Question

graph $f(x) = -\frac{1}{3}x^2 + 4$ on the set of axes below.

graph grid image

state the vertex of this function. state the equation of the axis of symmetry of this function.

Explanation:

Step1: Identify vertex form

The function $f(x) = -\frac{1}{3}x^2 + 4$ matches the vertex form $f(x)=a(x-h)^2+k$, where $h=0$, $k=4$, $a=-\frac{1}{3}$.

Step2: Find vertex coordinates

For $f(x)=a(x-h)^2+k$, vertex is $(h,k)$. Substitute $h=0,k=4$: Vertex = $(0,4)$.

Step3: Find axis of symmetry

Axis of symmetry is $x=h$. Substitute $h=0$: $x=0$.

Step4: Plot key points for graph

• When $x=0$, $f(0)=-\frac{1}{3}(0)^2+4=4$ (vertex)
• When $x=3$, $f(3)=-\frac{1}{3}(9)+4=-3+4=1$ → point $(3,1)$
• When $x=-3$, $f(-3)=-\frac{1}{3}(9)+4=-3+4=1$ → point $(-3,1)$
• When $x=6$, $f(6)=-\frac{1}{3}(36)+4=-12+4=-8$ → point $(6,-8)$
• When $x=-6$, $f(-6)=-\frac{1}{3}(36)+4=-12+4=-8$ → point $(-6,-8)$

Connect these symmetric points to form a downward-opening parabola.

Answer:

• Vertex of the function: $(0, 4)$
• Equation of the axis of symmetry: $x=0$
• (Graph: A downward-opening parabola with vertex at $(0,4)$, passing through points like $(\pm3,1)$ and $(\pm6,-8)$, symmetric about the y-axis)