Create a table of values for y = 2x² + 2x − 8.

To plot the graph, we need to calculate y-values for a range of x-values. We will choose x-values from $-4$ to 3 to capture the shape of the parabola, including its vertex and x-intercepts.

$$y = 2x^2 + 2x - 8$$

$x$	$x^2$	$2x^2$	$2x$	$-8$	$y = 2x^2 + 2x - 8$
$-4$	16	32	$-8$	$-8$	$32 - 8 - 8 = 16$
$-3$	9	18	$-6$	$-8$	$18 - 6 - 8 = 4$
$-2$	4	8	$-4$	$-8$	$8 - 4 - 8 = -4$
$-1$	1	2	$-2$	$-8$	$2 - 2 - 8 = -8$
$-0.5$	0.25	0.5	$-1$	$-8$	$0.5 - 1 - 8 = -8.5$
0	0	0	0	$-8$	$0 + 0 - 8 = -8$
1	1	2	2	$-8$	$2 + 2 - 8 = -4$
2	4	8	4	$-8$	$8 + 4 - 8 = 4$
3	9	18	6	$-8$	$18 + 6 - 8 = 16$

a)

1. Draw and label the x-axis and y-axis on the provided graph paper.
2. Mark the scale as specified: 1 unit for the y-axis and 1 unit for the x-axis. This means each major grid line represents 1 unit on both axes.
3. Plot the points from the table: $(-4, 16)$, $(-3, 4)$, $(-2, -4)$, $(-1, -8)$, $(-0.5, -8.5)$, $(0, -8)$, $(1, -4)$, $(2, 4)$, $(3, 16)$.
4. Draw a smooth curve through these points to form the parabola.

b)

i) Solve $2x^2 + 2x - 8 = 0$
This equation is equivalent to finding the x-intercepts of the graph $y = 2x^2 + 2x - 8$. These are the points where the parabola crosses the x-axis (where $y=0$).
By observing the graph, the parabola intersects the x-axis at approximately:
$x \approx -2.56$ and $x \approx 1.56$.
(For exact values: $x = \frac{-2 \pm \sqrt{2^2 - 4(2)(-8)}}{2(2)} = \frac{-2 \pm \sqrt{4+64}}{4} = \frac{-2 \pm \sqrt{68}}{4} = \frac{-2 \pm 2\sqrt{17}}{4} = \frac{-1 \pm \sqrt{17}}{2}$)
The solutions are $\boxed{x \approx -2.56, x \approx 1.56}$.

ii) Solve $2x^2 + x + 2 = 0$
To solve