Find Decreasing Intervals of y = -2x² - 12x - 16

To solve this problem, we'll follow these steps:

• Step 1: Compute the derivative of the function.
• Step 2: Find the critical point where the derivative is zero.
• Step 3: Determine where the function is decreasing by analyzing the sign of the derivative.

Now, let's work through each step:

Step 1: Compute the derivative of the function $y = -2x^2 - 12x - 16$. The derivative, $y'(x)$, is found by applying the power rule:

$y' = \frac{d}{dx}(-2x^2 - 12x - 16) = -4x - 12$

Step 2: Find the critical point by setting the derivative equal to zero and solving for $x$:

$-4x - 12 = 0$

$-4x = 12$

$x = -3$

This is the critical point where the function changes direction.

Step 3: Determine where the function is decreasing. A quadratic function, which is a parabola that opens downwards (since the leading coefficient $-2$ is negative), will be decreasing to the right of its vertex at $x = -3$. This means that the interval where the function is decreasing is when $x > -3$.

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