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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 $(a, b, c)$?

 Mar 18, 2020
 #1
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Help Cphill

 Mar 18, 2020
 #2
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If  the  point  (1,0) is  on the graph, then by  symmetry  with the veryex, the  point  (5,0)  is also on the  graph

 

Then  x  =1  and  x  = 5  are roots

 

So......we  have this form

 

y = a (x - 1) ( x - 5)

 

And since the point (3, 2)  is on the graph, we can solve for  "a" thusly

 

2 = a( 3 - 1) (3 - 5)     simplify

 

2 = a (2) (-2)

 

2= - 4a       divide  both sides  by  -4

 

2/- 4  =  a =   -1/2

 

So  we  have that

 

y = (-1/2)(x - 1) (x - 5)

 

y = (-1/2) (x^2  - 6x  + 5)

 

y = (-1/2)x^2  + 3x  - 5/2

 

(a, b, c) =   [ (-1/2) ,  3,  ( -5/2)  ]

 

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

 

 

cool cool cool

 Mar 18, 2020

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