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The graph of \(y=ax^2 + bx + c\) is a parabola with vertical axis of symmetry. The vertex of this parabola is \((2,3)\) and the parabola contains the point \((4,4)\). Find the value of \(y\) when \(x=6\).

 Jan 14, 2018
 #1
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y = ax^2 + bx  +  c

 

The x coordinate of the vertex  =  -b / [2a]  =  2  ⇒  b = - 4a    

 

We know that

 

a(2)^2 + b(2)  +  c  =  3  ⇒  4a  + 2b  +  c  =  3  ⇒  4a -  8a  + c  =  3  ⇒  -4a + c = 3    (1)

a(4)^2 + b(4)  +  c  =  4  ⇒ 16a  + 4b  + c  = 4  ⇒  16a - 16a + c  = 4  ⇒  c  = 4   (2)

 

Subbing (2)  into (1)  we have that   

-4a  +  4 =  3

-4a  =  -1

a  = -1/-4  =  1/4

 

And     b  =  -4(1/4)  =  -1

 

So......the function is 

 

y  = (1/4)x^2  -  x  +  4

 

So....when  x  = 6 

y  =  (1/4)6^2 - (6)  +  4  =  9  - 6  +  4   =   7

 

 

cool cool cool

 Jan 14, 2018

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