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Suppose that we have an equation y=ax^2+bx+c whose graph is a parabola with vertex (3,2), vertical axis of symmetry, and contains the point (1,0). What is (a, b, c)?

 Oct 10, 2019
 #1
avatar+103858 
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If   the vertex  is  (3,2)    and  it contains the point   (1,0).....then x  = 1 is a  root

 

So...by symmetry......(5, 0) is also a root

 

So....we can find  "a"  thusly

 

0  =  a(1 - 3)^2  + 2

 

-2  =  a(-2)^2

 

-2  = 4a

 

a  =  -2/4  =  -1/2

 

And by Vieta......

 

The sum of the roots   -b/a

So

1 + 5  = - b / ( -1/2)

6 = 2b

b = 3

 

And the product of the roots  = c/a

So

5*1  =  c /(-1/2)

5 = -2c

c = -5/2

 

So

 

(a,b , c)   =  ( -1/2, 3, -5/2)

 

Here's the graph :  https://www.desmos.com/calculator/yd99xbrpsg

 

 

cool cool cool

 
 Oct 10, 2019

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