Determine the Positive and Negative Domains for y = 3x² + 12x + 12

To solve for the positive and negative domains of the quadratic function $y=3x^2+12x+12$, we will follow these steps:

• Step 1: Calculate the roots using the quadratic formula.
• Step 2: Analyze the sign of the function between and beyond these roots.

Step 1: Find the roots of the quadratic function using the quadratic formula:

The given quadratic function is $y = 3x^2 + 12x + 12$. Identifying coefficients, we have $a = 3$, $b = 12$, and $c = 12$.

Using the quadratic formula:

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

Substitute the values of $a$, $b$, and $c$:

$x = \frac{-12 \pm \sqrt{12^2 - 4 \times 3 \times 12}}{2 \times 3}$ $x = \frac{-12 \pm \sqrt{144 - 144}}{6}$ $x = \frac{-12 \pm \sqrt{0}}{6}$ $x = \frac{-12}{6}$ $x = -2$

The quadratic function has a single root at $x = -2$, meaning it is a perfect square trinomial, and there is only one point where the function equals zero.

Step 2: Analyze the sign of the quadratic:

• Because the parabola opens upwards (since $a = 3 > 0$), it implies the function is positive for all $x$ except at the vertex $x = -2$.
• The function $y$ reaches its minimum value (vertex) at $x = -2$, where $y = 0$.
• Therefore, $y = 3x^2 + 12x + 12 > 0$ for all $x \neq -2$.

Therefore, the function is positive for all $x$ but not for $x = -2$, where it is zero. It never reaches negativity.

To summarize, the positive domain (where $y > 0$) is $x \neq -2$ and the negative domain (where $y < 0$) does not exist.

In terms of the choices given, the correct answer is:

x > 0 :x\ne-2

x < 0 : none