-1$	$\sqrt{-1}$	$-7$	$\left(\sqrt{-1},-7\right)$
$0$	$0$	$-5$	$\left(0,-5\right)$
$1$	$1$	$-3$	$\left(1,-3\right)$
$2$	$\sqrt{2}$	$-1$	$\left(\sqrt{2},-1\right)$
$3$	$\sqrt{3}$	$1$	$\left(\sqrt{3},1\right)$

Given information:

The given points are $-3,-2,-1,0,1,2,3$ and $x=\sqrt{t},y=2t-5$

Concept used:

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

Calculations:

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

$t$	$x=\sqrt{t}$	$y=2t-5$	$\left(x,y\right)$
$-3$	$\sqrt{-3}$	$-11$	$\left(\sqrt{-3},-11\right)$
$-2$	$\sqrt{-2}$	$-9$	$\left(\sqrt{-2},-9\right)$
$-1$	$\sqrt{-1}$	$-7$	$\left(\sqrt{-1},-7\right)$
$0$	$0$	$-5$	$\left(0,-5\right)$
$1$	$1$	$-3$	$\left(1,-3\right)$
$2$	$\sqrt{2}$	$-1$	$\left(\sqrt{2},-1\right)$
$3$	$\sqrt{3}$	$1$	$\left(\sqrt{3},1\right)$

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

  $x=\sqrt{t},y=2t-5$ .

The relation between $x\text{and}y\text{is}y=2x^2-5.$

Given information:

The given $x=\sqrt{t},y=2t-5$ .

Concept used :

By using the substitution, method can find the relation.

Calculations:

Here

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

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

  $y=2x^2-5$

So the relation between $x\text{and}y\text{is}y=\left(2x^2-5\right).$

c.

To plot: the relation between $x\text{and}y\text{is}y=\left(2x^2-5\right).$

Given information:

The given equation is $y=2x^2-5$ .

Plot the graph:

Graph the relation $y=2x^2-5$ is given below.