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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.

 Feb 9, 2020
 #1
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( 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)

 
 Feb 9, 2020

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