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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.

 Feb 17, 2020
 #1
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We have the form Ax^2 + Bx + C

 

The x coordinate of the vertex  is given by   -B / [2A]  =  -17/ [ 2(-5)]  =  17/10

 

Put this value back into the function to get the y coordinate of the vertex 

 

-5(17/10)^2 + 17(17/10) - 12  =

-5 (289/100) + 289/10 - 12   =   49/20

 

So.....the vertex  is  ( 17/10, 49/20)

 Feb 17, 2020

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