a. x -3 -2 -1 0 1 2 3 \(2x^2\) 18 8 2 0 2 8 -18 - x 3 2 1 0 - 1 - 2 - 3 - 4 - 4 - 4 - 4 -4 -4 -4 -4 y 17 6 -1 -4 -3 2 11 b. the graph above, solution to question 8b. ci. The roots of the equation from the graph are -1.2 and 1.7. ii. Let y = \(2x^2 - x - 4\) \(\frac{dy}{dx}\) = 4x - 1 At critical point \(\frac{dy}{dx}\) = 0 so that 4x - 1 = 0; x = \(\frac{1}{4}\) Using test points: 1. we use a test point less than \(\frac{1}{4}\), e.g x = 0 \(\frac{dy}{dx}\)(0) = 4(0) - 1 = -1 at this point, y = \(2x^2 - x - 4\) is decreasing 2. we use a test point greater than \(\frac{1}{4}\) e.g x = 1 \(\frac{dy}{dx}\)(1) = 4(1) - 1 = 4 - 1 = 3. at this point, y = \(2x^2 - x - 4\) is increasing. so, the function y = \(2x^2 - x - 4\) increases as x increases when x is greater than \(\frac{1}{4}\) Therefore, the value of x for which y increases are x is greater than \(\frac{1}{4}\) iii. The minimum point of y = -4.1.