Determining x When the Parabola y = x² + 10x + 16 Is Below Zero

The problem asks us to determine where the function $y = x^2 + 10x + 16$ is less than zero.

• Step 1: Find the roots of the equation $x^2 + 10x + 16 = 0$ using the quadratic formula.
• Step 2: Calculate the discriminant $\Delta = b^2 - 4ac = 10^2 - 4 \times 1 \times 16 = 100 - 64 = 36$.
• Step 3: Find the roots using $x = \frac{{-b \pm \sqrt{\Delta}}}{2a} = \frac{{-10 \pm \sqrt{36}}}{2}$.
• Step 4: Calculate the roots: $x = \frac{{-10 + 6}}{2} = -2$ $x = \frac{{-10 - 6}}{2} = -8$

The roots are $x = -2$ and $x = -8$. These roots will divide the number line into intervals.

• Step 5: Analyze the sign of the quadratic on the intervals $-\infty, -8$, $-8, -2$, and $-2, \infty$.
• Step 6: Choose test points: - For $x < -8$, choose $x = -9$, - For $-8 < x < -2$, choose $x = -5$, - For $x > -2$, choose $x = 0$.

Test each interval:

• For $x = -9$, $y = (-9)^2 + 10 \times (-9) + 16 = 81 - 90 + 16 = 7$ (positive).
• For $x = -5$, $y = (-5)^2 + 10 \times (-5) + 16 = 25 - 50 + 16 = -9$ (negative).
• For $x = 0$, $y = (0)^2 + 10 \times 0 + 16 = 16$ (positive).

The function is negative between $x = -8$ and $x = -2$. Therefore, the solution to $f(x) < 0$ is $-8 < x < -2$.

Therefore, the correct answer is $-8 < x < -2$.