Solve: Finding Positive Values for y = -1/6x² - 3⅓x

Look at the following function:

$y=-\frac{1}{6}x^2-3\frac{1}{9}x$

Determine for which values of $x$ the following is true:

$f(x) > 0$

To solve this problem, we'll follow these steps:

• Step 1: Identify the given equation in a standard form.
• Step 2: Calculate the roots using the quadratic formula.
• Step 3: Determine the sign of the function in different intervals determined by the roots.

Now, let's work through each step:

Step 1: The given quadratic function is $y = -\frac{1}{6}x^2 - \frac{10}{3}x$. Here, $a = -\frac{1}{6}$, $b = -\frac{10}{3}$, and $c = 0$.
We rewrite the function as $y = -\frac{1}{6}x^2 - \frac{10}{3}x$.

Step 2: To find the roots of the quadratic equation, apply the quadratic formula:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Plugging in the values, we get:
$x = \frac{-(-\frac{10}{3}) \pm \sqrt{(-\frac{10}{3})^2 - 4 \cdot (-\frac{1}{6}) \cdot 0}}{2 \cdot (-\frac{1}{6})}$
This simplifies to:
$x = \frac{\frac{10}{3}}{-\frac{1}{3}}$ (since the discriminant simplifies to zero as $c$ is zero)
$x = 0$ and $x = -18\frac{2}{3}$ (solving for the roots)

Step 3: Determining the intervals:
Because the parabola opens downwards (as $a = -\frac{1}{6}$ is negative), the quadratic is positive between the roots.
Thus, $f(x) > 0$ in the interval $-18\frac{2}{3} < x < 0$.

Therefore, the solution to the problem is $-18\frac{2}{3} < x < 0$.