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the axis of symmetry for the graph of the function is f(x)=1/4x^2+bx+10 is x=6. what is the value of b?

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hello :
note : x = b/2a an equation for The axis of symmetry when f(x) =ax²+bx+c
in this exercice : 6 = (b)/(2)(1/4)
6=(b)/0.5
b = 3
anathor solution :
f(x) = (1/4)x² +3x +10 = (1/4)(x² +12x )+10 =(1/4)(x²-2(6)(x)+6²-6²)+10
f(x) = (1/4)(x-6)² +1.....(vertex form the vertex is : (6 . 1)
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