Specific Quadratic Function Challenge: Analyze 3x² + 18x + 27

The task is to determine where the function $y=3x^2+18x+27$ is positive and negative. A quadratic function is upward-facing if the coefficient of $x^2$ is positive. Here, $a = 3 > 0$, indicating an upward parabola.

First, find the roots using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Calculate the discriminant:

$b^2 - 4ac = 18^2 - 4 \times 3 \times 27 = 324 - 324 = 0$

The discriminant is 0, implying one repeated root at:

$x = \frac{-18}{2 \times 3} = -3$

The vertex at $x = -3$ means the parabola touches the x-axis without crossing it, and there are no intervals where $y < 0$.

Since the parabola is always above the x-axis except at this point, for $x > -3$ and $x < -3$, $y > 0$ except at $x = -3$ where $y = 0$.

Therefore, the function is positive for $x \ne -3$.

The positive domain where $y > 0$ is:

$x > 0 : x \ne -3$

There is no negative domain (where $y < 0$).

$x < 0 :$ none

The correct choice is option 4:

\($x > 0 : x \ne -3$\) and \($x < 0 :$ none\).