Analyze the Domains of the Function: y = -1/3x² + 2/3x - 1/3

Find the positive and negative domains of the function below:

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

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

• Step 1: Identify the roots of the quadratic function.
• Step 2: Use the roots to determine intervals.
• Step 3: Test each interval to determine if the function is positive or negative.

Now, let's work through each step:
Step 1: The quadratic function is $y = -\frac{1}{3}x^2 + \frac{2}{3}x - \frac{1}{3}$. We apply the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to find the roots, where $a = -\frac{1}{3}$, $b = \frac{2}{3}$, and $c = -\frac{1}{3}$.

Calculating the discriminant $b^2 - 4ac$:

$\left(\frac{2}{3}\right)^2 - 4\left(-\frac{1}{3}\right)\left(-\frac{1}{3}\right) = \frac{4}{9} - \frac{4}{9} = 0$.

The discriminant is zero, indicating a repeated root at:

$x = \frac{-\frac{2}{3}}{2\left(-\frac{1}{3}\right)} = 1$.

Step 2: The repeated root is $x = 1$. For $x < 1$ and $x > 1$, evaluate the sign of the function.

Step 3: Testing intervals:

• For $x = 0$ (as a test point for $x < 1$), substitute into the original function:

$y = -\frac{1}{3}(0)^2 + \frac{2}{3}(0) - \frac{1}{3} = -\frac{1}{3}$. The function is negative.

• For $x > 1$, any positive $x$ will substitute into the function and confirm it remains negative, as the parabola opens downwards and cannot turn positive again.

Therefore, the solution to the problem is:

$x < 0 : x \ne 1$, and $x > 0$: none.