Use the point (0, -12) from the table and substitute x=0 and y=-12 into the relation y = px² − 5x + q.

Using the relation $y = 2x^2 - 5x - 12$, we calculate the missing y-values:

For $x=-1$:

$$y = 2(-1)^2 - 5(-1) - 12$$ $$y = 2(1) + 5 - 12$$ $$y = 2 + 5 - 12$$ $$y = 7 - 12 = -5$$

For $x=1$:

$$y = 2(1)^2 - 5(1) - 12$$ $$y = 2(1) - 5 - 12$$ $$y = 2 - 5 - 12$$ $$y = -3 - 12 = -15$$

For $x=2$:

$$y = 2(2)^2 - 5(2) - 12$$ $$y = 2(4) - 10 - 12$$ $$y = 8 - 10 - 12$$ $$y = -2 - 12 = -14$$

For $x=3$:

$$y = 2(3)^2 - 5(3) - 12$$ $$y = 2(9) - 15 - 12$$ $$y = 18 - 15 - 12$$ $$y = 3 - 12 = -9$$

The completed table is:

x	-3	-2	-1	0	1	2	3	4	5
y	21	6	-5	-12	-15	-14	-9	0	13

Use the graph to find

To find $y$ when $x=1.8$ from the graph, locate $x=1.8$ on the x-axis. Move vertically from $x=1.8$ to intersect the curve. From the intersection point, move horizontally to the y-axis to read the corresponding y-value.
Using the equation $y = 2x^2 - 5x - 12$:

$$y = 2(1.8)^2 - 5(1.8) - 12$$ $$y = 2(3.24) - 9 - 12$$ $$y = 6.48 - 9 - 12$$ $$y = 6.48 - 21$$ $$y = -14.52$$

From the graph, $y$ would be approximately $\boxed{-14.5}$.

To find $x$ when $y=-8$ from the graph, locate $y=-8$ on the y-axis. Move horizontally from $y=-8$ to intersect the curve. From the intersection points, move vertically to the x-axis to read the corresponding x-values.
Using the equation $y = 2x^2 - 5x - 12$:

$$-8 = 2x^2 - 5x - 12$$ $$0 = 2x^2 - 5x - 4$$

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$$x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-4)}}{2(2)}$$ $$x = \frac{5 \pm \sqrt{25 + 32}}{4}$$ $$x = \frac{5 \pm \sqrt{57}}{4}$$ $$x_1 = \frac{5 + \sqrt{57}}{4} \approx \frac{5 + 7.5498}{4} \approx \frac{12.5498}{4} \approx 3.137$$ $$x_2 = \frac{5 - \sqrt{57}}{4} \approx \frac{5 - 7.5498}{4} \approx \frac{-2.5498}{4} \approx -0.637$$

From the graph, $x$ would be approximately $\boxed{3.1 \text{ or } -0.6}$.