Question

state the axis of symmetry of f(x)=-(1)/(2)x^(2)+3

Answer 1

$f(x)=-\frac{1}{2}x^2+3$

The axis of symmetry is located at the x-coordinate of vertex of the function. For a function of the form:

$f(x)=ax^2+bx+c$

The coordinates of its vertex V(h,k) are given by:

$\begin{gathered} h=-\frac{b}{2a} \\ where \\ a=-\frac{1}{2} \\ b=0 \\ so\colon \\ h=-\frac{0}{2(-\frac{1}{2})}=0 \end{gathered}$

Therefore, the axis of symmetry is located at:

$x=0$