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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.

 Jan 14, 2020
 #1
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ignoring dollar sign glitch, 

 

your polynomials is -2[(x-(3+root5)(x-(3-root5)]

simplyfing  

p(x)=-2(x^2-6x+4)

vertex coordinates are (-b/2a, -d/4a)

 

so vertex is V(3,10)

 

 

 

 

CALCUALTION :  X = -b/2a=-(-6)/2=3

                             Y=   -D/4a=-(20)/2=10

 Jan 14, 2020

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