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Find $a+b+c$ if the graph of the equation $y=ax^2+bx+c$ is a parabola with vertex $(5,3)$, vertical axis of symmetry, and contains the point $(2,0)$.

 Dec 9, 2019
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Find $a+b+c$ if the graph of the equation $y=ax^2+bx+c$ is a parabola with vertex $(5,3)$, vertical axis of symmetry, and contains the point $(2,0)$.

 

We have the vertex form

 

y  = a(x - h)^2  +  k            where  (h, k)  =  ( 5, 3)

 

And since the point (2,0)  is on the graph, we can slove for "a"  thusly

 

0  =  a(2 - 5)^2  + 3      subract 3 from both sides    

 

-3  = a(-3)^2

 

-3  = 9a           divide both sides by  9

 

(-1/3)  =  a

 

 

So  we have

 

y = (-1/3) (x - 5)^2  +  3

 

y = (-1/3) [ x^2 - 10x + 25)  + 3

 

y =(-1/3)x^2  + (10/3)x - 25/3  +  9/3

 

y  = (-1/3)x^2 + (10/3)x -  16/3

 

So

 

a +  b  +  c   =

 

(-1/3)  + (10/3) +  (-16/3)  =

 

- 7 / 3

 

 

cool cool cool

 Dec 9, 2019

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