Answer:
Let f(x) = 0 and solve for x to get the x intercepts.
2x2 -3x -5 =0 Factor to get (2x -5) (x + 1) =0 then
2x - 5 = 0 and x = 2.5 x + 1 = 0 and x = -1 These are the x intercepts.
B: As the coefficient of the x^2 term is positive, the parabola open up, and the vertex is a minimum. The coordinates of the parabola are given by -b/2a, f( -b/2a) where a and b represent the coefficients of a parabola in the standard form ax^2 + bx +c = 0. In this case a = 2, be = -3, c = -5. So -b/2a = 3 / 4 f( -b/2a) = 2 * (3/4)^2 -3 * (3/4) - 5 = -6.125. The coordinates of the vertex are (3/4, -6.125)
C: To graph this function, first establish where the vertex will be from the coordinates found. Next plot the position of the x intercepts. You know it opens up, so you would draw it so the vertex is a minimum. You could also get the y intercept by letting x = 0, which gives -5 for the y intercept. This should give you a rough idea of what the graph looks like.
