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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.

 Oct 25, 2020
 #1
avatar+28021 
+1

y=(x-3+sqrt5)(x-3-sqrt5)   multiply this out :

y=x^2 -3x-sqrt5x - 3x +9+3sqrt5+ sqrt5x-3sqrt5 -5      simplify

y=x^2 -6x +4       multiply by -2

 

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

 

x value of vertex =  - b/2a      use this value of x in the equation to find the y value of the vertex

 Oct 25, 2020
 #2
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0

I am kinda confused on how to fine y

 Oct 25, 2020
 #3
avatar+28021 
+1

Y= -2x^2+12x-8      Using x = -b/2a        Where b= 12    a= -2

 Oct 26, 2020
 #4
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0

Thanks!!

 Oct 26, 2020

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