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The equation of a parabola is given.

 

y=1/4x^2−3x+18

 

What are the coordinates of the focus of the parabola?

Acceptfully  May 29, 2017
 #1
avatar+90001 
+2

 

y  =  (1/4)x^2  - 3x  + 18    multiply through by 4

 

4y  = x ^2 - 12x + 72       subtract 72 from both sides

 

4y - 72  =  x^2 - 12x       take (1/2) of 12  = 6, square it = 36  and add it to both sides

 

4y - 72 + 36  =  x^2 - 12x + 36    factor the right side, simplify the right

 

4y - 36   =  ( x - 6)^2          factor the left side as

 

4 ( y - 9)  =  ( x - 6)^2

 

And in the form

 

4p( y - k)  = ( x - h)^2       the vertex is ( h, k)  = (6, 9)  and p  = 1 

 

So..... the focus is given by  ( h, k + p)  =  ( 6, 9 + 1)  =  (6, 10 )

 

Here's the graph :  https://www.desmos.com/calculator/eddman8tqs

 

 

cool cool cool

CPhill  May 29, 2017

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