A quadratic of the form \(-2x^2+bx+c\) has roots of \(x=3+\sqrt5\) and \(x = 3-\sqrt5\). The graph of \(y=\) \(-2x^2+bx+c\) is a parabola. Find the vertex of this parabola.

Guest Jul 5, 2020

#1**+1 **

Here is the quadratic equation

(x -3 - sqrt5)(x-3+sqrt5) = 0 Multiplying it out

x^2 -3x +sqrt5 x -3x +9 -3sqrt5 - sqrt5 x +3 sqrt 5 - 5

x^2 -6x +4 = 0 to get it into the form asked about in the questiion , multiply through by -2

-2x^2 +12x -8 = 0

Now find 'x' coordinate of the vertex by - b /2a b = 12 a = -2

Use THIS 'x' value in the final equation to find the 'y' coordinate of the vertex....

ElectricPavlov Jul 5, 2020