Analyze the Quadratic Function: Domains of y = -2x² + 3x + 2

To solve the given problem, we will perform the following steps:

• Calculate the roots of the quadratic equation $y = -2x^2 + 3x + 2$ using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Given $a = -2$, $b = 3$, and $c = 2$, use the formula:

$x = \frac{-3 \pm \sqrt{3^2 - 4(-2)(2)}}{2(-2)} = \frac{-3 \pm \sqrt{9 + 16}}{-4} = \frac{-3 \pm \sqrt{25}}{-4}$

• The roots are $x = \frac{-3 + 5}{-4} = -\frac{1}{2}$ and $x = \frac{-3 - 5}{-4} = 2$.
• The roots $-\frac{1}{2}$ and $2$ divide the x-axis into three intervals: $(-\infty, -\frac{1}{2})$, $(-\frac{1}{2}, 2)$, and $(2, \infty)$.
• Test a point from each interval in the function to determine the sign of the function in those intervals:

Choose $x = -1$ for interval $(- \infty, -\frac{1}{2})$:
Substitute into the function: $y = -2(-1)^2 + 3(-1) + 2 = -2 - 3 + 2 = -3$ (negative).

Choose $x = 0$ for interval $(- \frac{1}{2}, 2)$:
Substitute into the function: $y = -2(0)^2 + 3(0) + 2 = 2$ (positive).

Choose $x = 3$ for interval $(2, \infty)$:
Substitute into the function: $y = -2(3)^2 + 3(3) + 2 = -18 + 9 + 2 = -7$ (negative).

Therefore, the positive domain where the function is positive is $-\frac{1}{2} < x < 2$, and the negative domains are $x < -\frac{1}{2}$ or $x > 2$.

The solution to the problem is:

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

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