Find the Domain of Function y=-6x²+27: Positive and Negative Regions

To find the positive and negative domains of the quadratic function $y = -6x^2 + 27$, we begin by finding the roots of the equation where $y = 0$.

Step 1: Set the function equal to zero: $-6x^2 + 27 = 0$.

Step 2: Solve for $x^2$: $-6x^2 + 27 = 0 \quad \Rightarrow \quad 6x^2 = 27 \quad \Rightarrow \quad x^2 = \frac{27}{6} \quad \Rightarrow \quad x^2 = \frac{9}{2}$

Step 3: Solve for $x$ by taking the square root: $x = \pm \sqrt{\frac{9}{2}} \quad \Rightarrow \quad x = \pm \frac{3}{\sqrt{2}}$

Step 4: These roots, $x = \frac{3}{\sqrt{2}}$ and $x = -\frac{3}{\sqrt{2}}$, divide the real number line into three intervals: $x < -\frac{3}{\sqrt{2}}$, $-\frac{3}{\sqrt{2}} < x < \frac{3}{\sqrt{2}}$, and $x > \frac{3}{\sqrt{2}}$.

Step 5: Test the sign of the function in each interval:

• For $x < -\frac{3}{\sqrt{2}}$, pick a test point like $x = -2$: $y = -6(-2)^2 + 27 = -24 + 27 = 3$ (positive)
• For $-\frac{3}{\sqrt{2}} < x < \frac{3}{\sqrt{2}}$, pick $x = 0$: $y = -6(0)^2 + 27 = 27$ (positive)
• For $x > \frac{3}{\sqrt{2}}$, pick $x = 2$: $y = -6(2)^2 + 27 = -24 + 27 = 3$ (positive)

The function is negative nowhere as the parabola opens downward (due to negative coefficient of $x^2$), it achieves maximum $y$ at its vertex, and beyond the roots, remains positive.

Therefore, the positive domain, where $y > 0$ is for $x < -\frac{3}{\sqrt{2}}$ and $-\frac{3}{\sqrt{2}} < x < \frac{3}{\sqrt{2}}$.

The function doesn’t change sign compared to standard expectations because of its formulation in this problem. The negative domain is non-existent.

The correct solution is then given by matching the described situation to the choice:

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

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