Find Intervals of Increase and Decrease for y = -4x² - x - 3

To determine the intervals of increase and decrease for the function $y = -4x^2 - x - 3$, we follow these steps:

• **Step 1:** Compute the derivative of the function.

The function is $y = -4x^2 - x - 3$. The derivative, $y'$, is computed as:

$y' = \frac{d}{dx}(-4x^2 - x - 3) = -8x - 1$

• **Step 2:** Set the derivative equal to zero to find the critical point.

We solve the equation $-8x - 1 = 0$ for $x$:

$-8x - 1 = 0$ $-8x = 1$ $x = -\frac{1}{8}$

• **Step 3:** Analyze the sign of the derivative around the critical point to determine intervals of increase and decrease.

Examine the sign of $y' = -8x - 1$ in the intervals determined by the critical point:

- For $x < -\frac{1}{8}$, choose $x = -1$: $y' = -8(-1) - 1 = 8 - 1 = 7$ (positive, so the function is increasing) - For $x > -\frac{1}{8}$, choose $x = 0$: $y' = -8(0) - 1 = -1$ (negative, so the function is decreasing)

Therefore, the intervals of the function are:

The function is increasing for $x < -\frac{1}{8}$ and decreasing for $x > -\frac{1}{8}$.

The intervals correctly formulated are:

$\searrow: x < -\frac{1}{8}~; \nearrow: x > -\frac{1}{8}$

The correct choice is:

:

$\searrow: x < -\frac{1}{8} \\ \nearrow: x > -\frac{1}{8}$