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A parabola with equation y=ax^2+bx+c has a vertical line of symmetry at x=2 and goes through the two points (1,1) and (5,-1).  Find (a,b,c).

 Dec 9, 2020
 #1
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If the line of symmetry  is  x =2.....then this is the x coordinate of the  vertex

 

We have the form

 

y =  a(x - h)^2  + k         where  h  is the  x oordinate of the vertex ....we need to find a  and k thusly

 

1  = a  (1 - 2)^2  +  k   →  1  = a  +  k

-1  = a  (5 - 2)^2  + k  → -1  = 9a + k

 

Subtract  the second equation from the first

 

2  = -8a

a = 2/-8  = -1/4

 

And   1 =  -1/4  +  k

k  = 1 + 1/4  = 5/4

 

So  we  have that

 

y  =(-1/4) ( x - 2)^2 +  5/4

 

y = (-1/4)  ( x^2  -4x  + 4) + 5/4

 

y = (-1/4)x^2  + x  -1  +  5/4

 

y  = (-1/4)x^2  + x  + 1/4

 

 

a = -1/4    b  =  1     c   =   1/4

 

And their sum is     1

 

 

cool cool cool

 Dec 9, 2020

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