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Vertex formula question help!

Guest Sep 25, 2017
 #1
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For f(x), write y

 

y  = ax^2 + bx + c         subtract c from both sides       

 

y - c  = a  [ x^2  + (b/a)x   ]       

Complete the square on x....take (1/2)  of (b/a)  =  b/[2a]....square it  =  b^2/[4a^2 ]........add to both sides.....don't forget that we're actually adding  ab^2/[4a^2 ]  to the left side

 

y - c + ab^2 / [4a^2]  =  a [ x^2 + (b/a)x  + b^2/(4a^2 )  ]       

Factor the right side........simplify the left

 

 

y - c + b^2 / (4a)  =  a [ x  +  b /(2a) ] ^2 

Add c, subtract b^2 / (4a)  to both sides

 

y  = a [ x  +  b/(2a) ] ^2  - b^2 / (4a) + c

 

y  = a [ x + b / (2a) ]^2   +    (  - b^2  / (4a)  + c  )

 

So.....the vertex   is given by     [ -b / (2a) ,     - b^2  / (4a)  + c  ]

 

Note that we can prove that this is correct.........

 

The x coordinate of the vertex is  -b / (2a)

 

Putting this into   ax^2 + bx + c   for x, we  can find the y coordinate of the vertex thusly :

 

y  = a [b^2 / (4a^2) ] + b [ -b/(2a)] + c

 

y  =  [b^2 / (4a)] - b^2 /(2a)  + c

 

y =  [ b^2  - 2b^2 ] / (4a)  + c

 

y  =  [-b^2 / (4a) ]  + c

 

 

 

cool cool cool

CPhill  Sep 25, 2017
edited by CPhill  Sep 25, 2017

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