Here, total distance covered $= x,km$

Total taxi fare $=$ Rs. $y$

Fare for the $1^{st} \,km = Rs. $Rs. $ 8$

Remaining distance $=( x -1)\, km$

$\therefore $ Fare for $( x -1) \,km =$ Rs. $ 5 \times( x -1)\, km$

Total taxi fare $=$ Rs. $8+$ Rs. $ 5( x -1)$

$\therefore $   According to the condition,

$\Rightarrow $        $y=8+5(x-1)$

$\Rightarrow $        $y=8+5 x-5$

$\Rightarrow $        $y=5 x+3$

which is the required linear equation representing the given information.

Graph: We have   $y =5 x +3$

$\therefore $ When $x=0$,        $y=5(0)+3$

$\Rightarrow $                            $y=3$

When $x=-\,1$,                          $y=5(-\,1)+3$

$\Rightarrow $                            $y=-\,2$

When $x=-\,2$,                          $y=5(-\,2)+3$

$\Rightarrow $                            $y=-\,7$

$\therefore $ We get the following table :

$x$	$0$	$-1$	$-2$
$y$	$3$	$-2$	$-7$

 Now, plotting the ordered pairs $(0,\,3),\,(-1,\,-2)$ and $(-2,\,-7)$ on a graph paper and joining them, get a straight line $PQ$.

Thus, $PQ$ is the required graph of the linear equation $y =5 x +3$.