Step 1: Create a table of values for $y = 2x^2 + 2x - 8$. 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. $$y = 2x^2 + 2x - 8$$ | $x$ | $x^2$ | $2x^2$ | $2x$ | $-8$ | $y = 2x^2 + 2x - 8$ | | :----- | :---- | :----- | :--- | :--- | :------------------ | | $-4$ | $16$ | $32$ | $-8$ | $-8$ | $32 - 8 - 8 = 16$ | | $-3$ | $9$ | $18$ | $-6$ | $-8$ | $18 - 6 - 8 = 4$ | | $-2$ | $4$ | $8$ | $-4$ | $-8$ | $8 - 4 - 8 = -4$ | | $-1$ | $1$ | $2$ | $-2$ | $-8$ | $2 - 2 - 8 = -8$ | | $-0.5$ | $0.25$| $0.5$ | $-1$ | $-8$ | $0.5 - 1 - 8 = -8.5$| | $0$ | $0$ | $0$ | $0$ | $-8$ | $0 + 0 - 8 = -8$ | | $1$ | $1$ | $2$ | $2$ | $-8$ | $2 + 2 - 8 = -4$ | | $2$ | $4$ | $8$ | $4$ | $-8$ | $8 + 4 - 8 = 4$ | | $3$ | $9$ | $18$ | $6$ | $-8$ | $18 + 6 - 8 = 16$ | Step 2: Plot the graph of $y = 2x^2 + 2x - 8$. 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. Step 3: Use the graph to solve the equations. 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