Analyzing Positive and Negative Domains: -x² - 6x - 9 Explained

To solve this problem, let's find the roots of the quadratic function by using the quadratic formula. The function is given as:

$y = -x^2 - 6x - 9$

The quadratic formula is:

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

Given $a = -1$, $b = -6$, and $c = -9$, let's calculate the discriminant:

$\Delta = b^2 - 4ac = (-6)^2 - 4(-1)(-9) = 36 - 36 = 0$

Since the discriminant is zero, there is exactly one real root, which is also the vertex of the parabola. Let's find this root:

$x = \frac{-(-6) \pm \sqrt{0}}{2(-1)} = \frac{6}{-2} = -3$

So, the vertex of the parabola is at $x = -3$. The quadratic function opens downwards ($a < 0$), which means the function will be zero at $x = -3$, negative for all other values of $x$. There is no positive domain because the parabola does not go above the x-axis.

Therefore, the negative domain of the function is:

$x < 0 : x \ne -3$

and there is no positive domain.

This matches choice 3.

Therefore, the solution to the problem is:

$x < 0 : x\ne-3$$x > 0 :$ none