Determine Positive X Values in the Quadratic y = -1/6x² + 11/3x

Look at the following function:

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

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

$f(x) > 0$

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

• Step 1: Write the function in standard quadratic form and identify coefficients.
• Step 2: Find the roots using the quadratic formula.
• Step 3: Determine the sign of the quadratic on intervals determined by the roots.
• Step 4: Identify intervals where the function is positive.

Step 1: The function given is $y = -\frac{1}{6}x^2 + 3\frac{2}{3}x$, or equivalently:

$y = -\frac{1}{6}x^2 + \frac{11}{3}x$ in standard form.

With coefficients $a = -\frac{1}{6}$, $b = \frac{11}{3}$, and $c = 0$.

Step 2: Apply the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to find roots:

The roots are given by:

$x = \frac{-\frac{11}{3} \pm \sqrt{\left(\frac{11}{3}\right)^2 - 4 \left(-\frac{1}{6}\right) \cdot 0}}{2 \cdot \left(-\frac{1}{6}\right)}$

Since $c = 0$, simplify to:

$x = \frac{-\frac{11}{3} \pm \sqrt{\left(\frac{11}{3}\right)^2}}{-\frac{1}{3}}$

Solve to get roots:

Roots are $x = 0$ and $x = 22$.

Step 3: Analyze the sign of $y$:

Since the parabola opens downwards (as $a < 0$), the function is positive between the roots:

$0 < x < 22$.

Therefore, the solution to the problem is where the function is positive:

$0 < x < 22$.