What are the x-intercepts and the vertex of the parabola represented by y = 2x2 - 5x -12?

To find the x-intercepts of the parabola, we need to set y = 0 and solve for x.

0 = 2x^2 - 5x - 12

This is a quadratic equation in the form ax^2 + bx + c = 0. We can solve it using the quadratic formula, x = [-b ± sqrt(b^2 - 4ac)] / (2a).

Here, a = 2, b = -5, and c = -12.

x = [5 ± sqrt((-5)^2 - 42(-12))] / (2*2) x = [5 ± sqrt(25 + 96)] / 4 x = [5 ± sqrt(121)] / 4 x = [5 ± 11] / 4

So, the x-intercepts are x = 4 and x = -1.5.

The vertex of a parabola y = ax^2 + bx + c is given by the point (h, k), where h = -b / (2a) and k = c - b^2 / (4a).

Here, h = -(-5) / (22) = 5/4 = 1.25 And, k = -12 - (-5)^2 / (42) = -12 - 25/8 = -12 - 3.125 = -15.125

So, the vertex of the parabola is (1.25, -15.125).