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The graph of the equation y = Ax^2 + Bx + C , where are A, B, C integers, is shown below. Find A + B + C.

 

 Nov 20, 2020
 #1
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The vertex  is  (-1, -1)

 

So we  have the form

 

4p ( y + 1)  = (x + 1)^2

 

And the point (1,3)  is on the  graph so 

 

4p ( 3 + 1)  = ( 1 + 1)^2

 

16p  = 4

 

p =  4/16 = 1/4

 

 

The equation  is

 

y  + 1 = ( x + 1)^2

 

y = x^2 + 2x + 1 - 1

 

y = x^2 + 2x

 

A = 1   B   = 2   C   = 0 

 

A + B + C =    3

 

 

cool cool cool

 Nov 20, 2020

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