over what interval is the function increasing...
over what interval is the function increasing, and over what interval is the function decreasing?\n|x|f(x)=-\frac{1}{3}x^{2}|(x,y)|\n|-6|-12|(-6, - 12)|\n|-3|-3|(-3, - 3)|\n|0|0|(0,0)|\n|3|-3|(3, - 3)|\n|6|-12|(6, - 12)|\nthe function f(x) is increasing over the interval . (simplify your answer. type an inequality.)
Answer
# Explanation:
## Step1: Identify the function type
The function $f(x)=-\frac{1}{3}x^{2}$ is a quadratic function in the form $y = ax^{2}+bx + c$ where $a=-\frac{1}{3}$, $b = 0$, $c = 0$. The graph of a quadratic function $y=ax^{2}+bx + c$ is a parabola, and its axis of symmetry is given by the formula $x=-\frac{b}{2a}$.
## Step2: Find the axis of symmetry
Substitute $a =-\frac{1}{3}$ and $b = 0$ into the axis - of - symmetry formula $x=-\frac{b}{2a}$. We get $x=-\frac{0}{2\times(-\frac{1}{3})}=0$.
## Step3: Determine increasing and decreasing intervals
Since $a=-\frac{1}{3}<0$, the parabola opens downwards. A parabola that opens downwards is increasing to the left of its axis of symmetry and decreasing to the right of its axis of symmetry.
# Answer:
$x<0$