A = 8
b = -5
c = 2
Work out the value of b^2-4ac

b²- 4ac

-5² - 4×8×2

25-64

-39

F(x) = 2x^2 − x − 10  part a: what are the x-intercepts of the graph of f(x)? show your work. (2 points)x-intercepts => f(x) = 0 or y = 0 (remember y = f(x) ).so, you need to solve 2x^2 - x - 10 = 0you can either use the quadratic formula or complete squares to solve it.i will complete squares: 2( x^2 - x/2) - 10 = 02 [(x - 1/4)^2 - 1/16] - 10 = 02(x -1/4)^2 - 1/8 - 10 = 02(x - 1/4)^2 - 81/8 = 0(x - 1/4)^2 = 81/16(x -1/4) = +/- √ (81/16) =+/- 9/4=> x = 1/4 +/- 9/4=> x = 1/4 + 9/4 = 10/4 = 2.5 and x = 1/4 - 9/4 = -8/4 = -2answer: -2 and 2.5part b: is the vertex of the graph of f(x) going to be a maximum or minimum?   given that the coefficient of the quadratic term (x^2) is positive, the parabola opens upward, so the vertex is a minimum.what are the coordinates of the vertex?   use the vertex form to conclude the coordinates of the vertex.the vertex form of the parabola is y = a (x - h)^2 + k, where the coordinates of the verex are (h,k).the equation of this parabola is 2(x - 1/4)^2 - 81/8, so h = 1/4 and k = -81/8 are the coordinates of the vertex.part c: what are the steps you would use to graph f(x)? justify that you can use the answers obtained in part a and part b to draw the graph. (5 points)1) put the x-intercepts2) put the vertex (h,k)3) draw a curve that pass through the three points: it is your parabola.