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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 \((2,0)\).

 Jul 16, 2019
 #1
avatar+111393 
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y = ax^2  + bx  +  c

 

The x coordinate of the vertex =  -b / (2a)   =  5  ⇒   -b = (2a)*5  ⇒  b = -10a

 

So  we have this system

 

a(5)^2  - 10a(5) + c  = 3   ⇒  25a - 50a + c = 3  ⇒  -25a + c  = 3     (1)

a(2)^2 - 10a(2) + c  =  0  ⇒  4a - 20a + c  =  0  ⇒  - 16a  + c  = 0  ⇒  16a - c  = 0   (2)

 

Add (1)  and (2)    and we get that

-9a  = 3

a  =  -1/3

b = -10 (-1/3) = 10/3

 

And using  (2)    16 (-1/3) - c  = 0   ⇒  -16/3  = c

 

So....

 

y  = (-1/3)x^2  + (10/3)x - 16/3

 

a  = -1/3      b  =  10/3     c  = -16/3

 

And their sum is    -7/3

 

 

cool cool cool

 Jul 16, 2019

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