Finding Positive Values in the Quadratic Function: y = 1/3x² + 2.33x

The function given is $y = \frac{1}{3}x^2 + \frac{7}{3}x$. Our goal is to determine when this function is greater than 0.

Firstly, we set the function equal to 0 to find the critical points:

$\frac{1}{3}x^2 + \frac{7}{3}x = 0$

Factor out $\frac{1}{3}x$ from the equation:

$\frac{1}{3}x(x + 7) = 0$

This gives us two roots: $x = 0$ and $x = -7$.

Now, consider the intervals determined by these roots: $x < -7$, $-7 < x < 0$, and $x > 0$. Analyze the sign of $y$ in each interval by selecting test points.

• Interval $x < -7$: Choose $x = -8$. Substituting into the function gives $\frac{1}{3}(-8)^2 + \frac{7}{3}(-8) = \frac{64}{3} - \frac{56}{3} = \frac{8}{3} > 0$
• Interval $-7 < x < 0$: Choose $x = -1$. Substituting into the function gives $\frac{1}{3}(-1)^2 + \frac{7}{3}(-1) = \frac{1}{3} - \frac{7}{3} = -\frac{6}{3} = -2 < 0$
• Interval $x > 0$: Choose $x = 1$. Substituting into the function gives $\frac{1}{3}(1)^2 + \frac{7}{3}(1) = \frac{1}{3} + \frac{7}{3} = \frac{8}{3} > 0$

From this analysis, the function $y$ is positive when $x < -7$ or $x > 0$. Thus, the solution is:

The function $f(x)$ is positive for $x > 0$ or $x < -7$.