Calculate the missing y values using the equation y = 2x2 + 5x − 3.

For $x = -3$: $y = 2(-3)^2 + 5(-3) - 3 = 2(9) - 15 - 3 = 18 - 15 - 3 = 0$.
For $x = -1.5$: $y = 2(-1.5)^2 + 5(-1.5) - 3 = 2(2.25) - 7.5 - 3 = 4.5 - 7.5 - 3 = -6$.
For $x = -0.5$: $y = 2(-0.5)^2 + 5(-0.5) - 3 = 2(0.25) - 2.5 - 3 = 0.5 - 2.5 - 3 = -5$.
For $x = 2$: $y = 2(2)^2 + 5(2) - 3 = 2(4) + 10 - 3 = 8 + 10 - 3 = 15$.

8. (b) Using a scale of 2cm to 1 unit on the $x$-axis and 2cm to 2 units on the $y$-axis, draw the graph of $y = 2x^2 + 5x - 3$, for the interval $-5 \le x \le 2.5$.

This part requires drawing a graph on paper using the specified scale and the values from the completed table.

8. (c) Using your graph in (b), find the truth set of:

• i) $2x^2 + 5x - 3 = 0$. This corresponds to the $x$-intercepts of the graph where $y=0$. From the table, $y=0$ when $x=-3$ and $x=-2$. The truth set is $\boxed{\{-3, -2\}}$.

• ii) $2x^2 + 6x - 3 = 0$. This equation can be rewritten as $2x^2 + 5x - 3 + x = 0$, which means $y + x = 0$, or $y = -x$. To solve this graphically, you would draw the line $y = -x$ on the same axes as $y = 2x^2 + 5x - 3$ and find the $x$-coordinates of their intersection points. Algebraically, we solve $2x^2 + 6x - 3 = 0$ using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

  $$x = \frac{-6 \pm \sqrt{6^2 - 4(2)(-3)}}{2(2)}$$ $$x = \frac{-6 \pm \sqrt{36 + 24}}{4}$$ $$x = \frac{-6 \pm \sqrt{60}}{4}$$ $$x = \frac{-6 \pm 2\sqrt{15}}{4}$$ $$x = \frac{-3 \pm \sqrt{15}}{2}$$

  Approximating $\sqrt{15} \approx 3.873$:

  $$x_1 = \frac{-3 + 3.873}{2} = \frac{0.873}{2} \approx 0.437$$ $$x_2 = \frac{-3 - 3.873}{2} = \frac{-6.873}{2} \approx -3.437$$

  The truth set is approximately $\boxed{\{-3.44, 0.44\}}$.

8. (d) The minimum value of the curve.

For $y = 2x^2 + 5x - 3$, we have $a=2$ and $b=5$.

The minimum value of the curve is $\boxed{-6.125}$.