determine the interval of increase and the interval of decrease for the function f(x)=-2x^2 - 8x - 4.

Explanation:

Step1: Find the derivative

The derivative of $f(x)=-2x^{2}-8x - 4$ using the power - rule $(x^n)'=nx^{n - 1}$ is $f'(x)=-4x-8$.

Step2: Set the derivative equal to zero

Solve $-4x - 8=0$. Add 8 to both sides: $-4x=8$. Then divide by - 4, getting $x=-2$.

Step3: Test intervals

Choose a test point in the interval $x\lt - 2$, say $x=-3$. Then $f'(-3)=-4\times(-3)-8 = 12 - 8=4>0$. So the function is increasing on the interval $(-\infty,-2)$. Choose a test point in the interval $x>-2$, say $x = 0$. Then $f'(0)=-4\times0-8=-8<0$. So the function is decreasing on the interval $(-2,\infty)$.

Answer:

The function $f(x)$ is increasing on the interval $(-\infty,-2)$ and decreasing on the interval $(-2,\infty)$.