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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.

 May 5, 2020
 #1
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\( x = 3 + \sqrt{5}$ and $x = 3 - \sqrt{5}.\)

 

By Vieta  the  sum of the roots  = -b/-2  = 6   ⇒   -b  = -12   ⇒ 12

And the product of the roots  [ i.e., 4]  =  c / -2   ⇒   c = -8 

By symmetry, the  x  coordinate of  the  vertex  =  3 

So......the y coordinate of the  vertex =  -2(3)^2  + 12(3) - 8  = 10

 

So.....the  vertex =  (3, 10)  

 

 

 

cool cool cool

 May 5, 2020
 #2
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Thank you very much CPhill!

 May 5, 2020
 #3
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 one-half divided by 3?

 May 5, 2020

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