Analyze the Function: Where Is -1/9x² + 1 2/3x Greater Than Zero?

Look at the following function:

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

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

$f(x) > 0$

To determine where the function $f(x) = -\frac{1}{9}x^2 + \frac{5}{3}x$ is positive, we follow these steps:

• Step 1: Convert the mixed fraction: The term $1\frac{2}{3}x$ can be written as $\frac{5}{3}x$.
• Step 2: The function can be expressed as $f(x) = -\frac{1}{9}x^2 + \frac{5}{3}x$.
• Step 3: Set the function equal to zero to find the roots: $-\frac{1}{9}x^2 + \frac{5}{3}x = 0$.
• Step 4: Factor out $x$: $x\left(-\frac{1}{9}x + \frac{5}{3}\right) = 0$.
• Step 5: Solve for $x$: From $x = 0$ and $-\frac{1}{9}x + \frac{5}{3} = 0$, find the second root:

Solving the second equation:

$\frac{5}{3} = \frac{1}{9}x$, which simplifies to:

$x = \frac{5}{3} \times 9 = 15$.

The roots are $x = 0$ and $x = 15$.

Since the parabola opens downwards (as indicated by the negative leading coefficient $-\frac{1}{9}$), the function will be positive between the roots.

Thus, $f(x) > 0$ for $0 < x < 15$.

Therefore, the values of $x$ such that $f(x) > 0$ are given by:

$0 < x < 15$.