Discover the Domains: Solving y=-3x²+5x-2 for Positive and Negative Values

To find the positive and negative domains of the function $y = -3x^2 + 5x - 2$, we follow these steps:

• Step 1: Find the roots of the equation by solving $-3x^2 + 5x - 2 = 0$.
• Step 2: The quadratic equation $ax^2 + bx + c = 0$ has roots $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Step 3: Calculate the discriminant, $b^2 - 4ac = 5^2 - 4(-3)(-2) = 25 - 24 = 1$.
• Step 4: Find the roots: $x = \frac{-5 \pm \sqrt{1}}{2(-3)} = \frac{-5 \pm 1}{-6}$.
• Step 5: This gives roots $x = \frac{-4}{-6} = \frac{2}{3}$ and $x = \frac{-6}{-6} = 1$.
• Step 6: Determine intervals: $(-\infty, \frac{2}{3})$, $(\frac{2}{3}, 1)$, $(1, \infty)$.
• Step 7: Test a point from each interval in the original equation to determine sign:
• For $x \in (-\infty, \frac{2}{3})$, choose $x = 0$: $y = -3(0)^2 + 5(0) - 2 = -2$ (negative).
• For $x \in (\frac{2}{3}, 1)$, choose $x = 0.8$: $y = -3(0.8)^2 + 5(0.8) - 2 = -1.28 + 4 - 2 = 0.72$ (positive).
• For $x \in (1, \infty)$, choose $x = 2$: $y = -3(2)^2 + 5(2) - 2 = -12 + 10 - 2 = -4$ (negative).

Therefore, the positive domains of the function are when $\frac{2}{3} < x < 1$, and the negative domains are when $x < \frac{2}{3}$ or $x > 1$.

Thus, the solution to the problem is:

$x > 1$ or $x < 0 : x < \frac{2}{3}$

$x > 0 : \frac{2}{3} < x < 1$