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A quadratic of the form $-2x^2 + bx + c$   has roots of x=3+sqrt5 and x=3-sqrt5 The graph of 

$y = -2x^2 + bx + c$

 is a parabola. Find the vertex of this parabola.

 

Hint-The coefficient of x^2 is not 1!

 Mar 12, 2020
 #1
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Please Help Electric Pavlov!

 Mar 12, 2020
 #2
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If the roots are  3 + sqrt(5)  and  3 - sqrt(5), then the equation is:

    y  =  [x - ( 3 + sqrt(5) ) ] · [x - ( 3 - sqrt(5) ) ]

 

If you expand the above equation, you will end with:  y  =  x2 - 6x + 4

 

Now, to find the vertex:           y  =  x2 - 6x + 4

                                          y - 4  =  x2 - 6x

Complete the square:  y - 4 + 9  =  x2 - 6x + 9

                                          y + 5  =  (x - 3)2

 

So, the vertex is:  (3, -5)

 Mar 12, 2020

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