Analyze Domains of a Quadratic Function: Exploring -2x² + 7x - 3

Find the positive and negative domains of the following function:

$y=-2x^2+7x-3$

To solve this problem, we'll find the intervals where the given quadratic function $y = -2x^2 + 7x - 3$ is greater than zero (positive) and less than zero (negative).

Step 1: Find the roots of the quadratic function.

The general form of the quadratic equation is $ax^2 + bx + c = 0$. Here, $a = -2$, $b = 7$, and $c = -3$.

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we calculate the roots.

First, calculate the discriminant:

$b^2 - 4ac = 7^2 - 4 \times (-2) \times (-3) = 49 - 24 = 25$.

Thus, the roots are:

$x = \frac{-7 \pm \sqrt{25}}{2 \times (-2)} = \frac{-7 \pm 5}{-4}$.

Calculating for the two roots:

• $x_1 = \frac{-7 + 5}{-4} = \frac{-2}{-4} = \frac{1}{2}$
• $x_2 = \frac{-7 - 5}{-4} = \frac{-12}{-4} = 3$

The roots are $x = \frac{1}{2}$ and $x = 3$.

Step 2: Determine the sign of the function in each interval.

The function is defined as:

$(-\infty, \frac{1}{2}), \left(\frac{1}{2}, 3\right), (3, \infty)$.

Test each interval to determine where the function is positive or negative:

• For $x < \frac{1}{2}$, choose $x = 0$:
  $y = -2(0)^2 + 7(0) - 3 = -3$ (negative)
• For $\frac{1}{2} < x < 3$, choose $x = 1$:
  $y = -2(1)^2 + 7(1) - 3 = 2$ (positive)
• For $x > 3$, choose $x = 4$:
  $y = -2(4)^2 + 7(4) - 3 = -3$ (negative)

Conclusion: The positive domain is $\frac{1}{2} < x < 3$, and the negative domain is $x < \frac{1}{2}$ or $x > 3$.

Therefore, the correct option is:

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

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