Exercise 8 Page 196

Let's begin by making a table of the residual values.

$x$	$y$	$y = - 0.5 x - 2$	$y -$ value From Model	Residual
$4$	$- 1$	$- 0.5 (4) - 2$	$- 4$	$- 1 - (- 4) = 3$
$6$	$- 3$	$- 0.5 (6) - 2$	$- 5$	$- 3 - (- 5) = 2$
$8$	$- 6$	$- 0.5 (8) - 2$	$- 6$	$- 6 - (- 6) = 0$
$10$	$- 8$	$- 0.5 (10) - 2$	$- 7$	$- 8 - (- 7) = - 1$
$12$	$- 10$	$- 0.5 (12) - 2$	$- 8$	$- 10 - (- 8) = - 2$
$14$	$- 10$	$- 0.5 (14) - 2$	$- 9$	$- 10 - (- 9) = - 1$
$16$	$- 10$	$- 0.5 (16) - 2$	$- 10$	$- 10 - (- 10) = 0$
$18$	$- 9$	$- 0.5 (18) - 2$	$- 11$	$- 9 - (- 11) = 2$
$20$	$- 9$	$- 0.5 (20) - 2$	$- 12$	$- 9 - (- 12) = 3$

Now we can create a scatter plot using the given $x -$ values and our residuals. Remember, if the model is a good fit for the data, the scatter plot will be evenly distributed above and below the $x -$ axis. Also, there will be no apparent patterns.

This line of fit does not model the data well. It is not evenly distributed above and below the $x -$ axis. We can see that the residual points form a U-shaped pattern, which suggests the data is not linear.