Determine the Positive and Negative Domains of y = 2x² + 7x - 9

To solve this problem, we need to find when the quadratic function changes from positive to negative and vice versa.

• Step 1: Find the roots using the quadratic formula. For $y = 2x^2 + 7x - 9$, identify $a = 2$, $b = 7$, and $c = -9$.
• Step 2: Apply the quadratic formula:
  $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
• Step 3: Calculate:
  $x = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 2 \cdot (-9)}}{2 \cdot 2}$
  $x = \frac{-7 \pm \sqrt{49 + 72}}{4}$
  $x = \frac{-7 \pm \sqrt{121}}{4}$
  $x = \frac{-7 \pm 11}{4}$
• Step 4: Solve for the roots:
  $x = \frac{-7 + 11}{4} = 1$
  $x = \frac{-7 - 11}{4} = -\frac{18}{4} = -4.5$
• Step 5: The roots are $x = 1$ and $x = -4.5$. These roots divide the real number line into intervals: $(-\infty, -4.5)$, $(-4.5, 1)$, and $(1, \infty)$.
• Step 6: Test each interval:
• Interval $(-\infty, -4.5)$: Choose a test point like $x = -5$.
  Calculate $y$ using $x = -5$:
  $y = 2(-5)^2 + 7(-5) - 9 = 50 - 35 - 9 = 6 \, (> 0)$. The function is positive in this interval.
• Interval $(-4.5, 1)$: Choose a test point like $x = 0$.
  Calculate $y$ using $x = 0$:
  $y = 2(0)^2 + 7(0) - 9 = -9 \, (< 0)$. The function is negative in this interval.
• Interval $(1, \infty)$: Choose a test point like $x = 2$.
  Calculate $y$ using $x = 2$:
  $y = 2(2)^2 + 7(2) - 9 = 8 + 14 - 9 = 13 \, (> 0)$. The function is positive in this interval.

The positive domain of $y$ is $(-\infty, -4.5) \cup (1, \infty)$.
The negative domain of $y$ is $(-4.5, 1)$.

Thus, the domains are as follows:
$x < 0 : -4\frac{1}{2} < x < 1$
$x > 1$ or $x > 0 : x < -4\frac{1}{2}$