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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. Help needed ASAP

 Jul 18, 2020
 #1
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Just use the quadratic formula!  It's easy.

 Jul 18, 2020
 #2
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y  =  -2x2 + bx + c          with roots     3 + sqrt(5)     and     3 - sqrt(5)

 

One answer is:  y  =  [ x - ( 3 + sqrt(5) ) ] · [ x - ( 3 - sqrt(5) ) ]

 

Let's multiply out the right-hand side:

     [ x - ( 3 + sqrt(5) ) ] · [ x - ( 3 - sqrt(5) ) ]     --->     [ x - 3 - sqrt(5) ] · [ x - 3 + sqrt(5) ]

                  --->   x2 - 3x + sqrt(5)·x - 3x + 9 - 3·sqrt(5) - sqrt(5)·x + 3·sqrt(5) - 5

                  --->   x2 - 6x + 4

 

However, since the original problem started with "-2x2", we'll have to multiply this answer by -2:

--->     y  =  -2x2 + 12x - 8

 

To find the vertex:                              y  =  -2x2 + 12x - 8

--->  add 8 to both sides:             y + 8  =  -2x2 + 12x

--->   factor out the -2:                 y + 8  =  -2(x2 - 6x)

--->   complete the square:    y + 8 - 18 =  -2(x2 - 6x + 9)

                                                    y - 10  =  -2(x - 3)2  

 

The vertex is  (3, 10)

 Jul 18, 2020

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