I have $f(x)=x^2-2bx+c$ If the minimum value of the function is 6, what is the possible value of c.
I tried$$f(x)=(x-b)^2-b^2+c$$ $$b^2=c-6$$ I couldn't solve for the value of c.
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abee 99
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1Any $c\ge 6$ will do, as you have shown. – BAI Apr 15 '18 at 04:24
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The vertex is where the min/max occurs. In this case, it's the min. Thus $x = -\dfrac{-2b}{2} = b \implies f_{\text{min}} = f(b) = b^2-2b\cdot b+c = c-b^2 = 6\implies c = b^2+6$.

DeepSea
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