Determine Where the Quadratic y = -x² - 6x - 8 is Negative: Solving Inequalities

To solve the problem, we need to find where the function $f(x) = -x^2 - 6x - 8$ is negative. Let's proceed with a step-by-step solution:

• Step 1: Find the roots of the equation $-x^2 - 6x - 8 = 0$ using the quadratic formula.
• Step 2: Determine the intervals formed by these roots.
• Step 3: Test the intervals to see where $f(x) < 0$.

Step 1: The quadratic formula is given as follows:

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

In our equation, $a = -1$, $b = -6$, and $c = -8$.

Plugging these values into the formula:

$x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(-1)(-8)}}{2(-1)}$

$x = \frac{6 \pm \sqrt{36 - 32}}{-2}$

$x = \frac{6 \pm \sqrt{4}}{-2}$

$x = \frac{6 \pm 2}{-2}$

This gives the roots:

• $x = \frac{6 + 2}{-2} = -4$
• $x = \frac{6 - 2}{-2} = -2$

Step 2: The roots divide the number line into three intervals: $x < -4$, $-4 < x < -2$, and $x > -2$.

Step 3: Test these intervals:

• For $x < -4$, pick $x = -5$: $f(-5) = -(-5)^2 - 6(-5) - 8 = -25 + 30 - 8 = -3$ (negative).
• For $-4 < x < -2$, pick $x = -3$: $f(-3) = -(-3)^2 - 6(-3) - 8 = -9 + 18 - 8 = 1$ (positive).
• For $x > -2$, pick $x = 0$: $f(0) = -(0)^2 - 6(0) - 8 = -8$ (negative).

Thus, $f(x) < 0$ when $x < -4$ or $x > -2$.

Therefore, the solution to the problem is $x > -2$ or $x < -4$.