Finding Intervals of Increase and Decrease: y = 25x² + 20x + 4

To solve for intervals of increase and decrease, we follow these detailed steps:

• Step 1: Differentiate the function.
• Step 2: Set the derivative to zero to find critical points.
• Step 3: Analyze sign changes around the critical points.

Let's begin:

Step 1: Differentiate the function $y = 25x^2 + 20x + 4$.

The derivative is:
$y' = \frac{d}{dx}(25x^2 + 20x + 4) = 50x + 20$.

Step 2: Set $y' = 0$ to find critical points:
$50x + 20 = 0$.

Solving for $x$:
$50x = -20$
$x = -\frac{20}{50} = -\frac{2}{5} = -0.4$.

Step 3: Test intervals around $x = -0.4$:

• For $x < -0.4$, choose a test point such as $x = -1$:
  $y'(-1) = 50(-1) + 20 = -50 + 20 = -30$, which is less than zero, indicating decrease.
• For $x > -0.4$, choose a test point such as $x = 0$:
  $y'(0) = 50(0) + 20 = 20$, which is greater than zero, indicating increase.

Thus, the function decreases on the interval $x < -0.4$ and increases on the interval $x > -0.4$.

The correct intervals of increase and decrease for the function are:
$\searrow: x < -0.4 \\ \nearrow: x > -0.4$.

The correct answer choice is:

$\searrow: x < -0.4 \\ \nearrow: x > -0.4$