Find Increasing Intervals for y = -2x² - 12x - 16: Quadratic Function Analysis

To determine where the function $y = -2x^2 - 12x - 16$ is increasing, we follow these steps:

• Step 1: Find the derivative of the function.
  Starting with $y = -2x^2 - 12x - 16$, the derivative is calculated as:
  $\frac{dy}{dx} = -4x - 12$.
• Step 2: Identify the critical points by setting the derivative to zero and solving for $x$:
  $-4x - 12 = 0$
  $-4x = 12$
  $x = -3$.
• Step 3: Determine the sign of the derivative in the intervals divided by the critical point. The critical point divides the real line into intervals: $(-\infty, -3)$ and $(-3, \infty)$.
• Step 4: Test an $x$-value from each interval to evaluate the sign of $\frac{dy}{dx}$:
  For $x = -4$ in the interval $(-\infty, -3)$,
  $\frac{dy}{dx} = -4(-4) - 12 = 16 - 12 = 4$ (positive).
  For $x = 0$ in the interval $(-3, \infty)$,
  $\frac{dy}{dx} = -4(0) - 12 = -12$ (negative).
• Conclusion: The function $y = -2x^2 - 12x - 16$ is increasing in the interval where the derivative is positive, which is $x < -3$.

Therefore, the solution to the problem is $x < -3$.