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

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 ≈ -2.56 and x ≈ 1.56. (For exact values: x = -2 ± sqrt(2^2 - 4(2)(-8))2(2) = -2 ± sqrt(4+64)4 = -2 ± sqrt(68)4 = -2 ± 2sqrt(17)4 = -1 ± sqrt(17)2) The solutions are x ≈ -2.56, x ≈ 1.56. ii) Solve 2x^2 + x + 2 = 0 To solve