1)
A quadratic of the form $-2x^2 + bx + c$ has roots of $x = 3 + \sqrt{5}$ and $x = 3 - \sqrt{5}.$ The graph of $y = -2x^2 + bx + c$ is a parabola. Find the vertex of this parabola.
We have the form Ax^2 + Bx + C
The x coordinate of the vertex is given by -B / [2A] = -17/ [ 2(-5)] = 17/10
Put this value back into the function to get the y coordinate of the vertex
-5(17/10)^2 + 17(17/10) - 12 =
-5 (289/100) + 289/10 - 12 = 49/20
So.....the vertex is ( 17/10, 49/20)