Domain Analysis of y = (1/3)x² + x + 2: Finding Valid Input Values

To determine the positive and negative domains of the quadratic function $y = \frac{1}{3}x^2 + x + 2$, we will follow these general steps:

• Find the roots of the equation $\frac{1}{3}x^2 + x + 2 = 0$.
• Analyze the sign of the quadratic between and beyond these roots.

First, let's identify the roots of the quadratic function:

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = \frac{1}{3}$, $b = 1$, and $c = 2$, the discriminant $\Delta = b^2 - 4ac = 1^2 - 4 \cdot \frac{1}{3} \cdot 2 = 1 - \frac{8}{3} = \frac{-5}{3}$.

Since the discriminant is negative, the quadratic equation $\frac{1}{3}x^2 + x + 2 = 0$ has no real roots.

Given that the coefficient of $x^2$ (i.e., $\frac{1}{3}$) is positive, the parabola opens upwards, meaning the entire function is greater than zero for all real values of $x$.

Therefore, the positive domain is all real numbers, and there is no negative domain.

Therefore, the solution is: $x > 0$: for all $x$; $x < 0$: none.