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.
\( x = 3 + \sqrt{5}$ and $x = 3 - \sqrt{5}.\)
By Vieta the sum of the roots = -b/-2 = 6 ⇒ -b = -12 ⇒ 12
And the product of the roots [ i.e., 4] = c / -2 ⇒ c = -8
By symmetry, the x coordinate of the vertex = 3
So......the y coordinate of the vertex = -2(3)^2 + 12(3) - 8 = 10
So.....the vertex = (3, 10)