Question

The axis of symmerty for the graph of the function f(x) = 1/4x^2 + bc +10 is x=6 what is the value of b

Answer 1

Completing the square gives

$\dfrac14x^2+bx+10=\dfrac14(x^2+4bx+40)=\dfrac14(x^2+4bx+4b^2+40-4b^2)$

$\implies\dfrac14x^2+bx+10=\dfrac14(x+2b)^2+10-b^2$

At the point on the graph along the axis of symmetry, the squared term vanishes, so that when $x=6$, we have $x+2b=0$. So

$6+2b=0\implies 2b=-6\implies\boxed{b=-3}$