Question

identify the vertex and the axis of symmetry for the function.
$f(x)=x^{2}-8$

the vertex of the function is $(0, -8)$
(type an ordered pair.)
the axis of symmetry of the function is
(type an equation.)

Explanation:

Step1: Recall the form of a parabola

The function \( f(x) = x^2 - 8 \) is a quadratic function in the form \( f(x)=ax^2 + bx + c \), or in vertex form \( f(x)=a(x - h)^2 + k \), where \((h,k)\) is the vertex. For \( f(x)=x^2 - 8 \), we can rewrite it as \( f(x)=1(x - 0)^2 - 8 \), so \( h = 0 \) and \( k=-8 \), which matches the given vertex \((0,-8)\).

Step2: Determine the axis of symmetry

For a quadratic function in the form \( f(x)=ax^2+bx + c \) (or vertex form \( f(x)=a(x - h)^2 + k \)), the axis of symmetry is the vertical line \( x = h \). Since the vertex is \((h,k)=(0,-8)\), the axis of symmetry is the line where \( x = 0 \).

Answer:

The axis of symmetry of the function is \( x = 0 \)