Question

Given g(x)=x^2-7x+1/4 show that the least possible value of g(x) is -12

Answer 1

   
$\displaystyle\\ g(x)=x^2-7x+\frac{1}{4} \\ \\ g(x)=x^2-7x+0.25\\ \\ \\ \\ x \text{ for } g_{min} = \frac{-b}{2a}=\frac{-(-7)}{2\times 1}= \frac{7}{2}=3.5 \\ \\ g_{min} = g(3.5) = (3.5)^2 -7\times 3.5 + 0.25= 12.25 - 24.5 +0.25 = \boxed{-12}$