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A quadratic of the form -2x2 + bx + c has roots of x = 3 + sqrt5 and x = 3 - sqrt5. The graph of y = -2x2 + bx + c is a parabola. Find the vertex of this parabola. P.S: This is the first time I'm asking a question :) P.S.P.S: Thanks for answering this whoever does. It will help me ALOT

 Feb 26, 2020
edited by PharaoCarl  Feb 26, 2020
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-2x^2  + bx  + c

 

By Vieta.....

 

The  sum of the roots =  6  =  -b/ -2  =  b/2  which implies  that  b  =  12

 

The product of the roots   =   c /-2  =  (3  + sqrt (5) ) * (3  -sqrt (5))  =   9 -  5  = 4

So      c  / -2  =  4    which implies  that  (c  = -8

 

So

 

We  have  that   

 

-2x^2  + 12x  - 8

 

The x coordinate of the  vertex =  (-b)/ (2a)  =  -12/ 2 (-2)  =  3    [ which we might have guessed ]

 

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

 

Here's  the graph  :

 

https://www.desmos.com/calculator/fsqrjynnxb

 

cool cool cool

 Feb 26, 2020
edited by CPhill  Feb 26, 2020

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