Find Intervals of Increase for y = -2x² - 8x - 10

Step 1: Differentiate the Function
First, find the derivative of $y$. The derivative $\frac{dy}{dx}$ of $y = -2x^2 - 8x - 10$ is:

$y' = \frac{d}{dx}(-2x^2 - 8x - 10) = -4x - 8$.

Step 2: Find the Critical Point
Set the derivative equal to zero to find the critical point:

$-4x - 8 = 0$.

Solving for $x$, we get:

$-4x = 8$.

$x = -2$.

Step 3: Analyze the Function Around the Critical Point
The critical point $x = -2$ is where the function changes from increasing to decreasing or vice versa. To find the intervals of increase, check the sign of the derivative on either side of $x = -2$. Since the leading coefficient (-2) in the quadratic function is negative, the parabola opens downwards.

Step 4: Establish the Intervals
- For $x < -2$, choose a test point (e.g., $x = -3$) and substitute into the derivative:

$y' = -4(-3) - 8 = 12 - 8 = 4$.

The derivative is positive, indicating that the function is increasing on the interval $x < -2$.

- For $x > -2$, choose a test point (e.g., $x = 0$) and substitute into the derivative:

$y' = -4(0) - 8 = -8$.

The derivative is negative, indicating that the function is decreasing on the interval $x > -2$.

Therefore, the function is increasing in the interval $x < -2$.

$x < -2$