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.

Guest Feb 9, 2020

#1**+1 **

( x -3 + sqrt5)(x-3-sqrt5) is the eqution given the roots supplied

Multiply all of this out to get

x^2-6x+4 now multiply by -2 to get the same form as the question

-2x^2+12x-8

this is a dome shaped parabola max value will be at x = -b/2a = -12/-4 = 3

plug this into the equation to find y -2(3^2)+12(3) - 8 = y = 10

Vertex = (3,10)

ElectricPavlov Feb 9, 2020