Solve: When is -1/6x² - 3⅑x Less Than Zero?

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, follow these steps:

• Step 1: Identify the coefficients from the quadratic function $y = -\frac{1}{6}x^2 - 3\frac{1}{9}x$. Here, $a = -\frac{1}{6}$, $b = -\frac{28}{9}$, and $c = 0$.
• Step 2: Find the roots using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Plugging in the values, we get:

$x = \frac{-\left(-\frac{28}{9}\right) \pm \sqrt{\left(-\frac{28}{9}\right)^2 - 4 \times \left(-\frac{1}{6}\right) \times 0}}{2 \times \left(-\frac{1}{6}\right)}$

Simplifying, we find:

$x = \frac{\frac{28}{9}}{-\frac{1}{3}}$

$x = \frac{28}{9} \times -3$

$x = -\frac{84}{9} = -9\frac{1}{3}$

• Step 3: Explore the intervals determined by these roots for where the function $y$ is negative:
• Consider intervals $(-\infty, -18\frac{2}{3})$, $(-18\frac{2}{3}, 0)$, and $(0, \infty)$.
• Evaluate the sign of $y$ over each interval:

For $(-\infty, -18\frac{2}{3})$: Choose a sample point like $x = -20$. Plug it in to see if $y < 0$. It turns out this interval is negative.

For $(-18\frac{2}{3}, 0)$: Choose a sample point like $x = -1$. Plug it in to see if $y > 0$. This interval turns out positive.

For $(0, \infty)$: Choose a sample point like $x = 1$. Plug it in to see if $y < 0$. This interval turns out negative.

Therefore, the solution is when $x > 0$ or $x < -18\frac{2}{3}$.

Thus, the quadratic function is negative outside the roots.

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